{"id":14195,"date":"2018-09-27T16:40:34","date_gmt":"2018-09-27T16:40:34","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/solving-trigonometric-equations-with-identities\/"},"modified":"2021-02-04T22:45:33","modified_gmt":"2021-02-04T22:45:33","slug":"solving-trigonometric-equations-with-identities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/pdx-precalculus\/chapter\/solving-trigonometric-equations-with-identities\/","title":{"raw":"Walkthrough of Unit 7: Trigonometric Identities","rendered":"Walkthrough of Unit 7: Trigonometric Identities"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"font-weight: 400\">Verify the fundamental trigonometric identities.<\/li>\r\n \t<li style=\"font-weight: 400\">Simplify trigonometric expressions using algebra and the identities.<\/li>\r\n \t<li style=\"font-weight: 400\">Use sum and difference formulas for cosine.<\/li>\r\n \t<li style=\"font-weight: 400\">Use sum and difference formulas for sine.<\/li>\r\n \t<li style=\"font-weight: 400\">Use sum and difference formulas for tangent.<\/li>\r\n \t<li style=\"font-weight: 400\">Use sum and difference formulas for cofunctions.<\/li>\r\n \t<li style=\"font-weight: 400\">Use sum and difference formulas to verify identities.<\/li>\r\n \t<li style=\"font-weight: 400\">Use double-angle formulas to find exact values.<\/li>\r\n \t<li style=\"font-weight: 400\">Use double-angle formulas to verify identities.<\/li>\r\n \t<li style=\"font-weight: 400\">Use reduction formulas to simplify an expression.<\/li>\r\n \t<li style=\"font-weight: 400\">Use half-angle formulas to find exact values.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Verify the fundamental trigonometric identities<\/h2>\r\nIdentities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.\r\n\r\nTo verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the <strong>Pythagorean identities<\/strong>, the even-odd identities, the reciprocal identities, and the quotient identities.\r\n\r\nWe will begin with the <strong>Pythagorean identities<\/strong>, which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.\r\n<table id=\"Table_07_01_01\" summary=\"&quot;Pythagorean Identities&quot; with three cells. First: sin(theta)^2 + cos(theta)^2 = 1. Second: 1 + cot(theta)^2 = csc(theta)^2. Third: 1 + tan(theta)^2 = sec(theta)^2.\"><colgroup> <col \/> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\" colspan=\"3\">Pythagorean Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1[\/latex]<\/td>\r\n<td>[latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex]<\/td>\r\n<td>[latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe second and third identities can be obtained by manipulating the first. The identity [latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta\\[\/latex] is found by rewriting the left side of the equation in terms of sine and cosine.\r\n\r\nProve: [latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta \/latex]\r\n<div style=\"text-align: center\">[latex]\\begin{align}1+{\\cot }^{2}\\theta&amp; =\\left(1+\\frac{{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)&amp;&amp; \\text{Rewrite the left side}. \\\\ &amp;=\\left(\\frac{{\\sin }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)+\\left(\\frac{{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)&amp;&amp; \\text{Write both terms with the common denominator}. \\\\ &amp;=\\frac{{\\sin }^{2}\\theta +{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta } \\\\ &amp;=\\frac{1}{{\\sin }^{2}\\theta } \\\\ &amp;={\\csc }^{2}\\theta \\end{align}[\/latex]<\/div>\r\nSimilarly, [latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex] can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives\r\n<div style=\"text-align: center\">[latex]\\begin{align}1+{\\tan }^{2}\\theta &amp;=1+{\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)}^{2}&amp;&amp; \\text{Rewrite left side}. \\\\ &amp;={\\left(\\frac{\\cos \\theta }{\\cos \\theta }\\right)}^{2}+{\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)}^{2}&amp;&amp; \\text{Write both terms with the common denominator}. \\\\ &amp;=\\frac{{\\cos }^{2}\\theta +{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &amp;=\\frac{1}{{\\cos }^{2}\\theta } \\\\ &amp;={\\sec }^{2}\\theta \\end{align}[\/latex]<\/div>\r\nThe next set of fundamental identities is the set of <strong>even-odd identities<\/strong>. The <strong>even-odd identities<\/strong> relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle and determine whether the identity is odd or even.\r\n<table id=\"Table_07_01_02\" summary=\"&quot;Even-Odd Identities&quot; with three cells. First: tan(-theta) = -tan(theta) and cot(-theta) = -cot(theta). Second: sin(-theta) = -sin(theta) and csc(-theta) = -csc(theta). Third: cos(-theta) = cos(theta) and sec(-theta) = sec(theta).\">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\" colspan=\"3\">Even-Odd Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{gathered}\\tan \\left(-\\theta \\right)=-\\tan \\theta\\\\ \\cot \\left(-\\theta \\right)=-\\cot \\theta \\end{gathered}[\/latex]<\/td>\r\n<td>[latex]\\begin{gathered}\\sin \\left(-\\theta \\right)=-\\sin \\theta\\\\ \\csc \\left(-\\theta \\right)=-\\csc \\theta\\end{gathered}[\/latex]<\/td>\r\n<td>[latex]\\begin{gathered}\\cos \\left(-\\theta \\right)=\\cos \\theta \\\\ \\sec \\left(-\\theta \\right)=\\sec \\theta \\end{gathered}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRecall that an <strong>odd function<\/strong> is one in which [latex]f\\left(-x\\right)= -f\\left(x\\right)[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. The <strong>sine<\/strong> function is an odd function because [latex]\\sin \\left(-\\theta \\right)=-\\sin \\theta[\/latex]. The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of [latex]\\frac{\\pi }{2}\\\\[\/latex] and [latex]-\\frac{\\pi }{2}[\/latex]. The output of [latex]\\sin \\left(\\frac{\\pi }{2}\\right)[\/latex] is opposite the output of [latex]\\sin \\left(-\\frac{\\pi }{2}\\right)[\/latex]. Thus,\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\frac{\\pi }{2}\\right)=1\\end{align}[\/latex] and [latex]\\begin{align} \\sin \\left(-\\frac{\\pi }{2}\\right)=-\\sin \\left(\\frac{\\pi }{2}\\right) =-1 \\end{align}[\/latex]<\/div>\r\nThis is shown in\u00a0Figure 2.\r\n\r\n<span id=\"fs-id1491023\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164029\/CNX_Precalc_Figure_07_01_0022.jpg\" alt=\"Graph of y=sin(theta) from -2pi to 2pi, showing in particular that it is symmetric about the origin. Points given are (pi\/2, 1) and (-pi\/2, -1).\" \/><\/span>\r\n<p style=\"text-align: center\"><strong>Figure 2.<\/strong>\u00a0Graph of [latex]y=\\sin \\theta[\/latex]<\/p>\r\nRecall that an <strong>even function<\/strong> is one in which\r\n<div style=\"text-align: center\">[latex]f\\left(-x\\right)=f\\left(x\\right)[\/latex] for all <em>x<\/em>\u00a0in the domain of <em>f<\/em>.<\/div>\r\nThe graph of an even function is symmetric about the <em>y-<\/em>axis. The cosine function is an even function because [latex]\\cos \\left(-\\theta \\right)=\\cos \\theta[\/latex].\r\nFor example, consider corresponding inputs [latex]\\frac{\\pi }{4}[\/latex] and [latex]-\\frac{\\pi }{4}[\/latex]. The output of [latex]\\cos \\left(\\frac{\\pi }{4}\\right)[\/latex] is the same as the output of [latex]\\cos \\left(-\\frac{\\pi }{4}\\right)[\/latex]. Thus,\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(-\\frac{\\pi }{4}\\right)=\\cos \\left(\\frac{\\pi }{4}\\right) \\approx 0.707 \\end{align}[\/latex]<\/div>\r\nSee Figure 3.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164031\/CNX_Precalc_Figure_07_01_0032.jpg\" alt=\"Graph of y=cos(theta) from -2pi to 2pi, showing in particular that it is symmetric about the y-axis. Points given are (-pi\/4, .707) and (pi\/4, .707). \" \/>\r\n<p style=\"text-align: center\"><strong>Figure 3.<\/strong>\u00a0Graph of [latex]y=\\cos \\theta[\/latex]<\/p>\r\nFor all [latex]\\theta[\/latex] in the domain of the sine and cosine functions, respectively, we can state the following:\r\n<ul id=\"fs-id2141576\">\r\n \t<li>Since [latex]\\sin \\left(-\\theta \\right)=-\\sin \\theta[\/latex], sine is an odd function.<\/li>\r\n \t<li>Since, [latex]\\cos \\left(-\\theta \\right)=\\cos \\theta[\/latex], cosine is an even function.<\/li>\r\n<\/ul>\r\nThe other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity, [latex]\\tan \\left(-\\theta \\right)=\\mathrm{-tan}\\theta[\/latex]. We can interpret the tangent of a negative angle as [latex]\\tan \\left(-\\theta \\right)=\\frac{\\sin \\left(-\\theta \\right)}{\\cos \\left(-\\theta \\right)}=\\frac{-\\sin \\theta }{\\cos \\theta }=-\\tan \\theta[\/latex]. Tangent is therefore an odd function, which means that [latex]\\tan \\left(-\\theta \\right)=-\\tan \\left(\\theta \\right)[\/latex] for all [latex]\\theta[\/latex] in the domain of the <strong>tangent function<\/strong>.\r\n\r\nThe cotangent identity, [latex]\\cot \\left(-\\theta \\right)=-\\cot \\theta[\/latex], also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as [latex]\\cot \\left(-\\theta \\right)=\\frac{\\cos \\left(-\\theta \\right)}{\\sin \\left(-\\theta \\right)}=\\frac{\\cos \\theta }{-\\sin \\theta }=-\\cot \\theta[\/latex]. Cotangent is therefore an odd function, which means that [latex]\\cot \\left(-\\theta \\right)=-\\cot \\left(\\theta \\right)[\/latex] for all [latex]\\theta[\/latex] in the domain of the <strong>cotangent function<\/strong>.\r\n\r\nThe <strong>cosecant function<\/strong> is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as [latex]\\csc \\left(-\\theta \\right)=\\frac{1}{\\sin \\left(-\\theta \\right)}=\\frac{1}{-\\sin \\theta }=-\\csc \\theta[\/latex]. The cosecant function is therefore odd.\r\n\r\nFinally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as [latex]\\sec \\left(-\\theta \\right)=\\frac{1}{\\cos \\left(-\\theta \\right)}=\\frac{1}{\\cos \\theta }=\\sec \\theta[\/latex]. The secant function is therefore even.\r\n\r\nTo sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.\r\n\r\nThe next set of fundamental identities is the set of <strong>reciprocal identities<\/strong>, which, as their name implies, relate trigonometric functions that are reciprocals of each other.\r\n<table id=\"fs-id2031263\" summary=\"Table labeled &quot;Reciprocal Identities.&quot; Three rows, two columns. The table has ordered pairs of these row values: (sin(theta) = 1\/csc(theta), csc(theta) = 1\/sin(theta)), (cos(theta) = 1\/sec(theta), sec(theta) = 1\/cos(theta)), (tan(theta) = 1\/cot(theta), cot(theta) = 1\/tan(theta)).\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\" colspan=\"2\">Reciprocal Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\sin \\theta =\\frac{1}{\\csc \\theta }[\/latex]<\/td>\r\n<td>[latex]\\csc \\theta =\\frac{1}{\\sin \\theta }[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\cos \\theta =\\frac{1}{\\sec \\theta }[\/latex]<\/td>\r\n<td>[latex]\\sec \\theta =\\frac{1}{\\cos \\theta }[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\tan \\theta =\\frac{1}{\\cot \\theta }[\/latex]<\/td>\r\n<td>[latex]\\cot \\theta =\\frac{1}{\\tan \\theta }[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe final set of identities is the set of <strong>quotient identities<\/strong>, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities.\r\n<table id=\"fs-id937819\" summary=\"Table labeled &quot;Quotient Identities.&quot; First cell: tan(theta) = sin(theta) \/ cos(theta). Second cell: cot(theta) = cos(theta) \/ sin(theta).\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\" colspan=\"2\">Quotient Identities<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\tan \\theta =\\frac{\\sin \\theta }{\\cos \\theta }[\/latex]<\/td>\r\n<td>[latex]\\cot \\theta =\\frac{\\cos \\theta }{\\sin \\theta }[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Summarizing Trigonometric Identities<\/h3>\r\nThe <strong>Pythagorean identities<\/strong> are based on the properties of a right triangle.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} {\\cos}^{2}\\theta + {\\sin}^{2}\\theta=1 \\\\ 1+{\\tan}^{2}\\theta={\\sec}^{2}\\theta \\\\ 1+{\\cot}^{2}\\theta={\\csc}^{2}\\theta\\end{gathered}[\/latex]<\/p>\r\nThe <strong>even-odd identities<\/strong> relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\cos(-\\theta)=\\cos(\\theta) \\\\\\sin(-\\theta)=-\\sin(\\theta) \\\\\\tan(-\\theta)=-\\tan(\\theta) \\\\\\cot(-\\theta)=-\\cot(\\theta) \\\\\\sec(-\\theta)=\\sec(\\theta) \\\\\\csc(-\\theta)=-\\csc(\\theta) \\end{gathered}[\/latex]<\/p>\r\nThe <strong>reciprocal identities<\/strong> define reciprocals of the trigonometric functions.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin\\theta=\\frac{1}{\\csc\\theta} \\\\ \\cos\\theta=\\frac{1}{\\sec\\theta} \\\\ \\tan\\theta=\\frac{1}{\\cot\\theta} \\\\ \\cot\\theta=\\frac{1}{\\tan\\theta} \\\\ \\sec\\theta=\\frac{1}{\\cos\\theta} \\\\ \\csc\\theta=\\frac{1}{\\sin\\theta}\\end{gathered}[\/latex]<\/p>\r\nThe <strong>quotient identities<\/strong> define the relationship among the trigonometric functions.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\tan\\theta=\\frac{\\sin\\theta}{\\cos\\theta} \\\\ \\cot\\theta=\\frac{\\cos\\theta}{\\sin\\theta} \\end{gathered}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Graphing the Equations of an Identity<\/h3>\r\nGraph both sides of the identity [latex]\\cot \\theta =\\frac{1}{\\tan \\theta }[\/latex]. In other words, on the graphing calculator, graph [latex]y=\\cot \\theta[\/latex] and [latex]y=\\frac{1}{\\tan \\theta }[\/latex].\r\n\r\n[reveal-answer q=\"943922\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"943922\"]\r\n<h3><span id=\"fs-id1353869\"><img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164033\/CNX_Precalc_Figure_07_01_0072.jpg\" alt=\"Graph of y = cot(theta) and y=1\/tan(theta) from -2pi to 2pi. They are the same!\" \/><\/span><\/h3>\r\n&nbsp;\r\n<h4>Analysis of the Solution<\/h4>\r\nWe see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trigonometric identity, verify that it is true.<strong>\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id2191946\">\r\n \t<li>Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.<\/li>\r\n \t<li>Look for opportunities to factor expressions, square a binomial, or add fractions.<\/li>\r\n \t<li>Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.<\/li>\r\n \t<li>If these steps do not yield the desired result, try converting all terms to sines and cosines.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Verifying a Trigonometric Identity<\/h3>\r\nVerify [latex]\\tan \\theta \\cos \\theta =\\sin \\theta[\/latex].\r\n\r\n[reveal-answer q=\"136260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"136260\"]\r\n\r\nWe will start on the left side, as it is the more complicated side:\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\theta \\cos \\theta &amp;=\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)\\cos \\theta \\\\ &amp;=\\left(\\frac{\\sin \\theta }{\\cancel{\\cos \\theta }}\\right)\\cancel{\\cos \\theta } \\\\ &amp;=\\sin \\theta \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis identity was fairly simple to verify, as it only required writing [latex]\\tan \\theta[\/latex] in terms of [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nVerify the identity [latex]\\csc \\theta \\cos \\theta \\tan \\theta =1[\/latex].\r\n\r\n[reveal-answer q=\"361361\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"361361\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\csc \\theta \\cos \\theta \\tan \\theta &amp;=\\left(\\frac{1}{\\sin \\theta }\\right)\\cos \\theta \\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right) \\\\ &amp;=\\frac{\\cos \\theta }{\\sin \\theta }\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right) \\\\ &amp;=\\frac{\\sin \\theta \\cos \\theta }{\\sin \\theta \\cos \\theta } \\\\ &amp;=1\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Verifying the Equivalency Using the Even-Odd Identities<\/h3>\r\nVerify the following equivalency using the even-odd identities:\r\n<p style=\"text-align: center\">[latex]\\left(1+\\sin x\\right)\\left[1+\\sin \\left(-x\\right)\\right]={\\cos }^{2}x[\/latex]<\/p>\r\n[reveal-answer q=\"208801\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"208801\"]\r\n\r\nWorking on the left side of the equation, we have\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\left(1+\\sin x\\right)\\left[1+\\sin \\left(-x\\right)\\right]&amp;=\\left(1+\\sin x\\right)\\left(1-\\sin x\\right)&amp;&amp; \\text{Since sin(-}x\\text{)=}-\\sin x \\\\ &amp;=1-{\\sin }^{2}x&amp;&amp; \\text{Difference of squares} \\\\ &amp;={\\cos }^{2}x&amp;&amp; {\\text{cos}}^{2}x=1-{\\sin }^{2}x\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Verifying a Trigonometric Identity Involving <em>sec<sup>2<\/sup>\u03b8<\/em><\/h3>\r\nVerify the identity [latex]\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }={\\sin }^{2}\\theta[\/latex]\r\n\r\n[reveal-answer q=\"565941\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"565941\"]\r\n\r\nAs the left side is more complicated, let\u2019s begin there.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&amp;=\\frac{\\left({\\tan }^{2}\\theta +1\\right)-1}{{\\sec }^{2}\\theta }&amp;&amp; {\\sec}^{2}\\theta ={\\tan }^{2}\\theta +1 \\\\ &amp;=\\frac{{\\tan }^{2}\\theta }{{\\sec }^{2}\\theta } \\\\ &amp;={\\tan }^{2}\\theta \\left(\\frac{1}{{\\sec }^{2}\\theta }\\right) \\\\ &amp;={\\tan }^{2}\\theta \\left({\\cos }^{2}\\theta \\right)&amp;&amp; {\\cos }^{2}\\theta =\\frac{1}{{\\sec }^{2}\\theta } \\\\ &amp;=\\left(\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta }\\right)\\left({\\cos }^{2}\\theta \\right)&amp;&amp; {\\tan}^{2}\\theta =\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &amp;=\\left(\\frac{{\\sin }^{2}\\theta }{\\cancel{{\\cos }^{2}\\theta}}\\right)\\left(\\cancel{{\\cos }^{2}\\theta} \\right) \\\\ &amp;={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\r\nThere is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&amp;=\\frac{{\\sec }^{2}\\theta }{{\\sec }^{2}\\theta }-\\frac{1}{{\\sec }^{2}\\theta } \\\\ &amp;=1-{\\cos }^{2}\\theta \\\\ &amp;={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<h3>Analysis<\/h3>\r\nIn the first method, we used the identity [latex]{\\sec }^{2}\\theta ={\\tan }^{2}\\theta +1\\\\[\/latex] and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nShow that [latex]\\frac{\\cot \\theta }{\\csc \\theta }=\\cos \\theta[\/latex].\r\n\r\n[reveal-answer q=\"813945\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"813945\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\cot \\theta }{\\csc \\theta }&amp;=\\frac{\\frac{\\cos \\theta }{\\sin \\theta }}{\\frac{1}{\\sin \\theta }} \\\\ &amp;=\\frac{\\cos \\theta }{\\sin \\theta }\\cdot \\frac{\\sin \\theta }{1} \\\\ &amp;=\\cos \\theta \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Creating and Verifying an Identity<\/h3>\r\nCreate an identity for the expression [latex]2\\tan \\theta \\sec \\theta[\/latex] by rewriting strictly in terms of sine.\r\n\r\n[reveal-answer q=\"482916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"482916\"]\r\n\r\nThere are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:\r\n<p style=\"text-align: center\">[latex]\\begin{align}2\\tan \\theta \\sec \\theta &amp;=2\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)\\left(\\frac{1}{\\cos \\theta }\\right) \\\\ &amp;=\\frac{2\\sin \\theta }{{\\cos }^{2}\\theta } \\\\ &amp;=\\frac{2\\sin \\theta }{1-{\\sin }^{2}\\theta }&amp;&amp; \\text{Substitute }1-{\\sin }^{2}\\theta \\text{ for }{\\cos }^{2}\\theta \\end{align}[\/latex]<\/p>\r\nThus,\r\n<p style=\"text-align: center\">[latex]2\\tan \\theta \\sec \\theta =\\frac{2\\sin \\theta }{1-{\\sin }^{2}\\theta }[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Verifying an Identity Using Algebra and Even\/Odd Identities<\/h3>\r\nVerify the identity:\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{{\\sin }^{2}\\left(-\\theta \\right)-{\\cos }^{2}\\left(-\\theta \\right)}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)}=\\cos \\theta -\\sin \\theta\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"689339\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689339\"]\r\n\r\nLet\u2019s start with the left side and simplify:\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{{\\sin }^{2}\\left(-\\theta \\right)-{\\cos }^{2}\\left(-\\theta \\right)}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)}&amp;=\\frac{{\\left[\\sin \\left(-\\theta \\right)\\right]}^{2}-{\\left[\\cos \\left(-\\theta \\right)\\right]}^{2}}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)} \\\\ &amp;=\\frac{{\\left(-\\sin \\theta \\right)}^{2}-{\\left(\\cos \\theta \\right)}^{2}}{-\\sin \\theta -\\cos \\theta }&amp;&amp; \\sin \\left(-x\\right)=-\\sin x\\text{ and }\\cos \\left(-x\\right)=\\cos x \\\\ &amp;=\\frac{{\\left(\\sin \\theta \\right)}^{2}-{\\left(\\cos \\theta \\right)}^{2}}{-\\sin \\theta -\\cos \\theta }&amp;&amp; \\text{Difference of squares} \\\\ &amp;=\\frac{\\left(\\sin \\theta -\\cos \\theta \\right)\\left(\\sin \\theta +\\cos \\theta \\right)}{-\\left(\\sin \\theta +\\cos \\theta \\right)} \\\\ &amp;=\\frac{\\left(\\sin \\theta -\\cos \\theta \\right)\\left(\\cancel{\\sin \\theta +\\cos \\theta }\\right)}{-\\left(\\cancel{\\sin \\theta +\\cos \\theta }\\right)} \\\\ &amp;=\\cos \\theta -\\sin \\theta\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nVerify the identity [latex]\\frac{{\\sin }^{2}\\theta -1}{\\tan \\theta \\sin \\theta -\\tan \\theta }=\\frac{\\sin \\theta +1}{\\tan \\theta }[\/latex].\r\n\r\n[reveal-answer q=\"307900\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"307900\"]\r\n\r\n[latex]\\begin{align}\\frac{{\\sin }^{2}\\theta -1}{\\tan \\theta \\sin \\theta -\\tan \\theta }&amp;=\\frac{\\left(\\sin \\theta +1\\right)\\left(\\sin \\theta -1\\right)}{\\tan \\theta \\left(\\sin \\theta -1\\right)}\\\\ &amp;=\\frac{\\sin \\theta +1}{\\tan \\theta }\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Verifying an Identity Involving Cosines and Cotangents<\/h3>\r\nVerify the identity: [latex]\\left(1-{\\cos }^{2}x\\right)\\left(1+{\\cot }^{2}x\\right)=1[\/latex].\r\n\r\n[reveal-answer q=\"690675\"]Show Solutions[\/reveal-answer]\r\n[hidden-answer a=\"690675\"]\r\n\r\nWe will work on the left side of the equation.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\left(1-{\\cos }^{2}x\\right)\\left(1+{\\cot }^{2}x\\right)&amp;=\\left(1-{\\cos }^{2}x\\right)\\left(1+\\frac{{\\cos }^{2}x}{{\\sin }^{2}x}\\right) \\\\ &amp;=\\left(1-{\\cos }^{2}x\\right)\\left(\\frac{{\\sin }^{2}x}{{\\sin }^{2}x}+\\frac{{\\cos }^{2}x}{{\\sin }^{2}x}\\right) &amp;&amp; \\text{Find the common denominator}. \\\\ &amp;=\\left(1-{\\cos }^{2}x\\right)\\left(\\frac{{\\sin }^{2}x+{\\cos }^{2}x}{{\\sin }^{2}x}\\right) \\\\ &amp;=\\left({\\sin }^{2}x\\right)\\left(\\frac{1}{{\\sin }^{2}x}\\right) \\\\ &amp;=1\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Simplify trigonometric expressions using algebra and the identities<\/h2>\r\nWe have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.\r\n\r\nFor example, the equation [latex]\\left(\\sin x+1\\right)\\left(\\sin x - 1\\right)=0[\/latex] resembles the equation [latex]\\left(x+1\\right)\\left(x - 1\\right)=0[\/latex], which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.\r\n\r\nAnother example is the difference of squares formula, [latex]{a}^{2}-{b}^{2}=\\left(a-b\\right)\\left(a+b\\right)[\/latex], which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Writing the Trigonometric Expression as an Algebraic Expression<\/h3>\r\nWrite the following trigonometric expression as an algebraic expression: [latex]2{\\cos }^{2}\\theta +\\cos \\theta -1[\/latex].\r\n\r\n[reveal-answer q=\"19423\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"19423\"]\r\n\r\nNotice that the pattern displayed has the same form as a standard quadratic expression, [latex]a{x}^{2}+bx+c[\/latex]. Letting [latex]\\cos \\theta =x[\/latex], we can rewrite the expression as follows:\r\n<p style=\"text-align: center\">[latex]2{x}^{2}+x - 1[\/latex]<\/p>\r\nThis expression can be factored as [latex]\\left(2x+1\\right)\\left(x - 1\\right)[\/latex]. If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for [latex]x[\/latex]. At this point, we would replace [latex]x[\/latex] with [latex]\\cos \\theta [\/latex] and solve for [latex]\\theta [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Rewriting a Trigonometric Expression Using the Difference of Squares<\/h3>\r\nRewrite the trigonometric expression: [latex]4{\\cos }^{2}\\theta -1[\/latex].\r\n\r\n[reveal-answer q=\"947607\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"947607\"]\r\n\r\nNotice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,\r\n<p style=\"text-align: center\">[latex]\\begin{align}4{\\cos }^{2}\\theta -1&amp;={\\left(2\\cos \\theta \\right)}^{2}-1 \\\\ &amp;=\\left(2\\cos \\theta -1\\right)\\left(2\\cos \\theta +1\\right) \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<h3>Analysis<\/h3>\r\nIf this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let [latex]\\cos \\theta =x[\/latex], rewrite the expression as [latex]4{x}^{2}-1[\/latex], and factor [latex]\\left(2x - 1\\right)\\left(2x+1\\right)[\/latex]. Then replace [latex]x[\/latex] with [latex]\\cos \\theta [\/latex] and solve for the angle.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nRewrite the trigonometric expression: [latex]25 - 9{\\sin }^{2}\\theta [\/latex].\r\n\r\n[reveal-answer q=\"979877\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979877\"]\r\n\r\nThis is a difference of squares formula: [latex]25 - 9{\\sin }^{2}\\theta =\\left(5 - 3\\sin \\theta \\right)\\left(5+3\\sin \\theta \\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Simplify by Rewriting and Using Substitution<\/h3>\r\nSimplify the expression by rewriting and using identities:\r\n<p style=\"text-align: center\">[latex]{\\csc }^{2}\\theta -{\\cot }^{2}\\theta [\/latex]<\/p>\r\n[reveal-answer q=\"120852\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"120852\"]\r\n\r\nWe can start with the Pythagorean identity.\r\n<p style=\"text-align: center\">[latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta [\/latex]<\/p>\r\nNow we can simplify by substituting [latex]1+{\\cot }^{2}\\theta [\/latex] for [latex]{\\csc }^{2}\\theta [\/latex]. We have\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\csc }^{2}\\theta -{\\cot }^{2}\\theta &amp;=1+{\\cot }^{2}\\theta -{\\cot }^{2}\\theta \\\\ &amp;=1\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]120496[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nUse algebraic techniques to verify the identity: [latex]\\frac{\\cos\\theta}{1+\\sin\\theta}=\\frac{1-\\sin\\theta}{\\cos\\theta}[\/latex].\r\n\r\n(Hint: Multiply the numerator and denominator on the left side by [latex]1-\\sin\\theta[\/latex]).\r\n\r\n[reveal-answer q=\"94618\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"94618\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\cos \\theta }{1+\\sin \\theta }\\left(\\frac{1-\\sin \\theta }{1-\\sin \\theta }\\right)&amp;=\\frac{\\cos \\theta \\left(1-\\sin \\theta \\right)}{1-{\\sin }^{2}\\theta } \\\\ &amp;=\\frac{\\cos \\theta \\left(1-\\sin \\theta \\right)}{{\\cos }^{2}\\theta } \\\\ &amp;=\\frac{1-\\sin \\theta }{\\cos \\theta } \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use sum and difference formulas for cosine<\/h2>\r\nFinding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the <strong>special angles<\/strong>, which we can review in the unit circle shown in Figure 2.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164036\/CNX_Precalc_Figure_07_01_0042.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B. \" width=\"975\" height=\"638\" \/> <b>Figure 4.<\/b> The Unit Circle[\/caption]\r\n\r\nWe will begin with the <strong>sum and difference formulas for cosine<\/strong>, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles.\r\n<table id=\"Table_07_02_01\" summary=\"Two rows, two columns. The table has ordered pairs of these row values: (Sum formula for cosine, cos(a+B) = cos(a)cos(B) - sin(a)sin(B)) and (Difference formula for cosine, cos(a-B) = cos(a)cos(B) + sin(a)sin(B)).\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Sum formula for cosine<\/strong><\/td>\r\n<td>[latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Difference formula for cosine<\/strong><\/td>\r\n<td>[latex]\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta [\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFirst, we will prove the difference formula for cosines. Let\u2019s consider two points on the unit circle. Point [latex]P[\/latex] is at an angle [latex]\\alpha [\/latex] from the positive <em>x-<\/em>axis with coordinates [latex]\\left(\\cos \\alpha ,\\sin \\alpha \\right)[\/latex] and point [latex]Q[\/latex] is at an angle of [latex]\\beta [\/latex] from the positive <em>x-<\/em>axis with coordinates [latex]\\left(\\cos \\beta ,\\sin \\beta \\right)[\/latex]. Note the measure of angle [latex]POQ[\/latex] is [latex]\\alpha -\\beta [\/latex].\r\n\r\nLabel two more points: [latex]A[\/latex] at an angle of [latex]\\left(\\alpha -\\beta \\right)[\/latex] from the positive <em>x-<\/em>axis with coordinates [latex]\\left(\\cos \\left(\\alpha -\\beta \\right),\\sin \\left(\\alpha -\\beta \\right)\\right)[\/latex]; and point [latex]B[\/latex] with coordinates [latex]\\left(1,0\\right)[\/latex]. Triangle [latex]POQ[\/latex] is a rotation of triangle [latex]AOB[\/latex] and thus the distance from [latex]P[\/latex] to [latex]Q[\/latex] is the same as the distance from [latex]A[\/latex] to [latex]B[\/latex].\r\n<figure id=\"Figure_07_02_002\" class=\"small\"><span id=\"fs-id1130636\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164038\/CNX_Precalc_Figure_07_02_0022.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" \/><\/span><\/figure>\r\n<p style=\"text-align: center\"><strong>Figure 5.\u00a0<\/strong>We can find the distance from [latex]P[\/latex] to [latex]Q[\/latex] using the <strong>distance formula<\/strong>.<\/p>\r\n\r\n<div><\/div>\r\n<div style=\"text-align: center\">[latex]\\begin{align}{d}_{PQ}&amp;=\\sqrt{{\\left(\\cos \\alpha -\\cos \\beta \\right)}^{2}+{\\left(\\sin \\alpha -\\sin \\beta \\right)}^{2}} \\\\ &amp;=\\sqrt{{\\cos }^{2}\\alpha -2\\cos \\alpha \\cos \\beta +{\\cos }^{2}\\beta +{\\sin }^{2}\\alpha -2\\sin \\alpha \\sin \\beta +{\\sin }^{2}\\beta } \\end{align}[\/latex]<\/div>\r\nThen we apply the <strong>Pythagorean identity<\/strong> and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align} &amp;=\\sqrt{\\left({\\cos }^{2}\\alpha +{\\sin }^{2}\\alpha \\right)+\\left({\\cos }^{2}\\beta +{\\sin }^{2}\\beta \\right)-2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\\\ &amp;=\\sqrt{1+1 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\\\ &amp;=\\sqrt{2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\end{align}[\/latex]<\/div>\r\nSimilarly, using the distance formula we can find the distance from [latex]A[\/latex] to [latex]B[\/latex].\r\n<div style=\"text-align: center\">[latex]\\begin{align}{d}_{AB}&amp;=\\sqrt{{\\left(\\cos \\left(\\alpha -\\beta \\right)-1\\right)}^{2}+{\\left(\\sin \\left(\\alpha -\\beta \\right)-0\\right)}^{2}} \\\\ &amp;=\\sqrt{{\\cos }^{2}\\left(\\alpha -\\beta \\right)-2\\cos \\left(\\alpha -\\beta \\right)+1+{\\sin }^{2}\\left(\\alpha -\\beta \\right)} \\end{align}[\/latex]<\/div>\r\nApplying the Pythagorean identity and simplifying we get:\r\n<div style=\"text-align: center\">[latex]\\begin{align} &amp;=\\sqrt{\\left({\\cos }^{2}\\left(\\alpha -\\beta \\right)+{\\sin }^{2}\\left(\\alpha -\\beta \\right)\\right)-2\\cos \\left(\\alpha -\\beta \\right)+1}\\\\ &amp;=\\sqrt{1 - 2\\cos \\left(\\alpha -\\beta \\right)+1}\\\\ &amp;=\\sqrt{2 - 2\\cos \\left(\\alpha -\\beta \\right)} \\end{align}[\/latex]<\/div>\r\nBecause the two distances are the same, we set them equal to each other and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align} \\sqrt{2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta }&amp;=\\sqrt{2 - 2\\cos \\left(\\alpha -\\beta \\right)} \\\\ 2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta &amp;=2 - 2\\cos \\left(\\alpha -\\beta \\right) \\end{align}[\/latex]<\/div>\r\nFinally we subtract [latex]2[\/latex] from both sides and divide both sides by [latex]-2[\/latex].\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta =\\cos \\left(\\alpha -\\beta \\right)\\end{align}[\/latex]<\/div>\r\nThus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Sum and Difference Formulas for Cosine<\/h3>\r\nThese formulas can be used to calculate the cosine of sums and differences of angles.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta\\end{align} [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta\\end{align} [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two angles, find the cosine of the difference between the angles.<strong>\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id1374215\">\r\n \t<li>Write the difference formula for cosine.<\/li>\r\n \t<li>Substitute the values of the given angles into the formula.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles<\/h3>\r\nUsing the formula for the cosine of the difference of two angles, find the exact value of [latex]\\cos \\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"956915\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"956915\"]\r\n\r\nUse the formula for the cosine of the difference of two angles. We have\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\alpha -\\beta \\right)&amp;=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\\\ \\cos \\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)&amp;=\\cos \\left(\\frac{5\\pi }{4}\\right)\\cos \\left(\\frac{\\pi }{6}\\right)+\\sin \\left(\\frac{5\\pi }{4}\\right)\\sin \\left(\\frac{\\pi }{6}\\right) \\\\ &amp;=\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)-\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{1}{2}\\right) \\\\ &amp;=-\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4} \\\\ &amp;=\\frac{-\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the exact value of [latex]\\cos \\left(\\frac{\\pi }{3}-\\frac{\\pi }{4}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"700317\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"700317\"]\r\n\r\n[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 12: Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine<\/h3>\r\nFind the exact value of [latex]\\cos \\left({75}^{\\circ }\\right)[\/latex].\r\n\r\n[reveal-answer q=\"617422\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"617422\"]\r\n\r\nAs [latex]{75}^{\\circ }={45}^{\\circ }+{30}^{\\circ }[\/latex], we can evaluate [latex]\\cos \\left({75}^{\\circ }\\right)[\/latex] as [latex]\\cos \\left({45}^{\\circ }+{30}^{\\circ }\\right)[\/latex]. Thus,\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left({45}^{\\circ }+{30}^{\\circ }\\right)&amp;=\\cos \\left({45}^{\\circ }\\right)\\cos \\left({30}^{\\circ }\\right)-\\sin \\left({45}^{\\circ }\\right)\\sin \\left({30}^{\\circ }\\right) \\\\ &amp;=\\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right) \\\\ &amp;=\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4} \\\\ &amp;=\\frac{\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the exact value of [latex]\\cos \\left({105}^{\\circ }\\right)[\/latex].\r\n\r\n[reveal-answer q=\"683323\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"683323\"]\r\n\r\n[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173445[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use sum and difference formulas for sine<\/h2>\r\nThe <strong>sum and difference formulas for sine<\/strong> can be derived in the same manner as those for cosine, and they resemble the cosine formulas.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Sum and Difference Formulas for Sine<\/h3>\r\nThese formulas can be used to calculate the sines of sums and differences of angles.\r\n<p style=\"text-align: center\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta [\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\sin \\left(\\alpha -\\beta \\right)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two angles, find the sine of the difference between the angles.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Write the difference formula for sine.<\/li>\r\n \t<li>Substitute the given angles into the formula.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 13: Using Sum and Difference Identities to Evaluate the Difference of Angles<\/h3>\r\nUse the sum and difference identities to evaluate the difference of the angles and show that part <em>a<\/em> equals part <em>b.<\/em>\r\n<ol>\r\n \t<li>[latex]\\sin \\left({45}^{\\circ }-{30}^{\\circ }\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\sin \\left({135}^{\\circ }-{120}^{\\circ }\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"191680\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"191680\"]\r\n<ol>\r\n \t<li>Let\u2019s begin by writing the formula and substitute the given angles.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\alpha -\\beta \\right)&amp;=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\\\ \\sin \\left({45}^{\\circ }-{30}^{\\circ }\\right)&amp;=\\sin \\left({45}^{\\circ }\\right)\\cos \\left({30}^{\\circ }\\right)-\\cos \\left({45}^{\\circ }\\right)\\sin \\left({30}^{\\circ }\\right) \\end{align}[\/latex]<\/div>\r\nNext, we need to find the values of the trigonometric expressions.\r\n<div style=\"text-align: center\">[latex]\\sin \\left({45}^{\\circ }\\right)=\\frac{\\sqrt{2}}{2},\\text{ }\\cos \\left({30}^{\\circ }\\right)=\\frac{\\sqrt{3}}{2},\\text{ }\\cos \\left({45}^{\\circ }\\right)=\\frac{\\sqrt{2}}{2},\\text{ }\\sin \\left({30}^{\\circ }\\right)=\\frac{1}{2}[\/latex]<\/div>\r\nNow we can substitute these values into the equation and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align} \\sin \\left({45}^{\\circ }-{30}^{\\circ }\\right)&amp;=\\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right) \\\\ &amp;=\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Again, we write the formula and substitute the given angles.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\alpha -\\beta \\right)&amp;=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta\\\\ \\sin \\left({135}^{\\circ }-{120}^{\\circ }\\right)&amp;=\\sin \\left({135}^{\\circ }\\right)\\cos \\left({120}^{\\circ }\\right)-\\cos \\left({135}^{\\circ }\\right)\\sin \\left({120}^{\\circ }\\right)\\end{align}[\/latex]<\/div>\r\nNext, we find the values of the trigonometric expressions.\r\n<div style=\"text-align: center\">[latex]\\sin \\left({135}^{\\circ }\\right)=\\frac{\\sqrt{2}}{2},\\cos \\left({120}^{\\circ }\\right)=-\\frac{1}{2},\\cos \\left({135}^{\\circ }\\right)=\\frac{\\sqrt{2}}{2},\\sin \\left({120}^{\\circ }\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/div>\r\nNow we can substitute these values into the equation and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left({135}^{\\circ }-{120}^{\\circ }\\right)&amp;=\\frac{\\sqrt{2}}{2}\\left(-\\frac{1}{2}\\right)-\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right) \\\\ &amp;=\\frac{-\\sqrt{2}+\\sqrt{6}}{4} \\\\ &amp;=\\frac{\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 14: Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function<\/h3>\r\nFind the exact value of [latex]\\sin \\left({\\cos }^{-1}\\frac{1}{2}+{\\sin }^{-1}\\frac{3}{5}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"954448\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"954448\"]\r\n\r\nThe pattern displayed in this problem is [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex]. Let [latex]\\alpha ={\\cos }^{-1}\\frac{1}{2}[\/latex] and [latex]\\beta ={\\sin }^{-1}\\frac{3}{5}[\/latex]. Then we can write\r\n<p style=\"text-align: center\">[latex]\\begin{align} \\cos \\alpha &amp;=\\frac{1}{2},0\\le \\alpha \\le \\pi\\\\ \\sin \\beta &amp;=\\frac{3}{5},-\\frac{\\pi }{2}\\le \\beta \\le \\frac{\\pi }{2} \\end{align}[\/latex]<\/p>\r\nWe will use the Pythagorean identities to find [latex]\\sin \\alpha [\/latex] and [latex]\\cos \\beta [\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\alpha &amp;=\\sqrt{1-{\\cos }^{2}\\alpha } \\\\ &amp;=\\sqrt{1-\\frac{1}{4}} \\\\ &amp;=\\sqrt{\\frac{3}{4}} \\\\ &amp;=\\frac{\\sqrt{3}}{2} \\\\ \\cos \\beta &amp;=\\sqrt{1-{\\sin }^{2}\\beta } \\\\ &amp;=\\sqrt{1-\\frac{9}{25}} \\\\ &amp;=\\sqrt{\\frac{16}{25}} \\\\ &amp;=\\frac{4}{5}\\end{align}[\/latex]<\/p>\r\nUsing the sum formula for sine,\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\left({\\cos }^{-1}\\frac{1}{2}+{\\sin }^{-1}\\frac{3}{5}\\right)&amp;=\\sin \\left(\\alpha +\\beta \\right) \\\\ &amp;=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\\\ &amp;=\\frac{\\sqrt{3}}{2}\\cdot \\frac{4}{5}+\\frac{1}{2}\\cdot \\frac{3}{5} \\\\ &amp;=\\frac{4\\sqrt{3}+3}{10}\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173443[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use sum and difference formulas for tangent<\/h2>\r\nFinding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.\r\n\r\nFinding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, [latex]\\tan x=\\frac{\\sin x}{\\cos x},\\cos x\\ne 0[\/latex].\r\n\r\nLet\u2019s derive the sum formula for tangent.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&amp;=\\frac{\\sin \\left(\\alpha +\\beta \\right)}{\\cos \\left(\\alpha +\\beta \\right)} \\\\[1mm] &amp;=\\frac{\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta } \\\\[1mm] &amp;=\\frac{\\frac{\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }}{\\frac{\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} &amp;&amp; \\text{Divide the numerator and denominator by cos}\\alpha \\text{cos}\\beta \\\\[1mm] &amp;=\\frac{\\frac{\\sin \\alpha \\cos \\beta }{\\cos \\alpha \\cos \\beta}+\\frac{\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }}{\\frac{\\cos \\alpha \\cos \\beta }{\\cos \\alpha \\cos \\beta }-\\frac{\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} &amp;&amp; \\text{Split the fractions.} \\\\[1mm] &amp;=\\frac{\\frac{\\sin \\alpha }{\\cos \\alpha }+\\frac{\\sin \\beta }{\\cos \\beta }}{1-\\frac{\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} &amp;&amp; \\text{Cancel.} \\\\[1mm] &amp;=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta } \\end{align}[\/latex]<\/div>\r\nWe can derive the difference formula for tangent in a similar way.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Sum and Difference Formulas for Tangent<\/h3>\r\nThe <strong>sum and difference formulas for tangent<\/strong> are:\r\n<p style=\"text-align: center\">[latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\tan \\left(\\alpha -\\beta \\right)=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two angles, find the tangent of the sum of the angles.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Write the sum formula for tangent.<\/li>\r\n \t<li>Substitute the given angles into the formula.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 15: Finding the Exact Value of an Expression Involving Tangent<\/h3>\r\nFind the exact value of [latex]\\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"796417\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"796417\"]\r\n\r\nLet\u2019s first write the sum formula for tangent and substitute the given angles into the formula.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&amp;=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)&amp;=\\frac{\\tan \\left(\\frac{\\pi }{6}\\right)+\\tan \\left(\\frac{\\pi }{4}\\right)}{1-\\left(\\tan \\left(\\frac{\\pi }{6}\\right)\\right)\\left(\\tan \\left(\\frac{\\pi }{4}\\right)\\right)} \\end{align}[\/latex]<\/p>\r\nNext, we determine the individual tangents within the formula:\r\n<p style=\"text-align: center\">[latex]\\tan \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{\\sqrt{3}},\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex]<\/p>\r\nSo we have\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)&amp;=\\frac{\\frac{1}{\\sqrt{3}}+1}{1-\\left(\\frac{1}{\\sqrt{3}}\\right)\\left(1\\right)} \\\\ &amp;=\\frac{\\frac{1+\\sqrt{3}}{\\sqrt{3}}}{\\frac{\\sqrt{3}-1}{\\sqrt{3}}} \\\\ &amp;=\\frac{1+\\sqrt{3}}{\\sqrt{3}}\\left(\\frac{\\sqrt{3}}{\\sqrt{3}-1}\\right) \\\\ &amp;=\\frac{\\sqrt{3}+1}{\\sqrt{3}-1} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the exact value of [latex]\\tan \\left(\\frac{2\\pi }{3}+\\frac{\\pi }{4}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"962695\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"962695\"]\r\n\r\n[latex]\\frac{1-\\sqrt{3}}{1+\\sqrt{3}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173454[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 16: Finding Multiple Sums and Differences of Angles<\/h3>\r\nGiven [latex]\\text{ }\\sin \\alpha =\\frac{3}{5},0&lt;\\alpha &lt;\\frac{\\pi }{2},\\cos \\beta =-\\frac{5}{13},\\pi &lt;\\beta &lt;\\frac{3\\pi }{2}[\/latex], find\r\n<ol>\r\n \t<li>[latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\tan \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\tan \\left(\\alpha -\\beta \\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"622511\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"622511\"]\r\n\r\nWe can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.\r\n<ol>\r\n \t<li>To find [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex], we begin with [latex]\\sin \\alpha =\\frac{3}{5}[\/latex] and [latex]0&lt;\\alpha &lt;\\frac{\\pi }{2}[\/latex]. The side opposite [latex]\\alpha [\/latex] has length 3, the hypotenuse has length 5, and [latex]\\alpha [\/latex] is in the first quadrant. Using the Pythagorean Theorem, we can find the length of side [latex]a:[\/latex]\r\n<div style=\"text-align: center\">\r\n\r\n[latex]\\begin{gathered}{a}^{2}+{3}^{2}={5}^{2} \\\\ {a}^{2}=16 \\\\ a=4 \\end{gathered}[\/latex]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164040\/CNX_Precalc_Figure_07_02_0032.jpg\" alt=\"Diagram of a triangle in the x,y plane. The vertices are at the origin, (4,0), and (4,3). The angle at the origin is alpha degrees, The angle formed by the x-axis and the side from (4,3) to (4,0) is a right angle. The side opposite the right angle has length 5.\" width=\"487\" height=\"252\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n<\/div>\r\nSince [latex]\\cos \\beta =-\\frac{5}{13}[\/latex] and [latex]\\pi &lt;\\beta &lt;\\frac{3\\pi }{2}[\/latex], the side adjacent to [latex]\\beta [\/latex] is [latex]-5[\/latex], the hypotenuse is 13, and [latex]\\beta [\/latex] is in the third quadrant. Again, using the Pythagorean Theorem, we have<\/li>\r\n \t<li>\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\left(-5\\right)}^{2}+{a}^{2}&amp;={13}^{2} \\\\ 25+{a}^{2}&amp;=169 \\\\ {a}^{2}&amp;=144\\\\ a&amp;=\\pm 12 \\end{align}[\/latex]<\/p>\r\nSince [latex]\\beta [\/latex] is in the third quadrant, [latex]a=-12[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164043\/CNX_Precalc_Figure_07_02_0042.jpg\" alt=\"Diagram of a triangle in the x,y plane. The vertices are at the origin, (-5,0), and (-5, -12). The angle at the origin is Beta degrees. The angle formed by the x axis and the side from (-5, -12) to (-5,0) is a right angle. The side opposite the right angle has length 13.\" width=\"487\" height=\"568\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\nThe next step is finding the cosine of [latex]\\alpha [\/latex] and the sine of [latex]\\beta [\/latex]. The cosine of [latex]\\alpha [\/latex] is the adjacent side over the hypotenuse. We can find it from the triangle in Figure 5:\u00a0[latex]\\cos \\alpha =\\frac{4}{5}[\/latex]. We can also find the sine of [latex]\\beta [\/latex] from the triangle in Figure 5, as opposite side over the hypotenuse: [latex]\\sin \\beta =-\\frac{12}{13}[\/latex]. Now we are ready to evaluate [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex].\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\alpha +\\beta \\right)&amp;=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\\\ &amp;=\\left(\\frac{3}{5}\\right)\\left(-\\frac{5}{13}\\right)+\\left(\\frac{4}{5}\\right)\\left(-\\frac{12}{13}\\right) \\\\ &amp;=-\\frac{15}{65}-\\frac{48}{65} \\\\ &amp;=-\\frac{63}{65} \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>We can find [latex]\\cos \\left(\\alpha +\\beta \\right)[\/latex] in a similar manner. We substitute the values according to the formula.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\alpha +\\beta \\right)&amp;=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\\\ &amp;=\\left(\\frac{4}{5}\\right)\\left(-\\frac{5}{13}\\right)-\\left(\\frac{3}{5}\\right)\\left(-\\frac{12}{13}\\right) \\\\ &amp;=-\\frac{20}{65}+\\frac{36}{65} \\\\ &amp;=\\frac{16}{65} \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>For [latex]\\tan \\left(\\alpha +\\beta \\right)[\/latex], if [latex]\\sin \\alpha =\\frac{3}{5}[\/latex] and [latex]\\cos \\alpha =\\frac{4}{5}[\/latex], then\r\n<div style=\"text-align: center\">[latex]\\tan \\alpha =\\frac{\\frac{3}{5}}{\\frac{4}{5}}=\\frac{3}{4}[\/latex]<\/div>\r\nIf [latex]\\sin \\beta =-\\frac{12}{13}[\/latex] and [latex]\\cos \\beta =-\\frac{5}{13}[\/latex],\r\nthen\r\n<div style=\"text-align: center\">[latex]\\tan \\beta =\\frac{\\frac{-12}{13}}{\\frac{-5}{13}}=\\frac{12}{5}[\/latex]<\/div>\r\nThen,\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&amp;=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta } \\\\ &amp;=\\frac{\\frac{3}{4}+\\frac{12}{5}}{1-\\frac{3}{4}\\left(\\frac{12}{5}\\right)} \\\\ &amp;=\\frac{\\text{ }\\frac{63}{20}}{-\\frac{16}{20}} \\\\ &amp;=-\\frac{63}{16} \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>To find [latex]\\tan \\left(\\alpha -\\beta \\right)[\/latex], we have the values we need. We can substitute them in and evaluate.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha -\\beta \\right)&amp;=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta } \\\\ &amp;=\\frac{\\frac{3}{4}-\\frac{12}{5}}{1+\\frac{3}{4}\\left(\\frac{12}{5}\\right)} \\\\ &amp;=\\frac{-\\frac{33}{20}}{\\frac{56}{20}} \\\\ &amp;=-\\frac{33}{56}\\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<h4>Analysis of the Solution<\/h4>\r\nA common mistake when addressing problems such as this one is that we may be tempted to think that [latex]\\alpha [\/latex] and [latex]\\beta [\/latex] are angles in the same triangle, which of course, they are not. Also note that\r\n<div style=\"text-align: center\">[latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\sin \\left(\\alpha +\\beta \\right)}{\\cos \\left(\\alpha +\\beta \\right)}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<div style=\"text-align: left\"><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600;text-align: left\">Use sum and difference formulas for cofunctions<\/span><\/div>\r\n<\/div>\r\nNow that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall that if the sum of two positive angles is [latex]\\frac{\\pi }{2}[\/latex], those two angles are complements, and the sum of the two acute angles in a right triangle is [latex]\\frac{\\pi }{2}[\/latex], so they are also complements. In Figure 6, notice that if one of the acute angles is labeled as [latex]\\theta [\/latex], then the other acute angle must be labeled [latex]\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164045\/CNX_Precalc_Figure_07_02_0072.jpg\" alt=\"Image of a right triangle. The remaining angles are labeled theta and pi\/2 - theta.\" width=\"487\" height=\"268\" \/> <b>Figure 8.<\/b> From these relationships, the cofunction identities are formed.[\/caption]\r\n\r\nNotice also that [latex]\\sin \\theta =\\cos \\left(\\frac{\\pi }{2}-\\theta \\right):[\/latex] opposite over hypotenuse. Thus, when two angles are complimentary, we can say that the sine of [latex]\\theta [\/latex] equals the <strong>cofunction<\/strong> of the complement of [latex]\\theta [\/latex]. Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions.<span id=\"fs-id2872065\">\r\n<\/span>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Cofunction Identities<\/h3>\r\nThe cofunction identities are summarized in the table below.\r\n<table id=\"Table_07_02_02\" summary=\"Three rows, two columns\/ The table has ordered pairs of these row values: (sin(theta) = cos(pi\/2 - theta), cos(theta) = sin(pi\/2 - theta)), (tan(theta) = cot(pi\/2 - theta), cot(theta) = tan(pi\/2 - theta)), and (sec(theta) = csc(pi\/2 - theta), csc(theta) = sec(pi\/2 - theta)).\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\sin \\theta =\\cos \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\r\n<td>[latex]\\cos \\theta =\\sin \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\tan \\theta =\\cot \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\r\n<td>[latex]\\cot \\theta =\\tan \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\sec \\theta =\\csc \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\r\n<td>[latex]\\csc \\theta =\\sec \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nNotice that the formulas in the table may also justified algebraically using the sum and difference formulas. For example, using\r\n<div style=\"text-align: center\">[latex]\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta [\/latex],<\/div>\r\nwe can write\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\frac{\\pi }{2}-\\theta \\right)&amp;=\\cos \\frac{\\pi }{2}\\cos \\theta +\\sin \\frac{\\pi }{2}\\sin \\theta \\\\ &amp;=\\left(0\\right)\\cos \\theta +\\left(1\\right)\\sin \\theta \\\\ &amp;=\\sin \\theta\\end{align}[\/latex]<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 17: Finding a Cofunction with the Same Value as the Given Expression<\/h3>\r\nWrite [latex]\\tan \\frac{\\pi }{9}[\/latex] in terms of its cofunction.\r\n\r\n[reveal-answer q=\"406106\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"406106\"]\r\n\r\nThe cofunction of [latex]\\tan \\theta =\\cot \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]. Thus,\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\frac{\\pi }{9}\\right)&amp;=\\cot \\left(\\frac{\\pi }{2}-\\frac{\\pi }{9}\\right) \\\\&amp; =\\cot \\left(\\frac{9\\pi }{18}-\\frac{2\\pi }{18}\\right) \\\\&amp; =\\cot \\left(\\frac{7\\pi }{18}\\right)\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite [latex]\\sin \\frac{\\pi }{7}[\/latex] in terms of its cofunction.\r\n\r\n[reveal-answer q=\"615431\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"615431\"]\r\n\r\n[latex]\\cos \\left(\\frac{5\\pi }{14}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173455[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use sum and difference formulas to verify identities<\/h2>\r\nVerifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an identity, verify using sum and difference formulas.<\/h3>\r\n<ol>\r\n \t<li>Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.<\/li>\r\n \t<li>Look for opportunities to use the sum and difference formulas.<\/li>\r\n \t<li>Rewrite sums or differences of quotients as single quotients.<\/li>\r\n \t<li>If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div>\r\n<div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 18: Verifying an Identity Involving Sine<\/h3>\r\nVerify the identity [latex]\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)=2\\sin \\alpha \\cos \\beta [\/latex].\r\n\r\n[reveal-answer q=\"464113\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"464113\"]\r\n\r\nWe see that the left side of the equation includes the sines of the sum and the difference of angles.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta\\\\ \\sin \\left(\\alpha -\\beta \\right)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\end{gathered}[\/latex]<\/p>\r\nWe can rewrite each using the sum and difference formulas.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)&amp;=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta +\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\\\ &amp;=2\\sin \\alpha \\cos \\beta \\end{align}[\/latex]<\/p>\r\nWe see that the identity is verified.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 19: Verifying an Identity Involving Tangent<\/h3>\r\nVerify the following identity.\r\n<p style=\"text-align: center\">[latex]\\frac{\\sin \\left(\\alpha -\\beta \\right)}{\\cos \\alpha \\cos \\beta }=\\tan \\alpha -\\tan \\beta [\/latex]<\/p>\r\n[reveal-answer q=\"605610\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"605610\"]\r\n\r\nWe can begin by rewriting the numerator on the left side of the equation.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\sin \\left(\\alpha -\\beta \\right)}{\\cos \\alpha \\cos \\beta }&amp;=\\frac{\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta } \\\\ &amp;=\\frac{\\sin \\alpha \\cos \\beta}{\\cos \\alpha \\cos \\beta}-\\frac{\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta } &amp;&amp; \\text{Rewrite using a common denominator}. \\\\ &amp;=\\frac{\\sin \\alpha }{\\cos \\alpha }-\\frac{\\sin \\beta }{\\cos \\beta }&amp;&amp; \\text{Cancel}. \\\\ &amp;=\\tan \\alpha -\\tan \\beta &amp;&amp; \\text{Rewrite in terms of tangent}.\\end{align}[\/latex]<\/p>\r\nWe see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nVerify the identity: [latex]\\tan \\left(\\pi -\\theta \\right)=-\\tan \\theta [\/latex].\r\n\r\n[reveal-answer q=\"184004\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"184004\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\pi -\\theta \\right)&amp;=\\frac{\\tan \\left(\\pi \\right)-\\tan \\theta }{1+\\tan \\left(\\pi \\right)\\tan \\theta } \\\\ &amp;=\\frac{0-\\tan \\theta }{1+0\\cdot \\tan \\theta } \\\\ &amp;=-\\tan \\theta \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 20: Using Sum and Difference Formulas to Solve an Application Problem<\/h3>\r\nLet [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex] denote two non-vertical intersecting lines, and let [latex]\\theta [\/latex] denote the acute angle between [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex]. Show that\r\n<p style=\"text-align: center\">[latex]\\tan \\theta =\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}[\/latex]<\/p>\r\nwhere [latex]{m}_{1}[\/latex] and [latex]{m}_{2}[\/latex] are the slopes of [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex] respectively. (<strong>Hint:<\/strong> Use the fact that [latex]\\tan {\\theta }_{1}={m}_{1}[\/latex] and [latex]\\tan {\\theta }_{2}={m}_{2}[\/latex]. )\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164047\/CNX_Precalc_Figure_07_02_0052.jpg\" alt=\"Diagram of two non-vertical intersecting lines L1 and L2 also intersecting the x-axis. The acute angle formed by the intersection of L1 and L2 is theta. The acute angle formed by L2 and the x-axis is theta 1, and the acute angle formed by the x-axis and L1 is theta 2. \" width=\"487\" height=\"289\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"429859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"429859\"]\r\n\r\nUsing the difference formula for tangent, this problem does not seem as daunting as it might.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\theta &amp;=\\tan \\left({\\theta }_{2}-{\\theta }_{1}\\right) \\\\ &amp;=\\frac{\\tan {\\theta }_{2}-\\tan {\\theta }_{1}}{1+\\tan {\\theta }_{1}\\tan {\\theta }_{2}} \\\\ &amp;=\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div><section id=\"fs-id2056229\" class=\"key-equations\">\r\n<h2>Using Double-Angle Formulas to Find Exact Values<\/h2>\r\nIn the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The <strong>double-angle formulas<\/strong> are a special case of the sum formulas, where [latex]\\alpha =\\beta [\/latex]. Deriving the double-angle formula for sine begins with the sum formula,\r\n<div style=\"text-align: center\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta [\/latex]<\/div>\r\nIf we let [latex]\\alpha =\\beta =\\theta [\/latex], then we have\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\theta +\\theta \\right)&amp;=\\sin \\theta \\cos \\theta +\\cos \\theta \\sin \\theta \\\\ \\sin \\left(2\\theta \\right)&amp;=2\\sin \\theta \\cos \\theta \\end{align}[\/latex]<\/div>\r\nDeriving the double-angle for cosine gives us three options. First, starting from the sum formula, [latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta [\/latex], and letting [latex]\\alpha =\\beta =\\theta [\/latex], we have\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\theta +\\theta \\right)&amp;=\\cos \\theta \\cos \\theta -\\sin \\theta \\sin \\theta \\\\ \\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\end{align}[\/latex]<\/div>\r\nUsing the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;=\\left(1-{\\sin }^{2}\\theta \\right)-{\\sin }^{2}\\theta \\\\ &amp;=1 - 2{\\sin }^{2}\\theta\\end{align}[\/latex]<\/div>\r\nThe second interpretation is:\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;={\\cos }^{2}\\theta -\\left(1-{\\cos }^{2}\\theta \\right) \\\\ &amp;=2{\\cos }^{2}\\theta -1\\end{align}[\/latex]<\/div>\r\nSimilarly, to derive the double-angle formula for tangent, replacing [latex]\\alpha =\\beta =\\theta [\/latex] in the sum formula gives\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&amp;=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\theta +\\theta \\right)&amp;=\\frac{\\tan \\theta +\\tan \\theta }{1-\\tan \\theta \\tan \\theta }\\\\ \\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Double-Angle Formulas<\/h3>\r\nThe <strong>double-angle formulas<\/strong> are summarized as follows:\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&amp;=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;=1 - 2{\\sin }^{2}\\theta \\\\ &amp;=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Draw a triangle to reflect the given information.<\/li>\r\n \t<li>Determine the correct double-angle formula.<\/li>\r\n \t<li>Substitute values into the formula based on the triangle.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 21: Using a Double-Angle Formula to Find the Exact Value Involving Tangent<\/h3>\r\nGiven that [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex] and [latex]\\theta [\/latex] is in quadrant II, find the following:\r\n<ol>\r\n \t<li>[latex]\\sin \\left(2\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\left(2\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\tan \\left(2\\theta \\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"477962\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"477962\"]\r\n\r\nIf we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\\tan \\theta =-\\frac{3}{4}[\/latex], such that [latex]\\theta [\/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\\theta [\/latex] is in the second quadrant, the adjacent side is on the <em>x<\/em>-axis and is negative. Use the <strong>Pythagorean Theorem<\/strong> to find the length of the hypotenuse:\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\left(-4\\right)}^{2}+{\\left(3\\right)}^{2}&amp;={c}^{2}\\\\ 16+9&amp;={c}^{2}\\\\ 25&amp;={c}^{2}\\\\ c&amp;=5\\end{align}[\/latex]<\/p>\r\nNow we can draw a triangle similar to the one shown in Figure 2.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164101\/CNX_Precalc_Figure_07_03_0022.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.\" width=\"487\" height=\"251\" \/> <b>Figure 10<\/b>[\/caption]\r\n<ol>\r\n \t<li>Let\u2019s begin by writing the double-angle formula for sine.\r\n<div style=\"text-align: center\">[latex]\\sin \\left(2\\theta \\right)=2\\sin \\theta \\cos \\theta [\/latex]<\/div>\r\nWe see that we to need to find [latex]\\sin \\theta [\/latex] and [latex]\\cos \\theta [\/latex]. Based on Figure 2, we see that the hypotenuse equals 5, so [latex]\\sin \\theta =\\frac{3}{5}[\/latex], and [latex]\\cos \\theta =\u2212\\frac{4}{5}[\/latex]. Substitute these values into the equation, and simplify.\r\nThus,\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&amp;=2\\left(\\frac{3}{5}\\right)\\left(\u2212\\frac{4}{5}\\right) \\\\ &amp;=\u2212\\frac{24}{25} \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Write the double-angle formula for cosine.\r\n<div style=\"text-align: center\">[latex]\\cos \\left(2\\theta \\right)={\\cos }^{2}\\theta -{\\sin }^{2}\\theta [\/latex]<\/div>\r\nAgain, substitute the values of the sine and cosine into the equation, and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\left(\u2212\\frac{4}{5}\\right)}^{2}\u2212{\\left(\\frac{3}{5}\\right)}^{2} \\\\ &amp;=\\frac{16}{25}\u2212\\frac{9}{25} \\\\ &amp;=\\frac{7}{25}\\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Write the double-angle formula for tangent.\r\n<div style=\"text-align: center\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2\\tan \\theta }{1\u2212{\\tan }^{2}\\theta }[\/latex]<\/div>\r\nIn this formula, we need the tangent, which we were given as [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex]. Substitute this value into the equation, and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&amp;=\\frac{2\\left(-\\frac{3}{4}\\right)}{1-{\\left(-\\frac{3}{4}\\right)}^{2}} \\\\ &amp;=\\frac{-\\frac{3}{2}}{1-\\frac{9}{16}} \\\\ &amp;=-\\frac{3}{2}\\left(\\frac{16}{7}\\right) \\\\ &amp;=-\\frac{24}{7} \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]\\sin \\alpha =\\frac{5}{8}[\/latex], with [latex]\\theta [\/latex] in quadrant I, find [latex]\\cos \\left(2\\alpha \\right)[\/latex].\r\n\r\n[reveal-answer q=\"827677\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"827677\"]\r\n\r\n[latex]\\cos \\left(2\\alpha \\right)=\\frac{7}{32}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]156232[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 22: Using the Double-Angle Formula for Cosine without Exact Values<\/h3>\r\nUse the double-angle formula for cosine to write [latex]\\cos \\left(6x\\right)[\/latex] in terms of [latex]\\cos \\left(3x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"676542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"676542\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(6x\\right)&amp;=\\cos \\left(3x+3x\\right) \\\\ &amp;=\\cos 3x\\cos 3x-\\sin 3x\\sin 3x \\\\ &amp;={\\cos }^{2}3x-{\\sin }^{2}3x \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using Double-Angle Formulas to Verify Identities<\/h2>\r\nEstablishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 23: Using the Double-Angle Formulas to Establish an Identity<\/h3>\r\nEstablish the following identity using double-angle formulas:\r\n<p style=\"text-align: center\">[latex]1+\\sin \\left(2\\theta \\right)={\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}[\/latex]<\/p>\r\n[reveal-answer q=\"600157\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"600157\"]\r\n\r\nWe will work on the right side of the equal sign and rewrite the expression until it matches the left side.\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}&amp;={\\sin }^{2}\\theta +2\\sin \\theta \\cos \\theta +{\\cos }^{2}\\theta \\\\ &amp;=\\left({\\sin }^{2}\\theta +{\\cos }^{2}\\theta \\right)+2\\sin \\theta \\cos \\theta \\\\ &amp;=1+2\\sin \\theta \\cos \\theta \\\\ &amp;=1+\\sin \\left(2\\theta \\right)\\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis process is not complicated, as long as we recall the perfect square formula from algebra:\r\n<p style=\"text-align: center\">[latex]{\\left(a\\pm b\\right)}^{2}={a}^{2}\\pm 2ab+{b}^{2}[\/latex]<\/p>\r\nwhere [latex]a=\\sin \\theta [\/latex] and [latex]b=\\cos \\theta [\/latex]. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEstablish the identity: [latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\cos \\left(2\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"178499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178499\"]\r\n\r\n[latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)=\\cos \\left(2\\theta \\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 24: Verifying a Double-Angle Identity for Tangent<\/h3>\r\nVerify the identity:\r\n<p style=\"text-align: center\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\r\n[reveal-answer q=\"395376\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"395376\"]\r\n\r\nIn this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }&amp;&amp; \\text{Double-angle formula} \\\\ &amp;=\\frac{2\\tan \\theta \\left(\\frac{1}{\\tan \\theta }\\right)}{\\left(1-{\\tan }^{2}\\theta \\right)\\left(\\frac{1}{\\tan \\theta }\\right)}&amp;&amp; \\text{Multiply by a term that results in desired numerator}. \\\\ &amp;=\\frac{2}{\\frac{1}{\\tan \\theta }-\\frac{{\\tan }^{2}\\theta }{\\tan \\theta }} \\\\ &amp;=\\frac{2}{\\cot \\theta -\\tan \\theta }&amp;&amp; \\text{Use reciprocal identity for }\\frac{1}{\\tan \\theta }.\\end{align}[\/latex]<\/p>\r\n\r\n<h3>Analysis of the Solution<\/h3>\r\nHere is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show\r\n<p style=\"text-align: center\">[latex]\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\r\nLet\u2019s work on the right side.\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{2}{\\cot \\theta -\\tan \\theta }&amp;=\\frac{2}{\\frac{1}{\\tan \\theta }-\\tan \\theta }\\left(\\frac{\\tan \\theta }{\\tan \\theta }\\right) \\\\ &amp;=\\frac{2\\tan \\theta }{\\frac{1}{\\cancel{\\tan \\theta }}\\left(\\cancel{\\tan \\theta }\\right)-\\tan \\theta \\left(\\tan \\theta \\right)} \\\\ &amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta } \\end{align}[\/latex]<\/p>\r\nWhen using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nVerify the identity: [latex]\\cos \\left(2\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta [\/latex].\r\n\r\n[reveal-answer q=\"584630\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"584630\"]\r\n\r\n[latex]\\cos \\left(2\\theta \\right)\\cos \\theta =\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use Reduction Formulas to Simplify an Expression<\/h2>\r\nThe double-angle formulas can be used to derive the <strong>reduction formulas<\/strong>, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.\r\n\r\nWe can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let\u2019s begin with [latex]\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta [\/latex]. Solve for [latex]{\\sin }^{2}\\theta :[\/latex]\r\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta \\\\ 2{\\sin }^{2}\\theta =1-\\cos \\left(2\\theta \\right) \\\\ {\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\end{gathered}[\/latex]<\/div>\r\nNext, we use the formula [latex]\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1[\/latex]. Solve for [latex]{\\cos }^{2}\\theta :[\/latex]\r\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1 \\\\ 1+\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta \\\\ \\frac{1+\\cos \\left(2\\theta \\right)}{2}={\\cos }^{2}\\theta \\end{gathered}[\/latex]<\/div>\r\nThe last reduction formula is derived by writing tangent in terms of sine and cosine:\r\n<div style=\"text-align: center\">[latex]\\begin{align}{\\tan }^{2}\\theta &amp;=\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &amp;=\\frac{\\frac{1-\\cos \\left(2\\theta \\right)}{2}}{\\frac{1+\\cos \\left(2\\theta \\right)}{2}}&amp;&amp; \\text{Substitute the reduction formulas.} \\\\ &amp;=\\left(\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\right)\\left(\\frac{2}{1+\\cos \\left(2\\theta \\right)}\\right) \\\\ &amp;=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Reduction Formulas<\/h3>\r\nThe <strong>reduction formulas<\/strong> are summarized as follows:\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &amp;{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &amp;{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 25: Writing an Equivalent Expression Not Containing Powers Greater Than 1<\/h3>\r\nWrite an equivalent expression for [latex]{\\cos }^{4}x[\/latex] that does not involve any powers of sine or cosine greater than 1.\r\n\r\n[reveal-answer q=\"109691\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"109691\"]\r\n\r\nWe will apply the reduction formula for cosine twice.\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\cos }^{4}x&amp;={\\left({\\cos }^{2}x\\right)}^{2} \\\\ &amp;={\\left(\\frac{1+\\cos \\left(2x\\right)}{2}\\right)}^{2}&amp;&amp; {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &amp;=\\frac{1}{4}\\left(1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right) \\\\ &amp;=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{4}\\left(\\frac{1+\\cos \\left(2\\left(2x\\right)\\right)}{2}\\right) &amp;&amp; {\\text{ Substitute reduction formula for cos}}^{2}x. \\\\ &amp;=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}+\\frac{1}{8}\\cos \\left(4x\\right) \\\\ &amp;=\\frac{3}{8}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThe solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 26: Using the Power-Reducing Formulas to Prove an Identity<\/h3>\r\nUse the power-reducing formulas to prove\r\n<p style=\"text-align: center\">[latex]{\\sin }^{3}\\left(2x\\right)=\\left[\\frac{1}{2}\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right][\/latex]<\/p>\r\n[reveal-answer q=\"828553\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"828553\"]\r\n\r\nWe will work on simplifying the left side of the equation:\r\n<p style=\"text-align: center\">[latex]\\begin{align}{\\sin }^{3}\\left(2x\\right)&amp;=\\left[\\sin \\left(2x\\right)\\right]\\left[{\\sin }^{2}\\left(2x\\right)\\right] \\\\ &amp;=\\sin \\left(2x\\right)\\left[\\frac{1-\\cos \\left(4x\\right)}{2}\\right]&amp;&amp; \\text{Substitute the power-reduction formula}. \\\\ &amp;=\\sin \\left(2x\\right)\\left(\\frac{1}{2}\\right)\\left[1-\\cos \\left(4x\\right)\\right] \\\\ &amp;=\\frac{1}{2}\\left[\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right] \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that in this example, we substituted\r\n<div style=\"text-align: center\">[latex]\\frac{1-\\cos \\left(4x\\right)}{2}[\/latex]<\/div>\r\nfor [latex]{\\sin }^{2}\\left(2x\\right)[\/latex]. The formula states\r\n<div style=\"text-align: center\">[latex]{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2}[\/latex]<\/div>\r\nWe let [latex]\\theta =2x[\/latex], so [latex]2\\theta =4x[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nUse the power-reducing formulas to prove that [latex]10{\\cos }^{4}x=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"387102\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"387102\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align}10{\\cos }^{4}x&amp;=10{\\left({\\cos }^{2}x\\right)}^{2} \\\\ &amp;=10{\\left[\\frac{1+\\cos \\left(2x\\right)}{2}\\right]}^{2}&amp;&amp; {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &amp;=\\frac{10}{4}\\left[1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right] \\\\ &amp;=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{4}\\left(\\frac{1+\\cos\\left( 2\\left(2x\\right)\\right)}{2}\\right)&amp;&amp; {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &amp;=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{8}+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &amp;=\\frac{30}{8}+5\\cos \\left(2x\\right)+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &amp;=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]4533[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using Half-Angle Formulas to Find Exact Values<\/h2>\r\nThe next set of identities is the set of <strong>half-angle formulas<\/strong>, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace [latex]\\theta [\/latex] with [latex]\\frac{\\alpha }{2}[\/latex], the half-angle formula for sine is found by simplifying the equation and solving for [latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]. Note that the half-angle formulas are preceded by a [latex]\\pm [\/latex] sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which [latex]\\frac{\\alpha }{2}[\/latex] terminates.\r\n\r\nThe half-angle formula for sine is derived as follows:\r\n<div style=\"text-align: center\">[latex]\\begin{align}{\\sin }^{2}\\theta &amp;=\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\\\ {\\sin }^{2}\\left(\\frac{\\alpha }{2}\\right)&amp;=\\frac{1-\\left(\\cos 2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &amp;=\\frac{1-\\cos \\alpha }{2} \\\\ \\sin \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\r\nTo derive the half-angle formula for cosine, we have\r\n<div style=\"text-align: center\">[latex]\\begin{align}{\\cos }^{2}\\theta &amp;=\\frac{1+\\cos \\left(2\\theta \\right)}{2}\\\\ {\\cos }^{2}\\left(\\frac{\\alpha }{2}\\right)&amp;=\\frac{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &amp;=\\frac{1+\\cos \\alpha }{2} \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\r\nFor the tangent identity, we have\r\n<div style=\"text-align: center\">[latex]\\begin{align}{\\tan }^{2}\\theta &amp;=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\\\ {\\tan }^{2}\\left(\\frac{\\alpha }{2}\\right)&amp;=\\frac{1-\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)} \\\\ &amp;=\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }\\hfill \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\end{align}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Half-Angle Formulas<\/h3>\r\nThe <strong>half-angle formulas<\/strong> are as follows:\r\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &amp;=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &amp;=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 27: Using a Half-Angle Formula to Find the Exact Value of a Sine Function<\/h3>\r\nFind [latex]\\sin \\left({15}^{\\circ }\\right)[\/latex] using a half-angle formula.\r\n\r\n[reveal-answer q=\"283155\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"283155\"]\r\n\r\nSince [latex]{15}^{\\circ }=\\frac{{30}^{\\circ }}{2}[\/latex], we use the half-angle formula for sine:\r\n<p style=\"text-align: center\">[latex]\\begin{align} \\sin \\frac{{30}^{\\circ }}{2}&amp;=\\sqrt{\\frac{1-\\cos {30}^{\\circ }}{2}} \\\\ &amp;=\\sqrt{\\frac{1-\\frac{\\sqrt{3}}{2}}{2}} \\\\ &amp;=\\sqrt{\\frac{\\frac{2-\\sqrt{3}}{2}}{2}} \\\\ &amp;=\\sqrt{\\frac{2-\\sqrt{3}}{4}} \\\\ &amp;=\\frac{\\sqrt{2-\\sqrt{3}}}{2} \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice that we used only the positive root because [latex]\\sin \\left({15}^{\\text{o}}\\right)[\/latex] is positive.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.<\/h3>\r\n<ol>\r\n \t<li>Draw a triangle to represent the given information.<\/li>\r\n \t<li>Determine the correct half-angle formula.<\/li>\r\n \t<li>Substitute values into the formula based on the triangle.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 28: Finding Exact Values Using Half-Angle Identities<\/h3>\r\nGiven that [latex]\\tan \\alpha =\\frac{8}{15}[\/latex] and [latex]\\alpha [\/latex] lies in quadrant III, find the exact value of the following:\r\n<ol>\r\n \t<li>[latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\tan \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"751659\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"751659\"]\r\n\r\nUsing the given information, we can draw the triangle shown in Figure 3. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate [latex]\\sin \\alpha =-\\frac{8}{17}[\/latex] and [latex]\\cos \\alpha =-\\frac{15}{17}[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164103\/CNX_Precalc_Figure_07_03_0032.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.\" width=\"487\" height=\"289\" \/> <b>Figure 11<\/b>[\/caption]\r\n<ol>\r\n \t<li>Before we start, we must remember that, if [latex]\\alpha [\/latex] is in quadrant III, then [latex]180^\\circ &lt;\\alpha &lt;270^\\circ [\/latex], so [latex]\\frac{180^\\circ }{2}&lt;\\frac{\\alpha }{2}&lt;\\frac{270^\\circ }{2}[\/latex]. This means that the terminal side of [latex]\\frac{\\alpha }{2}[\/latex] is in quadrant II, since [latex]90^\\circ &lt;\\frac{\\alpha }{2}&lt;135^\\circ [\/latex].To find [latex]\\sin \\frac{\\alpha }{2}[\/latex], we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure 3\u00a0and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align} \\sin \\frac{\\alpha }{2}&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{\\frac{32}{17}}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{32}{17}\\cdot \\frac{1}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{16}{17}} \\\\ &amp;=\\pm \\frac{4}{\\sqrt{17}} \\\\ &amp;=\\frac{4\\sqrt{17}}{17} \\end{align}[\/latex]<\/div>\r\nWe choose the positive value of [latex]\\sin \\frac{\\alpha }{2}[\/latex] because the angle terminates in quadrant II and sine is positive in quadrant II.<\/li>\r\n \t<li>To find [latex]\\cos \\frac{\\alpha }{2}[\/latex], we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in Figure 3, and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align} \\cos \\frac{\\alpha }{2}&amp;=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{1+\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{\\frac{2}{17}}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{2}{17}\\cdot \\frac{1}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{1}{17}} \\\\ &amp;=-\\frac{\\sqrt{17}}{17}\\end{align}[\/latex]<\/div>\r\nWe choose the negative value of [latex]\\cos \\frac{\\alpha }{2}[\/latex] because the angle is in quadrant II because cosine is negative in quadrant II.<\/li>\r\n \t<li>To find [latex]\\tan \\frac{\\alpha }{2}[\/latex], we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in Figure 3\u00a0and simplify.\r\n<div style=\"text-align: center\">[latex]\\begin{align} \\tan \\frac{\\alpha }{2}&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &amp;=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{1+\\left(-\\frac{15}{17}\\right)}} \\\\ &amp;=\\pm \\sqrt{\\frac{\\frac{32}{17}}{\\frac{2}{17}}} \\\\ &amp;=\\pm \\sqrt{\\frac{32}{2}} \\\\ &amp;=-\\sqrt{16} \\\\ &amp;=-4 \\end{align}[\/latex]<\/div>\r\nWe choose the negative value of [latex]\\tan \\frac{\\alpha }{2}[\/latex] because [latex]\\frac{\\alpha }{2}[\/latex] lies in quadrant II, and tangent is negative in quadrant II.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven that [latex]\\sin \\alpha =-\\frac{4}{5}[\/latex] and [latex]\\alpha [\/latex] lies in quadrant IV, find the exact value of [latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"406030\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"406030\"]\r\n\r\n[latex]-\\frac{2}{\\sqrt{5}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173569[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 29: Finding the Measurement of a Half Angle<\/h3>\r\nNow, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of [latex]\\theta [\/latex] formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for higher-level competition, what is the measurement of the angle for novice competition?\r\n\r\n[reveal-answer q=\"685675\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"685675\"]\r\n\r\nSince the angle for novice competition measures half the steepness of the angle for the high level competition, and [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for high competition, we can find [latex]\\cos \\theta [\/latex] from the right triangle and the Pythagorean theorem so that we can use the half-angle identities.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}{3}^{2}+{5}^{2}=34 \\\\ c=\\sqrt{34} \\end{gathered}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164106\/CNX_Precalc_Figure_07_03_0042.jpg\" alt=\"Image of a right triangle with sides 3, 5, and rad34. Rad 34 is the hypotenuse, and 3 is the base. The angle formed by the hypotenuse and base is theta. The angle between the side of length 3 and side of length 5 is a right angle.\" width=\"487\" height=\"210\" \/> <b>Figure 12<\/b>[\/caption]\r\n\r\nWe see that [latex]\\cos \\theta =\\frac{3}{\\sqrt{34}}=\\frac{3\\sqrt{34}}{34}[\/latex]. We can use the half-angle formula for tangent: [latex]\\tan \\frac{\\theta }{2}=\\sqrt{\\frac{1-\\cos \\theta }{1+\\cos \\theta }}[\/latex]. Since [latex]\\tan \\theta [\/latex] is in the first quadrant, so is [latex]\\tan \\frac{\\theta }{2}[\/latex]. Thus,\r\n<p style=\"text-align: center\">[latex]\\begin{align} \\tan \\frac{\\theta }{2}&amp;=\\sqrt{\\frac{1-\\frac{3\\sqrt{34}}{34}}{1+\\frac{3\\sqrt{34}}{34}}} \\\\ &amp;=\\sqrt{\\frac{\\frac{34 - 3\\sqrt{34}}{34}}{\\frac{34+3\\sqrt{34}}{34}}} \\\\ &amp;=\\sqrt{\\frac{34 - 3\\sqrt{34}}{34+3\\sqrt{34}}} \\\\ &amp;\\approx 0.57\\end{align}[\/latex]<\/p>\r\nWe can take the inverse tangent to find the angle: [latex]{\\tan }^{-1}\\left(0.57\\right)\\approx {29.7}^{\\circ }[\/latex]. So the angle of the ramp for novice competition is [latex]\\approx {29.7}^{\\circ }[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id2253485\" class=\"definition\"><\/dl>\r\n<dl id=\"fs-id2253499\" class=\"definition\">\r\n \t<dd id=\"fs-id2253503\"><\/dd>\r\n<\/dl>\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table id=\"fs-id2056235\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Pythagorean identities<\/td>\r\n<td>[latex]\\begin{gathered}{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1\\\\ 1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta \\\\ 1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta \\end{gathered}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Even-odd identities<\/td>\r\n<td>[latex]\\begin{gathered}\\tan \\left(-\\theta \\right)=-\\tan \\theta \\\\ \\cot \\left(-\\theta \\right)=-\\cot \\theta \\\\ \\sin \\left(-\\theta \\right)=-\\sin \\theta \\\\ \\csc \\left(-\\theta \\right)=-\\csc \\theta \\\\ \\cos \\left(-\\theta \\right)=\\cos \\theta \\\\ \\sec \\left(-\\theta \\right)=\\sec \\theta \\end{gathered}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Reciprocal identities<\/td>\r\n<td>[latex]\\begin{gathered}\\sin \\theta =\\frac{1}{\\csc \\theta }\\\\ \\cos \\theta =\\frac{1}{\\sec \\theta }\\\\ \\tan \\theta =\\frac{1}{\\cot \\theta }\\\\ \\csc \\theta =\\frac{1}{\\sin \\theta }\\\\ \\sec \\theta =\\frac{1}{\\cos \\theta }\\\\ \\cot \\theta =\\frac{1}{\\tan \\theta }\\end{gathered}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Quotient identities<\/td>\r\n<td>[latex]\\begin{gathered} \\tan \\theta =\\frac{\\sin \\theta }{\\cos \\theta }\\\\ \\cot \\theta =\\frac{\\cos \\theta }{\\sin \\theta } \\end{gathered}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section id=\"fs-id2081298\" class=\"key-concepts\">\r\n<table style=\"width: 743px\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 182.984px\">Sum Formula for Cosine<\/td>\r\n<td style=\"width: 533.984px\">[latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 182.984px\">Difference Formula for Cosine<\/td>\r\n<td style=\"width: 533.984px\">[latex]\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 182.984px\">Sum Formula for Sine<\/td>\r\n<td style=\"width: 533.984px\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 182.984px\">Difference Formula for Sine<\/td>\r\n<td style=\"width: 533.984px\">[latex]\\sin \\left(\\alpha -\\beta \\right)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 182.984px\">Sum Formula for Tangent<\/td>\r\n<td style=\"width: 533.984px\">[latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 182.984px\">Difference Formula for Tangent<\/td>\r\n<td style=\"width: 533.984px\">[latex]\\tan \\left(\\alpha -\\beta \\right)=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 182.984px\">Cofunction identities<\/td>\r\n<td style=\"width: 533.984px\">[latex]\\sin \\theta =\\cos \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]\r\n\r\n[latex]\\cos \\theta =\\sin \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]\r\n\r\n[latex]\\tan \\theta =\\cot \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]\r\n\r\n[latex]\\cot \\theta =\\tan \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]\r\n\r\n[latex]\\sec \\theta =\\csc \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]\r\n\r\n[latex]\\csc \\theta =\\sec \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table><colgroup> <col \/> <col \/> <\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong>Double-angle formulas<\/strong><\/td>\r\n<td>[latex]\\begin{align}\\sin \\left(2\\theta \\right)&amp;=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;=1 - 2{\\sin }^{2}\\theta \\\\ &amp;=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Reduction formulas<\/strong><\/td>\r\n<td>[latex]\\begin{align}&amp;{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &amp;{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &amp;{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Half-angle formulas<\/strong><\/td>\r\n<td>[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &amp;=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &amp;=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id2081305\">\r\n \t<li>There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.<\/li>\r\n \t<li>Graphing both sides of an identity will verify it.<\/li>\r\n \t<li>Simplifying one side of the equation to equal the other side is another method for verifying an identity.<\/li>\r\n \t<li>The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation.<\/li>\r\n \t<li>We can create an identity by simplifying an expression and then verifying it.<\/li>\r\n \t<li>Verifying an identity may involve algebra with the fundamental identities.<\/li>\r\n \t<li>Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics.<\/li>\r\n \t<li>The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles.<\/li>\r\n \t<li>The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle.<\/li>\r\n \t<li>The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle.<\/li>\r\n \t<li>The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions.<\/li>\r\n \t<li>The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles.<\/li>\r\n \t<li>The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles.<\/li>\r\n \t<li>The cofunction identities apply to complementary angles and pairs of reciprocal functions.<\/li>\r\n \t<li>Sum and difference formulas are useful in verifying identities.<\/li>\r\n \t<li>Application problems are often easier to solve by using sum and difference formulas.<\/li>\r\n \t<li>Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent.<\/li>\r\n \t<li>Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term.<\/li>\r\n \t<li>Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id2253485\" class=\"definition\">\r\n \t<dt>double-angle formulas<\/dt>\r\n \t<dd id=\"fs-id2253488\">identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal<\/dd>\r\n<\/dl>\r\n<strong>even-odd identities<\/strong>\r\n<dl id=\"fs-id1881971\" class=\"definition\">\r\n \t<dd id=\"fs-id1881976\">set of equations involving trigonometric functions such that if [latex]f\\left(-x\\right)=-f\\left(x\\right)[\/latex], the identity is odd, and if [latex]f\\left(-x\\right)=f\\left(x\\right)[\/latex], the identity is even<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id2253492\" class=\"definition\">\r\n \t<dt>half-angle formulas<\/dt>\r\n \t<dd id=\"fs-id2253495\">identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions<\/dd>\r\n<\/dl>\r\n<strong>Pythagorean identities<\/strong>\r\n<dl id=\"fs-id1882066\" class=\"definition\">\r\n \t<dd id=\"fs-id1882069\">set of equations involving trigonometric functions based on the right triangle properties<\/dd>\r\n<\/dl>\r\n<strong>quotient identities<\/strong>\r\n<dl id=\"fs-id1882073\" class=\"definition\">\r\n \t<dd id=\"fs-id1882076\">pair of identities based on the fact that tangent is the ratio of sine and cosine, and cotangent is the ratio of cosine and sine<\/dd>\r\n<\/dl>\r\n<strong>reciprocal identities<\/strong>\r\n<dl id=\"fs-id1882080\" class=\"definition\">\r\n \t<dd id=\"fs-id1882083\">set of equations involving the reciprocals of basic trigonometric definitions<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id2253499\" class=\"definition\">\r\n \t<dt>reduction formulas<\/dt>\r\n \t<dd id=\"fs-id2253503\">identities derived from the double-angle formulas and used to reduce the power of a trigonometric function<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1881971\" class=\"definition\">\r\n \t<dd id=\"fs-id1881976\"><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1882080\" class=\"definition\">\r\n \t<dd id=\"fs-id1882083\"><\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"font-weight: 400\">Verify the fundamental trigonometric identities.<\/li>\n<li style=\"font-weight: 400\">Simplify trigonometric expressions using algebra and the identities.<\/li>\n<li style=\"font-weight: 400\">Use sum and difference formulas for cosine.<\/li>\n<li style=\"font-weight: 400\">Use sum and difference formulas for sine.<\/li>\n<li style=\"font-weight: 400\">Use sum and difference formulas for tangent.<\/li>\n<li style=\"font-weight: 400\">Use sum and difference formulas for cofunctions.<\/li>\n<li style=\"font-weight: 400\">Use sum and difference formulas to verify identities.<\/li>\n<li style=\"font-weight: 400\">Use double-angle formulas to find exact values.<\/li>\n<li style=\"font-weight: 400\">Use double-angle formulas to verify identities.<\/li>\n<li style=\"font-weight: 400\">Use reduction formulas to simplify an expression.<\/li>\n<li style=\"font-weight: 400\">Use half-angle formulas to find exact values.<\/li>\n<\/ul>\n<\/div>\n<h2>Verify the fundamental trigonometric identities<\/h2>\n<p>Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.<\/p>\n<p>To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the <strong>Pythagorean identities<\/strong>, the even-odd identities, the reciprocal identities, and the quotient identities.<\/p>\n<p>We will begin with the <strong>Pythagorean identities<\/strong>, which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.<\/p>\n<table id=\"Table_07_01_01\" summary=\"&quot;Pythagorean Identities&quot; with three cells. First: sin(theta)^2 + cos(theta)^2 = 1. Second: 1 + cot(theta)^2 = csc(theta)^2. Third: 1 + tan(theta)^2 = sec(theta)^2.\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th style=\"text-align: center\" colspan=\"3\">Pythagorean Identities<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1[\/latex]<\/td>\n<td>[latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex]<\/td>\n<td>[latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The second and third identities can be obtained by manipulating the first. The identity [latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta\\[\/latex] is found by rewriting the left side of the equation in terms of sine and cosine.<\/p>\n<p>Prove: [latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta \/latex]  <\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}1+{\\cot }^{2}\\theta& =\\left(1+\\frac{{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)&& \\text{Rewrite the left side}. \\\\ &=\\left(\\frac{{\\sin }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)+\\left(\\frac{{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta }\\right)&& \\text{Write both terms with the common denominator}. \\\\ &=\\frac{{\\sin }^{2}\\theta +{\\cos }^{2}\\theta }{{\\sin }^{2}\\theta } \\\\ &=\\frac{1}{{\\sin }^{2}\\theta } \\\\ &={\\csc }^{2}\\theta \\end{align}[\/latex]<\/div>\n<p>Similarly, [latex]1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta[\/latex] can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}1+{\\tan }^{2}\\theta &=1+{\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)}^{2}&& \\text{Rewrite left side}. \\\\ &={\\left(\\frac{\\cos \\theta }{\\cos \\theta }\\right)}^{2}+{\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)}^{2}&& \\text{Write both terms with the common denominator}. \\\\ &=\\frac{{\\cos }^{2}\\theta +{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\frac{1}{{\\cos }^{2}\\theta } \\\\ &={\\sec }^{2}\\theta \\end{align}[\/latex]<\/div>\n<p>The next set of fundamental identities is the set of <strong>even-odd identities<\/strong>. The <strong>even-odd identities<\/strong> relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle and determine whether the identity is odd or even.<\/p>\n<table id=\"Table_07_01_02\" summary=\"&quot;Even-Odd Identities&quot; with three cells. First: tan(-theta) = -tan(theta) and cot(-theta) = -cot(theta). Second: sin(-theta) = -sin(theta) and csc(-theta) = -csc(theta). Third: cos(-theta) = cos(theta) and sec(-theta) = sec(theta).\">\n<thead>\n<tr>\n<th style=\"text-align: center\" colspan=\"3\">Even-Odd Identities<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\begin{gathered}\\tan \\left(-\\theta \\right)=-\\tan \\theta\\\\ \\cot \\left(-\\theta \\right)=-\\cot \\theta \\end{gathered}[\/latex]<\/td>\n<td>[latex]\\begin{gathered}\\sin \\left(-\\theta \\right)=-\\sin \\theta\\\\ \\csc \\left(-\\theta \\right)=-\\csc \\theta\\end{gathered}[\/latex]<\/td>\n<td>[latex]\\begin{gathered}\\cos \\left(-\\theta \\right)=\\cos \\theta \\\\ \\sec \\left(-\\theta \\right)=\\sec \\theta \\end{gathered}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Recall that an <strong>odd function<\/strong> is one in which [latex]f\\left(-x\\right)= -f\\left(x\\right)[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. The <strong>sine<\/strong> function is an odd function because [latex]\\sin \\left(-\\theta \\right)=-\\sin \\theta[\/latex]. The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of [latex]\\frac{\\pi }{2}\\\\[\/latex] and [latex]-\\frac{\\pi }{2}[\/latex]. The output of [latex]\\sin \\left(\\frac{\\pi }{2}\\right)[\/latex] is opposite the output of [latex]\\sin \\left(-\\frac{\\pi }{2}\\right)[\/latex]. Thus,<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\frac{\\pi }{2}\\right)=1\\end{align}[\/latex] and [latex]\\begin{align} \\sin \\left(-\\frac{\\pi }{2}\\right)=-\\sin \\left(\\frac{\\pi }{2}\\right) =-1 \\end{align}[\/latex]<\/div>\n<p>This is shown in\u00a0Figure 2.<\/p>\n<p><span id=\"fs-id1491023\"><br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164029\/CNX_Precalc_Figure_07_01_0022.jpg\" alt=\"Graph of y=sin(theta) from -2pi to 2pi, showing in particular that it is symmetric about the origin. Points given are (pi\/2, 1) and (-pi\/2, -1).\" \/><\/span><\/p>\n<p style=\"text-align: center\"><strong>Figure 2.<\/strong>\u00a0Graph of [latex]y=\\sin \\theta[\/latex]<\/p>\n<p>Recall that an <strong>even function<\/strong> is one in which<\/p>\n<div style=\"text-align: center\">[latex]f\\left(-x\\right)=f\\left(x\\right)[\/latex] for all <em>x<\/em>\u00a0in the domain of <em>f<\/em>.<\/div>\n<p>The graph of an even function is symmetric about the <em>y-<\/em>axis. The cosine function is an even function because [latex]\\cos \\left(-\\theta \\right)=\\cos \\theta[\/latex].<br \/>\nFor example, consider corresponding inputs [latex]\\frac{\\pi }{4}[\/latex] and [latex]-\\frac{\\pi }{4}[\/latex]. The output of [latex]\\cos \\left(\\frac{\\pi }{4}\\right)[\/latex] is the same as the output of [latex]\\cos \\left(-\\frac{\\pi }{4}\\right)[\/latex]. Thus,<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(-\\frac{\\pi }{4}\\right)=\\cos \\left(\\frac{\\pi }{4}\\right) \\approx 0.707 \\end{align}[\/latex]<\/div>\n<p>See Figure 3.<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164031\/CNX_Precalc_Figure_07_01_0032.jpg\" alt=\"Graph of y=cos(theta) from -2pi to 2pi, showing in particular that it is symmetric about the y-axis. Points given are (-pi\/4, .707) and (pi\/4, .707).\" \/><\/p>\n<p style=\"text-align: center\"><strong>Figure 3.<\/strong>\u00a0Graph of [latex]y=\\cos \\theta[\/latex]<\/p>\n<p>For all [latex]\\theta[\/latex] in the domain of the sine and cosine functions, respectively, we can state the following:<\/p>\n<ul id=\"fs-id2141576\">\n<li>Since [latex]\\sin \\left(-\\theta \\right)=-\\sin \\theta[\/latex], sine is an odd function.<\/li>\n<li>Since, [latex]\\cos \\left(-\\theta \\right)=\\cos \\theta[\/latex], cosine is an even function.<\/li>\n<\/ul>\n<p>The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity, [latex]\\tan \\left(-\\theta \\right)=\\mathrm{-tan}\\theta[\/latex]. We can interpret the tangent of a negative angle as [latex]\\tan \\left(-\\theta \\right)=\\frac{\\sin \\left(-\\theta \\right)}{\\cos \\left(-\\theta \\right)}=\\frac{-\\sin \\theta }{\\cos \\theta }=-\\tan \\theta[\/latex]. Tangent is therefore an odd function, which means that [latex]\\tan \\left(-\\theta \\right)=-\\tan \\left(\\theta \\right)[\/latex] for all [latex]\\theta[\/latex] in the domain of the <strong>tangent function<\/strong>.<\/p>\n<p>The cotangent identity, [latex]\\cot \\left(-\\theta \\right)=-\\cot \\theta[\/latex], also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as [latex]\\cot \\left(-\\theta \\right)=\\frac{\\cos \\left(-\\theta \\right)}{\\sin \\left(-\\theta \\right)}=\\frac{\\cos \\theta }{-\\sin \\theta }=-\\cot \\theta[\/latex]. Cotangent is therefore an odd function, which means that [latex]\\cot \\left(-\\theta \\right)=-\\cot \\left(\\theta \\right)[\/latex] for all [latex]\\theta[\/latex] in the domain of the <strong>cotangent function<\/strong>.<\/p>\n<p>The <strong>cosecant function<\/strong> is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as [latex]\\csc \\left(-\\theta \\right)=\\frac{1}{\\sin \\left(-\\theta \\right)}=\\frac{1}{-\\sin \\theta }=-\\csc \\theta[\/latex]. The cosecant function is therefore odd.<\/p>\n<p>Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as [latex]\\sec \\left(-\\theta \\right)=\\frac{1}{\\cos \\left(-\\theta \\right)}=\\frac{1}{\\cos \\theta }=\\sec \\theta[\/latex]. The secant function is therefore even.<\/p>\n<p>To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.<\/p>\n<p>The next set of fundamental identities is the set of <strong>reciprocal identities<\/strong>, which, as their name implies, relate trigonometric functions that are reciprocals of each other.<\/p>\n<table id=\"fs-id2031263\" summary=\"Table labeled &quot;Reciprocal Identities.&quot; Three rows, two columns. The table has ordered pairs of these row values: (sin(theta) = 1\/csc(theta), csc(theta) = 1\/sin(theta)), (cos(theta) = 1\/sec(theta), sec(theta) = 1\/cos(theta)), (tan(theta) = 1\/cot(theta), cot(theta) = 1\/tan(theta)).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th style=\"text-align: center\" colspan=\"2\">Reciprocal Identities<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\sin \\theta =\\frac{1}{\\csc \\theta }[\/latex]<\/td>\n<td>[latex]\\csc \\theta =\\frac{1}{\\sin \\theta }[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\cos \\theta =\\frac{1}{\\sec \\theta }[\/latex]<\/td>\n<td>[latex]\\sec \\theta =\\frac{1}{\\cos \\theta }[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\tan \\theta =\\frac{1}{\\cot \\theta }[\/latex]<\/td>\n<td>[latex]\\cot \\theta =\\frac{1}{\\tan \\theta }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The final set of identities is the set of <strong>quotient identities<\/strong>, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities.<\/p>\n<table id=\"fs-id937819\" summary=\"Table labeled &quot;Quotient Identities.&quot; First cell: tan(theta) = sin(theta) \/ cos(theta). Second cell: cot(theta) = cos(theta) \/ sin(theta).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th style=\"text-align: center\" colspan=\"2\">Quotient Identities<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\tan \\theta =\\frac{\\sin \\theta }{\\cos \\theta }[\/latex]<\/td>\n<td>[latex]\\cot \\theta =\\frac{\\cos \\theta }{\\sin \\theta }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Summarizing Trigonometric Identities<\/h3>\n<p>The <strong>Pythagorean identities<\/strong> are based on the properties of a right triangle.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} {\\cos}^{2}\\theta + {\\sin}^{2}\\theta=1 \\\\ 1+{\\tan}^{2}\\theta={\\sec}^{2}\\theta \\\\ 1+{\\cot}^{2}\\theta={\\csc}^{2}\\theta\\end{gathered}[\/latex]<\/p>\n<p>The <strong>even-odd identities<\/strong> relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\cos(-\\theta)=\\cos(\\theta) \\\\\\sin(-\\theta)=-\\sin(\\theta) \\\\\\tan(-\\theta)=-\\tan(\\theta) \\\\\\cot(-\\theta)=-\\cot(\\theta) \\\\\\sec(-\\theta)=\\sec(\\theta) \\\\\\csc(-\\theta)=-\\csc(\\theta) \\end{gathered}[\/latex]<\/p>\n<p>The <strong>reciprocal identities<\/strong> define reciprocals of the trigonometric functions.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin\\theta=\\frac{1}{\\csc\\theta} \\\\ \\cos\\theta=\\frac{1}{\\sec\\theta} \\\\ \\tan\\theta=\\frac{1}{\\cot\\theta} \\\\ \\cot\\theta=\\frac{1}{\\tan\\theta} \\\\ \\sec\\theta=\\frac{1}{\\cos\\theta} \\\\ \\csc\\theta=\\frac{1}{\\sin\\theta}\\end{gathered}[\/latex]<\/p>\n<p>The <strong>quotient identities<\/strong> define the relationship among the trigonometric functions.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\tan\\theta=\\frac{\\sin\\theta}{\\cos\\theta} \\\\ \\cot\\theta=\\frac{\\cos\\theta}{\\sin\\theta} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Graphing the Equations of an Identity<\/h3>\n<p>Graph both sides of the identity [latex]\\cot \\theta =\\frac{1}{\\tan \\theta }[\/latex]. In other words, on the graphing calculator, graph [latex]y=\\cot \\theta[\/latex] and [latex]y=\\frac{1}{\\tan \\theta }[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q943922\">Show Solution<\/span><\/p>\n<div id=\"q943922\" class=\"hidden-answer\" style=\"display: none\">\n<h3><span id=\"fs-id1353869\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164033\/CNX_Precalc_Figure_07_01_0072.jpg\" alt=\"Graph of y = cot(theta) and y=1\/tan(theta) from -2pi to 2pi. They are the same!\" \/><\/span><\/h3>\n<p>&nbsp;<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trigonometric identity, verify that it is true.<strong><br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id2191946\">\n<li>Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.<\/li>\n<li>Look for opportunities to factor expressions, square a binomial, or add fractions.<\/li>\n<li>Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.<\/li>\n<li>If these steps do not yield the desired result, try converting all terms to sines and cosines.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Verifying a Trigonometric Identity<\/h3>\n<p>Verify [latex]\\tan \\theta \\cos \\theta =\\sin \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q136260\">Show Solution<\/span><\/p>\n<div id=\"q136260\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will start on the left side, as it is the more complicated side:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\theta \\cos \\theta &=\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)\\cos \\theta \\\\ &=\\left(\\frac{\\sin \\theta }{\\cancel{\\cos \\theta }}\\right)\\cancel{\\cos \\theta } \\\\ &=\\sin \\theta \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This identity was fairly simple to verify, as it only required writing [latex]\\tan \\theta[\/latex] in terms of [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Verify the identity [latex]\\csc \\theta \\cos \\theta \\tan \\theta =1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q361361\">Show Solution<\/span><\/p>\n<div id=\"q361361\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}\\csc \\theta \\cos \\theta \\tan \\theta &=\\left(\\frac{1}{\\sin \\theta }\\right)\\cos \\theta \\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right) \\\\ &=\\frac{\\cos \\theta }{\\sin \\theta }\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right) \\\\ &=\\frac{\\sin \\theta \\cos \\theta }{\\sin \\theta \\cos \\theta } \\\\ &=1\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Verifying the Equivalency Using the Even-Odd Identities<\/h3>\n<p>Verify the following equivalency using the even-odd identities:<\/p>\n<p style=\"text-align: center\">[latex]\\left(1+\\sin x\\right)\\left[1+\\sin \\left(-x\\right)\\right]={\\cos }^{2}x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q208801\">Show Solution<\/span><\/p>\n<div id=\"q208801\" class=\"hidden-answer\" style=\"display: none\">\n<p>Working on the left side of the equation, we have<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\left(1+\\sin x\\right)\\left[1+\\sin \\left(-x\\right)\\right]&=\\left(1+\\sin x\\right)\\left(1-\\sin x\\right)&& \\text{Since sin(-}x\\text{)=}-\\sin x \\\\ &=1-{\\sin }^{2}x&& \\text{Difference of squares} \\\\ &={\\cos }^{2}x&& {\\text{cos}}^{2}x=1-{\\sin }^{2}x\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Verifying a Trigonometric Identity Involving <em>sec<sup>2<\/sup>\u03b8<\/em><\/h3>\n<p>Verify the identity [latex]\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }={\\sin }^{2}\\theta[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q565941\">Show Solution<\/span><\/p>\n<div id=\"q565941\" class=\"hidden-answer\" style=\"display: none\">\n<p>As the left side is more complicated, let\u2019s begin there.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&=\\frac{\\left({\\tan }^{2}\\theta +1\\right)-1}{{\\sec }^{2}\\theta }&& {\\sec}^{2}\\theta ={\\tan }^{2}\\theta +1 \\\\ &=\\frac{{\\tan }^{2}\\theta }{{\\sec }^{2}\\theta } \\\\ &={\\tan }^{2}\\theta \\left(\\frac{1}{{\\sec }^{2}\\theta }\\right) \\\\ &={\\tan }^{2}\\theta \\left({\\cos }^{2}\\theta \\right)&& {\\cos }^{2}\\theta =\\frac{1}{{\\sec }^{2}\\theta } \\\\ &=\\left(\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta }\\right)\\left({\\cos }^{2}\\theta \\right)&& {\\tan}^{2}\\theta =\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\left(\\frac{{\\sin }^{2}\\theta }{\\cancel{{\\cos }^{2}\\theta}}\\right)\\left(\\cancel{{\\cos }^{2}\\theta} \\right) \\\\ &={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\n<p>There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{{\\sec }^{2}\\theta -1}{{\\sec }^{2}\\theta }&=\\frac{{\\sec }^{2}\\theta }{{\\sec }^{2}\\theta }-\\frac{1}{{\\sec }^{2}\\theta } \\\\ &=1-{\\cos }^{2}\\theta \\\\ &={\\sin }^{2}\\theta \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<h3>Analysis<\/h3>\n<p>In the first method, we used the identity [latex]{\\sec }^{2}\\theta ={\\tan }^{2}\\theta +1\\\\[\/latex] and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Show that [latex]\\frac{\\cot \\theta }{\\csc \\theta }=\\cos \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q813945\">Show Solution<\/span><\/p>\n<div id=\"q813945\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\cot \\theta }{\\csc \\theta }&=\\frac{\\frac{\\cos \\theta }{\\sin \\theta }}{\\frac{1}{\\sin \\theta }} \\\\ &=\\frac{\\cos \\theta }{\\sin \\theta }\\cdot \\frac{\\sin \\theta }{1} \\\\ &=\\cos \\theta \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Creating and Verifying an Identity<\/h3>\n<p>Create an identity for the expression [latex]2\\tan \\theta \\sec \\theta[\/latex] by rewriting strictly in terms of sine.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q482916\">Show Solution<\/span><\/p>\n<div id=\"q482916\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}2\\tan \\theta \\sec \\theta &=2\\left(\\frac{\\sin \\theta }{\\cos \\theta }\\right)\\left(\\frac{1}{\\cos \\theta }\\right) \\\\ &=\\frac{2\\sin \\theta }{{\\cos }^{2}\\theta } \\\\ &=\\frac{2\\sin \\theta }{1-{\\sin }^{2}\\theta }&& \\text{Substitute }1-{\\sin }^{2}\\theta \\text{ for }{\\cos }^{2}\\theta \\end{align}[\/latex]<\/p>\n<p>Thus,<\/p>\n<p style=\"text-align: center\">[latex]2\\tan \\theta \\sec \\theta =\\frac{2\\sin \\theta }{1-{\\sin }^{2}\\theta }[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Verifying an Identity Using Algebra and Even\/Odd Identities<\/h3>\n<p>Verify the identity:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{{\\sin }^{2}\\left(-\\theta \\right)-{\\cos }^{2}\\left(-\\theta \\right)}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)}=\\cos \\theta -\\sin \\theta\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q689339\">Show Solution<\/span><\/p>\n<div id=\"q689339\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u2019s start with the left side and simplify:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{{\\sin }^{2}\\left(-\\theta \\right)-{\\cos }^{2}\\left(-\\theta \\right)}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)}&=\\frac{{\\left[\\sin \\left(-\\theta \\right)\\right]}^{2}-{\\left[\\cos \\left(-\\theta \\right)\\right]}^{2}}{\\sin \\left(-\\theta \\right)-\\cos \\left(-\\theta \\right)} \\\\ &=\\frac{{\\left(-\\sin \\theta \\right)}^{2}-{\\left(\\cos \\theta \\right)}^{2}}{-\\sin \\theta -\\cos \\theta }&& \\sin \\left(-x\\right)=-\\sin x\\text{ and }\\cos \\left(-x\\right)=\\cos x \\\\ &=\\frac{{\\left(\\sin \\theta \\right)}^{2}-{\\left(\\cos \\theta \\right)}^{2}}{-\\sin \\theta -\\cos \\theta }&& \\text{Difference of squares} \\\\ &=\\frac{\\left(\\sin \\theta -\\cos \\theta \\right)\\left(\\sin \\theta +\\cos \\theta \\right)}{-\\left(\\sin \\theta +\\cos \\theta \\right)} \\\\ &=\\frac{\\left(\\sin \\theta -\\cos \\theta \\right)\\left(\\cancel{\\sin \\theta +\\cos \\theta }\\right)}{-\\left(\\cancel{\\sin \\theta +\\cos \\theta }\\right)} \\\\ &=\\cos \\theta -\\sin \\theta\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Verify the identity [latex]\\frac{{\\sin }^{2}\\theta -1}{\\tan \\theta \\sin \\theta -\\tan \\theta }=\\frac{\\sin \\theta +1}{\\tan \\theta }[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q307900\">Show Solution<\/span><\/p>\n<div id=\"q307900\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}\\frac{{\\sin }^{2}\\theta -1}{\\tan \\theta \\sin \\theta -\\tan \\theta }&=\\frac{\\left(\\sin \\theta +1\\right)\\left(\\sin \\theta -1\\right)}{\\tan \\theta \\left(\\sin \\theta -1\\right)}\\\\ &=\\frac{\\sin \\theta +1}{\\tan \\theta }\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Verifying an Identity Involving Cosines and Cotangents<\/h3>\n<p>Verify the identity: [latex]\\left(1-{\\cos }^{2}x\\right)\\left(1+{\\cot }^{2}x\\right)=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q690675\">Show Solutions<\/span><\/p>\n<div id=\"q690675\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will work on the left side of the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\left(1-{\\cos }^{2}x\\right)\\left(1+{\\cot }^{2}x\\right)&=\\left(1-{\\cos }^{2}x\\right)\\left(1+\\frac{{\\cos }^{2}x}{{\\sin }^{2}x}\\right) \\\\ &=\\left(1-{\\cos }^{2}x\\right)\\left(\\frac{{\\sin }^{2}x}{{\\sin }^{2}x}+\\frac{{\\cos }^{2}x}{{\\sin }^{2}x}\\right) && \\text{Find the common denominator}. \\\\ &=\\left(1-{\\cos }^{2}x\\right)\\left(\\frac{{\\sin }^{2}x+{\\cos }^{2}x}{{\\sin }^{2}x}\\right) \\\\ &=\\left({\\sin }^{2}x\\right)\\left(\\frac{1}{{\\sin }^{2}x}\\right) \\\\ &=1\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Simplify trigonometric expressions using algebra and the identities<\/h2>\n<p>We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.<\/p>\n<p>For example, the equation [latex]\\left(\\sin x+1\\right)\\left(\\sin x - 1\\right)=0[\/latex] resembles the equation [latex]\\left(x+1\\right)\\left(x - 1\\right)=0[\/latex], which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.<\/p>\n<p>Another example is the difference of squares formula, [latex]{a}^{2}-{b}^{2}=\\left(a-b\\right)\\left(a+b\\right)[\/latex], which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 8: Writing the Trigonometric Expression as an Algebraic Expression<\/h3>\n<p>Write the following trigonometric expression as an algebraic expression: [latex]2{\\cos }^{2}\\theta +\\cos \\theta -1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q19423\">Show Solution<\/span><\/p>\n<div id=\"q19423\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that the pattern displayed has the same form as a standard quadratic expression, [latex]a{x}^{2}+bx+c[\/latex]. Letting [latex]\\cos \\theta =x[\/latex], we can rewrite the expression as follows:<\/p>\n<p style=\"text-align: center\">[latex]2{x}^{2}+x - 1[\/latex]<\/p>\n<p>This expression can be factored as [latex]\\left(2x+1\\right)\\left(x - 1\\right)[\/latex]. If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for [latex]x[\/latex]. At this point, we would replace [latex]x[\/latex] with [latex]\\cos \\theta[\/latex] and solve for [latex]\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Rewriting a Trigonometric Expression Using the Difference of Squares<\/h3>\n<p>Rewrite the trigonometric expression: [latex]4{\\cos }^{2}\\theta -1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q947607\">Show Solution<\/span><\/p>\n<div id=\"q947607\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}4{\\cos }^{2}\\theta -1&={\\left(2\\cos \\theta \\right)}^{2}-1 \\\\ &=\\left(2\\cos \\theta -1\\right)\\left(2\\cos \\theta +1\\right) \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<h3>Analysis<\/h3>\n<p>If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let [latex]\\cos \\theta =x[\/latex], rewrite the expression as [latex]4{x}^{2}-1[\/latex], and factor [latex]\\left(2x - 1\\right)\\left(2x+1\\right)[\/latex]. Then replace [latex]x[\/latex] with [latex]\\cos \\theta[\/latex] and solve for the angle.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Rewrite the trigonometric expression: [latex]25 - 9{\\sin }^{2}\\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q979877\">Show Solution<\/span><\/p>\n<div id=\"q979877\" class=\"hidden-answer\" style=\"display: none\">\n<p>This is a difference of squares formula: [latex]25 - 9{\\sin }^{2}\\theta =\\left(5 - 3\\sin \\theta \\right)\\left(5+3\\sin \\theta \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 10: Simplify by Rewriting and Using Substitution<\/h3>\n<p>Simplify the expression by rewriting and using identities:<\/p>\n<p style=\"text-align: center\">[latex]{\\csc }^{2}\\theta -{\\cot }^{2}\\theta[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q120852\">Show Solution<\/span><\/p>\n<div id=\"q120852\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can start with the Pythagorean identity.<\/p>\n<p style=\"text-align: center\">[latex]1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta[\/latex]<\/p>\n<p>Now we can simplify by substituting [latex]1+{\\cot }^{2}\\theta[\/latex] for [latex]{\\csc }^{2}\\theta[\/latex]. We have<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\csc }^{2}\\theta -{\\cot }^{2}\\theta &=1+{\\cot }^{2}\\theta -{\\cot }^{2}\\theta \\\\ &=1\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm120496\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=120496&theme=oea&iframe_resize_id=ohm120496\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Use algebraic techniques to verify the identity: [latex]\\frac{\\cos\\theta}{1+\\sin\\theta}=\\frac{1-\\sin\\theta}{\\cos\\theta}[\/latex].<\/p>\n<p>(Hint: Multiply the numerator and denominator on the left side by [latex]1-\\sin\\theta[\/latex]).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q94618\">Show Solution<\/span><\/p>\n<div id=\"q94618\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\cos \\theta }{1+\\sin \\theta }\\left(\\frac{1-\\sin \\theta }{1-\\sin \\theta }\\right)&=\\frac{\\cos \\theta \\left(1-\\sin \\theta \\right)}{1-{\\sin }^{2}\\theta } \\\\ &=\\frac{\\cos \\theta \\left(1-\\sin \\theta \\right)}{{\\cos }^{2}\\theta } \\\\ &=\\frac{1-\\sin \\theta }{\\cos \\theta } \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use sum and difference formulas for cosine<\/h2>\n<p>Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the <strong>special angles<\/strong>, which we can review in the unit circle shown in Figure 2.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164036\/CNX_Precalc_Figure_07_01_0042.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" width=\"975\" height=\"638\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4.<\/b> The Unit Circle<\/p>\n<\/div>\n<p>We will begin with the <strong>sum and difference formulas for cosine<\/strong>, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles.<\/p>\n<table id=\"Table_07_02_01\" summary=\"Two rows, two columns. The table has ordered pairs of these row values: (Sum formula for cosine, cos(a+B) = cos(a)cos(B) - sin(a)sin(B)) and (Difference formula for cosine, cos(a-B) = cos(a)cos(B) + sin(a)sin(B)).\">\n<tbody>\n<tr>\n<td><strong>Sum formula for cosine<\/strong><\/td>\n<td>[latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Difference formula for cosine<\/strong><\/td>\n<td>[latex]\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>First, we will prove the difference formula for cosines. Let\u2019s consider two points on the unit circle. Point [latex]P[\/latex] is at an angle [latex]\\alpha[\/latex] from the positive <em>x-<\/em>axis with coordinates [latex]\\left(\\cos \\alpha ,\\sin \\alpha \\right)[\/latex] and point [latex]Q[\/latex] is at an angle of [latex]\\beta[\/latex] from the positive <em>x-<\/em>axis with coordinates [latex]\\left(\\cos \\beta ,\\sin \\beta \\right)[\/latex]. Note the measure of angle [latex]POQ[\/latex] is [latex]\\alpha -\\beta[\/latex].<\/p>\n<p>Label two more points: [latex]A[\/latex] at an angle of [latex]\\left(\\alpha -\\beta \\right)[\/latex] from the positive <em>x-<\/em>axis with coordinates [latex]\\left(\\cos \\left(\\alpha -\\beta \\right),\\sin \\left(\\alpha -\\beta \\right)\\right)[\/latex]; and point [latex]B[\/latex] with coordinates [latex]\\left(1,0\\right)[\/latex]. Triangle [latex]POQ[\/latex] is a rotation of triangle [latex]AOB[\/latex] and thus the distance from [latex]P[\/latex] to [latex]Q[\/latex] is the same as the distance from [latex]A[\/latex] to [latex]B[\/latex].<\/p>\n<figure id=\"Figure_07_02_002\" class=\"small\"><span id=\"fs-id1130636\"><br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164038\/CNX_Precalc_Figure_07_02_0022.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" \/><\/span><\/figure>\n<p style=\"text-align: center\"><strong>Figure 5.\u00a0<\/strong>We can find the distance from [latex]P[\/latex] to [latex]Q[\/latex] using the <strong>distance formula<\/strong>.<\/p>\n<div><\/div>\n<div style=\"text-align: center\">[latex]\\begin{align}{d}_{PQ}&=\\sqrt{{\\left(\\cos \\alpha -\\cos \\beta \\right)}^{2}+{\\left(\\sin \\alpha -\\sin \\beta \\right)}^{2}} \\\\ &=\\sqrt{{\\cos }^{2}\\alpha -2\\cos \\alpha \\cos \\beta +{\\cos }^{2}\\beta +{\\sin }^{2}\\alpha -2\\sin \\alpha \\sin \\beta +{\\sin }^{2}\\beta } \\end{align}[\/latex]<\/div>\n<p>Then we apply the <strong>Pythagorean identity<\/strong> and simplify.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} &=\\sqrt{\\left({\\cos }^{2}\\alpha +{\\sin }^{2}\\alpha \\right)+\\left({\\cos }^{2}\\beta +{\\sin }^{2}\\beta \\right)-2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\\\ &=\\sqrt{1+1 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\\\ &=\\sqrt{2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta } \\end{align}[\/latex]<\/div>\n<p>Similarly, using the distance formula we can find the distance from [latex]A[\/latex] to [latex]B[\/latex].<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}{d}_{AB}&=\\sqrt{{\\left(\\cos \\left(\\alpha -\\beta \\right)-1\\right)}^{2}+{\\left(\\sin \\left(\\alpha -\\beta \\right)-0\\right)}^{2}} \\\\ &=\\sqrt{{\\cos }^{2}\\left(\\alpha -\\beta \\right)-2\\cos \\left(\\alpha -\\beta \\right)+1+{\\sin }^{2}\\left(\\alpha -\\beta \\right)} \\end{align}[\/latex]<\/div>\n<p>Applying the Pythagorean identity and simplifying we get:<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} &=\\sqrt{\\left({\\cos }^{2}\\left(\\alpha -\\beta \\right)+{\\sin }^{2}\\left(\\alpha -\\beta \\right)\\right)-2\\cos \\left(\\alpha -\\beta \\right)+1}\\\\ &=\\sqrt{1 - 2\\cos \\left(\\alpha -\\beta \\right)+1}\\\\ &=\\sqrt{2 - 2\\cos \\left(\\alpha -\\beta \\right)} \\end{align}[\/latex]<\/div>\n<p>Because the two distances are the same, we set them equal to each other and simplify.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} \\sqrt{2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta }&=\\sqrt{2 - 2\\cos \\left(\\alpha -\\beta \\right)} \\\\ 2 - 2\\cos \\alpha \\cos \\beta -2\\sin \\alpha \\sin \\beta &=2 - 2\\cos \\left(\\alpha -\\beta \\right) \\end{align}[\/latex]<\/div>\n<p>Finally we subtract [latex]2[\/latex] from both sides and divide both sides by [latex]-2[\/latex].<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta =\\cos \\left(\\alpha -\\beta \\right)\\end{align}[\/latex]<\/div>\n<p>Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Sum and Difference Formulas for Cosine<\/h3>\n<p>These formulas can be used to calculate the cosine of sums and differences of angles.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta\\end{align}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta\\end{align}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two angles, find the cosine of the difference between the angles.<strong><br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id1374215\">\n<li>Write the difference formula for cosine.<\/li>\n<li>Substitute the values of the given angles into the formula.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 11: Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles<\/h3>\n<p>Using the formula for the cosine of the difference of two angles, find the exact value of [latex]\\cos \\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q956915\">Show Solution<\/span><\/p>\n<div id=\"q956915\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the formula for the cosine of the difference of two angles. We have<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\alpha -\\beta \\right)&=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\\\ \\cos \\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)&=\\cos \\left(\\frac{5\\pi }{4}\\right)\\cos \\left(\\frac{\\pi }{6}\\right)+\\sin \\left(\\frac{5\\pi }{4}\\right)\\sin \\left(\\frac{\\pi }{6}\\right) \\\\ &=\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)-\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{1}{2}\\right) \\\\ &=-\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4} \\\\ &=\\frac{-\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the exact value of [latex]\\cos \\left(\\frac{\\pi }{3}-\\frac{\\pi }{4}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q700317\">Show Solution<\/span><\/p>\n<div id=\"q700317\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 12: Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine<\/h3>\n<p>Find the exact value of [latex]\\cos \\left({75}^{\\circ }\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q617422\">Show Solution<\/span><\/p>\n<div id=\"q617422\" class=\"hidden-answer\" style=\"display: none\">\n<p>As [latex]{75}^{\\circ }={45}^{\\circ }+{30}^{\\circ }[\/latex], we can evaluate [latex]\\cos \\left({75}^{\\circ }\\right)[\/latex] as [latex]\\cos \\left({45}^{\\circ }+{30}^{\\circ }\\right)[\/latex]. Thus,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left({45}^{\\circ }+{30}^{\\circ }\\right)&=\\cos \\left({45}^{\\circ }\\right)\\cos \\left({30}^{\\circ }\\right)-\\sin \\left({45}^{\\circ }\\right)\\sin \\left({30}^{\\circ }\\right) \\\\ &=\\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right) \\\\ &=\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4} \\\\ &=\\frac{\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the exact value of [latex]\\cos \\left({105}^{\\circ }\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q683323\">Show Solution<\/span><\/p>\n<div id=\"q683323\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173445\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173445&theme=oea&iframe_resize_id=ohm173445\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use sum and difference formulas for sine<\/h2>\n<p>The <strong>sum and difference formulas for sine<\/strong> can be derived in the same manner as those for cosine, and they resemble the cosine formulas.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Sum and Difference Formulas for Sine<\/h3>\n<p>These formulas can be used to calculate the sines of sums and differences of angles.<\/p>\n<p style=\"text-align: center\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\sin \\left(\\alpha -\\beta \\right)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two angles, find the sine of the difference between the angles.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Write the difference formula for sine.<\/li>\n<li>Substitute the given angles into the formula.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 13: Using Sum and Difference Identities to Evaluate the Difference of Angles<\/h3>\n<p>Use the sum and difference identities to evaluate the difference of the angles and show that part <em>a<\/em> equals part <em>b.<\/em><\/p>\n<ol>\n<li>[latex]\\sin \\left({45}^{\\circ }-{30}^{\\circ }\\right)[\/latex]<\/li>\n<li>[latex]\\sin \\left({135}^{\\circ }-{120}^{\\circ }\\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q191680\">Show Solution<\/span><\/p>\n<div id=\"q191680\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Let\u2019s begin by writing the formula and substitute the given angles.\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\alpha -\\beta \\right)&=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\\\ \\sin \\left({45}^{\\circ }-{30}^{\\circ }\\right)&=\\sin \\left({45}^{\\circ }\\right)\\cos \\left({30}^{\\circ }\\right)-\\cos \\left({45}^{\\circ }\\right)\\sin \\left({30}^{\\circ }\\right) \\end{align}[\/latex]<\/div>\n<p>Next, we need to find the values of the trigonometric expressions.<\/p>\n<div style=\"text-align: center\">[latex]\\sin \\left({45}^{\\circ }\\right)=\\frac{\\sqrt{2}}{2},\\text{ }\\cos \\left({30}^{\\circ }\\right)=\\frac{\\sqrt{3}}{2},\\text{ }\\cos \\left({45}^{\\circ }\\right)=\\frac{\\sqrt{2}}{2},\\text{ }\\sin \\left({30}^{\\circ }\\right)=\\frac{1}{2}[\/latex]<\/div>\n<p>Now we can substitute these values into the equation and simplify.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align} \\sin \\left({45}^{\\circ }-{30}^{\\circ }\\right)&=\\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right) \\\\ &=\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{align}[\/latex]<\/div>\n<\/li>\n<li>Again, we write the formula and substitute the given angles.\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\alpha -\\beta \\right)&=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta\\\\ \\sin \\left({135}^{\\circ }-{120}^{\\circ }\\right)&=\\sin \\left({135}^{\\circ }\\right)\\cos \\left({120}^{\\circ }\\right)-\\cos \\left({135}^{\\circ }\\right)\\sin \\left({120}^{\\circ }\\right)\\end{align}[\/latex]<\/div>\n<p>Next, we find the values of the trigonometric expressions.<\/p>\n<div style=\"text-align: center\">[latex]\\sin \\left({135}^{\\circ }\\right)=\\frac{\\sqrt{2}}{2},\\cos \\left({120}^{\\circ }\\right)=-\\frac{1}{2},\\cos \\left({135}^{\\circ }\\right)=\\frac{\\sqrt{2}}{2},\\sin \\left({120}^{\\circ }\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/div>\n<p>Now we can substitute these values into the equation and simplify.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left({135}^{\\circ }-{120}^{\\circ }\\right)&=\\frac{\\sqrt{2}}{2}\\left(-\\frac{1}{2}\\right)-\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right) \\\\ &=\\frac{-\\sqrt{2}+\\sqrt{6}}{4} \\\\ &=\\frac{\\sqrt{6}-\\sqrt{2}}{4} \\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 14: Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function<\/h3>\n<p>Find the exact value of [latex]\\sin \\left({\\cos }^{-1}\\frac{1}{2}+{\\sin }^{-1}\\frac{3}{5}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q954448\">Show Solution<\/span><\/p>\n<div id=\"q954448\" class=\"hidden-answer\" style=\"display: none\">\n<p>The pattern displayed in this problem is [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex]. Let [latex]\\alpha ={\\cos }^{-1}\\frac{1}{2}[\/latex] and [latex]\\beta ={\\sin }^{-1}\\frac{3}{5}[\/latex]. Then we can write<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\cos \\alpha &=\\frac{1}{2},0\\le \\alpha \\le \\pi\\\\ \\sin \\beta &=\\frac{3}{5},-\\frac{\\pi }{2}\\le \\beta \\le \\frac{\\pi }{2} \\end{align}[\/latex]<\/p>\n<p>We will use the Pythagorean identities to find [latex]\\sin \\alpha[\/latex] and [latex]\\cos \\beta[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\alpha &=\\sqrt{1-{\\cos }^{2}\\alpha } \\\\ &=\\sqrt{1-\\frac{1}{4}} \\\\ &=\\sqrt{\\frac{3}{4}} \\\\ &=\\frac{\\sqrt{3}}{2} \\\\ \\cos \\beta &=\\sqrt{1-{\\sin }^{2}\\beta } \\\\ &=\\sqrt{1-\\frac{9}{25}} \\\\ &=\\sqrt{\\frac{16}{25}} \\\\ &=\\frac{4}{5}\\end{align}[\/latex]<\/p>\n<p>Using the sum formula for sine,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\left({\\cos }^{-1}\\frac{1}{2}+{\\sin }^{-1}\\frac{3}{5}\\right)&=\\sin \\left(\\alpha +\\beta \\right) \\\\ &=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\\\ &=\\frac{\\sqrt{3}}{2}\\cdot \\frac{4}{5}+\\frac{1}{2}\\cdot \\frac{3}{5} \\\\ &=\\frac{4\\sqrt{3}+3}{10}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173443\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173443&theme=oea&iframe_resize_id=ohm173443\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use sum and difference formulas for tangent<\/h2>\n<p>Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.<\/p>\n<p>Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, [latex]\\tan x=\\frac{\\sin x}{\\cos x},\\cos x\\ne 0[\/latex].<\/p>\n<p>Let\u2019s derive the sum formula for tangent.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\sin \\left(\\alpha +\\beta \\right)}{\\cos \\left(\\alpha +\\beta \\right)} \\\\[1mm] &=\\frac{\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta } \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }}{\\frac{\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Divide the numerator and denominator by cos}\\alpha \\text{cos}\\beta \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha \\cos \\beta }{\\cos \\alpha \\cos \\beta}+\\frac{\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }}{\\frac{\\cos \\alpha \\cos \\beta }{\\cos \\alpha \\cos \\beta }-\\frac{\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Split the fractions.} \\\\[1mm] &=\\frac{\\frac{\\sin \\alpha }{\\cos \\alpha }+\\frac{\\sin \\beta }{\\cos \\beta }}{1-\\frac{\\sin \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta }} && \\text{Cancel.} \\\\[1mm] &=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta } \\end{align}[\/latex]<\/div>\n<p>We can derive the difference formula for tangent in a similar way.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Sum and Difference Formulas for Tangent<\/h3>\n<p>The <strong>sum and difference formulas for tangent<\/strong> are:<\/p>\n<p style=\"text-align: center\">[latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\tan \\left(\\alpha -\\beta \\right)=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two angles, find the tangent of the sum of the angles.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Write the sum formula for tangent.<\/li>\n<li>Substitute the given angles into the formula.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 15: Finding the Exact Value of an Expression Involving Tangent<\/h3>\n<p>Find the exact value of [latex]\\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q796417\">Show Solution<\/span><\/p>\n<div id=\"q796417\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u2019s first write the sum formula for tangent and substitute the given angles into the formula.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)&=\\frac{\\tan \\left(\\frac{\\pi }{6}\\right)+\\tan \\left(\\frac{\\pi }{4}\\right)}{1-\\left(\\tan \\left(\\frac{\\pi }{6}\\right)\\right)\\left(\\tan \\left(\\frac{\\pi }{4}\\right)\\right)} \\end{align}[\/latex]<\/p>\n<p>Next, we determine the individual tangents within the formula:<\/p>\n<p style=\"text-align: center\">[latex]\\tan \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{\\sqrt{3}},\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex]<\/p>\n<p>So we have<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)&=\\frac{\\frac{1}{\\sqrt{3}}+1}{1-\\left(\\frac{1}{\\sqrt{3}}\\right)\\left(1\\right)} \\\\ &=\\frac{\\frac{1+\\sqrt{3}}{\\sqrt{3}}}{\\frac{\\sqrt{3}-1}{\\sqrt{3}}} \\\\ &=\\frac{1+\\sqrt{3}}{\\sqrt{3}}\\left(\\frac{\\sqrt{3}}{\\sqrt{3}-1}\\right) \\\\ &=\\frac{\\sqrt{3}+1}{\\sqrt{3}-1} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the exact value of [latex]\\tan \\left(\\frac{2\\pi }{3}+\\frac{\\pi }{4}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q962695\">Show Solution<\/span><\/p>\n<div id=\"q962695\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{1-\\sqrt{3}}{1+\\sqrt{3}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173454\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173454&theme=oea&iframe_resize_id=ohm173454\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 16: Finding Multiple Sums and Differences of Angles<\/h3>\n<p>Given [latex]\\text{ }\\sin \\alpha =\\frac{3}{5},0<\\alpha <\\frac{\\pi }{2},\\cos \\beta =-\\frac{5}{13},\\pi <\\beta <\\frac{3\\pi }{2}[\/latex], find\n\n\n<ol>\n<li>[latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n<li>[latex]\\cos \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n<li>[latex]\\tan \\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n<li>[latex]\\tan \\left(\\alpha -\\beta \\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q622511\">Show Solution<\/span><\/p>\n<div id=\"q622511\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.<\/p>\n<ol>\n<li>To find [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex], we begin with [latex]\\sin \\alpha =\\frac{3}{5}[\/latex] and [latex]0<\\alpha <\\frac{\\pi }{2}[\/latex]. The side opposite [latex]\\alpha[\/latex] has length 3, the hypotenuse has length 5, and [latex]\\alpha[\/latex] is in the first quadrant. Using the Pythagorean Theorem, we can find the length of side [latex]a:[\/latex]\n\n\n<div style=\"text-align: center\">\n<p>[latex]\\begin{gathered}{a}^{2}+{3}^{2}={5}^{2} \\\\ {a}^{2}=16 \\\\ a=4 \\end{gathered}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164040\/CNX_Precalc_Figure_07_02_0032.jpg\" alt=\"Diagram of a triangle in the x,y plane. The vertices are at the origin, (4,0), and (4,3). The angle at the origin is alpha degrees, The angle formed by the x-axis and the side from (4,3) to (4,0) is a right angle. The side opposite the right angle has length 5.\" width=\"487\" height=\"252\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<p>Since [latex]\\cos \\beta =-\\frac{5}{13}[\/latex] and [latex]\\pi <\\beta <\\frac{3\\pi }{2}[\/latex], the side adjacent to [latex]\\beta[\/latex] is [latex]-5[\/latex], the hypotenuse is 13, and [latex]\\beta[\/latex] is in the third quadrant. Again, using the Pythagorean Theorem, we have<\/li>\n<li>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\left(-5\\right)}^{2}+{a}^{2}&={13}^{2} \\\\ 25+{a}^{2}&=169 \\\\ {a}^{2}&=144\\\\ a&=\\pm 12 \\end{align}[\/latex]<\/p>\n<p>Since [latex]\\beta[\/latex] is in the third quadrant, [latex]a=-12[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164043\/CNX_Precalc_Figure_07_02_0042.jpg\" alt=\"Diagram of a triangle in the x,y plane. The vertices are at the origin, (-5,0), and (-5, -12). The angle at the origin is Beta degrees. The angle formed by the x axis and the side from (-5, -12) to (-5,0) is a right angle. The side opposite the right angle has length 13.\" width=\"487\" height=\"568\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<p>The next step is finding the cosine of [latex]\\alpha[\/latex] and the sine of [latex]\\beta[\/latex]. The cosine of [latex]\\alpha[\/latex] is the adjacent side over the hypotenuse. We can find it from the triangle in Figure 5:\u00a0[latex]\\cos \\alpha =\\frac{4}{5}[\/latex]. We can also find the sine of [latex]\\beta[\/latex] from the triangle in Figure 5, as opposite side over the hypotenuse: [latex]\\sin \\beta =-\\frac{12}{13}[\/latex]. Now we are ready to evaluate [latex]\\sin \\left(\\alpha +\\beta \\right)[\/latex].<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\alpha +\\beta \\right)&=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\\\ &=\\left(\\frac{3}{5}\\right)\\left(-\\frac{5}{13}\\right)+\\left(\\frac{4}{5}\\right)\\left(-\\frac{12}{13}\\right) \\\\ &=-\\frac{15}{65}-\\frac{48}{65} \\\\ &=-\\frac{63}{65} \\end{align}[\/latex]<\/div>\n<\/li>\n<li>We can find [latex]\\cos \\left(\\alpha +\\beta \\right)[\/latex] in a similar manner. We substitute the values according to the formula.\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\alpha +\\beta \\right)&=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\\\ &=\\left(\\frac{4}{5}\\right)\\left(-\\frac{5}{13}\\right)-\\left(\\frac{3}{5}\\right)\\left(-\\frac{12}{13}\\right) \\\\ &=-\\frac{20}{65}+\\frac{36}{65} \\\\ &=\\frac{16}{65} \\end{align}[\/latex]<\/div>\n<\/li>\n<li>For [latex]\\tan \\left(\\alpha +\\beta \\right)[\/latex], if [latex]\\sin \\alpha =\\frac{3}{5}[\/latex] and [latex]\\cos \\alpha =\\frac{4}{5}[\/latex], then\n<div style=\"text-align: center\">[latex]\\tan \\alpha =\\frac{\\frac{3}{5}}{\\frac{4}{5}}=\\frac{3}{4}[\/latex]<\/div>\n<p>If [latex]\\sin \\beta =-\\frac{12}{13}[\/latex] and [latex]\\cos \\beta =-\\frac{5}{13}[\/latex],<br \/>\nthen<\/p>\n<div style=\"text-align: center\">[latex]\\tan \\beta =\\frac{\\frac{-12}{13}}{\\frac{-5}{13}}=\\frac{12}{5}[\/latex]<\/div>\n<p>Then,<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta } \\\\ &=\\frac{\\frac{3}{4}+\\frac{12}{5}}{1-\\frac{3}{4}\\left(\\frac{12}{5}\\right)} \\\\ &=\\frac{\\text{ }\\frac{63}{20}}{-\\frac{16}{20}} \\\\ &=-\\frac{63}{16} \\end{align}[\/latex]<\/div>\n<\/li>\n<li>To find [latex]\\tan \\left(\\alpha -\\beta \\right)[\/latex], we have the values we need. We can substitute them in and evaluate.\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha -\\beta \\right)&=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta } \\\\ &=\\frac{\\frac{3}{4}-\\frac{12}{5}}{1+\\frac{3}{4}\\left(\\frac{12}{5}\\right)} \\\\ &=\\frac{-\\frac{33}{20}}{\\frac{56}{20}} \\\\ &=-\\frac{33}{56}\\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\n<p>A common mistake when addressing problems such as this one is that we may be tempted to think that [latex]\\alpha[\/latex] and [latex]\\beta[\/latex] are angles in the same triangle, which of course, they are not. Also note that<\/p>\n<div style=\"text-align: center\">[latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\sin \\left(\\alpha +\\beta \\right)}{\\cos \\left(\\alpha +\\beta \\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div style=\"text-align: left\"><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600;text-align: left\">Use sum and difference formulas for cofunctions<\/span><\/div>\n<\/div>\n<p>Now that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall that if the sum of two positive angles is [latex]\\frac{\\pi }{2}[\/latex], those two angles are complements, and the sum of the two acute angles in a right triangle is [latex]\\frac{\\pi }{2}[\/latex], so they are also complements. In Figure 6, notice that if one of the acute angles is labeled as [latex]\\theta[\/latex], then the other acute angle must be labeled [latex]\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164045\/CNX_Precalc_Figure_07_02_0072.jpg\" alt=\"Image of a right triangle. The remaining angles are labeled theta and pi\/2 - theta.\" width=\"487\" height=\"268\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8.<\/b> From these relationships, the cofunction identities are formed.<\/p>\n<\/div>\n<p>Notice also that [latex]\\sin \\theta =\\cos \\left(\\frac{\\pi }{2}-\\theta \\right):[\/latex] opposite over hypotenuse. Thus, when two angles are complimentary, we can say that the sine of [latex]\\theta[\/latex] equals the <strong>cofunction<\/strong> of the complement of [latex]\\theta[\/latex]. Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions.<span id=\"fs-id2872065\"><br \/>\n<\/span><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Cofunction Identities<\/h3>\n<p>The cofunction identities are summarized in the table below.<\/p>\n<table id=\"Table_07_02_02\" summary=\"Three rows, two columns\/ The table has ordered pairs of these row values: (sin(theta) = cos(pi\/2 - theta), cos(theta) = sin(pi\/2 - theta)), (tan(theta) = cot(pi\/2 - theta), cot(theta) = tan(pi\/2 - theta)), and (sec(theta) = csc(pi\/2 - theta), csc(theta) = sec(pi\/2 - theta)).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>[latex]\\sin \\theta =\\cos \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\cos \\theta =\\sin \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\tan \\theta =\\cot \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\cot \\theta =\\tan \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\sec \\theta =\\csc \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\csc \\theta =\\sec \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>Notice that the formulas in the table may also justified algebraically using the sum and difference formulas. For example, using<\/p>\n<div style=\"text-align: center\">[latex]\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta[\/latex],<\/div>\n<p>we can write<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\frac{\\pi }{2}-\\theta \\right)&=\\cos \\frac{\\pi }{2}\\cos \\theta +\\sin \\frac{\\pi }{2}\\sin \\theta \\\\ &=\\left(0\\right)\\cos \\theta +\\left(1\\right)\\sin \\theta \\\\ &=\\sin \\theta\\end{align}[\/latex]<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 17: Finding a Cofunction with the Same Value as the Given Expression<\/h3>\n<p>Write [latex]\\tan \\frac{\\pi }{9}[\/latex] in terms of its cofunction.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q406106\">Show Solution<\/span><\/p>\n<div id=\"q406106\" class=\"hidden-answer\" style=\"display: none\">\n<p>The cofunction of [latex]\\tan \\theta =\\cot \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]. Thus,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\frac{\\pi }{9}\\right)&=\\cot \\left(\\frac{\\pi }{2}-\\frac{\\pi }{9}\\right) \\\\& =\\cot \\left(\\frac{9\\pi }{18}-\\frac{2\\pi }{18}\\right) \\\\& =\\cot \\left(\\frac{7\\pi }{18}\\right)\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write [latex]\\sin \\frac{\\pi }{7}[\/latex] in terms of its cofunction.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q615431\">Show Solution<\/span><\/p>\n<div id=\"q615431\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\left(\\frac{5\\pi }{14}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173455\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173455&theme=oea&iframe_resize_id=ohm173455\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use sum and difference formulas to verify identities<\/h2>\n<p>Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an identity, verify using sum and difference formulas.<\/h3>\n<ol>\n<li>Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.<\/li>\n<li>Look for opportunities to use the sum and difference formulas.<\/li>\n<li>Rewrite sums or differences of quotients as single quotients.<\/li>\n<li>If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.<\/li>\n<\/ol>\n<\/div>\n<div>\n<div>\n<div class=\"textbox shaded\">\n<h3>Example 18: Verifying an Identity Involving Sine<\/h3>\n<p>Verify the identity [latex]\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)=2\\sin \\alpha \\cos \\beta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q464113\">Show Solution<\/span><\/p>\n<div id=\"q464113\" class=\"hidden-answer\" style=\"display: none\">\n<p>We see that the left side of the equation includes the sines of the sum and the difference of angles.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta\\\\ \\sin \\left(\\alpha -\\beta \\right)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\end{gathered}[\/latex]<\/p>\n<p>We can rewrite each using the sum and difference formulas.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\alpha +\\beta \\right)+\\sin \\left(\\alpha -\\beta \\right)&=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta +\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\\\ &=2\\sin \\alpha \\cos \\beta \\end{align}[\/latex]<\/p>\n<p>We see that the identity is verified.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div>\n<div class=\"textbox shaded\">\n<h3>Example 19: Verifying an Identity Involving Tangent<\/h3>\n<p>Verify the following identity.<\/p>\n<p style=\"text-align: center\">[latex]\\frac{\\sin \\left(\\alpha -\\beta \\right)}{\\cos \\alpha \\cos \\beta }=\\tan \\alpha -\\tan \\beta[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q605610\">Show Solution<\/span><\/p>\n<div id=\"q605610\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can begin by rewriting the numerator on the left side of the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{\\sin \\left(\\alpha -\\beta \\right)}{\\cos \\alpha \\cos \\beta }&=\\frac{\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta } \\\\ &=\\frac{\\sin \\alpha \\cos \\beta}{\\cos \\alpha \\cos \\beta}-\\frac{\\cos \\alpha \\sin \\beta }{\\cos \\alpha \\cos \\beta } && \\text{Rewrite using a common denominator}. \\\\ &=\\frac{\\sin \\alpha }{\\cos \\alpha }-\\frac{\\sin \\beta }{\\cos \\beta }&& \\text{Cancel}. \\\\ &=\\tan \\alpha -\\tan \\beta && \\text{Rewrite in terms of tangent}.\\end{align}[\/latex]<\/p>\n<p>We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Verify the identity: [latex]\\tan \\left(\\pi -\\theta \\right)=-\\tan \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q184004\">Show Solution<\/span><\/p>\n<div id=\"q184004\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\pi -\\theta \\right)&=\\frac{\\tan \\left(\\pi \\right)-\\tan \\theta }{1+\\tan \\left(\\pi \\right)\\tan \\theta } \\\\ &=\\frac{0-\\tan \\theta }{1+0\\cdot \\tan \\theta } \\\\ &=-\\tan \\theta \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<h3>Example 20: Using Sum and Difference Formulas to Solve an Application Problem<\/h3>\n<p>Let [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex] denote two non-vertical intersecting lines, and let [latex]\\theta[\/latex] denote the acute angle between [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex]. Show that<\/p>\n<p style=\"text-align: center\">[latex]\\tan \\theta =\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}[\/latex]<\/p>\n<p>where [latex]{m}_{1}[\/latex] and [latex]{m}_{2}[\/latex] are the slopes of [latex]{L}_{1}[\/latex] and [latex]{L}_{2}[\/latex] respectively. (<strong>Hint:<\/strong> Use the fact that [latex]\\tan {\\theta }_{1}={m}_{1}[\/latex] and [latex]\\tan {\\theta }_{2}={m}_{2}[\/latex]. )<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164047\/CNX_Precalc_Figure_07_02_0052.jpg\" alt=\"Diagram of two non-vertical intersecting lines L1 and L2 also intersecting the x-axis. The acute angle formed by the intersection of L1 and L2 is theta. The acute angle formed by L2 and the x-axis is theta 1, and the acute angle formed by the x-axis and L1 is theta 2.\" width=\"487\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q429859\">Show Solution<\/span><\/p>\n<div id=\"q429859\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the difference formula for tangent, this problem does not seem as daunting as it might.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\theta &=\\tan \\left({\\theta }_{2}-{\\theta }_{1}\\right) \\\\ &=\\frac{\\tan {\\theta }_{2}-\\tan {\\theta }_{1}}{1+\\tan {\\theta }_{1}\\tan {\\theta }_{2}} \\\\ &=\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<section id=\"fs-id2056229\" class=\"key-equations\">\n<h2>Using Double-Angle Formulas to Find Exact Values<\/h2>\n<p>In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The <strong>double-angle formulas<\/strong> are a special case of the sum formulas, where [latex]\\alpha =\\beta[\/latex]. Deriving the double-angle formula for sine begins with the sum formula,<\/p>\n<div style=\"text-align: center\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta[\/latex]<\/div>\n<p>If we let [latex]\\alpha =\\beta =\\theta[\/latex], then we have<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\theta +\\theta \\right)&=\\sin \\theta \\cos \\theta +\\cos \\theta \\sin \\theta \\\\ \\sin \\left(2\\theta \\right)&=2\\sin \\theta \\cos \\theta \\end{align}[\/latex]<\/div>\n<p>Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, [latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta[\/latex], and letting [latex]\\alpha =\\beta =\\theta[\/latex], we have<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(\\theta +\\theta \\right)&=\\cos \\theta \\cos \\theta -\\sin \\theta \\sin \\theta \\\\ \\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\end{align}[\/latex]<\/div>\n<p>Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &=\\left(1-{\\sin }^{2}\\theta \\right)-{\\sin }^{2}\\theta \\\\ &=1 - 2{\\sin }^{2}\\theta\\end{align}[\/latex]<\/div>\n<p>The second interpretation is:<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &={\\cos }^{2}\\theta -\\left(1-{\\cos }^{2}\\theta \\right) \\\\ &=2{\\cos }^{2}\\theta -1\\end{align}[\/latex]<\/div>\n<p>Similarly, to derive the double-angle formula for tangent, replacing [latex]\\alpha =\\beta =\\theta[\/latex] in the sum formula gives<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\theta +\\theta \\right)&=\\frac{\\tan \\theta +\\tan \\theta }{1-\\tan \\theta \\tan \\theta }\\\\ \\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Double-Angle Formulas<\/h3>\n<p>The <strong>double-angle formulas<\/strong> are summarized as follows:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &=1 - 2{\\sin }^{2}\\theta \\\\ &=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Draw a triangle to reflect the given information.<\/li>\n<li>Determine the correct double-angle formula.<\/li>\n<li>Substitute values into the formula based on the triangle.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 21: Using a Double-Angle Formula to Find the Exact Value Involving Tangent<\/h3>\n<p>Given that [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex] and [latex]\\theta[\/latex] is in quadrant II, find the following:<\/p>\n<ol>\n<li>[latex]\\sin \\left(2\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\cos \\left(2\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\tan \\left(2\\theta \\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q477962\">Show Solution<\/span><\/p>\n<div id=\"q477962\" class=\"hidden-answer\" style=\"display: none\">\n<p>If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\\tan \\theta =-\\frac{3}{4}[\/latex], such that [latex]\\theta[\/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\\theta[\/latex] is in the second quadrant, the adjacent side is on the <em>x<\/em>-axis and is negative. Use the <strong>Pythagorean Theorem<\/strong> to find the length of the hypotenuse:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\left(-4\\right)}^{2}+{\\left(3\\right)}^{2}&={c}^{2}\\\\ 16+9&={c}^{2}\\\\ 25&={c}^{2}\\\\ c&=5\\end{align}[\/latex]<\/p>\n<p>Now we can draw a triangle similar to the one shown in Figure 2.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164101\/CNX_Precalc_Figure_07_03_0022.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.\" width=\"487\" height=\"251\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<ol>\n<li>Let\u2019s begin by writing the double-angle formula for sine.\n<div style=\"text-align: center\">[latex]\\sin \\left(2\\theta \\right)=2\\sin \\theta \\cos \\theta[\/latex]<\/div>\n<p>We see that we to need to find [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex]. Based on Figure 2, we see that the hypotenuse equals 5, so [latex]\\sin \\theta =\\frac{3}{5}[\/latex], and [latex]\\cos \\theta =\u2212\\frac{4}{5}[\/latex]. Substitute these values into the equation, and simplify.<br \/>\nThus,<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&=2\\left(\\frac{3}{5}\\right)\\left(\u2212\\frac{4}{5}\\right) \\\\ &=\u2212\\frac{24}{25} \\end{align}[\/latex]<\/div>\n<\/li>\n<li>Write the double-angle formula for cosine.\n<div style=\"text-align: center\">[latex]\\cos \\left(2\\theta \\right)={\\cos }^{2}\\theta -{\\sin }^{2}\\theta[\/latex]<\/div>\n<p>Again, substitute the values of the sine and cosine into the equation, and simplify.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\left(\u2212\\frac{4}{5}\\right)}^{2}\u2212{\\left(\\frac{3}{5}\\right)}^{2} \\\\ &=\\frac{16}{25}\u2212\\frac{9}{25} \\\\ &=\\frac{7}{25}\\end{align}[\/latex]<\/div>\n<\/li>\n<li>Write the double-angle formula for tangent.\n<div style=\"text-align: center\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2\\tan \\theta }{1\u2212{\\tan }^{2}\\theta }[\/latex]<\/div>\n<p>In this formula, we need the tangent, which we were given as [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex]. Substitute this value into the equation, and simplify.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&=\\frac{2\\left(-\\frac{3}{4}\\right)}{1-{\\left(-\\frac{3}{4}\\right)}^{2}} \\\\ &=\\frac{-\\frac{3}{2}}{1-\\frac{9}{16}} \\\\ &=-\\frac{3}{2}\\left(\\frac{16}{7}\\right) \\\\ &=-\\frac{24}{7} \\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given [latex]\\sin \\alpha =\\frac{5}{8}[\/latex], with [latex]\\theta[\/latex] in quadrant I, find [latex]\\cos \\left(2\\alpha \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q827677\">Show Solution<\/span><\/p>\n<div id=\"q827677\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\left(2\\alpha \\right)=\\frac{7}{32}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm156232\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=156232&theme=oea&iframe_resize_id=ohm156232\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 22: Using the Double-Angle Formula for Cosine without Exact Values<\/h3>\n<p>Use the double-angle formula for cosine to write [latex]\\cos \\left(6x\\right)[\/latex] in terms of [latex]\\cos \\left(3x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q676542\">Show Solution<\/span><\/p>\n<div id=\"q676542\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}\\cos \\left(6x\\right)&=\\cos \\left(3x+3x\\right) \\\\ &=\\cos 3x\\cos 3x-\\sin 3x\\sin 3x \\\\ &={\\cos }^{2}3x-{\\sin }^{2}3x \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using Double-Angle Formulas to Verify Identities<\/h2>\n<p>Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 23: Using the Double-Angle Formulas to Establish an Identity<\/h3>\n<p>Establish the following identity using double-angle formulas:<\/p>\n<p style=\"text-align: center\">[latex]1+\\sin \\left(2\\theta \\right)={\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q600157\">Show Solution<\/span><\/p>\n<div id=\"q600157\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will work on the right side of the equal sign and rewrite the expression until it matches the left side.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}&={\\sin }^{2}\\theta +2\\sin \\theta \\cos \\theta +{\\cos }^{2}\\theta \\\\ &=\\left({\\sin }^{2}\\theta +{\\cos }^{2}\\theta \\right)+2\\sin \\theta \\cos \\theta \\\\ &=1+2\\sin \\theta \\cos \\theta \\\\ &=1+\\sin \\left(2\\theta \\right)\\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This process is not complicated, as long as we recall the perfect square formula from algebra:<\/p>\n<p style=\"text-align: center\">[latex]{\\left(a\\pm b\\right)}^{2}={a}^{2}\\pm 2ab+{b}^{2}[\/latex]<\/p>\n<p>where [latex]a=\\sin \\theta[\/latex] and [latex]b=\\cos \\theta[\/latex]. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Establish the identity: [latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\cos \\left(2\\theta \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q178499\">Show Solution<\/span><\/p>\n<div id=\"q178499\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)=\\cos \\left(2\\theta \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 24: Verifying a Double-Angle Identity for Tangent<\/h3>\n<p>Verify the identity:<\/p>\n<p style=\"text-align: center\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q395376\">Show Solution<\/span><\/p>\n<div id=\"q395376\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }&& \\text{Double-angle formula} \\\\ &=\\frac{2\\tan \\theta \\left(\\frac{1}{\\tan \\theta }\\right)}{\\left(1-{\\tan }^{2}\\theta \\right)\\left(\\frac{1}{\\tan \\theta }\\right)}&& \\text{Multiply by a term that results in desired numerator}. \\\\ &=\\frac{2}{\\frac{1}{\\tan \\theta }-\\frac{{\\tan }^{2}\\theta }{\\tan \\theta }} \\\\ &=\\frac{2}{\\cot \\theta -\\tan \\theta }&& \\text{Use reciprocal identity for }\\frac{1}{\\tan \\theta }.\\end{align}[\/latex]<\/p>\n<h3>Analysis of the Solution<\/h3>\n<p>Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show<\/p>\n<p style=\"text-align: center\">[latex]\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\n<p>Let\u2019s work on the right side.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\frac{2}{\\cot \\theta -\\tan \\theta }&=\\frac{2}{\\frac{1}{\\tan \\theta }-\\tan \\theta }\\left(\\frac{\\tan \\theta }{\\tan \\theta }\\right) \\\\ &=\\frac{2\\tan \\theta }{\\frac{1}{\\cancel{\\tan \\theta }}\\left(\\cancel{\\tan \\theta }\\right)-\\tan \\theta \\left(\\tan \\theta \\right)} \\\\ &=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta } \\end{align}[\/latex]<\/p>\n<p>When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Verify the identity: [latex]\\cos \\left(2\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q584630\">Show Solution<\/span><\/p>\n<div id=\"q584630\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\left(2\\theta \\right)\\cos \\theta =\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use Reduction Formulas to Simplify an Expression<\/h2>\n<p>The double-angle formulas can be used to derive the <strong>reduction formulas<\/strong>, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.<\/p>\n<p>We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let\u2019s begin with [latex]\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta[\/latex]. Solve for [latex]{\\sin }^{2}\\theta :[\/latex]<\/p>\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta \\\\ 2{\\sin }^{2}\\theta =1-\\cos \\left(2\\theta \\right) \\\\ {\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\end{gathered}[\/latex]<\/div>\n<p>Next, we use the formula [latex]\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1[\/latex]. Solve for [latex]{\\cos }^{2}\\theta :[\/latex]<\/p>\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1 \\\\ 1+\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta \\\\ \\frac{1+\\cos \\left(2\\theta \\right)}{2}={\\cos }^{2}\\theta \\end{gathered}[\/latex]<\/div>\n<p>The last reduction formula is derived by writing tangent in terms of sine and cosine:<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}{\\tan }^{2}\\theta &=\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\frac{\\frac{1-\\cos \\left(2\\theta \\right)}{2}}{\\frac{1+\\cos \\left(2\\theta \\right)}{2}}&& \\text{Substitute the reduction formulas.} \\\\ &=\\left(\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\right)\\left(\\frac{2}{1+\\cos \\left(2\\theta \\right)}\\right) \\\\ &=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Reduction Formulas<\/h3>\n<p>The <strong>reduction formulas<\/strong> are summarized as follows:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 25: Writing an Equivalent Expression Not Containing Powers Greater Than 1<\/h3>\n<p>Write an equivalent expression for [latex]{\\cos }^{4}x[\/latex] that does not involve any powers of sine or cosine greater than 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q109691\">Show Solution<\/span><\/p>\n<div id=\"q109691\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will apply the reduction formula for cosine twice.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\cos }^{4}x&={\\left({\\cos }^{2}x\\right)}^{2} \\\\ &={\\left(\\frac{1+\\cos \\left(2x\\right)}{2}\\right)}^{2}&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{1}{4}\\left(1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right) \\\\ &=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{4}\\left(\\frac{1+\\cos \\left(2\\left(2x\\right)\\right)}{2}\\right) && {\\text{ Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}+\\frac{1}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{3}{8}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 26: Using the Power-Reducing Formulas to Prove an Identity<\/h3>\n<p>Use the power-reducing formulas to prove<\/p>\n<p style=\"text-align: center\">[latex]{\\sin }^{3}\\left(2x\\right)=\\left[\\frac{1}{2}\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q828553\">Show Solution<\/span><\/p>\n<div id=\"q828553\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will work on simplifying the left side of the equation:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}{\\sin }^{3}\\left(2x\\right)&=\\left[\\sin \\left(2x\\right)\\right]\\left[{\\sin }^{2}\\left(2x\\right)\\right] \\\\ &=\\sin \\left(2x\\right)\\left[\\frac{1-\\cos \\left(4x\\right)}{2}\\right]&& \\text{Substitute the power-reduction formula}. \\\\ &=\\sin \\left(2x\\right)\\left(\\frac{1}{2}\\right)\\left[1-\\cos \\left(4x\\right)\\right] \\\\ &=\\frac{1}{2}\\left[\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right] \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that in this example, we substituted<\/p>\n<div style=\"text-align: center\">[latex]\\frac{1-\\cos \\left(4x\\right)}{2}[\/latex]<\/div>\n<p>for [latex]{\\sin }^{2}\\left(2x\\right)[\/latex]. The formula states<\/p>\n<div style=\"text-align: center\">[latex]{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2}[\/latex]<\/div>\n<p>We let [latex]\\theta =2x[\/latex], so [latex]2\\theta =4x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Use the power-reducing formulas to prove that [latex]10{\\cos }^{4}x=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q387102\">Show Solution<\/span><\/p>\n<div id=\"q387102\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align}10{\\cos }^{4}x&=10{\\left({\\cos }^{2}x\\right)}^{2} \\\\ &=10{\\left[\\frac{1+\\cos \\left(2x\\right)}{2}\\right]}^{2}&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{10}{4}\\left[1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right] \\\\ &=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{4}\\left(\\frac{1+\\cos\\left( 2\\left(2x\\right)\\right)}{2}\\right)&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{8}+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{30}{8}+5\\cos \\left(2x\\right)+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4533\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4533&theme=oea&iframe_resize_id=ohm4533\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using Half-Angle Formulas to Find Exact Values<\/h2>\n<p>The next set of identities is the set of <strong>half-angle formulas<\/strong>, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace [latex]\\theta[\/latex] with [latex]\\frac{\\alpha }{2}[\/latex], the half-angle formula for sine is found by simplifying the equation and solving for [latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]. Note that the half-angle formulas are preceded by a [latex]\\pm[\/latex] sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which [latex]\\frac{\\alpha }{2}[\/latex] terminates.<\/p>\n<p>The half-angle formula for sine is derived as follows:<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}{\\sin }^{2}\\theta &=\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\\\ {\\sin }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1-\\left(\\cos 2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &=\\frac{1-\\cos \\alpha }{2} \\\\ \\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\n<p>To derive the half-angle formula for cosine, we have<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}{\\cos }^{2}\\theta &=\\frac{1+\\cos \\left(2\\theta \\right)}{2}\\\\ {\\cos }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &=\\frac{1+\\cos \\alpha }{2} \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\n<p>For the tangent identity, we have<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}{\\tan }^{2}\\theta &=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\\\ {\\tan }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1-\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)} \\\\ &=\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }\\hfill \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\end{align}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Half-Angle Formulas<\/h3>\n<p>The <strong>half-angle formulas<\/strong> are as follows:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 27: Using a Half-Angle Formula to Find the Exact Value of a Sine Function<\/h3>\n<p>Find [latex]\\sin \\left({15}^{\\circ }\\right)[\/latex] using a half-angle formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q283155\">Show Solution<\/span><\/p>\n<div id=\"q283155\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since [latex]{15}^{\\circ }=\\frac{{30}^{\\circ }}{2}[\/latex], we use the half-angle formula for sine:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\sin \\frac{{30}^{\\circ }}{2}&=\\sqrt{\\frac{1-\\cos {30}^{\\circ }}{2}} \\\\ &=\\sqrt{\\frac{1-\\frac{\\sqrt{3}}{2}}{2}} \\\\ &=\\sqrt{\\frac{\\frac{2-\\sqrt{3}}{2}}{2}} \\\\ &=\\sqrt{\\frac{2-\\sqrt{3}}{4}} \\\\ &=\\frac{\\sqrt{2-\\sqrt{3}}}{2} \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice that we used only the positive root because [latex]\\sin \\left({15}^{\\text{o}}\\right)[\/latex] is positive.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.<\/h3>\n<ol>\n<li>Draw a triangle to represent the given information.<\/li>\n<li>Determine the correct half-angle formula.<\/li>\n<li>Substitute values into the formula based on the triangle.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 28: Finding Exact Values Using Half-Angle Identities<\/h3>\n<p>Given that [latex]\\tan \\alpha =\\frac{8}{15}[\/latex] and [latex]\\alpha[\/latex] lies in quadrant III, find the exact value of the following:<\/p>\n<ol>\n<li>[latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\n<li>[latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\n<li>[latex]\\tan \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q751659\">Show Solution<\/span><\/p>\n<div id=\"q751659\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the given information, we can draw the triangle shown in Figure 3. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate [latex]\\sin \\alpha =-\\frac{8}{17}[\/latex] and [latex]\\cos \\alpha =-\\frac{15}{17}[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164103\/CNX_Precalc_Figure_07_03_0032.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.\" width=\"487\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<ol>\n<li>Before we start, we must remember that, if [latex]\\alpha[\/latex] is in quadrant III, then [latex]180^\\circ <\\alpha <270^\\circ[\/latex], so [latex]\\frac{180^\\circ }{2}<\\frac{\\alpha }{2}<\\frac{270^\\circ }{2}[\/latex]. This means that the terminal side of [latex]\\frac{\\alpha }{2}[\/latex] is in quadrant II, since [latex]90^\\circ <\\frac{\\alpha }{2}<135^\\circ[\/latex].To find [latex]\\sin \\frac{\\alpha }{2}[\/latex], we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure 3\u00a0and simplify.\n\n\n<div style=\"text-align: center\">[latex]\\begin{align} \\sin \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ &=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{32}{17}}{2}} \\\\ &=\\pm \\sqrt{\\frac{32}{17}\\cdot \\frac{1}{2}} \\\\ &=\\pm \\sqrt{\\frac{16}{17}} \\\\ &=\\pm \\frac{4}{\\sqrt{17}} \\\\ &=\\frac{4\\sqrt{17}}{17} \\end{align}[\/latex]<\/div>\n<p>We choose the positive value of [latex]\\sin \\frac{\\alpha }{2}[\/latex] because the angle terminates in quadrant II and sine is positive in quadrant II.<\/li>\n<li>To find [latex]\\cos \\frac{\\alpha }{2}[\/latex], we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in Figure 3, and simplify.\n<div style=\"text-align: center\">[latex]\\begin{align} \\cos \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ &=\\pm \\sqrt{\\frac{1+\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{2}{17}}{2}} \\\\ &=\\pm \\sqrt{\\frac{2}{17}\\cdot \\frac{1}{2}} \\\\ &=\\pm \\sqrt{\\frac{1}{17}} \\\\ &=-\\frac{\\sqrt{17}}{17}\\end{align}[\/latex]<\/div>\n<p>We choose the negative value of [latex]\\cos \\frac{\\alpha }{2}[\/latex] because the angle is in quadrant II because cosine is negative in quadrant II.<\/li>\n<li>To find [latex]\\tan \\frac{\\alpha }{2}[\/latex], we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in Figure 3\u00a0and simplify.\n<div style=\"text-align: center\">[latex]\\begin{align} \\tan \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{1+\\left(-\\frac{15}{17}\\right)}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{32}{17}}{\\frac{2}{17}}} \\\\ &=\\pm \\sqrt{\\frac{32}{2}} \\\\ &=-\\sqrt{16} \\\\ &=-4 \\end{align}[\/latex]<\/div>\n<p>We choose the negative value of [latex]\\tan \\frac{\\alpha }{2}[\/latex] because [latex]\\frac{\\alpha }{2}[\/latex] lies in quadrant II, and tangent is negative in quadrant II.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given that [latex]\\sin \\alpha =-\\frac{4}{5}[\/latex] and [latex]\\alpha[\/latex] lies in quadrant IV, find the exact value of [latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q406030\">Show Solution<\/span><\/p>\n<div id=\"q406030\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\frac{2}{\\sqrt{5}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173569\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173569&theme=oea&iframe_resize_id=ohm173569\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 29: Finding the Measurement of a Half Angle<\/h3>\n<p>Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of [latex]\\theta[\/latex] formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for higher-level competition, what is the measurement of the angle for novice competition?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q685675\">Show Solution<\/span><\/p>\n<div id=\"q685675\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the angle for novice competition measures half the steepness of the angle for the high level competition, and [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for high competition, we can find [latex]\\cos \\theta[\/latex] from the right triangle and the Pythagorean theorem so that we can use the half-angle identities.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}{3}^{2}+{5}^{2}=34 \\\\ c=\\sqrt{34} \\end{gathered}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164106\/CNX_Precalc_Figure_07_03_0042.jpg\" alt=\"Image of a right triangle with sides 3, 5, and rad34. Rad 34 is the hypotenuse, and 3 is the base. The angle formed by the hypotenuse and base is theta. The angle between the side of length 3 and side of length 5 is a right angle.\" width=\"487\" height=\"210\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12<\/b><\/p>\n<\/div>\n<p>We see that [latex]\\cos \\theta =\\frac{3}{\\sqrt{34}}=\\frac{3\\sqrt{34}}{34}[\/latex]. We can use the half-angle formula for tangent: [latex]\\tan \\frac{\\theta }{2}=\\sqrt{\\frac{1-\\cos \\theta }{1+\\cos \\theta }}[\/latex]. Since [latex]\\tan \\theta[\/latex] is in the first quadrant, so is [latex]\\tan \\frac{\\theta }{2}[\/latex]. Thus,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} \\tan \\frac{\\theta }{2}&=\\sqrt{\\frac{1-\\frac{3\\sqrt{34}}{34}}{1+\\frac{3\\sqrt{34}}{34}}} \\\\ &=\\sqrt{\\frac{\\frac{34 - 3\\sqrt{34}}{34}}{\\frac{34+3\\sqrt{34}}{34}}} \\\\ &=\\sqrt{\\frac{34 - 3\\sqrt{34}}{34+3\\sqrt{34}}} \\\\ &\\approx 0.57\\end{align}[\/latex]<\/p>\n<p>We can take the inverse tangent to find the angle: [latex]{\\tan }^{-1}\\left(0.57\\right)\\approx {29.7}^{\\circ }[\/latex]. So the angle of the ramp for novice competition is [latex]\\approx {29.7}^{\\circ }[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id2253485\" class=\"definition\"><\/dl>\n<dl id=\"fs-id2253499\" class=\"definition\">\n<dd id=\"fs-id2253503\"><\/dd>\n<\/dl>\n<\/div>\n<h2>Key Equations<\/h2>\n<table id=\"fs-id2056235\" summary=\"..\">\n<tbody>\n<tr>\n<td>Pythagorean identities<\/td>\n<td>[latex]\\begin{gathered}{\\sin }^{2}\\theta +{\\cos }^{2}\\theta =1\\\\ 1+{\\cot }^{2}\\theta ={\\csc }^{2}\\theta \\\\ 1+{\\tan }^{2}\\theta ={\\sec }^{2}\\theta \\end{gathered}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Even-odd identities<\/td>\n<td>[latex]\\begin{gathered}\\tan \\left(-\\theta \\right)=-\\tan \\theta \\\\ \\cot \\left(-\\theta \\right)=-\\cot \\theta \\\\ \\sin \\left(-\\theta \\right)=-\\sin \\theta \\\\ \\csc \\left(-\\theta \\right)=-\\csc \\theta \\\\ \\cos \\left(-\\theta \\right)=\\cos \\theta \\\\ \\sec \\left(-\\theta \\right)=\\sec \\theta \\end{gathered}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Reciprocal identities<\/td>\n<td>[latex]\\begin{gathered}\\sin \\theta =\\frac{1}{\\csc \\theta }\\\\ \\cos \\theta =\\frac{1}{\\sec \\theta }\\\\ \\tan \\theta =\\frac{1}{\\cot \\theta }\\\\ \\csc \\theta =\\frac{1}{\\sin \\theta }\\\\ \\sec \\theta =\\frac{1}{\\cos \\theta }\\\\ \\cot \\theta =\\frac{1}{\\tan \\theta }\\end{gathered}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Quotient identities<\/td>\n<td>[latex]\\begin{gathered} \\tan \\theta =\\frac{\\sin \\theta }{\\cos \\theta }\\\\ \\cot \\theta =\\frac{\\cos \\theta }{\\sin \\theta } \\end{gathered}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section id=\"fs-id2081298\" class=\"key-concepts\">\n<table style=\"width: 743px\">\n<tbody>\n<tr>\n<td style=\"width: 182.984px\">Sum Formula for Cosine<\/td>\n<td style=\"width: 533.984px\">[latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 182.984px\">Difference Formula for Cosine<\/td>\n<td style=\"width: 533.984px\">[latex]\\cos \\left(\\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 182.984px\">Sum Formula for Sine<\/td>\n<td style=\"width: 533.984px\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 182.984px\">Difference Formula for Sine<\/td>\n<td style=\"width: 533.984px\">[latex]\\sin \\left(\\alpha -\\beta \\right)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 182.984px\">Sum Formula for Tangent<\/td>\n<td style=\"width: 533.984px\">[latex]\\tan \\left(\\alpha +\\beta \\right)=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 182.984px\">Difference Formula for Tangent<\/td>\n<td style=\"width: 533.984px\">[latex]\\tan \\left(\\alpha -\\beta \\right)=\\frac{\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 182.984px\">Cofunction identities<\/td>\n<td style=\"width: 533.984px\">[latex]\\sin \\theta =\\cos \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/p>\n<p>[latex]\\cos \\theta =\\sin \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/p>\n<p>[latex]\\tan \\theta =\\cot \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/p>\n<p>[latex]\\cot \\theta =\\tan \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/p>\n<p>[latex]\\sec \\theta =\\csc \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/p>\n<p>[latex]\\csc \\theta =\\sec \\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table>\n<colgroup>\n<col \/>\n<col \/> <\/colgroup>\n<tbody>\n<tr>\n<td><strong>Double-angle formulas<\/strong><\/td>\n<td>[latex]\\begin{align}\\sin \\left(2\\theta \\right)&=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &=1 - 2{\\sin }^{2}\\theta \\\\ &=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Reduction formulas<\/strong><\/td>\n<td>[latex]\\begin{align}&{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Half-angle formulas<\/strong><\/td>\n<td>[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id2081305\">\n<li>There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.<\/li>\n<li>Graphing both sides of an identity will verify it.<\/li>\n<li>Simplifying one side of the equation to equal the other side is another method for verifying an identity.<\/li>\n<li>The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation.<\/li>\n<li>We can create an identity by simplifying an expression and then verifying it.<\/li>\n<li>Verifying an identity may involve algebra with the fundamental identities.<\/li>\n<li>Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics.<\/li>\n<li>The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles.<\/li>\n<li>The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle.<\/li>\n<li>The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle.<\/li>\n<li>The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions.<\/li>\n<li>The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles.<\/li>\n<li>The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles.<\/li>\n<li>The cofunction identities apply to complementary angles and pairs of reciprocal functions.<\/li>\n<li>Sum and difference formulas are useful in verifying identities.<\/li>\n<li>Application problems are often easier to solve by using sum and difference formulas.<\/li>\n<li>Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent.<\/li>\n<li>Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term.<\/li>\n<li>Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not.<\/li>\n<\/ul>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id2253485\" class=\"definition\">\n<dt>double-angle formulas<\/dt>\n<dd id=\"fs-id2253488\">identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal<\/dd>\n<\/dl>\n<p><strong>even-odd identities<\/strong><\/p>\n<dl id=\"fs-id1881971\" class=\"definition\">\n<dd id=\"fs-id1881976\">set of equations involving trigonometric functions such that if [latex]f\\left(-x\\right)=-f\\left(x\\right)[\/latex], the identity is odd, and if [latex]f\\left(-x\\right)=f\\left(x\\right)[\/latex], the identity is even<\/dd>\n<\/dl>\n<dl id=\"fs-id2253492\" class=\"definition\">\n<dt>half-angle formulas<\/dt>\n<dd id=\"fs-id2253495\">identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions<\/dd>\n<\/dl>\n<p><strong>Pythagorean identities<\/strong><\/p>\n<dl id=\"fs-id1882066\" class=\"definition\">\n<dd id=\"fs-id1882069\">set of equations involving trigonometric functions based on the right triangle properties<\/dd>\n<\/dl>\n<p><strong>quotient identities<\/strong><\/p>\n<dl id=\"fs-id1882073\" class=\"definition\">\n<dd id=\"fs-id1882076\">pair of identities based on the fact that tangent is the ratio of sine and cosine, and cotangent is the ratio of cosine and sine<\/dd>\n<\/dl>\n<p><strong>reciprocal identities<\/strong><\/p>\n<dl id=\"fs-id1882080\" class=\"definition\">\n<dd id=\"fs-id1882083\">set of equations involving the reciprocals of basic trigonometric definitions<\/dd>\n<\/dl>\n<dl id=\"fs-id2253499\" class=\"definition\">\n<dt>reduction formulas<\/dt>\n<dd id=\"fs-id2253503\">identities derived from the double-angle formulas and used to reduce the power of a trigonometric function<\/dd>\n<\/dl>\n<dl id=\"fs-id1881971\" class=\"definition\">\n<dd id=\"fs-id1881976\"><\/dd>\n<\/dl>\n<dl id=\"fs-id1882080\" class=\"definition\">\n<dd id=\"fs-id1882083\"><\/dd>\n<\/dl>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-14195\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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