{"id":1936,"date":"2023-10-12T00:36:07","date_gmt":"2023-10-12T00:36:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/double-angle-half-angle-and-reduction-formulas\/"},"modified":"2024-03-11T18:30:50","modified_gmt":"2024-03-11T18:30:50","slug":"double-angle-half-angle-and-reduction-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/double-angle-half-angle-and-reduction-formulas\/","title":{"raw":"Double Angle, Half Angle, and Reduction Formulas","rendered":"Double Angle, Half Angle, and Reduction Formulas"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Use double-angle formulas to find exact values.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use double-angle formulas to verify identities.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use reduction formulas to simplify an expression.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use half-angle formulas to find exact values.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Using Double-Angle Formulas to Find Exact Values<\/h2>\r\nIn the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The <strong>double-angle formulas<\/strong> are a special case of the sum formulas, where [latex]\\alpha =\\beta [\/latex]. Deriving the double-angle formula for sine begins with the sum formula,\r\n<div style=\"text-align: center;\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta [\/latex]<\/div>\r\nIf we let [latex]\\alpha =\\beta =\\theta [\/latex], then we have\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(\\theta +\\theta \\right)&amp;=\\sin \\theta \\cos \\theta +\\cos \\theta \\sin \\theta \\\\ \\sin \\left(2\\theta \\right)&amp;=2\\sin \\theta \\cos \\theta \\end{align}[\/latex]<\/div>\r\nDeriving the double-angle for cosine gives us three options. First, starting from the sum formula, [latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta [\/latex], and letting [latex]\\alpha =\\beta =\\theta [\/latex], we have\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(\\theta +\\theta \\right)&amp;=\\cos \\theta \\cos \\theta -\\sin \\theta \\sin \\theta \\\\ \\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\end{align}[\/latex]<\/div>\r\nUsing the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;=\\left(1-{\\sin }^{2}\\theta \\right)-{\\sin }^{2}\\theta \\\\ &amp;=1 - 2{\\sin }^{2}\\theta\\end{align}[\/latex]<\/div>\r\nThe second interpretation is:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;={\\cos }^{2}\\theta -\\left(1-{\\cos }^{2}\\theta \\right) \\\\ &amp;=2{\\cos }^{2}\\theta -1\\end{align}[\/latex]<\/div>\r\nSimilarly, to derive the double-angle formula for tangent, replacing [latex]\\alpha =\\beta =\\theta [\/latex] in the sum formula gives\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&amp;=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\theta +\\theta \\right)&amp;=\\frac{\\tan \\theta +\\tan \\theta }{1-\\tan \\theta \\tan \\theta }\\\\ \\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Double-Angle Formulas<\/h3>\r\nThe <strong>double-angle formulas<\/strong> are summarized as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&amp;=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;=1 - 2{\\sin }^{2}\\theta \\\\ &amp;=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/p>\r\n\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question sameseed=1 hide_question_numbers=1]18740[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n[ohm_question sameseed=1 hide_question_numbers=1]132474[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Draw a triangle to reflect the given information.<\/li>\r\n \t<li>Determine the correct double-angle formula.<\/li>\r\n \t<li>Substitute values into the formula based on the triangle.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Using a Double-Angle Formula to Find the Exact Value Involving Tangent<\/h3>\r\nGiven that [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex] and [latex]\\theta [\/latex] is in quadrant II, find the following:\r\n<ol>\r\n \t<li>[latex]\\sin \\left(2\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\left(2\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\tan \\left(2\\theta \\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"477962\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"477962\"]\r\n\r\nIf we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\\tan \\theta =-\\frac{3}{4}[\/latex], such that [latex]\\theta [\/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\\theta [\/latex] is in the second quadrant, the adjacent side is on the <em>x<\/em>-axis and is negative. Use the <strong>Pythagorean Theorem<\/strong> to find the length of the hypotenuse:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\left(-4\\right)}^{2}+{\\left(3\\right)}^{2}&amp;={c}^{2}\\\\ 16+9&amp;={c}^{2}\\\\ 25&amp;={c}^{2}\\\\ c&amp;=5\\end{align}[\/latex]<\/p>\r\nNow we can draw a triangle similar to the one shown in Figure 2.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164101\/CNX_Precalc_Figure_07_03_0022.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.\" width=\"487\" height=\"251\" \/> <b>Figure 2<\/b>[\/caption]\r\n<ol>\r\n \t<li>Let\u2019s begin by writing the double-angle formula for sine.\r\n<div style=\"text-align: center;\">[latex]\\sin \\left(2\\theta \\right)=2\\sin \\theta \\cos \\theta [\/latex]<\/div>\r\nWe see that we to need to find [latex]\\sin \\theta [\/latex] and [latex]\\cos \\theta [\/latex]. Based on Figure 2, we see that the hypotenuse equals 5, so [latex]\\sin \\theta =\\frac{3}{5}[\/latex], and [latex]\\cos \\theta =\u2212\\frac{4}{5}[\/latex]. Substitute these values into the equation, and simplify.\r\nThus,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&amp;=2\\left(\\frac{3}{5}\\right)\\left(\u2212\\frac{4}{5}\\right) \\\\ &amp;=\u2212\\frac{24}{25} \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Write the double-angle formula for cosine.\r\n<div style=\"text-align: center;\">[latex]\\cos \\left(2\\theta \\right)={\\cos }^{2}\\theta -{\\sin }^{2}\\theta [\/latex]<\/div>\r\nAgain, substitute the values of the sine and cosine into the equation, and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&amp;={\\left(\u2212\\frac{4}{5}\\right)}^{2}\u2212{\\left(\\frac{3}{5}\\right)}^{2} \\\\ &amp;=\\frac{16}{25}\u2212\\frac{9}{25} \\\\ &amp;=\\frac{7}{25}\\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Write the double-angle formula for tangent.\r\n<div style=\"text-align: center;\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2\\tan \\theta }{1\u2212{\\tan }^{2}\\theta }[\/latex]<\/div>\r\nIn this formula, we need the tangent, which we were given as [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex]. Substitute this value into the equation, and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&amp;=\\frac{2\\left(-\\frac{3}{4}\\right)}{1-{\\left(-\\frac{3}{4}\\right)}^{2}} \\\\ &amp;=\\frac{-\\frac{3}{2}}{1-\\frac{9}{16}} \\\\ &amp;=-\\frac{3}{2}\\left(\\frac{16}{7}\\right) \\\\ &amp;=-\\frac{24}{7} \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]\\sin \\alpha =\\frac{5}{8}[\/latex], with [latex]\\theta [\/latex] in quadrant I, find [latex]\\cos \\left(2\\alpha \\right)[\/latex].\r\n\r\n[reveal-answer q=\"827677\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"827677\"]\r\n\r\n[latex]\\cos \\left(2\\alpha \\right)=\\frac{7}{32}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]156232[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Using the Double-Angle Formula for Cosine without Exact Values<\/h3>\r\nUse the double-angle formula for cosine to write [latex]\\cos \\left(6x\\right)[\/latex] in terms of [latex]\\cos \\left(3x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"676542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"676542\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(6x\\right)&amp;=\\cos \\left(3x+3x\\right) \\\\ &amp;=\\cos 3x\\cos 3x-\\sin 3x\\sin 3x \\\\ &amp;={\\cos }^{2}3x-{\\sin }^{2}3x \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using Double-Angle Formulas to Verify Identities<\/h2>\r\nEstablishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Using the Double-Angle Formulas to Establish an Identity<\/h3>\r\nEstablish the following identity using double-angle formulas:\r\n<p style=\"text-align: center;\">[latex]1+\\sin \\left(2\\theta \\right)={\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}[\/latex]<\/p>\r\n[reveal-answer q=\"600157\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"600157\"]\r\n\r\nWe will work on the right side of the equal sign and rewrite the expression until it matches the left side.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}&amp;={\\sin }^{2}\\theta +2\\sin \\theta \\cos \\theta +{\\cos }^{2}\\theta \\\\ &amp;=\\left({\\sin }^{2}\\theta +{\\cos }^{2}\\theta \\right)+2\\sin \\theta \\cos \\theta \\\\ &amp;=1+2\\sin \\theta \\cos \\theta \\\\ &amp;=1+\\sin \\left(2\\theta \\right)\\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis process is not complicated, as long as we recall the perfect square formula from algebra:\r\n<p style=\"text-align: center;\">[latex]{\\left(a\\pm b\\right)}^{2}={a}^{2}\\pm 2ab+{b}^{2}[\/latex]<\/p>\r\nwhere [latex]a=\\sin \\theta [\/latex] and [latex]b=\\cos \\theta [\/latex]. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEstablish the identity: [latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\cos \\left(2\\theta \\right)[\/latex].\r\n\r\n[reveal-answer q=\"178499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178499\"]\r\n\r\n[latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)=\\cos \\left(2\\theta \\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Verifying a Double-Angle Identity for Tangent<\/h3>\r\nVerify the identity:\r\n<p style=\"text-align: center;\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\r\n[reveal-answer q=\"395376\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"395376\"]\r\n\r\nIn this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }&amp;&amp; \\text{Double-angle formula} \\\\ &amp;=\\frac{2\\tan \\theta \\left(\\frac{1}{\\tan \\theta }\\right)}{\\left(1-{\\tan }^{2}\\theta \\right)\\left(\\frac{1}{\\tan \\theta }\\right)}&amp;&amp; \\text{Multiply by a term that results in desired numerator}. \\\\ &amp;=\\frac{2}{\\frac{1}{\\tan \\theta }-\\frac{{\\tan }^{2}\\theta }{\\tan \\theta }} \\\\ &amp;=\\frac{2}{\\cot \\theta -\\tan \\theta }&amp;&amp; \\text{Use reciprocal identity for }\\frac{1}{\\tan \\theta }.\\end{align}[\/latex]<\/p>\r\n\r\n<h3>Analysis of the Solution<\/h3>\r\nHere is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show\r\n<p style=\"text-align: center;\">[latex]\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\r\nLet\u2019s work on the right side.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{2}{\\cot \\theta -\\tan \\theta }&amp;=\\frac{2}{\\frac{1}{\\tan \\theta }-\\tan \\theta }\\left(\\frac{\\tan \\theta }{\\tan \\theta }\\right) \\\\ &amp;=\\frac{2\\tan \\theta }{\\frac{1}{\\cancel{\\tan \\theta }}\\left(\\cancel{\\tan \\theta }\\right)-\\tan \\theta \\left(\\tan \\theta \\right)} \\\\ &amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta } \\end{align}[\/latex]<\/p>\r\nWhen using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nVerify the identity: [latex]\\cos \\left(2\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta [\/latex].\r\n\r\n[reveal-answer q=\"584630\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"584630\"]\r\n\r\n[latex]\\cos \\left(2\\theta \\right)\\cos \\theta =\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use Reduction Formulas to Simplify an Expression<\/h2>\r\nThe double-angle formulas can be used to derive the <strong>reduction formulas<\/strong>, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.\r\n\r\nWe can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let\u2019s begin with [latex]\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta [\/latex]. Solve for [latex]{\\sin }^{2}\\theta :[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta \\\\ 2{\\sin }^{2}\\theta =1-\\cos \\left(2\\theta \\right) \\\\ {\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\end{gathered}[\/latex]<\/div>\r\nNext, we use the formula [latex]\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1[\/latex]. Solve for [latex]{\\cos }^{2}\\theta :[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1 \\\\ 1+\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta \\\\ \\frac{1+\\cos \\left(2\\theta \\right)}{2}={\\cos }^{2}\\theta \\end{gathered}[\/latex]<\/div>\r\nThe last reduction formula is derived by writing tangent in terms of sine and cosine:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\tan }^{2}\\theta &amp;=\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &amp;=\\frac{\\frac{1-\\cos \\left(2\\theta \\right)}{2}}{\\frac{1+\\cos \\left(2\\theta \\right)}{2}}&amp;&amp; \\text{Substitute the reduction formulas.} \\\\ &amp;=\\left(\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\right)\\left(\\frac{2}{1+\\cos \\left(2\\theta \\right)}\\right) \\\\ &amp;=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Reduction Formulas<\/h3>\r\nThe <strong>reduction formulas<\/strong> are summarized as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &amp;{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &amp;{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Writing an Equivalent Expression Not Containing Powers Greater Than 1<\/h3>\r\nWrite an equivalent expression for [latex]{\\cos }^{4}x[\/latex] that does not involve any powers of sine or cosine greater than 1.\r\n\r\n[reveal-answer q=\"109691\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"109691\"]\r\n\r\nWe will apply the reduction formula for cosine twice.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\cos }^{4}x&amp;={\\left({\\cos }^{2}x\\right)}^{2} \\\\ &amp;={\\left(\\frac{1+\\cos \\left(2x\\right)}{2}\\right)}^{2}&amp;&amp; {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &amp;=\\frac{1}{4}\\left(1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right) \\\\ &amp;=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{4}\\left(\\frac{1+\\cos \\left(2\\left(2x\\right)\\right)}{2}\\right) &amp;&amp; {\\text{ Substitute reduction formula for cos}}^{2}x. \\\\ &amp;=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}+\\frac{1}{8}\\cos \\left(4x\\right) \\\\ &amp;=\\frac{3}{8}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThe solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Using the Power-Reducing Formulas to Prove an Identity<\/h3>\r\nUse the power-reducing formulas to prove\r\n<p style=\"text-align: center;\">[latex]{\\sin }^{3}\\left(2x\\right)=\\left[\\frac{1}{2}\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right][\/latex]<\/p>\r\n[reveal-answer q=\"828553\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"828553\"]\r\n\r\nWe will work on simplifying the left side of the equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\sin }^{3}\\left(2x\\right)&amp;=\\left[\\sin \\left(2x\\right)\\right]\\left[{\\sin }^{2}\\left(2x\\right)\\right] \\\\ &amp;=\\sin \\left(2x\\right)\\left[\\frac{1-\\cos \\left(4x\\right)}{2}\\right]&amp;&amp; \\text{Substitute the power-reduction formula}. \\\\ &amp;=\\sin \\left(2x\\right)\\left(\\frac{1}{2}\\right)\\left[1-\\cos \\left(4x\\right)\\right] \\\\ &amp;=\\frac{1}{2}\\left[\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right] \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that in this example, we substituted\r\n<div style=\"text-align: center;\">[latex]\\frac{1-\\cos \\left(4x\\right)}{2}[\/latex]<\/div>\r\nfor [latex]{\\sin }^{2}\\left(2x\\right)[\/latex]. The formula states\r\n<div style=\"text-align: center;\">[latex]{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2}[\/latex]<\/div>\r\nWe let [latex]\\theta =2x[\/latex], so [latex]2\\theta =4x[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nUse the power-reducing formulas to prove that [latex]10{\\cos }^{4}x=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"387102\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"387102\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}10{\\cos }^{4}x&amp;=10{\\left({\\cos }^{2}x\\right)}^{2} \\\\ &amp;=10{\\left[\\frac{1+\\cos \\left(2x\\right)}{2}\\right]}^{2}&amp;&amp; {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &amp;=\\frac{10}{4}\\left[1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right] \\\\ &amp;=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{4}\\left(\\frac{1+\\cos\\left( 2\\left(2x\\right)\\right)}{2}\\right)&amp;&amp; {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &amp;=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{8}+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &amp;=\\frac{30}{8}+5\\cos \\left(2x\\right)+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &amp;=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]4533[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using Half-Angle Formulas to Find Exact Values<\/h2>\r\nThe next set of identities is the set of <strong>half-angle formulas<\/strong>, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace [latex]\\theta [\/latex] with [latex]\\frac{\\alpha }{2}[\/latex], the half-angle formula for sine is found by simplifying the equation and solving for [latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]. Note that the half-angle formulas are preceded by a [latex]\\pm [\/latex] sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which [latex]\\frac{\\alpha }{2}[\/latex] terminates.\r\n\r\nThe half-angle formula for sine is derived as follows:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\sin }^{2}\\theta &amp;=\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\\\ {\\sin }^{2}\\left(\\frac{\\alpha }{2}\\right)&amp;=\\frac{1-\\left(\\cos 2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &amp;=\\frac{1-\\cos \\alpha }{2} \\\\ \\sin \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\r\nTo derive the half-angle formula for cosine, we have\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\cos }^{2}\\theta &amp;=\\frac{1+\\cos \\left(2\\theta \\right)}{2}\\\\ {\\cos }^{2}\\left(\\frac{\\alpha }{2}\\right)&amp;=\\frac{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &amp;=\\frac{1+\\cos \\alpha }{2} \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\r\nFor the tangent identity, we have\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\tan }^{2}\\theta &amp;=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\\\ {\\tan }^{2}\\left(\\frac{\\alpha }{2}\\right)&amp;=\\frac{1-\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)} \\\\ &amp;=\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }\\hfill \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\end{align}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Half-Angle Formulas<\/h3>\r\nThe <strong>half-angle formulas<\/strong> are as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &amp;=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &amp;=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Using a Half-Angle Formula to Find the Exact Value of a Sine Function<\/h3>\r\nFind [latex]\\sin \\left({15}^{\\circ }\\right)[\/latex] using a half-angle formula.\r\n\r\n[reveal-answer q=\"283155\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"283155\"]\r\n\r\nSince [latex]{15}^{\\circ }=\\frac{{30}^{\\circ }}{2}[\/latex], we use the half-angle formula for sine:\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\sin \\frac{{30}^{\\circ }}{2}&amp;=\\sqrt{\\frac{1-\\cos {30}^{\\circ }}{2}} \\\\ &amp;=\\sqrt{\\frac{1-\\frac{\\sqrt{3}}{2}}{2}} \\\\ &amp;=\\sqrt{\\frac{\\frac{2-\\sqrt{3}}{2}}{2}} \\\\ &amp;=\\sqrt{\\frac{2-\\sqrt{3}}{4}} \\\\ &amp;=\\frac{\\sqrt{2-\\sqrt{3}}}{2} \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice that we used only the positive root because [latex]\\sin \\left({15}^{\\text{o}}\\right)[\/latex] is positive.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.<\/h3>\r\n<ol>\r\n \t<li>Draw a triangle to represent the given information.<\/li>\r\n \t<li>Determine the correct half-angle formula.<\/li>\r\n \t<li>Substitute values into the formula based on the triangle.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Finding Exact Values Using Half-Angle Identities<\/h3>\r\nGiven that [latex]\\tan \\alpha =\\frac{8}{15}[\/latex] and [latex]\\alpha [\/latex] lies in quadrant III, find the exact value of the following:\r\n<ol>\r\n \t<li>[latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\tan \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"751659\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"751659\"]\r\n\r\nUsing the given information, we can draw the triangle shown in Figure 3. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate [latex]\\sin \\alpha =-\\frac{8}{17}[\/latex] and [latex]\\cos \\alpha =-\\frac{15}{17}[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164103\/CNX_Precalc_Figure_07_03_0032.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.\" width=\"487\" height=\"289\" \/> <b>Figure 3<\/b>[\/caption]\r\n<ol>\r\n \t<li>Before we start, we must remember that, if [latex]\\alpha [\/latex] is in quadrant III, then [latex]180^\\circ &lt;\\alpha &lt;270^\\circ [\/latex], so [latex]\\frac{180^\\circ }{2}&lt;\\frac{\\alpha }{2}&lt;\\frac{270^\\circ }{2}[\/latex]. This means that the terminal side of [latex]\\frac{\\alpha }{2}[\/latex] is in quadrant II, since [latex]90^\\circ &lt;\\frac{\\alpha }{2}&lt;135^\\circ [\/latex].To find [latex]\\sin \\frac{\\alpha }{2}[\/latex], we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure 3\u00a0and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align} \\sin \\frac{\\alpha }{2}&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{\\frac{32}{17}}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{32}{17}\\cdot \\frac{1}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{16}{17}} \\\\ &amp;=\\pm \\frac{4}{\\sqrt{17}} \\\\ &amp;=\\frac{4\\sqrt{17}}{17} \\end{align}[\/latex]<\/div>\r\nWe choose the positive value of [latex]\\sin \\frac{\\alpha }{2}[\/latex] because the angle terminates in quadrant II and sine is positive in quadrant II.<\/li>\r\n \t<li>To find [latex]\\cos \\frac{\\alpha }{2}[\/latex], we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in Figure 3, and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align} \\cos \\frac{\\alpha }{2}&amp;=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{1+\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{\\frac{2}{17}}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{2}{17}\\cdot \\frac{1}{2}} \\\\ &amp;=\\pm \\sqrt{\\frac{1}{17}} \\\\ &amp;=-\\frac{\\sqrt{17}}{17}\\end{align}[\/latex]<\/div>\r\nWe choose the negative value of [latex]\\cos \\frac{\\alpha }{2}[\/latex] because the angle is in quadrant II because cosine is negative in quadrant II.<\/li>\r\n \t<li>To find [latex]\\tan \\frac{\\alpha }{2}[\/latex], we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in Figure 3\u00a0and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align} \\tan \\frac{\\alpha }{2}&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &amp;=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{1+\\left(-\\frac{15}{17}\\right)}} \\\\ &amp;=\\pm \\sqrt{\\frac{\\frac{32}{17}}{\\frac{2}{17}}} \\\\ &amp;=\\pm \\sqrt{\\frac{32}{2}} \\\\ &amp;=-\\sqrt{16} \\\\ &amp;=-4 \\end{align}[\/latex]<\/div>\r\nWe choose the negative value of [latex]\\tan \\frac{\\alpha }{2}[\/latex] because [latex]\\frac{\\alpha }{2}[\/latex] lies in quadrant II, and tangent is negative in quadrant II.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven that [latex]\\sin \\alpha =-\\frac{4}{5}[\/latex] and [latex]\\alpha [\/latex] lies in quadrant IV, find the exact value of [latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"406030\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"406030\"]\r\n\r\n[latex]-\\frac{2}{\\sqrt{5}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173569[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Finding the Measurement of a Half Angle<\/h3>\r\nNow, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of [latex]\\theta [\/latex] formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for higher-level competition, what is the measurement of the angle for novice competition?\r\n\r\n[reveal-answer q=\"685675\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"685675\"]\r\n\r\nSince the angle for novice competition measures half the steepness of the angle for the high level competition, and [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for high competition, we can find [latex]\\cos \\theta [\/latex] from the right triangle and the Pythagorean theorem so that we can use the half-angle identities.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{3}^{2}+{5}^{2}=34 \\\\ c=\\sqrt{34} \\end{gathered}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164106\/CNX_Precalc_Figure_07_03_0042.jpg\" alt=\"Image of a right triangle with sides 3, 5, and rad34. Rad 34 is the hypotenuse, and 3 is the base. The angle formed by the hypotenuse and base is theta. The angle between the side of length 3 and side of length 5 is a right angle.\" width=\"487\" height=\"210\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\nWe see that [latex]\\cos \\theta =\\frac{3}{\\sqrt{34}}=\\frac{3\\sqrt{34}}{34}[\/latex]. We can use the half-angle formula for tangent: [latex]\\tan \\frac{\\theta }{2}=\\sqrt{\\frac{1-\\cos \\theta }{1+\\cos \\theta }}[\/latex]. Since [latex]\\tan \\theta [\/latex] is in the first quadrant, so is [latex]\\tan \\frac{\\theta }{2}[\/latex]. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\tan \\frac{\\theta }{2}&amp;=\\sqrt{\\frac{1-\\frac{3\\sqrt{34}}{34}}{1+\\frac{3\\sqrt{34}}{34}}} \\\\ &amp;=\\sqrt{\\frac{\\frac{34 - 3\\sqrt{34}}{34}}{\\frac{34+3\\sqrt{34}}{34}}} \\\\ &amp;=\\sqrt{\\frac{34 - 3\\sqrt{34}}{34+3\\sqrt{34}}} \\\\ &amp;\\approx 0.57\\end{align}[\/latex]<\/p>\r\nWe can take the inverse tangent to find the angle: [latex]{\\tan }^{-1}\\left(0.57\\right)\\approx {29.7}^{\\circ }[\/latex]. So the angle of the ramp for novice competition is [latex]\\approx {29.7}^{\\circ }[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table><colgroup> <col \/> <col \/> <\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong>Double-angle formulas<\/strong><\/td>\r\n<td>[latex]\\begin{align}\\sin \\left(2\\theta \\right)&amp;=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&amp;={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &amp;=1 - 2{\\sin }^{2}\\theta \\\\ &amp;=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&amp;=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Reduction formulas<\/strong><\/td>\r\n<td>[latex]\\begin{align}&amp;{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &amp;{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &amp;{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Half-angle formulas<\/strong><\/td>\r\n<td>[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&amp;=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &amp;=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &amp;=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent.<\/li>\r\n \t<li>Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term.<\/li>\r\n \t<li>Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not.<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id2253485\" class=\"definition\">\r\n \t<dt>double-angle formulas<\/dt>\r\n \t<dd id=\"fs-id2253488\">identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id2253492\" class=\"definition\">\r\n \t<dt>half-angle formulas<\/dt>\r\n \t<dd id=\"fs-id2253495\">identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id2253499\" class=\"definition\">\r\n \t<dt>reduction formulas<\/dt>\r\n \t<dd id=\"fs-id2253503\">identities derived from the double-angle formulas and used to reduce the power of a trigonometric function<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"font-weight: 400;\">Use double-angle formulas to find exact values.<\/li>\n<li style=\"font-weight: 400;\">Use double-angle formulas to verify identities.<\/li>\n<li style=\"font-weight: 400;\">Use reduction formulas to simplify an expression.<\/li>\n<li style=\"font-weight: 400;\">Use half-angle formulas to find exact values.<\/li>\n<\/ul>\n<\/div>\n<h2>Using Double-Angle Formulas to Find Exact Values<\/h2>\n<p>In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The <strong>double-angle formulas<\/strong> are a special case of the sum formulas, where [latex]\\alpha =\\beta[\/latex]. Deriving the double-angle formula for sine begins with the sum formula,<\/p>\n<div style=\"text-align: center;\">[latex]\\sin \\left(\\alpha +\\beta \\right)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta[\/latex]<\/div>\n<p>If we let [latex]\\alpha =\\beta =\\theta[\/latex], then we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(\\theta +\\theta \\right)&=\\sin \\theta \\cos \\theta +\\cos \\theta \\sin \\theta \\\\ \\sin \\left(2\\theta \\right)&=2\\sin \\theta \\cos \\theta \\end{align}[\/latex]<\/div>\n<p>Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, [latex]\\cos \\left(\\alpha +\\beta \\right)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta[\/latex], and letting [latex]\\alpha =\\beta =\\theta[\/latex], we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(\\theta +\\theta \\right)&=\\cos \\theta \\cos \\theta -\\sin \\theta \\sin \\theta \\\\ \\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\end{align}[\/latex]<\/div>\n<p>Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more interpretations. The first one is:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &=\\left(1-{\\sin }^{2}\\theta \\right)-{\\sin }^{2}\\theta \\\\ &=1 - 2{\\sin }^{2}\\theta\\end{align}[\/latex]<\/div>\n<p>The second interpretation is:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &={\\cos }^{2}\\theta -\\left(1-{\\cos }^{2}\\theta \\right) \\\\ &=2{\\cos }^{2}\\theta -1\\end{align}[\/latex]<\/div>\n<p>Similarly, to derive the double-angle formula for tangent, replacing [latex]\\alpha =\\beta =\\theta[\/latex] in the sum formula gives<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(\\alpha +\\beta \\right)&=\\frac{\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }\\\\ \\tan \\left(\\theta +\\theta \\right)&=\\frac{\\tan \\theta +\\tan \\theta }{1-\\tan \\theta \\tan \\theta }\\\\ \\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Double-Angle Formulas<\/h3>\n<p>The <strong>double-angle formulas<\/strong> are summarized as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &=1 - 2{\\sin }^{2}\\theta \\\\ &=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/p>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm18740\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=18740&theme=oea&iframe_resize_id=ohm18740&sameseed=1\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm132474\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=132474&theme=oea&iframe_resize_id=ohm132474&sameseed=1\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Draw a triangle to reflect the given information.<\/li>\n<li>Determine the correct double-angle formula.<\/li>\n<li>Substitute values into the formula based on the triangle.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Using a Double-Angle Formula to Find the Exact Value Involving Tangent<\/h3>\n<p>Given that [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex] and [latex]\\theta[\/latex] is in quadrant II, find the following:<\/p>\n<ol>\n<li>[latex]\\sin \\left(2\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\cos \\left(2\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\tan \\left(2\\theta \\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q477962\">Show Solution<\/span><\/p>\n<div id=\"q477962\" class=\"hidden-answer\" style=\"display: none\">\n<p>If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\\tan \\theta =-\\frac{3}{4}[\/latex], such that [latex]\\theta[\/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\\theta[\/latex] is in the second quadrant, the adjacent side is on the <em>x<\/em>-axis and is negative. Use the <strong>Pythagorean Theorem<\/strong> to find the length of the hypotenuse:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\left(-4\\right)}^{2}+{\\left(3\\right)}^{2}&={c}^{2}\\\\ 16+9&={c}^{2}\\\\ 25&={c}^{2}\\\\ c&=5\\end{align}[\/latex]<\/p>\n<p>Now we can draw a triangle similar to the one shown in Figure 2.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164101\/CNX_Precalc_Figure_07_03_0022.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-4,0), and (-4,3). The angle at the origin is theta. The angle formed by the side (-4,3) to (-4,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 5.\" width=\"487\" height=\"251\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<ol>\n<li>Let\u2019s begin by writing the double-angle formula for sine.\n<div style=\"text-align: center;\">[latex]\\sin \\left(2\\theta \\right)=2\\sin \\theta \\cos \\theta[\/latex]<\/div>\n<p>We see that we to need to find [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex]. Based on Figure 2, we see that the hypotenuse equals 5, so [latex]\\sin \\theta =\\frac{3}{5}[\/latex], and [latex]\\cos \\theta =\u2212\\frac{4}{5}[\/latex]. Substitute these values into the equation, and simplify.<br \/>\nThus,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(2\\theta \\right)&=2\\left(\\frac{3}{5}\\right)\\left(\u2212\\frac{4}{5}\\right) \\\\ &=\u2212\\frac{24}{25} \\end{align}[\/latex]<\/div>\n<\/li>\n<li>Write the double-angle formula for cosine.\n<div style=\"text-align: center;\">[latex]\\cos \\left(2\\theta \\right)={\\cos }^{2}\\theta -{\\sin }^{2}\\theta[\/latex]<\/div>\n<p>Again, substitute the values of the sine and cosine into the equation, and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(2\\theta \\right)&={\\left(\u2212\\frac{4}{5}\\right)}^{2}\u2212{\\left(\\frac{3}{5}\\right)}^{2} \\\\ &=\\frac{16}{25}\u2212\\frac{9}{25} \\\\ &=\\frac{7}{25}\\end{align}[\/latex]<\/div>\n<\/li>\n<li>Write the double-angle formula for tangent.\n<div style=\"text-align: center;\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2\\tan \\theta }{1\u2212{\\tan }^{2}\\theta }[\/latex]<\/div>\n<p>In this formula, we need the tangent, which we were given as [latex]\\tan \\theta =\u2212\\frac{3}{4}[\/latex]. Substitute this value into the equation, and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&=\\frac{2\\left(-\\frac{3}{4}\\right)}{1-{\\left(-\\frac{3}{4}\\right)}^{2}} \\\\ &=\\frac{-\\frac{3}{2}}{1-\\frac{9}{16}} \\\\ &=-\\frac{3}{2}\\left(\\frac{16}{7}\\right) \\\\ &=-\\frac{24}{7} \\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given [latex]\\sin \\alpha =\\frac{5}{8}[\/latex], with [latex]\\theta[\/latex] in quadrant I, find [latex]\\cos \\left(2\\alpha \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q827677\">Show Solution<\/span><\/p>\n<div id=\"q827677\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\left(2\\alpha \\right)=\\frac{7}{32}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm156232\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=156232&theme=oea&iframe_resize_id=ohm156232\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Using the Double-Angle Formula for Cosine without Exact Values<\/h3>\n<p>Use the double-angle formula for cosine to write [latex]\\cos \\left(6x\\right)[\/latex] in terms of [latex]\\cos \\left(3x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q676542\">Show Solution<\/span><\/p>\n<div id=\"q676542\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}\\cos \\left(6x\\right)&=\\cos \\left(3x+3x\\right) \\\\ &=\\cos 3x\\cos 3x-\\sin 3x\\sin 3x \\\\ &={\\cos }^{2}3x-{\\sin }^{2}3x \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using Double-Angle Formulas to Verify Identities<\/h2>\n<p>Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 3: Using the Double-Angle Formulas to Establish an Identity<\/h3>\n<p>Establish the following identity using double-angle formulas:<\/p>\n<p style=\"text-align: center;\">[latex]1+\\sin \\left(2\\theta \\right)={\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q600157\">Show Solution<\/span><\/p>\n<div id=\"q600157\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will work on the right side of the equal sign and rewrite the expression until it matches the left side.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\left(\\sin \\theta +\\cos \\theta \\right)}^{2}&={\\sin }^{2}\\theta +2\\sin \\theta \\cos \\theta +{\\cos }^{2}\\theta \\\\ &=\\left({\\sin }^{2}\\theta +{\\cos }^{2}\\theta \\right)+2\\sin \\theta \\cos \\theta \\\\ &=1+2\\sin \\theta \\cos \\theta \\\\ &=1+\\sin \\left(2\\theta \\right)\\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This process is not complicated, as long as we recall the perfect square formula from algebra:<\/p>\n<p style=\"text-align: center;\">[latex]{\\left(a\\pm b\\right)}^{2}={a}^{2}\\pm 2ab+{b}^{2}[\/latex]<\/p>\n<p>where [latex]a=\\sin \\theta[\/latex] and [latex]b=\\cos \\theta[\/latex]. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Establish the identity: [latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\cos \\left(2\\theta \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q178499\">Show Solution<\/span><\/p>\n<div id=\"q178499\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\cos }^{4}\\theta -{\\sin }^{4}\\theta =\\left({\\cos }^{2}\\theta +{\\sin }^{2}\\theta \\right)\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)=\\cos \\left(2\\theta \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Verifying a Double-Angle Identity for Tangent<\/h3>\n<p>Verify the identity:<\/p>\n<p style=\"text-align: center;\">[latex]\\tan \\left(2\\theta \\right)=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q395376\">Show Solution<\/span><\/p>\n<div id=\"q395376\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }&& \\text{Double-angle formula} \\\\ &=\\frac{2\\tan \\theta \\left(\\frac{1}{\\tan \\theta }\\right)}{\\left(1-{\\tan }^{2}\\theta \\right)\\left(\\frac{1}{\\tan \\theta }\\right)}&& \\text{Multiply by a term that results in desired numerator}. \\\\ &=\\frac{2}{\\frac{1}{\\tan \\theta }-\\frac{{\\tan }^{2}\\theta }{\\tan \\theta }} \\\\ &=\\frac{2}{\\cot \\theta -\\tan \\theta }&& \\text{Use reciprocal identity for }\\frac{1}{\\tan \\theta }.\\end{align}[\/latex]<\/p>\n<h3>Analysis of the Solution<\/h3>\n<p>Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }=\\frac{2}{\\cot \\theta -\\tan \\theta }[\/latex]<\/p>\n<p>Let\u2019s work on the right side.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{2}{\\cot \\theta -\\tan \\theta }&=\\frac{2}{\\frac{1}{\\tan \\theta }-\\tan \\theta }\\left(\\frac{\\tan \\theta }{\\tan \\theta }\\right) \\\\ &=\\frac{2\\tan \\theta }{\\frac{1}{\\cancel{\\tan \\theta }}\\left(\\cancel{\\tan \\theta }\\right)-\\tan \\theta \\left(\\tan \\theta \\right)} \\\\ &=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta } \\end{align}[\/latex]<\/p>\n<p>When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Verify the identity: [latex]\\cos \\left(2\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q584630\">Show Solution<\/span><\/p>\n<div id=\"q584630\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\left(2\\theta \\right)\\cos \\theta =\\left({\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\right)\\cos \\theta ={\\cos }^{3}\\theta -\\cos \\theta {\\sin }^{2}\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use Reduction Formulas to Simplify an Expression<\/h2>\n<p>The double-angle formulas can be used to derive the <strong>reduction formulas<\/strong>, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.<\/p>\n<p>We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let\u2019s begin with [latex]\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta[\/latex]. Solve for [latex]{\\sin }^{2}\\theta :[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=1 - 2{\\sin }^{2}\\theta \\\\ 2{\\sin }^{2}\\theta =1-\\cos \\left(2\\theta \\right) \\\\ {\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\end{gathered}[\/latex]<\/div>\n<p>Next, we use the formula [latex]\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1[\/latex]. Solve for [latex]{\\cos }^{2}\\theta :[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta -1 \\\\ 1+\\cos \\left(2\\theta \\right)=2{\\cos }^{2}\\theta \\\\ \\frac{1+\\cos \\left(2\\theta \\right)}{2}={\\cos }^{2}\\theta \\end{gathered}[\/latex]<\/div>\n<p>The last reduction formula is derived by writing tangent in terms of sine and cosine:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\tan }^{2}\\theta &=\\frac{{\\sin }^{2}\\theta }{{\\cos }^{2}\\theta } \\\\ &=\\frac{\\frac{1-\\cos \\left(2\\theta \\right)}{2}}{\\frac{1+\\cos \\left(2\\theta \\right)}{2}}&& \\text{Substitute the reduction formulas.} \\\\ &=\\left(\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\right)\\left(\\frac{2}{1+\\cos \\left(2\\theta \\right)}\\right) \\\\ &=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Reduction Formulas<\/h3>\n<p>The <strong>reduction formulas<\/strong> are summarized as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Writing an Equivalent Expression Not Containing Powers Greater Than 1<\/h3>\n<p>Write an equivalent expression for [latex]{\\cos }^{4}x[\/latex] that does not involve any powers of sine or cosine greater than 1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q109691\">Show Solution<\/span><\/p>\n<div id=\"q109691\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will apply the reduction formula for cosine twice.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\cos }^{4}x&={\\left({\\cos }^{2}x\\right)}^{2} \\\\ &={\\left(\\frac{1+\\cos \\left(2x\\right)}{2}\\right)}^{2}&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{1}{4}\\left(1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right) \\\\ &=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{4}\\left(\\frac{1+\\cos \\left(2\\left(2x\\right)\\right)}{2}\\right) && {\\text{ Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{1}{4}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}+\\frac{1}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{3}{8}+\\frac{1}{2}\\cos \\left(2x\\right)+\\frac{1}{8}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Using the Power-Reducing Formulas to Prove an Identity<\/h3>\n<p>Use the power-reducing formulas to prove<\/p>\n<p style=\"text-align: center;\">[latex]{\\sin }^{3}\\left(2x\\right)=\\left[\\frac{1}{2}\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right][\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q828553\">Show Solution<\/span><\/p>\n<div id=\"q828553\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will work on simplifying the left side of the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\sin }^{3}\\left(2x\\right)&=\\left[\\sin \\left(2x\\right)\\right]\\left[{\\sin }^{2}\\left(2x\\right)\\right] \\\\ &=\\sin \\left(2x\\right)\\left[\\frac{1-\\cos \\left(4x\\right)}{2}\\right]&& \\text{Substitute the power-reduction formula}. \\\\ &=\\sin \\left(2x\\right)\\left(\\frac{1}{2}\\right)\\left[1-\\cos \\left(4x\\right)\\right] \\\\ &=\\frac{1}{2}\\left[\\sin \\left(2x\\right)\\right]\\left[1-\\cos \\left(4x\\right)\\right] \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that in this example, we substituted<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{1-\\cos \\left(4x\\right)}{2}[\/latex]<\/div>\n<p>for [latex]{\\sin }^{2}\\left(2x\\right)[\/latex]. The formula states<\/p>\n<div style=\"text-align: center;\">[latex]{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2}[\/latex]<\/div>\n<p>We let [latex]\\theta =2x[\/latex], so [latex]2\\theta =4x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Use the power-reducing formulas to prove that [latex]10{\\cos }^{4}x=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q387102\">Show Solution<\/span><\/p>\n<div id=\"q387102\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}10{\\cos }^{4}x&=10{\\left({\\cos }^{2}x\\right)}^{2} \\\\ &=10{\\left[\\frac{1+\\cos \\left(2x\\right)}{2}\\right]}^{2}&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{10}{4}\\left[1+2\\cos \\left(2x\\right)+{\\cos }^{2}\\left(2x\\right)\\right] \\\\ &=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{4}\\left(\\frac{1+\\cos\\left( 2\\left(2x\\right)\\right)}{2}\\right)&& {\\text{Substitute reduction formula for cos}}^{2}x. \\\\ &=\\frac{10}{4}+\\frac{10}{2}\\cos \\left(2x\\right)+\\frac{10}{8}+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{30}{8}+5\\cos \\left(2x\\right)+\\frac{10}{8}\\cos \\left(4x\\right) \\\\ &=\\frac{15}{4}+5\\cos \\left(2x\\right)+\\frac{5}{4}\\cos \\left(4x\\right) \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4533\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4533&theme=oea&iframe_resize_id=ohm4533\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using Half-Angle Formulas to Find Exact Values<\/h2>\n<p>The next set of identities is the set of <strong>half-angle formulas<\/strong>, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace [latex]\\theta[\/latex] with [latex]\\frac{\\alpha }{2}[\/latex], the half-angle formula for sine is found by simplifying the equation and solving for [latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]. Note that the half-angle formulas are preceded by a [latex]\\pm[\/latex] sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which [latex]\\frac{\\alpha }{2}[\/latex] terminates.<\/p>\n<p>The half-angle formula for sine is derived as follows:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\sin }^{2}\\theta &=\\frac{1-\\cos \\left(2\\theta \\right)}{2}\\\\ {\\sin }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1-\\left(\\cos 2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &=\\frac{1-\\cos \\alpha }{2} \\\\ \\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\n<p>To derive the half-angle formula for cosine, we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\cos }^{2}\\theta &=\\frac{1+\\cos \\left(2\\theta \\right)}{2}\\\\ {\\cos }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{2} \\\\ &=\\frac{1+\\cos \\alpha }{2} \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\end{align}[\/latex]<\/div>\n<p>For the tangent identity, we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\tan }^{2}\\theta &=\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\\\ {\\tan }^{2}\\left(\\frac{\\alpha }{2}\\right)&=\\frac{1-\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)}{1+\\cos \\left(2\\cdot \\frac{\\alpha }{2}\\right)} \\\\ &=\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }\\hfill \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\end{align}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Half-Angle Formulas<\/h3>\n<p>The <strong>half-angle formulas<\/strong> are as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Using a Half-Angle Formula to Find the Exact Value of a Sine Function<\/h3>\n<p>Find [latex]\\sin \\left({15}^{\\circ }\\right)[\/latex] using a half-angle formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q283155\">Show Solution<\/span><\/p>\n<div id=\"q283155\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since [latex]{15}^{\\circ }=\\frac{{30}^{\\circ }}{2}[\/latex], we use the half-angle formula for sine:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\sin \\frac{{30}^{\\circ }}{2}&=\\sqrt{\\frac{1-\\cos {30}^{\\circ }}{2}} \\\\ &=\\sqrt{\\frac{1-\\frac{\\sqrt{3}}{2}}{2}} \\\\ &=\\sqrt{\\frac{\\frac{2-\\sqrt{3}}{2}}{2}} \\\\ &=\\sqrt{\\frac{2-\\sqrt{3}}{4}} \\\\ &=\\frac{\\sqrt{2-\\sqrt{3}}}{2} \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice that we used only the positive root because [latex]\\sin \\left({15}^{\\text{o}}\\right)[\/latex] is positive.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.<\/h3>\n<ol>\n<li>Draw a triangle to represent the given information.<\/li>\n<li>Determine the correct half-angle formula.<\/li>\n<li>Substitute values into the formula based on the triangle.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Finding Exact Values Using Half-Angle Identities<\/h3>\n<p>Given that [latex]\\tan \\alpha =\\frac{8}{15}[\/latex] and [latex]\\alpha[\/latex] lies in quadrant III, find the exact value of the following:<\/p>\n<ol>\n<li>[latex]\\sin \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\n<li>[latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\n<li>[latex]\\tan \\left(\\frac{\\alpha }{2}\\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q751659\">Show Solution<\/span><\/p>\n<div id=\"q751659\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the given information, we can draw the triangle shown in Figure 3. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate [latex]\\sin \\alpha =-\\frac{8}{17}[\/latex] and [latex]\\cos \\alpha =-\\frac{15}{17}[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164103\/CNX_Precalc_Figure_07_03_0032.jpg\" alt=\"Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.\" width=\"487\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<ol>\n<li>Before we start, we must remember that, if [latex]\\alpha[\/latex] is in quadrant III, then [latex]180^\\circ <\\alpha <270^\\circ[\/latex], so [latex]\\frac{180^\\circ }{2}<\\frac{\\alpha }{2}<\\frac{270^\\circ }{2}[\/latex]. This means that the terminal side of [latex]\\frac{\\alpha }{2}[\/latex] is in quadrant II, since [latex]90^\\circ <\\frac{\\alpha }{2}<135^\\circ[\/latex].To find [latex]\\sin \\frac{\\alpha }{2}[\/latex], we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure 3\u00a0and simplify.\n\n\n<div style=\"text-align: center;\">[latex]\\begin{align} \\sin \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ &=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{32}{17}}{2}} \\\\ &=\\pm \\sqrt{\\frac{32}{17}\\cdot \\frac{1}{2}} \\\\ &=\\pm \\sqrt{\\frac{16}{17}} \\\\ &=\\pm \\frac{4}{\\sqrt{17}} \\\\ &=\\frac{4\\sqrt{17}}{17} \\end{align}[\/latex]<\/div>\n<p>We choose the positive value of [latex]\\sin \\frac{\\alpha }{2}[\/latex] because the angle terminates in quadrant II and sine is positive in quadrant II.<\/li>\n<li>To find [latex]\\cos \\frac{\\alpha }{2}[\/latex], we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in Figure 3, and simplify.\n<div style=\"text-align: center;\">[latex]\\begin{align} \\cos \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ &=\\pm \\sqrt{\\frac{1+\\left(-\\frac{15}{17}\\right)}{2}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{2}{17}}{2}} \\\\ &=\\pm \\sqrt{\\frac{2}{17}\\cdot \\frac{1}{2}} \\\\ &=\\pm \\sqrt{\\frac{1}{17}} \\\\ &=-\\frac{\\sqrt{17}}{17}\\end{align}[\/latex]<\/div>\n<p>We choose the negative value of [latex]\\cos \\frac{\\alpha }{2}[\/latex] because the angle is in quadrant II because cosine is negative in quadrant II.<\/li>\n<li>To find [latex]\\tan \\frac{\\alpha }{2}[\/latex], we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in Figure 3\u00a0and simplify.\n<div style=\"text-align: center;\">[latex]\\begin{align} \\tan \\frac{\\alpha }{2}&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &=\\pm \\sqrt{\\frac{1-\\left(-\\frac{15}{17}\\right)}{1+\\left(-\\frac{15}{17}\\right)}} \\\\ &=\\pm \\sqrt{\\frac{\\frac{32}{17}}{\\frac{2}{17}}} \\\\ &=\\pm \\sqrt{\\frac{32}{2}} \\\\ &=-\\sqrt{16} \\\\ &=-4 \\end{align}[\/latex]<\/div>\n<p>We choose the negative value of [latex]\\tan \\frac{\\alpha }{2}[\/latex] because [latex]\\frac{\\alpha }{2}[\/latex] lies in quadrant II, and tangent is negative in quadrant II.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given that [latex]\\sin \\alpha =-\\frac{4}{5}[\/latex] and [latex]\\alpha[\/latex] lies in quadrant IV, find the exact value of [latex]\\cos \\left(\\frac{\\alpha }{2}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q406030\">Show Solution<\/span><\/p>\n<div id=\"q406030\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\frac{2}{\\sqrt{5}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173569\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173569&theme=oea&iframe_resize_id=ohm173569\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Finding the Measurement of a Half Angle<\/h3>\n<p>Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of [latex]\\theta[\/latex] formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for higher-level competition, what is the measurement of the angle for novice competition?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q685675\">Show Solution<\/span><\/p>\n<div id=\"q685675\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the angle for novice competition measures half the steepness of the angle for the high level competition, and [latex]\\tan \\theta =\\frac{5}{3}[\/latex] for high competition, we can find [latex]\\cos \\theta[\/latex] from the right triangle and the Pythagorean theorem so that we can use the half-angle identities.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{3}^{2}+{5}^{2}=34 \\\\ c=\\sqrt{34} \\end{gathered}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164106\/CNX_Precalc_Figure_07_03_0042.jpg\" alt=\"Image of a right triangle with sides 3, 5, and rad34. Rad 34 is the hypotenuse, and 3 is the base. The angle formed by the hypotenuse and base is theta. The angle between the side of length 3 and side of length 5 is a right angle.\" width=\"487\" height=\"210\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<p>We see that [latex]\\cos \\theta =\\frac{3}{\\sqrt{34}}=\\frac{3\\sqrt{34}}{34}[\/latex]. We can use the half-angle formula for tangent: [latex]\\tan \\frac{\\theta }{2}=\\sqrt{\\frac{1-\\cos \\theta }{1+\\cos \\theta }}[\/latex]. Since [latex]\\tan \\theta[\/latex] is in the first quadrant, so is [latex]\\tan \\frac{\\theta }{2}[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\tan \\frac{\\theta }{2}&=\\sqrt{\\frac{1-\\frac{3\\sqrt{34}}{34}}{1+\\frac{3\\sqrt{34}}{34}}} \\\\ &=\\sqrt{\\frac{\\frac{34 - 3\\sqrt{34}}{34}}{\\frac{34+3\\sqrt{34}}{34}}} \\\\ &=\\sqrt{\\frac{34 - 3\\sqrt{34}}{34+3\\sqrt{34}}} \\\\ &\\approx 0.57\\end{align}[\/latex]<\/p>\n<p>We can take the inverse tangent to find the angle: [latex]{\\tan }^{-1}\\left(0.57\\right)\\approx {29.7}^{\\circ }[\/latex]. So the angle of the ramp for novice competition is [latex]\\approx {29.7}^{\\circ }[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Equations<\/h2>\n<table>\n<colgroup>\n<col \/>\n<col \/> <\/colgroup>\n<tbody>\n<tr>\n<td><strong>Double-angle formulas<\/strong><\/td>\n<td>[latex]\\begin{align}\\sin \\left(2\\theta \\right)&=2\\sin \\theta \\cos \\theta\\\\\\text{ }\\\\ \\cos \\left(2\\theta \\right)&={\\cos }^{2}\\theta -{\\sin }^{2}\\theta \\\\ &=1 - 2{\\sin }^{2}\\theta \\\\ &=2{\\cos }^{2}\\theta -1 \\\\\\text{ }\\\\ \\tan \\left(2\\theta \\right)&=\\frac{2\\tan \\theta }{1-{\\tan }^{2}\\theta }\\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Reduction formulas<\/strong><\/td>\n<td>[latex]\\begin{align}&{\\sin }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\cos }^{2}\\theta =\\frac{1+\\cos \\left(2\\theta \\right)}{2} \\\\ &{\\tan }^{2}\\theta =\\frac{1-\\cos \\left(2\\theta \\right)}{1+\\cos \\left(2\\theta \\right)} \\end{align}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Half-angle formulas<\/strong><\/td>\n<td>[latex]\\begin{align}\\sin \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\cos \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1+\\cos \\alpha }{2}} \\\\ \\text{ } \\\\ \\tan \\left(\\frac{\\alpha }{2}\\right)&=\\pm \\sqrt{\\frac{1-\\cos \\alpha }{1+\\cos \\alpha }} \\\\ &=\\frac{\\sin \\alpha }{1+\\cos \\alpha } \\\\ &=\\frac{1-\\cos \\alpha }{\\sin \\alpha }\\end{align}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent.<\/li>\n<li>Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term.<\/li>\n<li>Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not.<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id2253485\" class=\"definition\">\n<dt>double-angle formulas<\/dt>\n<dd id=\"fs-id2253488\">identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal<\/dd>\n<\/dl>\n<dl id=\"fs-id2253492\" class=\"definition\">\n<dt>half-angle formulas<\/dt>\n<dd id=\"fs-id2253495\">identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions<\/dd>\n<\/dl>\n<dl id=\"fs-id2253499\" class=\"definition\">\n<dt>reduction formulas<\/dt>\n<dd id=\"fs-id2253503\">identities derived from the double-angle formulas and used to reduce the power of a trigonometric function<\/dd>\n<\/dl>\n<\/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-1936\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":708740,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Precalculus\",\"author\":\"OpenStax 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