{"id":1148,"date":"2021-11-11T17:37:24","date_gmt":"2021-11-11T17:37:24","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-area-and-arc-length-in-polar-form\/"},"modified":"2022-11-09T16:26:53","modified_gmt":"2022-11-09T16:26:53","slug":"skills-review-for-area-and-arc-length-in-polar-form","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-area-and-arc-length-in-polar-form\/","title":{"raw":"Skills Review for Area and Arc Length in Polar Coordinates","rendered":"Skills Review for Area and Arc Length in Polar Coordinates"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify reference angles for angles measured in both radians and degrees<\/li>\r\n \t<li>Evaluate trigonometric functions using the unit circle<\/li>\r\n \t<li>Solve trigonometric equations<\/li>\r\n \t<li>Apply reduction formulas<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Area and Arc Length in Polar Coordinates section, rather than use the rectangular coordinate system to calculate area under a curve and arc length, we will use the polar coordinate system. Here we will review how to evaluate sine and cosine functions at specific angle measures, solve trigonometric equations, and use reduction formulas.\r\n<h2>Find Reference Angles<\/h2>\r\n<strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-polar-coordinates\/\" target=\"_blank\" rel=\"noopener\">Module 1, Skills Review for Polar Coordinates<\/a>)<\/em><\/strong>\r\n<h2>Evaluate Trigonometric Functions Using the Unit Circle<\/h2>\r\n<strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-polar-coordinates\/\" target=\"_blank\" rel=\"noopener\">Module 1, Skills Review for Polar Coordinates<\/a>)<\/em><\/strong>\r\n<h2>Solve Trigonometric Equations<\/h2>\r\nTrigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The <strong>period<\/strong> of both the sine function and the cosine function is [latex]2\\pi [\/latex]. In other words, every [latex]2\\pi [\/latex] units, the <em>y-<\/em>values repeat. If we need to find all possible solutions, then we must add [latex]2\\pi k[\/latex], where [latex]k[\/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\\pi :[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\sin \\theta =\\sin \\left(\\theta \\pm 2k\\pi \\right)[\/latex]<\/div>\r\nThere are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Linear Trigonometric Equation Containing Cosine<\/h3>\r\nFind all possible exact solutions for the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex].\r\n\r\n[reveal-answer q=\"84784\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"84784\"]\r\n\r\nFrom the <strong>unit circle<\/strong>, we know that\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{3},\\frac{5\\pi }{3} \\end{gathered}[\/latex]<\/p>\r\nThese are the solutions in the interval [latex]\\left[0,2\\pi \\right][\/latex]. All possible solutions are given by\r\n<p style=\"text-align: center;\">[latex]\\theta =\\frac{\\pi }{3}\\pm 2k\\pi \\text{ and }\\theta =\\frac{5\\pi }{3}\\pm 2k\\pi [\/latex]<\/p>\r\nwhere [latex]k[\/latex] is an integer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE: Solving a Linear Trigonometric Equation Containing Sine<\/h3>\r\nFind all possible exact solutions for the equation [latex]\\sin t=\\frac{1}{2}[\/latex].\r\n\r\n[reveal-answer q=\"535703\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"535703\"]\r\n\r\nSolving for all possible values of <em>t<\/em> means that solutions include angles beyond the period of [latex]2\\pi [\/latex]. From the unit circle, we can see that the solutions are [latex]t=\\frac{\\pi }{6}[\/latex] and [latex]t=\\frac{5\\pi }{6}[\/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is\r\n<p style=\"text-align: center;\">[latex]t=\\frac{\\pi }{6}\\pm 2\\pi k\\text{ and }t=\\frac{5\\pi }{6}\\pm 2\\pi k[\/latex]<\/p>\r\nwhere [latex]k[\/latex] is an integer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE: Solving a Linear Trigonometric Equation Containing Cosine<\/h3>\r\nSolve the equation exactly: [latex]2\\cos \\theta -3=-5,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"939405\"]Show Solution[\/reveal-answer]\r\n<p style=\"text-align: left;\">[hidden-answer a=\"939405\"]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2\\cos \\theta -3=-5 \\\\ \\cos \\theta =-2 \\\\ \\cos \\theta =-1 \\\\ \\theta =\\pi \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve exactly the following linear equation on the interval [latex]\\left[0,2\\pi \\right):2\\sin x+1=0[\/latex].\r\n\r\n[reveal-answer q=\"500341\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"500341\"]\r\n\r\n[latex]x=\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/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]149871[\/ohm_question]\r\n\r\n<\/div>\r\nWhen we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\\pi [\/latex], not [latex]2\\pi [\/latex]. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\\frac{\\pi }{2}[\/latex], unless, of course, a problem places its own restrictions on the domain.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE: Solving a Trigonometric Equation<\/h3>\r\nSolve the problem exactly: [latex]2{\\sin }^{2}\\theta -1=0,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"467313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"467313\"]\r\n\r\nAs this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\\sin \\theta [\/latex]. Then we will find the angles.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -1=0 \\\\ 2{\\sin }^{2}\\theta =1 \\\\ {\\sin }^{2}\\theta =\\frac{1}{2} \\\\ \\sqrt{{\\sin }^{2}\\theta }=\\pm \\sqrt{\\frac{1}{2}} \\\\ \\sin \\theta =\\pm \\frac{1}{\\sqrt{2}}=\\pm \\frac{\\sqrt{2}}{2} \\\\ \\theta =\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Trigonometric Equation<\/h3>\r\nSolve the following equation exactly: [latex]\\csc \\theta =-2,0\\le \\theta &lt;4\\pi [\/latex].\r\n\r\n[reveal-answer q=\"605306\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"605306\"]\r\n\r\nWe want all values of [latex]\\theta [\/latex] for which [latex]\\csc \\theta =-2[\/latex] over the interval [latex]0\\le \\theta &lt;4\\pi [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\csc \\theta =-2 \\\\ \\frac{1}{\\sin \\theta }=-2 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6},\\frac{19\\pi }{6},\\frac{23\\pi }{6} \\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nAs [latex]\\sin \\theta =-\\frac{1}{2}[\/latex], notice that all four solutions are in the third and fourth quadrants.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Trigonometric Equation<\/h3>\r\nSolve the equation exactly: [latex]\\tan \\left(\\theta -\\frac{\\pi }{2}\\right)=1,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"484899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"484899\"]\r\n\r\nRecall that the tangent function has a period of [latex]\\pi [\/latex]. On the interval [latex]\\left[0,\\pi \\right)[\/latex], and at the angle of [latex]\\frac{\\pi }{4}[\/latex], the tangent has a value of 1. However, the angle we want is [latex]\\left(\\theta -\\frac{\\pi }{2}\\right)[\/latex]. Thus, if [latex]\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex], then\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\theta -\\frac{\\pi }{2}=\\frac{\\pi }{4}\\\\ \\theta =\\frac{3\\pi }{4}\\pm k\\pi \\end{gathered}[\/latex]<\/p>\r\nOver the interval [latex]\\left[0,2\\pi \\right)[\/latex], we have two solutions:\r\n<p style=\"text-align: center;\">[latex]\\theta =\\frac{3\\pi }{4}\\text{ and }\\theta =\\frac{3\\pi }{4}+\\pi =\\frac{7\\pi }{4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind all solutions for [latex]\\tan x=\\sqrt{3}[\/latex].\r\n\r\n[reveal-answer q=\"629684\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"629684\"]\r\n\r\n[latex]\\frac{\\pi }{3}\\pm \\pi k[\/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]173739[\/ohm_question]\r\n\r\n<\/div>\r\nSometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as [latex]\\sin \\left(2x\\right)[\/latex] or [latex]\\cos \\left(3x\\right)[\/latex]. When confronted with these equations, recall that [latex]y=\\sin \\left(2x\\right)[\/latex] is a <strong>horizontal compression<\/strong> by a factor of 2 of the function [latex]y=\\sin x[\/latex]. On an interval of [latex]2\\pi [\/latex], we can graph two periods of [latex]y=\\sin \\left(2x\\right)[\/latex], as opposed to one cycle of [latex]y=\\sin x[\/latex]. This compression of the graph leads us to believe there may be twice as many <em>x<\/em>-intercepts or solutions to [latex]\\sin \\left(2x\\right)=0[\/latex] compared to [latex]\\sin x=0[\/latex]. This information will help us solve the equation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Multiple Angle Trigonometric Equation<\/h3>\r\nSolve exactly: [latex]\\cos \\left(2x\\right)=\\frac{1}{2}[\/latex] on [latex]\\left[0,2\\pi \\right)[\/latex].\r\n\r\n[reveal-answer q=\"299198\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"299198\"]\r\n\r\nWe can see that this equation is the standard equation with a multiple of an angle. If [latex]\\cos \\left(\\alpha \\right)=\\frac{1}{2}[\/latex], we know [latex]\\alpha [\/latex] is in quadrants I and IV. While [latex]\\theta ={\\cos }^{-1}\\frac{1}{2}[\/latex] will only yield solutions in quadrants I and II, we recognize that the solutions to the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex] will be in quadrants I and IV.\r\n\r\nTherefore, the possible angles are [latex]\\theta =\\frac{\\pi }{3}[\/latex] and [latex]\\theta =\\frac{5\\pi }{3}[\/latex]. So, [latex]2x=\\frac{\\pi }{3}[\/latex] or [latex]2x=\\frac{5\\pi }{3}[\/latex], which means that [latex]x=\\frac{\\pi }{6}[\/latex] or [latex]x=\\frac{5\\pi }{6}[\/latex]. Does this make sense? Yes, because [latex]\\cos \\left(2\\left(\\frac{\\pi }{6}\\right)\\right)=\\cos \\left(\\frac{\\pi }{3}\\right)=\\frac{1}{2}[\/latex].\r\n\r\nAre there any other possible answers? Let us return to our first step.\r\n\r\nIn quadrant I, [latex]2x=\\frac{\\pi }{3}[\/latex], so [latex]x=\\frac{\\pi }{6}[\/latex] as noted. Let us revolve around the circle again:\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x&amp;=\\frac{\\pi }{3}+2\\pi \\\\ &amp;=\\frac{\\pi }{3}+\\frac{6\\pi }{3} \\\\ &amp;=\\frac{7\\pi }{3} \\end{align}[\/latex]<\/p>\r\nso [latex]x=\\frac{7\\pi }{6}[\/latex].\r\n\r\nOne more rotation yields\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x&amp;=\\frac{\\pi }{3}+4\\pi \\\\ &amp;=\\frac{\\pi }{3}+\\frac{12\\pi }{3} \\\\ &amp;=\\frac{13\\pi }{3} \\end{align}[\/latex]<\/p>\r\n[latex]x=\\frac{13\\pi }{6}&gt;2\\pi [\/latex], so this value for [latex]x[\/latex] is larger than [latex]2\\pi [\/latex], so it is not a solution on [latex]\\left[0,2\\pi \\right)[\/latex].\r\n\r\nIn quadrant IV, [latex]2x=\\frac{5\\pi }{3}[\/latex], so [latex]x=\\frac{5\\pi }{6}[\/latex] as noted. Let us revolve around the circle again:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x&amp;=\\frac{5\\pi }{3}+2\\pi \\\\ &amp;=\\frac{5\\pi }{3}+\\frac{6\\pi }{3} \\\\ &amp;=\\frac{11\\pi }{3}\\end{align}[\/latex]<\/p>\r\nso [latex]x=\\frac{11\\pi }{6}[\/latex].\r\n\r\nOne more rotation yields\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x&amp;=\\frac{5\\pi }{3}+4\\pi \\\\ &amp;=\\frac{5\\pi }{3}+\\frac{12\\pi }{3} \\\\ &amp;=\\frac{17\\pi }{3} \\end{align}[\/latex]<\/p>\r\n[latex]x=\\frac{17\\pi }{6}&gt;2\\pi [\/latex], so this value for [latex]x[\/latex] is larger than [latex]2\\pi [\/latex], so it is not a solution on [latex]\\left[0,2\\pi \\right)[\/latex].\r\n\r\nOur solutions are [latex]x=\\frac{\\pi }{6},\\frac{5\\pi }{6},\\frac{7\\pi }{6},\\text{and }\\frac{11\\pi }{6}[\/latex]. Note that whenever we solve a problem in the form of [latex]\\sin \\left(nx\\right)=c[\/latex], we must go around the unit circle [latex]n[\/latex] times.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use Reduction Formulas<\/h2>\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 exercises\">\r\n<h3>Example: Using reduction formulas to reduce powers<\/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 exercises\">\r\n<h3>Example: Using Reduction formulas To Reduce Powers<\/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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify reference angles for angles measured in both radians and degrees<\/li>\n<li>Evaluate trigonometric functions using the unit circle<\/li>\n<li>Solve trigonometric equations<\/li>\n<li>Apply reduction formulas<\/li>\n<\/ul>\n<\/div>\n<p>In the Area and Arc Length in Polar Coordinates section, rather than use the rectangular coordinate system to calculate area under a curve and arc length, we will use the polar coordinate system. Here we will review how to evaluate sine and cosine functions at specific angle measures, solve trigonometric equations, and use reduction formulas.<\/p>\n<h2>Find Reference Angles<\/h2>\n<p><strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-polar-coordinates\/\" target=\"_blank\" rel=\"noopener\">Module 1, Skills Review for Polar Coordinates<\/a>)<\/em><\/strong><\/p>\n<h2>Evaluate Trigonometric Functions Using the Unit Circle<\/h2>\n<p><strong><em>(See <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-polar-coordinates\/\" target=\"_blank\" rel=\"noopener\">Module 1, Skills Review for Polar Coordinates<\/a>)<\/em><\/strong><\/p>\n<h2>Solve Trigonometric Equations<\/h2>\n<p>Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The <strong>period<\/strong> of both the sine function and the cosine function is [latex]2\\pi[\/latex]. In other words, every [latex]2\\pi[\/latex] units, the <em>y-<\/em>values repeat. If we need to find all possible solutions, then we must add [latex]2\\pi k[\/latex], where [latex]k[\/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\\pi :[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\sin \\theta =\\sin \\left(\\theta \\pm 2k\\pi \\right)[\/latex]<\/div>\n<p>There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Linear Trigonometric Equation Containing Cosine<\/h3>\n<p>Find all possible exact solutions for the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q84784\">Show Solution<\/span><\/p>\n<div id=\"q84784\" class=\"hidden-answer\" style=\"display: none\">\n<p>From the <strong>unit circle<\/strong>, we know that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{3},\\frac{5\\pi }{3} \\end{gathered}[\/latex]<\/p>\n<p>These are the solutions in the interval [latex]\\left[0,2\\pi \\right][\/latex]. All possible solutions are given by<\/p>\n<p style=\"text-align: center;\">[latex]\\theta =\\frac{\\pi }{3}\\pm 2k\\pi \\text{ and }\\theta =\\frac{5\\pi }{3}\\pm 2k\\pi[\/latex]<\/p>\n<p>where [latex]k[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE: Solving a Linear Trigonometric Equation Containing Sine<\/h3>\n<p>Find all possible exact solutions for the equation [latex]\\sin t=\\frac{1}{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q535703\">Show Solution<\/span><\/p>\n<div id=\"q535703\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solving for all possible values of <em>t<\/em> means that solutions include angles beyond the period of [latex]2\\pi[\/latex]. From the unit circle, we can see that the solutions are [latex]t=\\frac{\\pi }{6}[\/latex] and [latex]t=\\frac{5\\pi }{6}[\/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{\\pi }{6}\\pm 2\\pi k\\text{ and }t=\\frac{5\\pi }{6}\\pm 2\\pi k[\/latex]<\/p>\n<p>where [latex]k[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE: Solving a Linear Trigonometric Equation Containing Cosine<\/h3>\n<p>Solve the equation exactly: [latex]2\\cos \\theta -3=-5,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q939405\">Show Solution<\/span><\/p>\n<p style=\"text-align: left;\">\n<div id=\"q939405\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2\\cos \\theta -3=-5 \\\\ \\cos \\theta =-2 \\\\ \\cos \\theta =-1 \\\\ \\theta =\\pi \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve exactly the following linear equation on the interval [latex]\\left[0,2\\pi \\right):2\\sin x+1=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q500341\">Show Solution<\/span><\/p>\n<div id=\"q500341\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/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=\"ohm149871\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149871&theme=oea&iframe_resize_id=ohm149871\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\\pi[\/latex], not [latex]2\\pi[\/latex]. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\\frac{\\pi }{2}[\/latex], unless, of course, a problem places its own restrictions on the domain.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE: Solving a Trigonometric Equation<\/h3>\n<p>Solve the problem exactly: [latex]2{\\sin }^{2}\\theta -1=0,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q467313\">Show Solution<\/span><\/p>\n<div id=\"q467313\" class=\"hidden-answer\" style=\"display: none\">\n<p>As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\\sin \\theta[\/latex]. Then we will find the angles.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -1=0 \\\\ 2{\\sin }^{2}\\theta =1 \\\\ {\\sin }^{2}\\theta =\\frac{1}{2} \\\\ \\sqrt{{\\sin }^{2}\\theta }=\\pm \\sqrt{\\frac{1}{2}} \\\\ \\sin \\theta =\\pm \\frac{1}{\\sqrt{2}}=\\pm \\frac{\\sqrt{2}}{2} \\\\ \\theta =\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Trigonometric Equation<\/h3>\n<p>Solve the following equation exactly: [latex]\\csc \\theta =-2,0\\le \\theta <4\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q605306\">Show Solution<\/span><\/p>\n<div id=\"q605306\" class=\"hidden-answer\" style=\"display: none\">\n<p>We want all values of [latex]\\theta[\/latex] for which [latex]\\csc \\theta =-2[\/latex] over the interval [latex]0\\le \\theta <4\\pi[\/latex].\n\n\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\csc \\theta =-2 \\\\ \\frac{1}{\\sin \\theta }=-2 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6},\\frac{19\\pi }{6},\\frac{23\\pi }{6} \\end{gathered}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>As [latex]\\sin \\theta =-\\frac{1}{2}[\/latex], notice that all four solutions are in the third and fourth quadrants.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Trigonometric Equation<\/h3>\n<p>Solve the equation exactly: [latex]\\tan \\left(\\theta -\\frac{\\pi }{2}\\right)=1,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q484899\">Show Solution<\/span><\/p>\n<div id=\"q484899\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall that the tangent function has a period of [latex]\\pi[\/latex]. On the interval [latex]\\left[0,\\pi \\right)[\/latex], and at the angle of [latex]\\frac{\\pi }{4}[\/latex], the tangent has a value of 1. However, the angle we want is [latex]\\left(\\theta -\\frac{\\pi }{2}\\right)[\/latex]. Thus, if [latex]\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\theta -\\frac{\\pi }{2}=\\frac{\\pi }{4}\\\\ \\theta =\\frac{3\\pi }{4}\\pm k\\pi \\end{gathered}[\/latex]<\/p>\n<p>Over the interval [latex]\\left[0,2\\pi \\right)[\/latex], we have two solutions:<\/p>\n<p style=\"text-align: center;\">[latex]\\theta =\\frac{3\\pi }{4}\\text{ and }\\theta =\\frac{3\\pi }{4}+\\pi =\\frac{7\\pi }{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find all solutions for [latex]\\tan x=\\sqrt{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q629684\">Show Solution<\/span><\/p>\n<div id=\"q629684\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi }{3}\\pm \\pi k[\/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=\"ohm173739\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173739&theme=oea&iframe_resize_id=ohm173739\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as [latex]\\sin \\left(2x\\right)[\/latex] or [latex]\\cos \\left(3x\\right)[\/latex]. When confronted with these equations, recall that [latex]y=\\sin \\left(2x\\right)[\/latex] is a <strong>horizontal compression<\/strong> by a factor of 2 of the function [latex]y=\\sin x[\/latex]. On an interval of [latex]2\\pi[\/latex], we can graph two periods of [latex]y=\\sin \\left(2x\\right)[\/latex], as opposed to one cycle of [latex]y=\\sin x[\/latex]. This compression of the graph leads us to believe there may be twice as many <em>x<\/em>-intercepts or solutions to [latex]\\sin \\left(2x\\right)=0[\/latex] compared to [latex]\\sin x=0[\/latex]. This information will help us solve the equation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Multiple Angle Trigonometric Equation<\/h3>\n<p>Solve exactly: [latex]\\cos \\left(2x\\right)=\\frac{1}{2}[\/latex] on [latex]\\left[0,2\\pi \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q299198\">Show Solution<\/span><\/p>\n<div id=\"q299198\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can see that this equation is the standard equation with a multiple of an angle. If [latex]\\cos \\left(\\alpha \\right)=\\frac{1}{2}[\/latex], we know [latex]\\alpha[\/latex] is in quadrants I and IV. While [latex]\\theta ={\\cos }^{-1}\\frac{1}{2}[\/latex] will only yield solutions in quadrants I and II, we recognize that the solutions to the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex] will be in quadrants I and IV.<\/p>\n<p>Therefore, the possible angles are [latex]\\theta =\\frac{\\pi }{3}[\/latex] and [latex]\\theta =\\frac{5\\pi }{3}[\/latex]. So, [latex]2x=\\frac{\\pi }{3}[\/latex] or [latex]2x=\\frac{5\\pi }{3}[\/latex], which means that [latex]x=\\frac{\\pi }{6}[\/latex] or [latex]x=\\frac{5\\pi }{6}[\/latex]. Does this make sense? Yes, because [latex]\\cos \\left(2\\left(\\frac{\\pi }{6}\\right)\\right)=\\cos \\left(\\frac{\\pi }{3}\\right)=\\frac{1}{2}[\/latex].<\/p>\n<p>Are there any other possible answers? Let us return to our first step.<\/p>\n<p>In quadrant I, [latex]2x=\\frac{\\pi }{3}[\/latex], so [latex]x=\\frac{\\pi }{6}[\/latex] as noted. Let us revolve around the circle again:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x&=\\frac{\\pi }{3}+2\\pi \\\\ &=\\frac{\\pi }{3}+\\frac{6\\pi }{3} \\\\ &=\\frac{7\\pi }{3} \\end{align}[\/latex]<\/p>\n<p>so [latex]x=\\frac{7\\pi }{6}[\/latex].<\/p>\n<p>One more rotation yields<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x&=\\frac{\\pi }{3}+4\\pi \\\\ &=\\frac{\\pi }{3}+\\frac{12\\pi }{3} \\\\ &=\\frac{13\\pi }{3} \\end{align}[\/latex]<\/p>\n<p>[latex]x=\\frac{13\\pi }{6}>2\\pi[\/latex], so this value for [latex]x[\/latex] is larger than [latex]2\\pi[\/latex], so it is not a solution on [latex]\\left[0,2\\pi \\right)[\/latex].<\/p>\n<p>In quadrant IV, [latex]2x=\\frac{5\\pi }{3}[\/latex], so [latex]x=\\frac{5\\pi }{6}[\/latex] as noted. Let us revolve around the circle again:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x&=\\frac{5\\pi }{3}+2\\pi \\\\ &=\\frac{5\\pi }{3}+\\frac{6\\pi }{3} \\\\ &=\\frac{11\\pi }{3}\\end{align}[\/latex]<\/p>\n<p>so [latex]x=\\frac{11\\pi }{6}[\/latex].<\/p>\n<p>One more rotation yields<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x&=\\frac{5\\pi }{3}+4\\pi \\\\ &=\\frac{5\\pi }{3}+\\frac{12\\pi }{3} \\\\ &=\\frac{17\\pi }{3} \\end{align}[\/latex]<\/p>\n<p>[latex]x=\\frac{17\\pi }{6}>2\\pi[\/latex], so this value for [latex]x[\/latex] is larger than [latex]2\\pi[\/latex], so it is not a solution on [latex]\\left[0,2\\pi \\right)[\/latex].<\/p>\n<p>Our solutions are [latex]x=\\frac{\\pi }{6},\\frac{5\\pi }{6},\\frac{7\\pi }{6},\\text{and }\\frac{11\\pi }{6}[\/latex]. Note that whenever we solve a problem in the form of [latex]\\sin \\left(nx\\right)=c[\/latex], we must go around the unit circle [latex]n[\/latex] times.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use Reduction Formulas<\/h2>\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 exercises\">\n<h3>Example: Using reduction formulas to reduce powers<\/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 exercises\">\n<h3>Example: Using Reduction formulas To Reduce Powers<\/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\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-1148\">\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>Calculus Volume 1. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus1\/\">https:\/\/courses.lumenlearning.com\/calculus1\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculus Volume 2. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus2\/\">https:\/\/courses.lumenlearning.com\/calculus2\/<\/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":349141,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"\",\"organization\":\"Lumen 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