{"id":762,"date":"2019-03-07T13:42:45","date_gmt":"2019-03-07T13:42:45","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/solving-trigonometric-equations-with-identities\/"},"modified":"2020-02-18T19:54:24","modified_gmt":"2020-02-18T19:54:24","slug":"solving-trigonometric-equations-with-identities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/solving-trigonometric-equations-with-identities\/","title":{"raw":"3.7 Trigonometric  Identities","rendered":"3.7 Trigonometric  Identities"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Verify the fundamental trigonometric identities.<\/li>\r\n \t<li>Simplify trigonometric expressions using algebra and the identities.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"Figure_07_01_006\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07134234\/CNX_Precalc_Figure_07_01_006.jpg\" alt=\"Photo of international passports.\" width=\"488\" height=\"366\" \/> <strong>Figure 1.\u00a0<\/strong>International passports and travel documents[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id2479925\">In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports\u2014there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation.<\/p>\r\n<p id=\"fs-id1186261\">In this section, we will review some trigonometric identities that we have already seen in earlier sections, create some new ones and show how we can use identities, along with basic tools of algebra, to simplify trigonometric expressions.<\/p>\r\n\r\n<div id=\"fs-id2429534\" class=\"bc-section section\">\r\n<h3>Some Fundamental Trigonometric Identities<\/h3>\r\n<p id=\"fs-id1981287\">We have previously discussed the set of <strong>even-odd identities. <\/strong>The <strong><span class=\"no-emphasis\">even-odd identities<\/span><\/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. (See <a class=\"autogenerated-content\" href=\"#Table_07_01_02\">Table 1<\/a>).<\/p>\r\n\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).\"><caption><strong>Table 1<\/strong><\/caption>\r\n<tbody>\r\n<tr>\r\n<th class=\"border\" style=\"text-align: center\" colspan=\"3\">Even-Odd Identities<\/th>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\begin{array}{l}\\mathrm{tan}\\left(-\\theta \\right)=-\\mathrm{tan}\\text{ }\\left(\\theta\\right)\\hfill \\\\ \\mathrm{cot}\\left(-\\theta \\right)=-\\mathrm{cot}\\text{ }\\left(\\theta\\right) \\hfill \\end{array}[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\begin{array}{l}\\mathrm{sin}\\left(-\\theta \\right)=-\\mathrm{sin}\\text{ }\\left(\\theta\\right) \\hfill \\\\ \\mathrm{csc}\\left(-\\theta \\right)=-\\mathrm{csc}\\text{ }\\left(\\theta\\right) \\hfill \\end{array}[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\begin{array}{l}\\mathrm{cos}\\left(-\\theta \\right)=\\mathrm{cos}\\text{ }\\left(\\theta\\right) \\hfill \\\\ \\mathrm{sec}\\left(-\\theta \\right)=\\mathrm{sec}\\text{ }\\left(\\theta\\right) \\hfill \\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1419861\">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. See <a class=\"autogenerated-content\" href=\"#fs-id2031263\">Table 2<\/a>.<\/p>\r\n\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><caption><strong>Table 2<\/strong><\/caption>\r\n<tbody>\r\n<tr>\r\n<th class=\"border\" style=\"text-align: center\" colspan=\"2\">Reciprocal Identities<\/th>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{sin}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{csc}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{csc}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{sin}\\text{ }\\left(\\theta\\right)} [\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{cos}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{sec}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{sec}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{cos}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{tan}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{cot}\\text{ }\\left(\\theta\\right) }[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{cot}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{tan}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1354190\">Another set of identities is the set of <strong><span class=\"no-emphasis\">quotient identities<\/span><\/strong>, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See <a class=\"autogenerated-content\" href=\"#fs-id937819\">Table 3<\/a>.<\/p>\r\n\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).\"><caption>Table 3<\/caption><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<th class=\"border\" style=\"text-align: center\" colspan=\"2\">Quotient Identities<\/th>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{tan}\\left(\\theta\\right)=\\frac{\\mathrm{sin}\\left(\\theta\\right) }{\\mathrm{cos}\\left(\\theta\\right)}[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{cot}\\left(\\theta\\right)=\\frac{\\mathrm{cos}\\left(\\theta\\right) }{\\mathrm{sin}\\left(\\theta\\right) }[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1165137732163\" class=\"bc-section section\">\r\n<h4>Alternate Forms of the Pythagorean Identity<\/h4>\r\n<p id=\"fs-id1165135203709\">We can use these fundamental identities to derive alternative forms of the <span class=\"no-emphasis\">Pythagorean Identity<\/span>,<\/p>\r\n<p style=\"text-align: center\">[latex]{\\mathrm{cos}}^{2}\\left(\\theta\\right)+{\\mathrm{sin}}^{2}\\left(\\theta\\right)=1.[\/latex]<\/p>\r\n\r\n<div>\r\n<div id=\"fs-id2429534\" class=\"bc-section section\">\r\n<p id=\"fs-id2113436\"><span style=\"font-size: 1rem;text-align: initial\">You should recall that the identity shown above is a direct result of our definition of the sine and cosine functions in terms of the coordinates of points on the unit circle.\u00a0\u00a0We can derive two more identities using the methods shown below.<\/span><\/p>\r\n<p id=\"fs-id1313011\">The identity [latex]1+\\mathrm{cot}^{2}\\left(\\theta\\right) =\\mathrm{csc}^{2}\\left(\\theta\\right) [\/latex] is found by dividing each term of the first identity by [latex]\\mathrm{sin}^{2}\\left(\\theta\\right)[\/latex], and then rewriting each part of the equation using the identities we have already discussed in earlier sections.<\/p>\r\n<p id=\"fs-id1286496\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sin}^{2}\\left(\\theta \\right)}{\\mathrm{sin}^{2}\\left(\\theta\\right)}+\\frac{\\mathrm{cos}^{2}\\left(\\theta\\right)}{\\mathrm{sin}^{2}\\left(\\theta\\right)}&amp; =\\frac{1}{\\mathrm{sin}^{2}\\left(\\theta\\right)}\\end{align*}[\/latex]<\/p>\r\nWe can then use our earlier quotient and reciprocal identities to rewrite the expression in this equation as shown below.\r\n<p id=\"fs-id1286496\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}1+\\mathrm{cot}^{2}\\left(\\theta\\right)&amp; =\\mathrm{csc}^{2}\\left(\\theta\\right)\\end{align*}[\/latex]<\/p>\r\nSimilarly, [latex]1+\\mathrm{tan}^{2}\\left(\\theta\\right) =\\mathrm{sec}^{2}\\left(\\theta\\right) [\/latex] can be obtained by dividing all terms in the first identity by [latex]\\mathrm{cos}^{2}\\left(\\theta\\right)[\/latex], and then rewriting each part of the equation using the quotient and reciprocal identities .\r\n<p style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sin}^{2}\\left(\\theta \\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right)}+\\frac{\\mathrm{cos}^{2}\\left(\\theta\\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right)}&amp; =\\frac{1}{\\mathrm{cos}^{2}\\left(\\theta\\right)}\\end{align*}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<p style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{tan}^{2}\\left(\\theta\\right)+1&amp; =\\mathrm{sec}^{2}\\left(\\theta\\right)\\end{align*}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id2113436\">We now have the three<span style=\"font-size: 1rem;text-align: initial\">\u00a0<\/span><strong style=\"font-size: 1rem;text-align: initial\">Pythagorean identities<\/strong>\u00a0shown in <a class=\"autogenerated-content\" style=\"font-size: 1rem;text-align: initial\" href=\"#Table_07_01_01\">Table 4<\/a><span style=\"font-size: 1rem;text-align: initial\">.\u00a0\u00a0<\/span><\/p>\r\n\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.\"><caption>Table 4<\/caption><colgroup> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<th class=\"border\" style=\"text-align: center\" colspan=\"3\">Pythagorean Identities<\/th>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{sin}^{2}\\left(\\theta\\right) +\\mathrm{cos}^{2}\\left(\\theta\\right) =1[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]1+{\\mathrm{cot}}^{2}\\left(\\theta\\right) ={\\mathrm{csc}}^{2}\\left(\\theta\\right) [\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]1+{\\mathrm{tan}}^{2}\\left(\\theta\\right) ={\\mathrm{sec}}^{2}\\left(\\theta\\right) [\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id2429534\" class=\"bc-section section\" style=\"text-align: center\">\r\n<div id=\"Example_07_01_01\" class=\"textbox examples\">\r\n<h3>Example 1:\u00a0 Graphing the Equations of an Identity<\/h3>\r\n<p id=\"fs-id1693457\">Graph both sides of the identity [latex]\\mathrm{cot}\\left(\\theta\\right) =\\frac{1}{\\mathrm{tan}\\left(\\theta\\right)}.[\/latex] In other words, on the graphing calculator, graph [latex]y=\\mathrm{cot}\\left(\\theta\\right) [\/latex] and [latex]y=\\frac{1}{\\mathrm{tan}\\left(\\theta\\right)}.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1208452\">[reveal-answer q=\"fs-id1208452\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1208452\"]\r\n<p id=\"fs-id1859063\">See <a class=\"autogenerated-content\" href=\"#Figure_07_01_007\">Figure 1<\/a>.<\/p>\r\n\r\n<div id=\"Figure_07_01_007\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"387\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07134244\/CNX_Precalc_Figure_07_01_007.jpg\" alt=\"Graph of y = cot(theta) and y=1\/tan(theta) from -2pi to 2pi. They are the same!\" width=\"387\" height=\"300\" \/> <strong>Figure 1.<\/strong>[\/caption]\r\n<h3>Analysis<\/h3>\r\n<p id=\"fs-id3063508\">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 confirm an identity verified with analytical means. If both expressions give the same graph, then they are most likely identities.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2467864\" class=\"precalculus howto examples\">\r\n<h3>How To:<\/h3>\r\n<p id=\"fs-id693742\"><strong>Given a trigonometric identity, verify that it is true. <\/strong><\/p>\r\n\r\n<ol>\r\n \t<li style=\"text-align: left\">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 style=\"text-align: left\">Look for opportunities to factor expressions, square a binomial, or add fractions.<\/li>\r\n \t<li style=\"text-align: left\">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 style=\"text-align: left\">If these steps do not yield the desired result, try converting all terms to sines and cosines.<\/li>\r\n \t<li style=\"text-align: left\">Note the values not in the domain of the expression on the left and right as the identity does not hold for those values.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_07_01_02\" class=\"textbox examples\">\r\n<div id=\"fs-id3057901\">\r\n<div id=\"fs-id2291660\">\r\n<h3>Example 2:\u00a0 Verifying a Trigonometric Identity<\/h3>\r\n<p id=\"fs-id1947320\">Verify [latex]\\mathrm{tan}\\left(\\theta\\right) \\mathrm{cos}\\left(\\theta\\right)=\\mathrm{sin}\\left(\\theta\\right) .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1334614\">[reveal-answer q=\"fs-id1334614\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1334614\"]\r\nWe will start on the left side, as it is the more complicated side:\r\n<div>\r\n<div><\/div>\r\n<div>[latex]\\begin{align*}\\mathrm{tan}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right)&amp;=\\left(\\frac{\\mathrm{sin}\\text{ }\\left(\\theta\\right)}{\\mathrm{cos}\\left(\\theta\\right)}\\right)\\mathrm{cos}\\left(\\theta\\right)\\\\&amp;=\\mathrm{sin}\\text{ }\\left(\\theta\\right)\\end{align*}[\/latex]<\/div>\r\n<div>\r\n<h3>Analysis<\/h3>\r\n<p id=\"fs-id1266264\">This identity was fairly simple to verify, as it only required writing [latex]\\mathrm{tan}\\left(\\theta\\right) [\/latex] in terms of [latex]\\mathrm{sin}\\left(\\theta\\right) [\/latex] and [latex]\\mathrm{cos}\\left(\\theta\\right) .[\/latex]\u00a0 When determining identities we need to also consider the domains of the expressions on the left and right sides of the equation.\u00a0 The identity is only valid where both expressions are defined.\u00a0 For this problem, the domain of\u00a0[latex]\\mathrm{tan}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right)[\/latex] is all real numbers except [latex]\\frac{\\pi}{2}\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.\u00a0 The expression\u00a0[latex]\\frac{\\pi}{2}\\pm n\\pi[\/latex] means that we will have breaks in the domain starting at [latex]\\frac{\\pi}{2}[\/latex] and then every time we add or subtract [latex]\\pi[\/latex] from there.\u00a0 This means the identity will not be valid at these points.\u00a0 We also need to consider the left hand side of the equation.\u00a0 However, since the domain of [latex]\\mathrm{sin}\\left(\\theta\\right)[\/latex] is all real numbers there are no additional places that the identity is not valid.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id820368\" class=\"precalculus tryit\">\r\n<h3>Try it #1<\/h3>\r\n<div id=\"ti_07_01_01\">\r\n<div id=\"fs-id2335257\">\r\n<p id=\"fs-id2335258\">Verify the identity [latex]\\mathrm{csc}\\left(\\theta\\right)\\text{ } \\mathrm{cos}\\left(\\theta\\right)\\text{ } \\mathrm{tan}\\left(\\theta\\right) =1.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1237705\" style=\"text-align: left\">[reveal-answer q=\"fs-id1237705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1237705\"]\r\n<div style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{csc}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right) \\mathrm{tan}\\left(\\theta\\right) &amp;=\\left(\\frac{1}{\\mathrm{sin}\\left(\\theta\\right) }\\right)\\mathrm{cos}\\left(\\theta\\right) \\left(\\frac{\\mathrm{sin}\\left(\\theta\\right) }{\\mathrm{cos}\\left(\\theta\\right) }\\right) \\\\ &amp;=\\frac{\\mathrm{cos}\\left(\\theta\\right)}{\\mathrm{sin}\\left(\\theta\\right) }\\left(\\frac{\\mathrm{sin}\\left(\\theta\\right) }{\\mathrm{cos}\\left(\\theta\\right) }\\right) \\\\ &amp;=\\frac{\\mathrm{sin}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right) }{\\mathrm{sin}\\left(\\theta\\right) \\mathrm{cos}\\left(\\theta\\right)} \\\\ &amp;=1\\end{align*}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>This identity is not valid where [latex]\\mathrm{csc}\\left(\\theta\\right)[\/latex] or [latex]\\mathrm{tan}\\left(\\theta\\right)[\/latex] are not defined.\u00a0 This means that it is not valid at [latex]... -\\pi, \\frac{-\\pi}{2}, 0, \\frac{\\pi}{2}, \\pi, ...[\/latex].<\/div>\r\n<div><\/div>\r\n<div><span style=\"text-align: center;font-size: 1em\">We can capture these values with the expression\u00a0[latex]\\frac{\\pi}{2}\\pm\\frac{n\\pi}{2}[\/latex]\u00a0where [latex]n[\/latex] is an integer.\u00a0<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div id=\"ti_07_01_01\">\r\n<div id=\"fs-id1237705\" style=\"text-align: left\">\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_01_03\" class=\"textbox examples\">\r\n<div id=\"fs-id2066731\">\r\n<div id=\"fs-id1128511\">\r\n<h3>Example 3:\u00a0 Verifying the Equivalency Using the Even-Odd Identities<\/h3>\r\n<p id=\"fs-id937635\">Verify the following equivalency using the even-odd identities:<\/p>\r\n\r\n<div id=\"fs-id2463271\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\left(1+\\mathrm{sin}\\text{ }\\left(x\\right)\\right)\\left(1+\\mathrm{sin}\\left(-x\\right)\\right)={\\mathrm{cos}}^{2}\\left(x\\right).[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1492841\">[reveal-answer q=\"fs-id1492841\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1492841\"]\r\n<p id=\"fs-id1137525\">Working on the left side of the equation, we have<\/p>\r\n\r\n<div id=\"fs-id1277815\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\left(1+\\mathrm{sin}\\left(x\\right)\\right)\\left(1+\\mathrm{sin}\\left(-x\\right)\\right)&amp;=\\left(1+\\mathrm{sin}\\left(x\\right)\\right)\\left(1-\\mathrm{sin}\\left(x\\right)\\right)&amp;&amp; \\text{Since sin}\\left(-x\\right)=-\\mathrm{sin}\\left(x\\right).\\\\ &amp;=1-\\mathrm{sin}^{2}\\left(x\\right)&amp;&amp;\\text{Difference of squares}.\\\\ &amp;=\\mathrm{cos}^{2}\\left(x\\right)&amp;&amp; \\mathrm{cos}^{2}\\left(x\\right)=1-\\mathrm{sin}^{2}\\left(x\\right)\\text{ from Pythagorean Identity}.\\end{align*}[\/latex]<\/div>\r\n<div>This identity is valid for all real numbers since [latex]\\mathrm{sin}\\left(\\theta\\right) [\/latex] and [latex]\\mathrm{cos}\\left(\\theta\\right) [\/latex] have a domain of all real numbers.<\/div>\r\n<\/div>\r\n<div id=\"fs-id1492841\">\r\n<div><\/div>\r\n<div class=\"unnumbered\" style=\"text-align: center\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_01_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1758470\">\r\n<div id=\"fs-id1233021\">\r\n<h3>Example 4:\u00a0 Verifying a Trigonometric Identity Involving <em>sec<sup>2<\/sup>(\u03b8)<\/em><\/h3>\r\n<p id=\"fs-id2814564\">Verify the identity [latex]\\frac{{\\mathrm{sec}}^{2}\\left(\\theta\\right) -1}{{\\mathrm{sec}}^{2}\\left(\\theta\\right) }={\\mathrm{sin}}^{2}\\left(\\theta\\right). [\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2123498\">[reveal-answer q=\"fs-id2123498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2123498\"]\r\n<p id=\"fs-id2123500\">As the left side is more complicated, let\u2019s begin there.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sec}^{2}\\left(\\theta\\right) -1}{\\mathrm{sec}^{2}\\left(\\theta\\right) }&amp;=\\frac{\\left(\\mathrm{tan}^{2}\\left(\\theta\\right) +1\\right)-1}{\\mathrm{sec}^{2}\\left(\\theta\\right)}&amp;&amp;\\text{Because }\\mathrm{sec}^{2}\\left(\\theta\\right) =\\mathrm{tan}^{2}\\left(\\theta\\right) +1. \\\\ &amp;=\\frac{\\mathrm{tan}^{2}\\left(\\theta\\right)}{\\mathrm{sec}^{2}\\left(\\theta\\right)}\\\\ &amp;=\\mathrm{tan}^{2}\\left(\\theta\\right) \\left(\\frac{1}{\\mathrm{sec}^{2}\\left(\\theta\\right)} \\right)\\\\ &amp;=\\mathrm{tan}^{2}\\left(\\theta\\right)\\mathrm{cos}^{2}\\left(\\theta\\right)&amp;&amp;\\text{Because }\\mathrm{cos}^{2}\\left(\\theta\\right) =\\frac{1}{\\mathrm{sec}^{2}\\left(\\theta\\right)}.\\\\ &amp;=\\frac{\\mathrm{sin}^{2}\\left(\\theta\\right)} {\\mathrm{cos}^{2}\\left(\\theta\\right)}\\mathrm{cos}^{2}\\left(\\theta\\right) &amp;&amp; \\text{Because }\\mathrm{tan}^{2}\\left(\\theta\\right) =\\frac{\\mathrm{sin}^{2}\\left(\\theta\\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right)}.\\\\ &amp;=\\mathrm{sin}^{2}\\left(\\theta\\right)\\end{align*}[\/latex]<\/p>\r\n<p id=\"fs-id1244586\">There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.<\/p>\r\n\r\n<div id=\"fs-id2107026\" class=\"unnumbered\" style=\"text-align: left\">\r\n<p style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sec}^{2}\\left(\\theta\\right) -1}{\\mathrm{sec}^{2}\\left(\\theta\\right) }&amp;=\\frac{\\mathrm{sec}^{2}\\left(\\theta\\right) }{\\mathrm{sec}^{2}\\left(\\theta\\right) }-\\frac{1}{\\mathrm{sec}^{2}\\left(\\theta\\right) }\\\\&amp;=1-\\mathrm{cos}^{2}\\left(\\theta\\right) \\\\ &amp;=\\mathrm{sin}^{2}\\left(\\theta\\right)\\end{align*}[\/latex]<\/p>\r\nThis identity is valid on the domain of [latex]\\mathrm{sec}\\left(\\theta\\right).[\/latex]\u00a0 This means that we need to exclude values where\u00a0[latex]\\mathrm{cos}\\left(\\theta\\right)=0.[\/latex]\r\n\r\nThis would give us all real numbers except [latex]\\frac{\\pi}{2}\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.\r\n<h3>Analysis<\/h3>\r\n<p id=\"fs-id1698743\">In the first method, we used the identity [latex]\\mathrm{sec}^{2}\\left(\\theta\\right) = \\mathrm{tan}^{2}\\left(\\theta\\right) +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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2017058\" class=\"precalculus tryit\">\r\n<h3>Try it #2<\/h3>\r\n<div id=\"ti_07_01_02\">\r\n<div id=\"fs-id1185914\">\r\n<p id=\"fs-id1185915\">Show that [latex]\\frac{\\mathrm{cot}\\text{ }\\left(\\theta\\right) }{\\mathrm{csc}\\left(\\theta\\right)}=\\mathrm{cos}\\left(\\theta\\right) .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1506268\">[reveal-answer q=\"fs-id1506268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1506268\"]\r\n<div id=\"fs-id1506270\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{cot}\\left(\\theta\\right)}{\\mathrm{csc}\\left(\\theta\\right)}&amp;=\\frac{\\frac{\\mathrm{cos}\\left(\\theta\\right) }{\\mathrm{sin}\\left(\\theta\\right) }}{\\frac{1}{\\mathrm{sin}\\left(\\theta\\right)}} &amp;&amp; \\text{This is a complex fraction.}\\\\ &amp;=\\frac{\\mathrm{cos}\\left(\\theta\\right)}{\\mathrm{sin}\\left(\\theta\\right) }\\cdot \\frac{\\mathrm{sin}\\left(\\theta\\right)}{ 1}&amp;&amp; \\text{Multiple the numerator by the reciprocal of the denominator.}\\\\ &amp;=\\mathrm{cos}\\left(\\theta\\right)\\end{align*}[\/latex]<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: left\">\r\n\r\nThis identity is valid on the domain of [latex]\\mathrm{cot}\\left(\\theta\\right)[\/latex] and [latex]\\mathrm{csc}\\left(\\theta\\right).[\/latex]\u00a0 This means that the identity is not valid for the values [latex]... -2\\pi, -\\pi, 0, \\pi, 2\\pi, ...[\/latex]\r\n\r\nThis would give us all real numbers except [latex]\\pi\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_01_05\" class=\"textbox examples\">\r\n<div id=\"fs-id2122140\">\r\n<div id=\"fs-id1970227\">\r\n<h3>Example 5: Creating and Verifying an Identity<\/h3>\r\n<p id=\"fs-id1970233\">Create an identity for the expression [latex]2\\text{ }\\mathrm{tan}\\left(\\theta\\right) \\mathrm{sec}\\left(\\theta\\right) [\/latex] by rewriting strictly in terms of sine.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1429666\">[reveal-answer q=\"fs-id1429666\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1429666\"]\r\n<p id=\"fs-id1429668\">There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:<\/p>\r\n\r\n<div id=\"fs-id2338806\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\text{ }\\mathrm{tan}\\left(\\theta\\right)\\mathrm{sec}\\left(\\theta\\right)&amp;=2\\left(\\frac{\\mathrm{sin}\\left(\\theta\\right)}{\\mathrm{cos}\\left(\\theta\\right)}\\right)\\left(\\frac{1}{\\mathrm{cos}\\left(\\theta\\right)}\\right)\\\\ &amp;=\\frac{2\\text{ }\\mathrm{sin}\\left(\\theta\\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right)}\\\\ &amp;=\\frac{2\\text{ }\\mathrm{sin}\\left(\\theta\\right)}{1-\\mathrm{sin}^{2}\\left(\\theta\\right)}&amp;&amp;\\text{Substitute }1-\\mathrm{sin}^{2}\\left(\\theta\\right)\\text{ for }\\mathrm{cos}^{2}\\left(\\theta\\right).\\\\\r\n2\\mathrm{tan}\\left(\\theta\\right)\\mathrm{sec}\\left(\\theta\\right)&amp;=\\frac{2\\text{ }\\mathrm{sin}\\left(\\theta\\right)}{1-\\mathrm{sin}^{2}\\left(\\theta\\right)}\\end{align*}[\/latex]<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: left\">This identity holds on values in both of the domains of [latex]\\mathrm{tan}\\left(\\theta\\right)[\/latex] and [latex]\\mathrm{sec}\\left(\\theta\\right).[\/latex]\u00a0 This means it holds for all real numbers except [latex]... -\\frac{3\\pi}{2}, -\\frac{\\pi}{2}, \\frac{\\pi}{2}, \\frac{3\\pi}{2}...[\/latex]<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: left\">This would give us all real numbers except [latex]\\frac{\\pi}{2}\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: left\">The expression on the left also needs to be considered for restrictions but the denominator is zero exactly when the left side of the expression is undefined so no more values need to be excluded.[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_01_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1219228\">\r\n<div id=\"fs-id1219230\">\r\n<h3>Example 6: Verifying an Identity Using Algebra and Even\/Odd Identities<\/h3>\r\n<p id=\"fs-id2430931\">Verify the identity:<\/p>\r\n\r\n<div id=\"fs-id1583958\" class=\"unnumbered\">[latex]\\frac{\\mathrm{sin}^{2}\\left(-\\theta \\right)-\\mathrm{cos}^{2}\\left(-\\theta \\right)}{\\mathrm{sin}\\left(-\\theta \\right)-\\mathrm{cos}\\left(-\\theta \\right)}=\\mathrm{cos}\\left(\\theta \\right) -\\mathrm{sin}\\left(\\theta \\right) [\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<div id=\"fs-id2345275\">[reveal-answer q=\"fs-id2345275\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2345275\"]\r\n<p id=\"fs-id2345277\">Let\u2019s start with the left side and simplify:<\/p>\r\n\r\n<div id=\"fs-id1540225\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sin}^{2}\\left(-\\theta\\right)-\\mathrm{cos}^{2}\\left(-\\theta \\right)}{\\mathrm{sin}\\left(-\\theta\\right)-\\mathrm{cos}\\left(-\\theta \\right)}&amp;=\\frac{\\left(\\mathrm{sin}\\left(-\\theta \\right)-\\mathrm{cos}\\left(-\\theta \\right)\\right)\\left(\\mathrm{sin}\\left(-\\theta \\right)+\\mathrm{cos}\\left(-\\theta \\right)\\right)}{\\mathrm{sin}\\left(-\\theta \\right)-\\mathrm{cos}\\left(-\\theta \\right)}&amp;&amp; \\text{Difference of squares.}\\\\&amp;=\\frac{\\left(-\\mathrm{sin}\\left(\\theta \\right)-\\mathrm{cos}\\left(\\theta \\right)\\right)\\left(-\\mathrm{sin}\\left(\\theta \\right)+\\mathrm{cos}\\left(\\theta \\right)\\right)}{-\\mathrm{sin}\\left(\\theta \\right)-\\mathrm{cos}\\left(\\theta \\right)}&amp;&amp;\\text{Odd-Even Identities.}\\\\&amp;=\\mathrm{-sin}\\left(\\theta \\right)+\\mathrm{cos}\\left(\\theta \\right)&amp;&amp;\\text{Cancel the common factor.}\\\\&amp;=\\mathrm{cos}\\left(\\theta \\right)\\mathrm{-sin}\\left(\\theta \\right)&amp;&amp;\\text{Reorder terms.}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: left\">This identity will be valid when [latex]\\mathrm{sin}\\left(-\\theta \\right)-\\mathrm{cos}\\left(-\\theta \\right)\\ne 0.[\/latex]\u00a0 Solving this type of equation is discussed in section 3.8 Solving Trigonometric Equations.[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2137172\" class=\"precalculus tryit\">\r\n<h3>Try it #3<\/h3>\r\n<div id=\"ti_07_01_03\">\r\n<div id=\"fs-id2261720\">\r\n<p id=\"fs-id2261721\">Verify the identity [latex]\\frac{\\mathrm{sin}^{2}\\left(\\theta \\right) -1}{\\mathrm{tan}\\left(\\theta \\right)\\mathrm{sin}\\left(\\theta\\right) -\\mathrm{tan}\\left(\\theta\\right)}=\\frac{\\mathrm{sin}\\left(\\theta\\right) +1}{\\mathrm{tan}\\left(\\theta \\right)}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2701905\">[reveal-answer q=\"fs-id2701905\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2701905\"]\r\n<p id=\"fs-id2701906\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sin}^{2}\\left(\\theta \\right) -1}{\\mathrm{tan}\\left(\\theta \\right) \\mathrm{sin}\\left(\\theta \\right) -\\mathrm{tan}\\left(\\theta \\right)}&amp;=\\frac{\\left(\\mathrm{sin}\\left(\\theta \\right) +1\\right)\\left(\\mathrm{sin}\\left(\\theta \\right) -1\\right)}{\\mathrm{tan}\\left(\\theta \\right)\\left(\\mathrm{sin}\\left(\\theta \\right) -1\\right)}\\\\ &amp;=\\frac{\\mathrm{sin}\\left(\\theta \\right)+1}{\\mathrm{tan}\\left(\\theta \\right)}\\end{align*}[\/latex]<\/p>\r\nThis equation is valid on the domain of tangent and where we don't get a divide by zero.\u00a0 Determining when the denominator in the expression on the left requires equation solving discussed in section 3.8 Solving Trigonometric Equations.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_01_07\" class=\"textbox examples\">\r\n<div id=\"fs-id1664723\">\r\n<div id=\"fs-id1664725\">\r\n<h3>Example 7: Verifying an Identity Involving Cosines and Cotangents<\/h3>\r\n<p id=\"fs-id1664730\">Verify the identity: [latex]\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)\\left(1+{\\mathrm{cot}}^{2}\\left(x\\right)\\right)=1.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1913600\">[reveal-answer q=\"fs-id1913600\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1913600\"]\r\n<p id=\"fs-id1913602\">We will work on the left side of the equation.<\/p>\r\n\r\n<div id=\"fs-id1913605\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)&amp;\\left(1+\\mathrm{cot}^{2}\\left(x\\right)\\right)\\\\&amp;=\\left(1-\\mathrm{cos}^{2}\\left(x\\right)\\right)\\left(1+\\frac{\\mathrm{cos}^{2}\\left(x\\right)}{\\mathrm{sin}^{2}\\left(x\\right)}\\right)&amp;&amp;\\text{Rewrite }\\mathrm{cot}^{2}\\left(x\\right). \\\\ &amp;=\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)\\left(\\frac{{\\mathrm{sin}}^{2}\\left(x\\right)}{{\\mathrm{sin}}^{2}x}+\\frac{{\\mathrm{cos}}^{2}x}{{\\mathrm{sin}}^{2}\\left(x\\right)}\\right) &amp;&amp;\\text{Find the common denominator.}\\\\ &amp;=\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)\\left(\\frac{\\mathrm{sin}^{2}\\left(x\\right)+\\mathrm{cos}^{2}\\left(x\\right)}{\\mathrm{sin}^{2}\\left(x\\right)}\\right) \\\\ &amp;=\\left({\\mathrm{sin}}^{2}\\left(x\\right)\\right)\\left(\\frac{\\mathrm{sin}^{2}\\left(x\\right)+\\mathrm{cos}^{2}\\left(x\\right)}{\\mathrm{sin}^{2}\\left(x\\right)}\\right)&amp;&amp;\\text{Because }1-\\mathrm{cos}^{2}\\left(x\\right) =\\mathrm{sin}^{2}\\left(x\\right).\\\\&amp;=\\left({\\mathrm{sin}}^{2}\\left(x\\right)\\right)\\left(\\frac{1}{{\\mathrm{sin}}^{2}\\left(x\\right)}\\right)&amp;&amp; \\text{Because } \\mathrm{sin}^{2}\\left(x\\right)+\\mathrm{cos}^{2}\\left(x\\right)=1.\\\\ &amp;=1\\end{align*}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div>This identity is valid on the domain of cotangent.\u00a0\u00a0This would give us all real numbers except [latex]\\pi\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.<\/div>\r\n<div><\/div>\r\n<div>An alternate way to verify this identity is to begin by multiplying out the expression on the left and simplifying.<\/div>\r\n<div style=\"text-align: center\">[latex]\\begin{align*}\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)\\left(1+\\mathrm{cot}^{2}\\left(x\\right)\\right)&amp;=1-\\mathrm{cos}^{2}\\left(x\\right)+\\mathrm{cot}^{2}\\left(x\\right)-\\mathrm{cot}^{2}\\left(x\\right)\\mathrm{cos}^{2}\\left(x\\right)\\\\&amp;=1-\\mathrm{cos}^{2}\\left(x\\right)+\\mathrm{cot}^{2}\\left(x\\right)\\left(1-\\mathrm{cos}^{2}\\left(x\\right)\\right)\\\\&amp;=1-\\mathrm{cos}^{2}\\left(x\\right)+\\frac{\\mathrm{cos}^{2}\\left(x\\right)}{\\mathrm{sin}^{2}\\left(x\\right)}\\mathrm{sin}^{2}\\left(x\\right)\\\\&amp;=1-\\mathrm{cos}^{2}\\left(x\\right)+\\mathrm{cos}^{2}\\left(x\\right)\\\\&amp;=1\\end{align*}[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<div id=\"fs-id1913600\">\r\n<div class=\"unnumbered\" style=\"text-align: center\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2485952\" class=\"bc-section section\">\r\n<h3>Using Algebra to Simplify Trigonometric Expressions<\/h3>\r\n<p id=\"fs-id2485958\">We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions when we are solving equations. 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>\r\nAn 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\r\n<p id=\"fs-id2485964\">For example, the expression [latex]\\mathrm{sin}^2\\left(x\\right)-1[\/latex] resembles the difference of squares [latex]x^2-1.[\/latex] Recognizing that [latex]x^2-1[\/latex] can be factored as [latex]\\left(x+1\\right)\\left(x-1\\right)[\/latex] helps us quickly recognize that\u00a0 [latex]\\mathrm{sin}^2\\left(x\\right)-1[\/latex] can be factored as [latex]\\left(\\mathrm{sin}\\left(x\\right)+1\\right)\\left(\\mathrm{sin}\\left(x\\right)-1\\right).[\/latex]<\/p>\r\n<p id=\"fs-id1814865\">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 expressions and equations easier to work with.<\/p>\r\n\r\n<div id=\"Example_07_01_08\" class=\"textbox examples\">\r\n<div id=\"fs-id1814944\">\r\n<div id=\"fs-id2578086\">\r\n<h3>Example 8: Writing the Trigonometric Expression as an Algebraic Expression<\/h3>\r\n<p id=\"fs-id2578091\">Write the following trigonometric expression as an algebraic expression: [latex]2\\mathrm{cos}^{2}\\left(\\theta\\right) +\\mathrm{cos}\\left(\\theta\\right) -1.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2578132\">[reveal-answer q=\"fs-id2578132\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2578132\"]\r\n<p id=\"fs-id2578134\">Notice that the pattern displayed has the same form as a standard quadratic expression, [latex]a{x}^{2}+bx+c.[\/latex] Letting [latex]\\mathrm{cos}\\left(\\theta\\right) =x,[\/latex] we can rewrite the expression as follows:<\/p>\r\n\r\n<div id=\"fs-id2121948\" class=\"unnumbered\" style=\"text-align: center\">[latex]2{x}^{2}+x-1[\/latex]<\/div>\r\n<p id=\"fs-id2121978\">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]\\mathrm{cos}\\left(\\theta\\right) [\/latex] and solve for [latex]\\theta .[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_05_09\" class=\"textbox examples\">\r\n<div id=\"fs-id2721228\">\r\n<div id=\"fs-id2721230\">\r\n<h3>Example 9: Rewriting a Trigonometric Expression Using the Difference of Squares<\/h3>\r\n<p id=\"fs-id1777385\">Rewrite the trigonometric expression: [latex]4\\mathrm{cos}^{2}\\left(\\theta\\right) -1.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1777420\">[reveal-answer q=\"fs-id1777420\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1777420\"]\r\n<p id=\"fs-id1777422\">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>\r\n\r\n<div id=\"fs-id1777427\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}4\\text{ }\\mathrm{cos}^{2}\\left(\\theta\\right) -1&amp;=\\left(2 \\mathrm{cos}\\left(\\theta\\right)-1 \\right)\\left(2\\text{ }\\mathrm{cos}\\left(\\theta\\right) +1\\right)\\end{align*}[\/latex]<\/div>\r\n<div id=\"fs-id1739635\">\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]\\mathrm{cos}\\left(\\theta\\right) =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]\\mathrm{cos}\\left(\\theta\\right) [\/latex] and solve for the angle.\r\n<div id=\"fs-id1777427\" class=\"unnumbered\" style=\"text-align: center\">[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2438482\" class=\"precalculus tryit\">\r\n<h3>Try it #4<\/h3>\r\n<div id=\"ti_07_01_04\">\r\n<div id=\"fs-id2438492\">\r\n<p id=\"fs-id2438493\">Rewrite the trigonometric expression: [latex]25-9\\text{ }\\mathrm{sin}^{2}\\left(\\theta\\right) .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2438529\">[reveal-answer q=\"fs-id2438529\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2438529\"]\r\n<p id=\"fs-id2438532\">This is a difference of squares formula: [latex]25-9\\text{ }\\mathrm{sin}^{2}\\left(\\theta\\right)=\\left(5-3\\text{ }\\mathrm{sin}\\left(\\theta\\right) \\right)\\left(5+3\\text{ }\\mathrm{sin}\\left(\\theta\\right)\\right).[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_01_10\" class=\"textbox examples\">\r\n<div id=\"fs-id2917234\">\r\n<div id=\"fs-id2917236\">\r\n<h3>Example 10: Simplify by Rewriting and Using Substitution<\/h3>\r\n<p id=\"fs-id2917242\">Simplify the expression by rewriting and using identities:<\/p>\r\n\r\n<div id=\"fs-id2917245\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\mathrm{csc}^{2}\\left(\\theta\\right) -\\mathrm{cot}^{2}\\left(\\theta\\right) [\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<div id=\"fs-id2483936\">[reveal-answer q=\"fs-id2483936\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2483936\"]\r\n<p id=\"fs-id2483938\">We can start with the Pythagorean identity.\u00a0 We know:\u00a0[latex]1+\\mathrm{cot}^{2}\\left(\\theta\\right)=\\mathrm{csc}^{2}\\left(\\theta\\right)[\/latex]<\/p>\r\n\r\n<div id=\"fs-id2483941\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\r\n\\mathrm{csc}^{2}\\left(\\theta\\right)-\\mathrm{cot}^{2}\\left(\\theta\\right)&amp;=1+\\mathrm{cot}^{2}\\left(\\theta\\right)-\\mathrm{cot}^{2}\\left(\\theta\\right)&amp;&amp;\\text{Substitute expression in for }\\mathrm{csc}^{2}\\left(\\theta\\right).\\\\ &amp;=1\\end{align*}[\/latex]<\/div>\r\n<div>This identity is valid on the domains of cotangent and cosecant.\u00a0\u00a0This would give us all real numbers except [latex]\\pi\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: center\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2607392\" class=\"precalculus tryit\">\r\n<h3>Try it #5<\/h3>\r\n<div id=\"ti_07_01_05\">\r\n<div id=\"fs-id2607401\">\r\n<p id=\"fs-id2607402\">Use algebraic techniques to verify the identity: [latex]\\frac{\\mathrm{cos}\\left(\\theta\\right) }{1+\\mathrm{sin}\\left(\\theta\\right) }=\\frac{1-\\mathrm{sin}\\left(\\theta\\right) }{\\mathrm{cos}\\left(\\theta\\right)}.[\/latex]<\/p>\r\n<p id=\"fs-id2780700\">(Hint: Multiply the numerator and denominator on the left side by [latex]1-\\mathrm{sin}\\left(\\theta\\right) .)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2780726\">[reveal-answer q=\"fs-id2780726\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2780726\"]\r\n<div id=\"fs-id2780728\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{cos}\\left(\\theta\\right)}{1+\\mathrm{sin}\\left(\\theta\\right)}\\left(\\frac{1-\\mathrm{sin}\\left(\\theta\\right)}{1-\\mathrm{sin}\\left(\\theta\\right)}\\right)&amp;=\\frac{\\mathrm{cos}\\left(\\theta\\right)\\left(1-\\mathrm{sin}\\left(\\theta\\right)\\right)}{1-\\mathrm{sin}^{2}\\left(\\theta\\right)}\\\\ &amp;=\\frac{\\mathrm{cos}\\left(\\theta\\right)\\left(1-\\mathrm{sin}\\left(\\theta\\right)\\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right))}&amp;&amp;\\text{We know}\\text{ } 1-\\mathrm{sin}^{2}\\left(\\theta\\right)=\\mathrm{cos}^{2}\\left(\\theta\\right).\\\\ &amp;=\\frac{1-\\mathrm{sin}\\left(\\theta\\right)}{\\mathrm{cos}\\left(\\theta\\right)}&amp;&amp;\\text{Cancel a factor of}\\text{ }\\mathrm{cos}\\left(\\theta\\right).\\end{align*}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1646590\" class=\"precalculus media\">\r\n<div class=\"textbox shaded\">\r\n<h3 id=\"fs-id1646597\">Media<\/h3>\r\nAccess these online resources for additional instruction and practice with the fundamental trigonometric identities.\r\n<ul id=\"fs-id1646601\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/funtrigiden\">Fundamental Trigonometric Identities<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/verifytrigiden\">Verifying Trigonometric Identities<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2056229\" class=\"key-equations\">\r\n<h3>Key Equations<\/h3>\r\n<table id=\"fs-id2056235\" style=\"height: 256px\" summary=\"..\">\r\n<tbody>\r\n<tr style=\"height: 40px\">\r\n<td class=\"border\" style=\"height: 40px;width: 78.5px\"><strong>Pythagorean identities<\/strong><\/td>\r\n<td class=\"border\" style=\"text-align: center;height: 40px;width: 548.5px\">[latex]\\begin{align*}{\\mathrm{sin}}^{2}\\left(\\theta\\right) +{\\mathrm{cos}}^{2}\\left(\\theta\\right)&amp;=1\\\\ 1+{\\mathrm{cot}}^{2}\\left(\\theta\\right) &amp;={\\mathrm{csc}}^{2}\\left(\\theta\\right)\\\\ 1+{\\mathrm{tan}}^{2}\\left(\\theta\\right) &amp;={\\mathrm{sec}}^{2}\\left(\\theta\\right) \\end{align*}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 68px\">\r\n<td class=\"border\" style=\"height: 68px;width: 78.5px\"><strong>Even-odd identities<\/strong><\/td>\r\n<td class=\"border\" style=\"text-align: center;height: 68px;width: 548.5px\">[latex]\\begin{align*}\\mathrm{tan}\\left(-\\theta \\right)&amp;=-\\mathrm{tan}\\left(\\theta\\right) \\\\ \\mathrm{cot}\\left(-\\theta \\right)&amp;=-\\mathrm{cot}\\left(\\theta\\right) \\\\ \\mathrm{sin}\\left(-\\theta \\right)&amp;=-\\mathrm{sin}\\left(\\theta\\right)\\\\ \\mathrm{csc}\\left(-\\theta \\right)&amp;=-\\mathrm{csc}\\left(\\theta\\right) \\\\ \\mathrm{cos}\\left(-\\theta \\right)&amp;=\\mathrm{cos}\\left(\\theta\\right) \\\\ \\mathrm{sec}\\left(-\\theta \\right)&amp;=\\mathrm{sec}\\left(\\theta\\right) \\end{align*}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 82px\">\r\n<td class=\"border\" style=\"height: 82px;width: 78.5px\"><strong>Reciprocal identities<\/strong><\/td>\r\n<td class=\"border\" style=\"text-align: center;height: 82px;width: 548.5px\">[latex]\\begin{align*}\\mathrm{sin}\\left(\\theta\\right)&amp;=\\frac{1}{\\mathrm{csc}\\left(\\theta\\right) }\\\\ \\mathrm{cos}\\left(\\theta\\right)&amp;=\\frac{1}{\\mathrm{sec}\\left(\\theta\\right)}\\\\ \\mathrm{tan}\\left(\\theta\\right) &amp;=\\frac{1}{\\mathrm{cot}\\left(\\theta\\right) }\\\\ \\mathrm{csc}\\left(\\theta\\right)&amp;=\\frac{1}{\\mathrm{sin}\\left(\\theta\\right) }\\\\ \\mathrm{sec}\\left(\\theta\\right) &amp;=\\frac{1}{\\mathrm{cos}\\left(\\theta\\right)}\\\\ \\mathrm{cot}\\left(\\theta\\right)&amp;=\\frac{1}{\\mathrm{tan}\\left(\\theta\\right)}\\end{align*}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 40px\">\r\n<td class=\"border\" style=\"height: 40px;width: 78.5px\"><strong>Quotient identities<\/strong><\/td>\r\n<td class=\"border\" style=\"text-align: center;height: 40px;width: 548.5px\">[latex]\\begin{align*} \\mathrm{tan}\\left(\\theta\\right) &amp;=\\frac{\\mathrm{sin}\\left(\\theta\\right)}{\\mathrm{cos}\\left(\\theta\\right) }\\\\ \\mathrm{cot}\\left(\\theta\\right) &amp;=\\frac{\\mathrm{cos}\\left(\\theta\\right) }{\\mathrm{sin}\\left(\\theta\\right) }\\end{align*}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 26px\">\r\n<td class=\"border\" style=\"height: 40px;width: 78.5px\"><strong>Shift Identities<\/strong><\/td>\r\n<td class=\"border\" style=\"text-align: center;height: 40px;width: 548.5px\">[latex]\\begin{align*}\\mathrm{sin}\\left(\\theta\\right)&amp;=\\mathrm{cos}\\left(\\theta-\\frac{\\pi}{2}\\right)\\\\ \\mathrm{cos}\\left(\\theta\\right)&amp;=\\mathrm{sin}\\left(\\theta-\\frac{\\pi}{2}\\right)\\end{align*}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id2081298\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\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<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1881971\">\r\n \t<dt>even-odd identities<\/dt>\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-id1882066\">\r\n \t<dt>Pythagorean identities<\/dt>\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<dl id=\"fs-id1882073\">\r\n \t<dt>quotient identities<\/dt>\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<dl id=\"fs-id1882080\">\r\n \t<dt>reciprocal identities<\/dt>\r\n \t<dd id=\"fs-id1882083\">set of equations involving the reciprocals of basic trigonometric definitions<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Verify the fundamental trigonometric identities.<\/li>\n<li>Simplify trigonometric expressions using algebra and the identities.<\/li>\n<\/ul>\n<\/div>\n<div id=\"Figure_07_01_006\" class=\"small\">\n<div style=\"width: 498px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07134234\/CNX_Precalc_Figure_07_01_006.jpg\" alt=\"Photo of international passports.\" width=\"488\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.\u00a0<\/strong>International passports and travel documents<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id2479925\">In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports\u2014there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation.<\/p>\n<p id=\"fs-id1186261\">In this section, we will review some trigonometric identities that we have already seen in earlier sections, create some new ones and show how we can use identities, along with basic tools of algebra, to simplify trigonometric expressions.<\/p>\n<div id=\"fs-id2429534\" class=\"bc-section section\">\n<h3>Some Fundamental Trigonometric Identities<\/h3>\n<p id=\"fs-id1981287\">We have previously discussed the set of <strong>even-odd identities. <\/strong>The <strong><span class=\"no-emphasis\">even-odd identities<\/span><\/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. (See <a class=\"autogenerated-content\" href=\"#Table_07_01_02\">Table 1<\/a>).<\/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<caption><strong>Table 1<\/strong><\/caption>\n<tbody>\n<tr>\n<th class=\"border\" style=\"text-align: center\" colspan=\"3\">Even-Odd Identities<\/th>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center\">[latex]\\begin{array}{l}\\mathrm{tan}\\left(-\\theta \\right)=-\\mathrm{tan}\\text{ }\\left(\\theta\\right)\\hfill \\\\ \\mathrm{cot}\\left(-\\theta \\right)=-\\mathrm{cot}\\text{ }\\left(\\theta\\right) \\hfill \\end{array}[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]\\begin{array}{l}\\mathrm{sin}\\left(-\\theta \\right)=-\\mathrm{sin}\\text{ }\\left(\\theta\\right) \\hfill \\\\ \\mathrm{csc}\\left(-\\theta \\right)=-\\mathrm{csc}\\text{ }\\left(\\theta\\right) \\hfill \\end{array}[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]\\begin{array}{l}\\mathrm{cos}\\left(-\\theta \\right)=\\mathrm{cos}\\text{ }\\left(\\theta\\right) \\hfill \\\\ \\mathrm{sec}\\left(-\\theta \\right)=\\mathrm{sec}\\text{ }\\left(\\theta\\right) \\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1419861\">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. See <a class=\"autogenerated-content\" href=\"#fs-id2031263\">Table 2<\/a>.<\/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<caption><strong>Table 2<\/strong><\/caption>\n<tbody>\n<tr>\n<th class=\"border\" style=\"text-align: center\" colspan=\"2\">Reciprocal Identities<\/th>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{sin}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{csc}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{csc}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{sin}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{cos}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{sec}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{sec}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{cos}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{tan}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{cot}\\text{ }\\left(\\theta\\right) }[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{cot}\\text{ }\\left(\\theta\\right) =\\frac{1}{\\mathrm{tan}\\text{ }\\left(\\theta\\right)}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1354190\">Another set of identities is the set of <strong><span class=\"no-emphasis\">quotient identities<\/span><\/strong>, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities. See <a class=\"autogenerated-content\" href=\"#fs-id937819\">Table 3<\/a>.<\/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<caption>Table 3<\/caption>\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<th class=\"border\" style=\"text-align: center\" colspan=\"2\">Quotient Identities<\/th>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{tan}\\left(\\theta\\right)=\\frac{\\mathrm{sin}\\left(\\theta\\right) }{\\mathrm{cos}\\left(\\theta\\right)}[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{cot}\\left(\\theta\\right)=\\frac{\\mathrm{cos}\\left(\\theta\\right) }{\\mathrm{sin}\\left(\\theta\\right) }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1165137732163\" class=\"bc-section section\">\n<h4>Alternate Forms of the Pythagorean Identity<\/h4>\n<p id=\"fs-id1165135203709\">We can use these fundamental identities to derive alternative forms of the <span class=\"no-emphasis\">Pythagorean Identity<\/span>,<\/p>\n<p style=\"text-align: center\">[latex]{\\mathrm{cos}}^{2}\\left(\\theta\\right)+{\\mathrm{sin}}^{2}\\left(\\theta\\right)=1.[\/latex]<\/p>\n<div>\n<div id=\"fs-id2429534\" class=\"bc-section section\">\n<p id=\"fs-id2113436\"><span style=\"font-size: 1rem;text-align: initial\">You should recall that the identity shown above is a direct result of our definition of the sine and cosine functions in terms of the coordinates of points on the unit circle.\u00a0\u00a0We can derive two more identities using the methods shown below.<\/span><\/p>\n<p id=\"fs-id1313011\">The identity [latex]1+\\mathrm{cot}^{2}\\left(\\theta\\right) =\\mathrm{csc}^{2}\\left(\\theta\\right)[\/latex] is found by dividing each term of the first identity by [latex]\\mathrm{sin}^{2}\\left(\\theta\\right)[\/latex], and then rewriting each part of the equation using the identities we have already discussed in earlier sections.<\/p>\n<p id=\"fs-id1286496\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sin}^{2}\\left(\\theta \\right)}{\\mathrm{sin}^{2}\\left(\\theta\\right)}+\\frac{\\mathrm{cos}^{2}\\left(\\theta\\right)}{\\mathrm{sin}^{2}\\left(\\theta\\right)}& =\\frac{1}{\\mathrm{sin}^{2}\\left(\\theta\\right)}\\end{align*}[\/latex]<\/p>\n<p>We can then use our earlier quotient and reciprocal identities to rewrite the expression in this equation as shown below.<\/p>\n<p id=\"fs-id1286496\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}1+\\mathrm{cot}^{2}\\left(\\theta\\right)& =\\mathrm{csc}^{2}\\left(\\theta\\right)\\end{align*}[\/latex]<\/p>\n<p>Similarly, [latex]1+\\mathrm{tan}^{2}\\left(\\theta\\right) =\\mathrm{sec}^{2}\\left(\\theta\\right)[\/latex] can be obtained by dividing all terms in the first identity by [latex]\\mathrm{cos}^{2}\\left(\\theta\\right)[\/latex], and then rewriting each part of the equation using the quotient and reciprocal identities .<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sin}^{2}\\left(\\theta \\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right)}+\\frac{\\mathrm{cos}^{2}\\left(\\theta\\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right)}& =\\frac{1}{\\mathrm{cos}^{2}\\left(\\theta\\right)}\\end{align*}[\/latex]<\/p>\n<\/div>\n<p style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{tan}^{2}\\left(\\theta\\right)+1& =\\mathrm{sec}^{2}\\left(\\theta\\right)\\end{align*}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id2113436\">We now have the three<span style=\"font-size: 1rem;text-align: initial\">\u00a0<\/span><strong style=\"font-size: 1rem;text-align: initial\">Pythagorean identities<\/strong>\u00a0shown in <a class=\"autogenerated-content\" style=\"font-size: 1rem;text-align: initial\" href=\"#Table_07_01_01\">Table 4<\/a><span style=\"font-size: 1rem;text-align: initial\">.\u00a0\u00a0<\/span><\/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<caption>Table 4<\/caption>\n<colgroup>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<th class=\"border\" style=\"text-align: center\" colspan=\"3\">Pythagorean Identities<\/th>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center\">[latex]\\mathrm{sin}^{2}\\left(\\theta\\right) +\\mathrm{cos}^{2}\\left(\\theta\\right) =1[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]1+{\\mathrm{cot}}^{2}\\left(\\theta\\right) ={\\mathrm{csc}}^{2}\\left(\\theta\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]1+{\\mathrm{tan}}^{2}\\left(\\theta\\right) ={\\mathrm{sec}}^{2}\\left(\\theta\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id2429534\" class=\"bc-section section\" style=\"text-align: center\">\n<div id=\"Example_07_01_01\" class=\"textbox examples\">\n<h3>Example 1:\u00a0 Graphing the Equations of an Identity<\/h3>\n<p id=\"fs-id1693457\">Graph both sides of the identity [latex]\\mathrm{cot}\\left(\\theta\\right) =\\frac{1}{\\mathrm{tan}\\left(\\theta\\right)}.[\/latex] In other words, on the graphing calculator, graph [latex]y=\\mathrm{cot}\\left(\\theta\\right)[\/latex] and [latex]y=\\frac{1}{\\mathrm{tan}\\left(\\theta\\right)}.[\/latex]<\/p>\n<div id=\"fs-id1208452\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1208452\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1208452\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1859063\">See <a class=\"autogenerated-content\" href=\"#Figure_07_01_007\">Figure 1<\/a>.<\/p>\n<div id=\"Figure_07_01_007\" class=\"small\">\n<div style=\"width: 397px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07134244\/CNX_Precalc_Figure_07_01_007.jpg\" alt=\"Graph of y = cot(theta) and y=1\/tan(theta) from -2pi to 2pi. They are the same!\" width=\"387\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.<\/strong><\/p>\n<\/div>\n<h3>Analysis<\/h3>\n<p id=\"fs-id3063508\">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 confirm an identity verified with analytical means. If both expressions give the same graph, then they are most likely identities.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2467864\" class=\"precalculus howto examples\">\n<h3>How To:<\/h3>\n<p id=\"fs-id693742\"><strong>Given a trigonometric identity, verify that it is true. <\/strong><\/p>\n<ol>\n<li style=\"text-align: left\">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 style=\"text-align: left\">Look for opportunities to factor expressions, square a binomial, or add fractions.<\/li>\n<li style=\"text-align: left\">Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.<\/li>\n<li style=\"text-align: left\">If these steps do not yield the desired result, try converting all terms to sines and cosines.<\/li>\n<li style=\"text-align: left\">Note the values not in the domain of the expression on the left and right as the identity does not hold for those values.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_07_01_02\" class=\"textbox examples\">\n<div id=\"fs-id3057901\">\n<div id=\"fs-id2291660\">\n<h3>Example 2:\u00a0 Verifying a Trigonometric Identity<\/h3>\n<p id=\"fs-id1947320\">Verify [latex]\\mathrm{tan}\\left(\\theta\\right) \\mathrm{cos}\\left(\\theta\\right)=\\mathrm{sin}\\left(\\theta\\right) .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1334614\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1334614\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1334614\" class=\"hidden-answer\" style=\"display: none\">\nWe will start on the left side, as it is the more complicated side:<\/p>\n<div>\n<div><\/div>\n<div>[latex]\\begin{align*}\\mathrm{tan}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right)&=\\left(\\frac{\\mathrm{sin}\\text{ }\\left(\\theta\\right)}{\\mathrm{cos}\\left(\\theta\\right)}\\right)\\mathrm{cos}\\left(\\theta\\right)\\\\&=\\mathrm{sin}\\text{ }\\left(\\theta\\right)\\end{align*}[\/latex]<\/div>\n<div>\n<h3>Analysis<\/h3>\n<p id=\"fs-id1266264\">This identity was fairly simple to verify, as it only required writing [latex]\\mathrm{tan}\\left(\\theta\\right)[\/latex] in terms of [latex]\\mathrm{sin}\\left(\\theta\\right)[\/latex] and [latex]\\mathrm{cos}\\left(\\theta\\right) .[\/latex]\u00a0 When determining identities we need to also consider the domains of the expressions on the left and right sides of the equation.\u00a0 The identity is only valid where both expressions are defined.\u00a0 For this problem, the domain of\u00a0[latex]\\mathrm{tan}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right)[\/latex] is all real numbers except [latex]\\frac{\\pi}{2}\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.\u00a0 The expression\u00a0[latex]\\frac{\\pi}{2}\\pm n\\pi[\/latex] means that we will have breaks in the domain starting at [latex]\\frac{\\pi}{2}[\/latex] and then every time we add or subtract [latex]\\pi[\/latex] from there.\u00a0 This means the identity will not be valid at these points.\u00a0 We also need to consider the left hand side of the equation.\u00a0 However, since the domain of [latex]\\mathrm{sin}\\left(\\theta\\right)[\/latex] is all real numbers there are no additional places that the identity is not valid.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id820368\" class=\"precalculus tryit\">\n<h3>Try it #1<\/h3>\n<div id=\"ti_07_01_01\">\n<div id=\"fs-id2335257\">\n<p id=\"fs-id2335258\">Verify the identity [latex]\\mathrm{csc}\\left(\\theta\\right)\\text{ } \\mathrm{cos}\\left(\\theta\\right)\\text{ } \\mathrm{tan}\\left(\\theta\\right) =1.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1237705\" style=\"text-align: left\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1237705\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1237705\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{csc}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right) \\mathrm{tan}\\left(\\theta\\right) &=\\left(\\frac{1}{\\mathrm{sin}\\left(\\theta\\right) }\\right)\\mathrm{cos}\\left(\\theta\\right) \\left(\\frac{\\mathrm{sin}\\left(\\theta\\right) }{\\mathrm{cos}\\left(\\theta\\right) }\\right) \\\\ &=\\frac{\\mathrm{cos}\\left(\\theta\\right)}{\\mathrm{sin}\\left(\\theta\\right) }\\left(\\frac{\\mathrm{sin}\\left(\\theta\\right) }{\\mathrm{cos}\\left(\\theta\\right) }\\right) \\\\ &=\\frac{\\mathrm{sin}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right) }{\\mathrm{sin}\\left(\\theta\\right) \\mathrm{cos}\\left(\\theta\\right)} \\\\ &=1\\end{align*}[\/latex]<\/div>\n<div><\/div>\n<div>This identity is not valid where [latex]\\mathrm{csc}\\left(\\theta\\right)[\/latex] or [latex]\\mathrm{tan}\\left(\\theta\\right)[\/latex] are not defined.\u00a0 This means that it is not valid at [latex]... -\\pi, \\frac{-\\pi}{2}, 0, \\frac{\\pi}{2}, \\pi, ...[\/latex].<\/div>\n<div><\/div>\n<div><span style=\"text-align: center;font-size: 1em\">We can capture these values with the expression\u00a0[latex]\\frac{\\pi}{2}\\pm\\frac{n\\pi}{2}[\/latex]\u00a0where [latex]n[\/latex] is an integer.\u00a0<\/span><\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"ti_07_01_01\">\n<div id=\"fs-id1237705\" style=\"text-align: left\">\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_01_03\" class=\"textbox examples\">\n<div id=\"fs-id2066731\">\n<div id=\"fs-id1128511\">\n<h3>Example 3:\u00a0 Verifying the Equivalency Using the Even-Odd Identities<\/h3>\n<p id=\"fs-id937635\">Verify the following equivalency using the even-odd identities:<\/p>\n<div id=\"fs-id2463271\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\left(1+\\mathrm{sin}\\text{ }\\left(x\\right)\\right)\\left(1+\\mathrm{sin}\\left(-x\\right)\\right)={\\mathrm{cos}}^{2}\\left(x\\right).[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1492841\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1492841\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1492841\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1137525\">Working on the left side of the equation, we have<\/p>\n<div id=\"fs-id1277815\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\left(1+\\mathrm{sin}\\left(x\\right)\\right)\\left(1+\\mathrm{sin}\\left(-x\\right)\\right)&=\\left(1+\\mathrm{sin}\\left(x\\right)\\right)\\left(1-\\mathrm{sin}\\left(x\\right)\\right)&& \\text{Since sin}\\left(-x\\right)=-\\mathrm{sin}\\left(x\\right).\\\\ &=1-\\mathrm{sin}^{2}\\left(x\\right)&&\\text{Difference of squares}.\\\\ &=\\mathrm{cos}^{2}\\left(x\\right)&& \\mathrm{cos}^{2}\\left(x\\right)=1-\\mathrm{sin}^{2}\\left(x\\right)\\text{ from Pythagorean Identity}.\\end{align*}[\/latex]<\/div>\n<div>This identity is valid for all real numbers since [latex]\\mathrm{sin}\\left(\\theta\\right)[\/latex] and [latex]\\mathrm{cos}\\left(\\theta\\right)[\/latex] have a domain of all real numbers.<\/div>\n<\/div>\n<div id=\"fs-id1492841\">\n<div><\/div>\n<div class=\"unnumbered\" style=\"text-align: center\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_01_04\" class=\"textbox examples\">\n<div id=\"fs-id1758470\">\n<div id=\"fs-id1233021\">\n<h3>Example 4:\u00a0 Verifying a Trigonometric Identity Involving <em>sec<sup>2<\/sup>(\u03b8)<\/em><\/h3>\n<p id=\"fs-id2814564\">Verify the identity [latex]\\frac{{\\mathrm{sec}}^{2}\\left(\\theta\\right) -1}{{\\mathrm{sec}}^{2}\\left(\\theta\\right) }={\\mathrm{sin}}^{2}\\left(\\theta\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2123498\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2123498\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2123498\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2123500\">As the left side is more complicated, let\u2019s begin there.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sec}^{2}\\left(\\theta\\right) -1}{\\mathrm{sec}^{2}\\left(\\theta\\right) }&=\\frac{\\left(\\mathrm{tan}^{2}\\left(\\theta\\right) +1\\right)-1}{\\mathrm{sec}^{2}\\left(\\theta\\right)}&&\\text{Because }\\mathrm{sec}^{2}\\left(\\theta\\right) =\\mathrm{tan}^{2}\\left(\\theta\\right) +1. \\\\ &=\\frac{\\mathrm{tan}^{2}\\left(\\theta\\right)}{\\mathrm{sec}^{2}\\left(\\theta\\right)}\\\\ &=\\mathrm{tan}^{2}\\left(\\theta\\right) \\left(\\frac{1}{\\mathrm{sec}^{2}\\left(\\theta\\right)} \\right)\\\\ &=\\mathrm{tan}^{2}\\left(\\theta\\right)\\mathrm{cos}^{2}\\left(\\theta\\right)&&\\text{Because }\\mathrm{cos}^{2}\\left(\\theta\\right) =\\frac{1}{\\mathrm{sec}^{2}\\left(\\theta\\right)}.\\\\ &=\\frac{\\mathrm{sin}^{2}\\left(\\theta\\right)} {\\mathrm{cos}^{2}\\left(\\theta\\right)}\\mathrm{cos}^{2}\\left(\\theta\\right) && \\text{Because }\\mathrm{tan}^{2}\\left(\\theta\\right) =\\frac{\\mathrm{sin}^{2}\\left(\\theta\\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right)}.\\\\ &=\\mathrm{sin}^{2}\\left(\\theta\\right)\\end{align*}[\/latex]<\/p>\n<p id=\"fs-id1244586\">There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.<\/p>\n<div id=\"fs-id2107026\" class=\"unnumbered\" style=\"text-align: left\">\n<p style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sec}^{2}\\left(\\theta\\right) -1}{\\mathrm{sec}^{2}\\left(\\theta\\right) }&=\\frac{\\mathrm{sec}^{2}\\left(\\theta\\right) }{\\mathrm{sec}^{2}\\left(\\theta\\right) }-\\frac{1}{\\mathrm{sec}^{2}\\left(\\theta\\right) }\\\\&=1-\\mathrm{cos}^{2}\\left(\\theta\\right) \\\\ &=\\mathrm{sin}^{2}\\left(\\theta\\right)\\end{align*}[\/latex]<\/p>\n<p>This identity is valid on the domain of [latex]\\mathrm{sec}\\left(\\theta\\right).[\/latex]\u00a0 This means that we need to exclude values where\u00a0[latex]\\mathrm{cos}\\left(\\theta\\right)=0.[\/latex]<\/p>\n<p>This would give us all real numbers except [latex]\\frac{\\pi}{2}\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.<\/p>\n<h3>Analysis<\/h3>\n<p id=\"fs-id1698743\">In the first method, we used the identity [latex]\\mathrm{sec}^{2}\\left(\\theta\\right) = \\mathrm{tan}^{2}\\left(\\theta\\right) +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>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2017058\" class=\"precalculus tryit\">\n<h3>Try it #2<\/h3>\n<div id=\"ti_07_01_02\">\n<div id=\"fs-id1185914\">\n<p id=\"fs-id1185915\">Show that [latex]\\frac{\\mathrm{cot}\\text{ }\\left(\\theta\\right) }{\\mathrm{csc}\\left(\\theta\\right)}=\\mathrm{cos}\\left(\\theta\\right) .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1506268\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1506268\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1506268\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1506270\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{cot}\\left(\\theta\\right)}{\\mathrm{csc}\\left(\\theta\\right)}&=\\frac{\\frac{\\mathrm{cos}\\left(\\theta\\right) }{\\mathrm{sin}\\left(\\theta\\right) }}{\\frac{1}{\\mathrm{sin}\\left(\\theta\\right)}} && \\text{This is a complex fraction.}\\\\ &=\\frac{\\mathrm{cos}\\left(\\theta\\right)}{\\mathrm{sin}\\left(\\theta\\right) }\\cdot \\frac{\\mathrm{sin}\\left(\\theta\\right)}{ 1}&& \\text{Multiple the numerator by the reciprocal of the denominator.}\\\\ &=\\mathrm{cos}\\left(\\theta\\right)\\end{align*}[\/latex]<\/div>\n<div class=\"unnumbered\" style=\"text-align: left\">\n<p>This identity is valid on the domain of [latex]\\mathrm{cot}\\left(\\theta\\right)[\/latex] and [latex]\\mathrm{csc}\\left(\\theta\\right).[\/latex]\u00a0 This means that the identity is not valid for the values [latex]... -2\\pi, -\\pi, 0, \\pi, 2\\pi, ...[\/latex]<\/p>\n<p>This would give us all real numbers except [latex]\\pi\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_01_05\" class=\"textbox examples\">\n<div id=\"fs-id2122140\">\n<div id=\"fs-id1970227\">\n<h3>Example 5: Creating and Verifying an Identity<\/h3>\n<p id=\"fs-id1970233\">Create an identity for the expression [latex]2\\text{ }\\mathrm{tan}\\left(\\theta\\right) \\mathrm{sec}\\left(\\theta\\right)[\/latex] by rewriting strictly in terms of sine.<\/p>\n<\/div>\n<div id=\"fs-id1429666\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1429666\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1429666\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1429668\">There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:<\/p>\n<div id=\"fs-id2338806\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\text{ }\\mathrm{tan}\\left(\\theta\\right)\\mathrm{sec}\\left(\\theta\\right)&=2\\left(\\frac{\\mathrm{sin}\\left(\\theta\\right)}{\\mathrm{cos}\\left(\\theta\\right)}\\right)\\left(\\frac{1}{\\mathrm{cos}\\left(\\theta\\right)}\\right)\\\\ &=\\frac{2\\text{ }\\mathrm{sin}\\left(\\theta\\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right)}\\\\ &=\\frac{2\\text{ }\\mathrm{sin}\\left(\\theta\\right)}{1-\\mathrm{sin}^{2}\\left(\\theta\\right)}&&\\text{Substitute }1-\\mathrm{sin}^{2}\\left(\\theta\\right)\\text{ for }\\mathrm{cos}^{2}\\left(\\theta\\right).\\\\  2\\mathrm{tan}\\left(\\theta\\right)\\mathrm{sec}\\left(\\theta\\right)&=\\frac{2\\text{ }\\mathrm{sin}\\left(\\theta\\right)}{1-\\mathrm{sin}^{2}\\left(\\theta\\right)}\\end{align*}[\/latex]<\/div>\n<div class=\"unnumbered\" style=\"text-align: left\">This identity holds on values in both of the domains of [latex]\\mathrm{tan}\\left(\\theta\\right)[\/latex] and [latex]\\mathrm{sec}\\left(\\theta\\right).[\/latex]\u00a0 This means it holds for all real numbers except [latex]... -\\frac{3\\pi}{2}, -\\frac{\\pi}{2}, \\frac{\\pi}{2}, \\frac{3\\pi}{2}...[\/latex]<\/div>\n<div class=\"unnumbered\" style=\"text-align: left\">This would give us all real numbers except [latex]\\frac{\\pi}{2}\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.<\/div>\n<div class=\"unnumbered\" style=\"text-align: left\">The expression on the left also needs to be considered for restrictions but the denominator is zero exactly when the left side of the expression is undefined so no more values need to be excluded.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_01_06\" class=\"textbox examples\">\n<div id=\"fs-id1219228\">\n<div id=\"fs-id1219230\">\n<h3>Example 6: Verifying an Identity Using Algebra and Even\/Odd Identities<\/h3>\n<p id=\"fs-id2430931\">Verify the identity:<\/p>\n<div id=\"fs-id1583958\" class=\"unnumbered\">[latex]\\frac{\\mathrm{sin}^{2}\\left(-\\theta \\right)-\\mathrm{cos}^{2}\\left(-\\theta \\right)}{\\mathrm{sin}\\left(-\\theta \\right)-\\mathrm{cos}\\left(-\\theta \\right)}=\\mathrm{cos}\\left(\\theta \\right) -\\mathrm{sin}\\left(\\theta \\right)[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<div id=\"fs-id2345275\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2345275\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2345275\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2345277\">Let\u2019s start with the left side and simplify:<\/p>\n<div id=\"fs-id1540225\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sin}^{2}\\left(-\\theta\\right)-\\mathrm{cos}^{2}\\left(-\\theta \\right)}{\\mathrm{sin}\\left(-\\theta\\right)-\\mathrm{cos}\\left(-\\theta \\right)}&=\\frac{\\left(\\mathrm{sin}\\left(-\\theta \\right)-\\mathrm{cos}\\left(-\\theta \\right)\\right)\\left(\\mathrm{sin}\\left(-\\theta \\right)+\\mathrm{cos}\\left(-\\theta \\right)\\right)}{\\mathrm{sin}\\left(-\\theta \\right)-\\mathrm{cos}\\left(-\\theta \\right)}&& \\text{Difference of squares.}\\\\&=\\frac{\\left(-\\mathrm{sin}\\left(\\theta \\right)-\\mathrm{cos}\\left(\\theta \\right)\\right)\\left(-\\mathrm{sin}\\left(\\theta \\right)+\\mathrm{cos}\\left(\\theta \\right)\\right)}{-\\mathrm{sin}\\left(\\theta \\right)-\\mathrm{cos}\\left(\\theta \\right)}&&\\text{Odd-Even Identities.}\\\\&=\\mathrm{-sin}\\left(\\theta \\right)+\\mathrm{cos}\\left(\\theta \\right)&&\\text{Cancel the common factor.}\\\\&=\\mathrm{cos}\\left(\\theta \\right)\\mathrm{-sin}\\left(\\theta \\right)&&\\text{Reorder terms.}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div class=\"unnumbered\" style=\"text-align: left\">This identity will be valid when [latex]\\mathrm{sin}\\left(-\\theta \\right)-\\mathrm{cos}\\left(-\\theta \\right)\\ne 0.[\/latex]\u00a0 Solving this type of equation is discussed in section 3.8 Solving Trigonometric Equations.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2137172\" class=\"precalculus tryit\">\n<h3>Try it #3<\/h3>\n<div id=\"ti_07_01_03\">\n<div id=\"fs-id2261720\">\n<p id=\"fs-id2261721\">Verify the identity [latex]\\frac{\\mathrm{sin}^{2}\\left(\\theta \\right) -1}{\\mathrm{tan}\\left(\\theta \\right)\\mathrm{sin}\\left(\\theta\\right) -\\mathrm{tan}\\left(\\theta\\right)}=\\frac{\\mathrm{sin}\\left(\\theta\\right) +1}{\\mathrm{tan}\\left(\\theta \\right)}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2701905\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2701905\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2701905\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2701906\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{sin}^{2}\\left(\\theta \\right) -1}{\\mathrm{tan}\\left(\\theta \\right) \\mathrm{sin}\\left(\\theta \\right) -\\mathrm{tan}\\left(\\theta \\right)}&=\\frac{\\left(\\mathrm{sin}\\left(\\theta \\right) +1\\right)\\left(\\mathrm{sin}\\left(\\theta \\right) -1\\right)}{\\mathrm{tan}\\left(\\theta \\right)\\left(\\mathrm{sin}\\left(\\theta \\right) -1\\right)}\\\\ &=\\frac{\\mathrm{sin}\\left(\\theta \\right)+1}{\\mathrm{tan}\\left(\\theta \\right)}\\end{align*}[\/latex]<\/p>\n<p>This equation is valid on the domain of tangent and where we don&#8217;t get a divide by zero.\u00a0 Determining when the denominator in the expression on the left requires equation solving discussed in section 3.8 Solving Trigonometric Equations.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_01_07\" class=\"textbox examples\">\n<div id=\"fs-id1664723\">\n<div id=\"fs-id1664725\">\n<h3>Example 7: Verifying an Identity Involving Cosines and Cotangents<\/h3>\n<p id=\"fs-id1664730\">Verify the identity: [latex]\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)\\left(1+{\\mathrm{cot}}^{2}\\left(x\\right)\\right)=1.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1913600\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1913600\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1913600\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1913602\">We will work on the left side of the equation.<\/p>\n<div id=\"fs-id1913605\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)&\\left(1+\\mathrm{cot}^{2}\\left(x\\right)\\right)\\\\&=\\left(1-\\mathrm{cos}^{2}\\left(x\\right)\\right)\\left(1+\\frac{\\mathrm{cos}^{2}\\left(x\\right)}{\\mathrm{sin}^{2}\\left(x\\right)}\\right)&&\\text{Rewrite }\\mathrm{cot}^{2}\\left(x\\right). \\\\ &=\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)\\left(\\frac{{\\mathrm{sin}}^{2}\\left(x\\right)}{{\\mathrm{sin}}^{2}x}+\\frac{{\\mathrm{cos}}^{2}x}{{\\mathrm{sin}}^{2}\\left(x\\right)}\\right) &&\\text{Find the common denominator.}\\\\ &=\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)\\left(\\frac{\\mathrm{sin}^{2}\\left(x\\right)+\\mathrm{cos}^{2}\\left(x\\right)}{\\mathrm{sin}^{2}\\left(x\\right)}\\right) \\\\ &=\\left({\\mathrm{sin}}^{2}\\left(x\\right)\\right)\\left(\\frac{\\mathrm{sin}^{2}\\left(x\\right)+\\mathrm{cos}^{2}\\left(x\\right)}{\\mathrm{sin}^{2}\\left(x\\right)}\\right)&&\\text{Because }1-\\mathrm{cos}^{2}\\left(x\\right) =\\mathrm{sin}^{2}\\left(x\\right).\\\\&=\\left({\\mathrm{sin}}^{2}\\left(x\\right)\\right)\\left(\\frac{1}{{\\mathrm{sin}}^{2}\\left(x\\right)}\\right)&& \\text{Because } \\mathrm{sin}^{2}\\left(x\\right)+\\mathrm{cos}^{2}\\left(x\\right)=1.\\\\ &=1\\end{align*}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div>This identity is valid on the domain of cotangent.\u00a0\u00a0This would give us all real numbers except [latex]\\pi\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.<\/div>\n<div><\/div>\n<div>An alternate way to verify this identity is to begin by multiplying out the expression on the left and simplifying.<\/div>\n<div style=\"text-align: center\">[latex]\\begin{align*}\\left(1-{\\mathrm{cos}}^{2}\\left(x\\right)\\right)\\left(1+\\mathrm{cot}^{2}\\left(x\\right)\\right)&=1-\\mathrm{cos}^{2}\\left(x\\right)+\\mathrm{cot}^{2}\\left(x\\right)-\\mathrm{cot}^{2}\\left(x\\right)\\mathrm{cos}^{2}\\left(x\\right)\\\\&=1-\\mathrm{cos}^{2}\\left(x\\right)+\\mathrm{cot}^{2}\\left(x\\right)\\left(1-\\mathrm{cos}^{2}\\left(x\\right)\\right)\\\\&=1-\\mathrm{cos}^{2}\\left(x\\right)+\\frac{\\mathrm{cos}^{2}\\left(x\\right)}{\\mathrm{sin}^{2}\\left(x\\right)}\\mathrm{sin}^{2}\\left(x\\right)\\\\&=1-\\mathrm{cos}^{2}\\left(x\\right)+\\mathrm{cos}^{2}\\left(x\\right)\\\\&=1\\end{align*}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<div id=\"fs-id1913600\">\n<div class=\"unnumbered\" style=\"text-align: center\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2485952\" class=\"bc-section section\">\n<h3>Using Algebra to Simplify Trigonometric Expressions<\/h3>\n<p id=\"fs-id2485958\">We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions when we are solving equations. 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>An 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<\/p>\n<p id=\"fs-id2485964\">For example, the expression [latex]\\mathrm{sin}^2\\left(x\\right)-1[\/latex] resembles the difference of squares [latex]x^2-1.[\/latex] Recognizing that [latex]x^2-1[\/latex] can be factored as [latex]\\left(x+1\\right)\\left(x-1\\right)[\/latex] helps us quickly recognize that\u00a0 [latex]\\mathrm{sin}^2\\left(x\\right)-1[\/latex] can be factored as [latex]\\left(\\mathrm{sin}\\left(x\\right)+1\\right)\\left(\\mathrm{sin}\\left(x\\right)-1\\right).[\/latex]<\/p>\n<p id=\"fs-id1814865\">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 expressions and equations easier to work with.<\/p>\n<div id=\"Example_07_01_08\" class=\"textbox examples\">\n<div id=\"fs-id1814944\">\n<div id=\"fs-id2578086\">\n<h3>Example 8: Writing the Trigonometric Expression as an Algebraic Expression<\/h3>\n<p id=\"fs-id2578091\">Write the following trigonometric expression as an algebraic expression: [latex]2\\mathrm{cos}^{2}\\left(\\theta\\right) +\\mathrm{cos}\\left(\\theta\\right) -1.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2578132\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2578132\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2578132\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2578134\">Notice that the pattern displayed has the same form as a standard quadratic expression, [latex]a{x}^{2}+bx+c.[\/latex] Letting [latex]\\mathrm{cos}\\left(\\theta\\right) =x,[\/latex] we can rewrite the expression as follows:<\/p>\n<div id=\"fs-id2121948\" class=\"unnumbered\" style=\"text-align: center\">[latex]2{x}^{2}+x-1[\/latex]<\/div>\n<p id=\"fs-id2121978\">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]\\mathrm{cos}\\left(\\theta\\right)[\/latex] and solve for [latex]\\theta .[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_05_09\" class=\"textbox examples\">\n<div id=\"fs-id2721228\">\n<div id=\"fs-id2721230\">\n<h3>Example 9: Rewriting a Trigonometric Expression Using the Difference of Squares<\/h3>\n<p id=\"fs-id1777385\">Rewrite the trigonometric expression: [latex]4\\mathrm{cos}^{2}\\left(\\theta\\right) -1.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1777420\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1777420\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1777420\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1777422\">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<div id=\"fs-id1777427\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}4\\text{ }\\mathrm{cos}^{2}\\left(\\theta\\right) -1&=\\left(2 \\mathrm{cos}\\left(\\theta\\right)-1 \\right)\\left(2\\text{ }\\mathrm{cos}\\left(\\theta\\right) +1\\right)\\end{align*}[\/latex]<\/div>\n<div id=\"fs-id1739635\">\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]\\mathrm{cos}\\left(\\theta\\right) =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]\\mathrm{cos}\\left(\\theta\\right)[\/latex] and solve for the angle.<\/p>\n<div id=\"fs-id1777427\" class=\"unnumbered\" style=\"text-align: center\"><\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2438482\" class=\"precalculus tryit\">\n<h3>Try it #4<\/h3>\n<div id=\"ti_07_01_04\">\n<div id=\"fs-id2438492\">\n<p id=\"fs-id2438493\">Rewrite the trigonometric expression: [latex]25-9\\text{ }\\mathrm{sin}^{2}\\left(\\theta\\right) .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2438529\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2438529\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2438529\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2438532\">This is a difference of squares formula: [latex]25-9\\text{ }\\mathrm{sin}^{2}\\left(\\theta\\right)=\\left(5-3\\text{ }\\mathrm{sin}\\left(\\theta\\right) \\right)\\left(5+3\\text{ }\\mathrm{sin}\\left(\\theta\\right)\\right).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_01_10\" class=\"textbox examples\">\n<div id=\"fs-id2917234\">\n<div id=\"fs-id2917236\">\n<h3>Example 10: Simplify by Rewriting and Using Substitution<\/h3>\n<p id=\"fs-id2917242\">Simplify the expression by rewriting and using identities:<\/p>\n<div id=\"fs-id2917245\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\mathrm{csc}^{2}\\left(\\theta\\right) -\\mathrm{cot}^{2}\\left(\\theta\\right)[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<div id=\"fs-id2483936\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2483936\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2483936\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2483938\">We can start with the Pythagorean identity.\u00a0 We know:\u00a0[latex]1+\\mathrm{cot}^{2}\\left(\\theta\\right)=\\mathrm{csc}^{2}\\left(\\theta\\right)[\/latex]<\/p>\n<div id=\"fs-id2483941\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}  \\mathrm{csc}^{2}\\left(\\theta\\right)-\\mathrm{cot}^{2}\\left(\\theta\\right)&=1+\\mathrm{cot}^{2}\\left(\\theta\\right)-\\mathrm{cot}^{2}\\left(\\theta\\right)&&\\text{Substitute expression in for }\\mathrm{csc}^{2}\\left(\\theta\\right).\\\\ &=1\\end{align*}[\/latex]<\/div>\n<div>This identity is valid on the domains of cotangent and cosecant.\u00a0\u00a0This would give us all real numbers except [latex]\\pi\\pm n\\pi[\/latex] where [latex]n[\/latex] is an integer.<\/div>\n<div class=\"unnumbered\" style=\"text-align: center\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2607392\" class=\"precalculus tryit\">\n<h3>Try it #5<\/h3>\n<div id=\"ti_07_01_05\">\n<div id=\"fs-id2607401\">\n<p id=\"fs-id2607402\">Use algebraic techniques to verify the identity: [latex]\\frac{\\mathrm{cos}\\left(\\theta\\right) }{1+\\mathrm{sin}\\left(\\theta\\right) }=\\frac{1-\\mathrm{sin}\\left(\\theta\\right) }{\\mathrm{cos}\\left(\\theta\\right)}.[\/latex]<\/p>\n<p id=\"fs-id2780700\">(Hint: Multiply the numerator and denominator on the left side by [latex]1-\\mathrm{sin}\\left(\\theta\\right) .)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2780726\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2780726\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2780726\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id2780728\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\frac{\\mathrm{cos}\\left(\\theta\\right)}{1+\\mathrm{sin}\\left(\\theta\\right)}\\left(\\frac{1-\\mathrm{sin}\\left(\\theta\\right)}{1-\\mathrm{sin}\\left(\\theta\\right)}\\right)&=\\frac{\\mathrm{cos}\\left(\\theta\\right)\\left(1-\\mathrm{sin}\\left(\\theta\\right)\\right)}{1-\\mathrm{sin}^{2}\\left(\\theta\\right)}\\\\ &=\\frac{\\mathrm{cos}\\left(\\theta\\right)\\left(1-\\mathrm{sin}\\left(\\theta\\right)\\right)}{\\mathrm{cos}^{2}\\left(\\theta\\right))}&&\\text{We know}\\text{ } 1-\\mathrm{sin}^{2}\\left(\\theta\\right)=\\mathrm{cos}^{2}\\left(\\theta\\right).\\\\ &=\\frac{1-\\mathrm{sin}\\left(\\theta\\right)}{\\mathrm{cos}\\left(\\theta\\right)}&&\\text{Cancel a factor of}\\text{ }\\mathrm{cos}\\left(\\theta\\right).\\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1646590\" class=\"precalculus media\">\n<div class=\"textbox shaded\">\n<h3 id=\"fs-id1646597\">Media<\/h3>\n<p>Access these online resources for additional instruction and practice with the fundamental trigonometric identities.<\/p>\n<ul id=\"fs-id1646601\">\n<li><a href=\"http:\/\/openstax.org\/l\/funtrigiden\">Fundamental Trigonometric Identities<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/verifytrigiden\">Verifying Trigonometric Identities<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2056229\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"fs-id2056235\" style=\"height: 256px\" summary=\"..\">\n<tbody>\n<tr style=\"height: 40px\">\n<td class=\"border\" style=\"height: 40px;width: 78.5px\"><strong>Pythagorean identities<\/strong><\/td>\n<td class=\"border\" style=\"text-align: center;height: 40px;width: 548.5px\">[latex]\\begin{align*}{\\mathrm{sin}}^{2}\\left(\\theta\\right) +{\\mathrm{cos}}^{2}\\left(\\theta\\right)&=1\\\\ 1+{\\mathrm{cot}}^{2}\\left(\\theta\\right) &={\\mathrm{csc}}^{2}\\left(\\theta\\right)\\\\ 1+{\\mathrm{tan}}^{2}\\left(\\theta\\right) &={\\mathrm{sec}}^{2}\\left(\\theta\\right) \\end{align*}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 68px\">\n<td class=\"border\" style=\"height: 68px;width: 78.5px\"><strong>Even-odd identities<\/strong><\/td>\n<td class=\"border\" style=\"text-align: center;height: 68px;width: 548.5px\">[latex]\\begin{align*}\\mathrm{tan}\\left(-\\theta \\right)&=-\\mathrm{tan}\\left(\\theta\\right) \\\\ \\mathrm{cot}\\left(-\\theta \\right)&=-\\mathrm{cot}\\left(\\theta\\right) \\\\ \\mathrm{sin}\\left(-\\theta \\right)&=-\\mathrm{sin}\\left(\\theta\\right)\\\\ \\mathrm{csc}\\left(-\\theta \\right)&=-\\mathrm{csc}\\left(\\theta\\right) \\\\ \\mathrm{cos}\\left(-\\theta \\right)&=\\mathrm{cos}\\left(\\theta\\right) \\\\ \\mathrm{sec}\\left(-\\theta \\right)&=\\mathrm{sec}\\left(\\theta\\right) \\end{align*}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 82px\">\n<td class=\"border\" style=\"height: 82px;width: 78.5px\"><strong>Reciprocal identities<\/strong><\/td>\n<td class=\"border\" style=\"text-align: center;height: 82px;width: 548.5px\">[latex]\\begin{align*}\\mathrm{sin}\\left(\\theta\\right)&=\\frac{1}{\\mathrm{csc}\\left(\\theta\\right) }\\\\ \\mathrm{cos}\\left(\\theta\\right)&=\\frac{1}{\\mathrm{sec}\\left(\\theta\\right)}\\\\ \\mathrm{tan}\\left(\\theta\\right) &=\\frac{1}{\\mathrm{cot}\\left(\\theta\\right) }\\\\ \\mathrm{csc}\\left(\\theta\\right)&=\\frac{1}{\\mathrm{sin}\\left(\\theta\\right) }\\\\ \\mathrm{sec}\\left(\\theta\\right) &=\\frac{1}{\\mathrm{cos}\\left(\\theta\\right)}\\\\ \\mathrm{cot}\\left(\\theta\\right)&=\\frac{1}{\\mathrm{tan}\\left(\\theta\\right)}\\end{align*}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 40px\">\n<td class=\"border\" style=\"height: 40px;width: 78.5px\"><strong>Quotient identities<\/strong><\/td>\n<td class=\"border\" style=\"text-align: center;height: 40px;width: 548.5px\">[latex]\\begin{align*} \\mathrm{tan}\\left(\\theta\\right) &=\\frac{\\mathrm{sin}\\left(\\theta\\right)}{\\mathrm{cos}\\left(\\theta\\right) }\\\\ \\mathrm{cot}\\left(\\theta\\right) &=\\frac{\\mathrm{cos}\\left(\\theta\\right) }{\\mathrm{sin}\\left(\\theta\\right) }\\end{align*}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 26px\">\n<td class=\"border\" style=\"height: 40px;width: 78.5px\"><strong>Shift Identities<\/strong><\/td>\n<td class=\"border\" style=\"text-align: center;height: 40px;width: 548.5px\">[latex]\\begin{align*}\\mathrm{sin}\\left(\\theta\\right)&=\\mathrm{cos}\\left(\\theta-\\frac{\\pi}{2}\\right)\\\\ \\mathrm{cos}\\left(\\theta\\right)&=\\mathrm{sin}\\left(\\theta-\\frac{\\pi}{2}\\right)\\end{align*}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id2081298\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\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<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1881971\">\n<dt>even-odd identities<\/dt>\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-id1882066\">\n<dt>Pythagorean identities<\/dt>\n<dd id=\"fs-id1882069\">set of equations involving trigonometric functions based on the right triangle properties<\/dd>\n<\/dl>\n<dl id=\"fs-id1882073\">\n<dt>quotient identities<\/dt>\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<dl id=\"fs-id1882080\">\n<dt>reciprocal identities<\/dt>\n<dd id=\"fs-id1882083\">set of equations involving the reciprocals of basic trigonometric definitions<\/dd>\n<\/dl>\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-762\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Solving Trigonometric Equations with Identities. <strong>Authored by<\/strong>: Douglas Hoffman. <strong>Provided by<\/strong>: Openstax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/8si1Yf2B@2.21:K0AU6oK2@8\/Solving-Trigonometric-Equations-with-Identities\">https:\/\/cnx.org\/contents\/8si1Yf2B@2.21:K0AU6oK2@8\/Solving-Trigonometric-Equations-with-Identities<\/a>. <strong>Project<\/strong>: Essential Precalcus, Part 2. <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 <\/section>","protected":false},"author":311,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Solving Trigonometric Equations with Identities\",\"author\":\"Douglas Hoffman\",\"organization\":\"Openstax\",\"url\":\"https:\/\/cnx.org\/contents\/8si1Yf2B@2.21:K0AU6oK2@8\/Solving-Trigonometric-Equations-with-Identities\",\"project\":\"Essential Precalcus, Part 2\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-762","chapter","type-chapter","status-publish","hentry"],"part":478,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/762","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":51,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/762\/revisions"}],"predecessor-version":[{"id":3241,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/762\/revisions\/3241"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/parts\/478"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/762\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/media?parent=762"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=762"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/contributor?post=762"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/license?post=762"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}