{"id":17059,"date":"2020-04-14T00:25:11","date_gmt":"2020-04-14T00:25:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/?post_type=chapter&#038;p=17059"},"modified":"2021-09-12T06:03:41","modified_gmt":"2021-09-12T06:03:41","slug":"reference-angles","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/chapter\/reference-angles\/","title":{"raw":"Section 6.3: Properties of the Trigonometric Functions; Reference Angles","rendered":"Section 6.3: Properties of the Trigonometric Functions; Reference Angles"},"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 reference angles to evaluate trigonometric functions.<\/li>\r\n \t<li>Use even-odd properties to find the exact values of the trigonometric functions.<\/li>\r\n \t<li>Use periodic properties to find the exact values of the trigonometric functions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Reference Angles<\/h2>\r\nAn angle\u2019s <strong>reference angle<\/strong> is the measure of the smallest, positive, acute angle [latex]t[\/latex] formed by the terminal side of the angle [latex]t[\/latex] and the horizontal axis. Thus positive reference angles have terminal sides that lie in the first quadrant and can be used as models for angles in other quadrants. See Figure 1 for examples of reference angles for angles in different quadrants.<span id=\"fs-id1165137542464\">\u00a0<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/923\/2015\/04\/25180227\/CNX_Precalc_Figure_05_01_0194.jpg\" alt=\"Four side by side graphs. First graph shows an angle of t in quadrant 1 in it's normal position. Second graph shows an angle of t in quadrant 2 due to a rotation of pi minus t. Third graph shows an angle of t in quadrant 3 due to a rotation of t minus pi. Fourth graph shows an angle of t in quadrant 4 due to a rotation of two pi minus t.\" width=\"975\" height=\"331\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>A GENERAL NOTE: REFERENCE ANGLES<\/h3>\r\nAn angle\u2019s <strong>reference angle<\/strong> is the size of the smallest acute angle, [latex]{t}^{\\prime }[\/latex], formed by the terminal side of the angle [latex]t[\/latex]\u00a0and the horizontal axis.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an angle between [latex]0[\/latex] and [latex]2\\pi [\/latex], find its reference angle.<\/h3>\r\n<ol>\r\n \t<li>An angle in the first quadrant is its own reference angle.<\/li>\r\n \t<li>For an angle in the second or third quadrant, the reference angle is [latex]|\\pi -t|[\/latex] or [latex]|180^\\circ \\mathrm{-t}|[\/latex].<\/li>\r\n \t<li>For an angle in the fourth quadrant, the reference angle is [latex]2\\pi -t[\/latex] or [latex]360^\\circ \\mathrm{-t}[\/latex].<\/li>\r\n \t<li>If an angle is less than [latex]0[\/latex] or greater than [latex]2\\pi [\/latex], add or subtract [latex]2\\pi [\/latex] as many times as needed to find an equivalent angle between [latex]0[\/latex] and [latex]2\\pi [\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Finding a Reference Angle<\/h3>\r\nFind the reference angle of [latex]225^\\circ [\/latex] as 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\/27003606\/CNX_Precalc_Figure_05_02_0162.jpg\" alt=\"Graph of circle with 225 degree angle inscribed.\" width=\"487\" height=\"383\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"770468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"770468\"]\r\n\r\nBecause [latex]225^\\circ [\/latex] is in the third quadrant, the reference angle is\r\n<p style=\"text-align: center;\">[latex]|\\left(180^\\circ -225^\\circ \\right)|=|-45^\\circ |=45^\\circ [\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the reference angle of [latex]\\frac{5\\pi }{3}[\/latex].\r\n\r\n[reveal-answer q=\"227547\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"227547\"]\r\n\r\n[latex]\\frac{\\pi }{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Using Reference Angles<\/span>\r\n\r\nReference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find [latex]\\left(x,y\\right)[\/latex] coordinates for those angles. We will use the <strong>reference angle<\/strong> of the angle of rotation combined with the quadrant in which the terminal side of the angle lies.\u00a0 We can find the exact trig value of any angle in any quadrant if we apply the trig function to the reference angle.\u00a0 The sign depends on the quadrant of the original angle.\r\n\r\nThe trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by <em>x<\/em>- and <em>y<\/em>-values in the original quadrant. Figure 3\u00a0shows which functions are positive in which quadrant.\r\n\r\nTo help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase \"All Students Take Calculus\" Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is \"<strong>A<\/strong>,\" <strong><u>a<\/u><\/strong>ll of the six trigonometric functions are positive. In quadrant II, \"<strong>S<\/strong>tudents,\" only <strong><u>s<\/u><\/strong>ine and its reciprocal function, cosecant, are positive. In quadrant III, \"<strong>T<\/strong>ake,\" only <strong><u>t<\/u><\/strong>angent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, \"<strong>C<\/strong>alculus\" only <strong><u>c<\/u><\/strong>osine and its reciprocal function, secant, are positive.\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\/27003705\/CNX_Precalc_Figure_05_03_0042.jpg\" alt=\"Graph of circle with each quadrant labeled. Under quadrant 1, labels fro sin t, cos t, tan t, sec t, csc t, and cot t. Under quadrant 2, labels for sin t and csc t. Under quadrant 3, labels for tan t and cot t. Under quadrant 4, labels for cos t, sec t.\" width=\"487\" height=\"363\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>How To: Find the trigonometric value for any angle<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Measure the angle between the terminal side of the given angle and the horizontal axis. This is the reference angle.<\/li>\r\n \t<li>Apply the trig function to the reference angle.<\/li>\r\n \t<li>Apply the appropriate sign using the table above.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Using Reference Angles to Find Sine and Cosine<\/h3>\r\n<ol>\r\n \t<li>Using a reference angle, find the exact value of [latex]\\cos \\left(150^\\circ \\right)[\/latex] and [latex]\\text{sin}\\left(150^\\circ \\right)[\/latex].<\/li>\r\n \t<li>Using the reference angle, find [latex]\\cos \\frac{5\\pi }{4}[\/latex] and [latex]\\sin \\frac{5\\pi }{4}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"233925\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"233925\"]\r\n<ol>\r\n \t<li>150\u00b0 is located in the second quadrant. The angle it makes with the <em>x<\/em>-axis is 180\u00b0 \u2212 150\u00b0 = 30\u00b0, so the reference angle is 30\u00b0. This tells us that 150\u00b0 has the same sine and cosine values as 30\u00b0, except for the sign. We know that\r\n<div style=\"text-align: center;\">[latex]\\cos \\left(30^\\circ \\right)=\\frac{\\sqrt{3}}{2}\\text{ and }\\sin \\left(30^\\circ \\right)=\\frac{1}{2}[\/latex].<\/div>\r\nSince 150\u00b0 is in the second quadrant, the <em>x<\/em>-coordinate of the point on the circle is negative, so the cosine value is negative. The <em>y<\/em>-coordinate is positive, so the sine value is positive.\r\n<div style=\"text-align: center;\">[latex]\\cos \\left(150^\\circ \\right)=-\\frac{\\sqrt{3}}{2}\\text{ and }\\sin \\left(150^\\circ \\right)=\\frac{1}{2}[\/latex]<\/div><\/li>\r\n \t<li>[latex]\\frac{5\\pi }{4}[\/latex] is in the third quadrant. Its reference angle is [latex]\\frac{5\\pi }{4}-\\pi =\\frac{\\pi }{4}[\/latex]. The cosine and sine of [latex]\\frac{\\pi }{4}[\/latex] are both [latex]\\frac{\\sqrt{2}}{2}[\/latex]. In the third quadrant, both [latex]x[\/latex] and [latex]y[\/latex] are negative, so:\r\n<div style=\"text-align: center;\">[latex]\\cos \\frac{5\\pi }{4}=-\\frac{\\sqrt{2}}{2}\\text{ and }\\sin \\frac{5\\pi }{4}=-\\frac{\\sqrt{2}}{2}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Using Reference Angles to Find Tangent and Cotangent<\/h3>\r\n<ol>\r\n \t<li>Using a reference angle, find the exact value of [latex]\\tan \\left(240^\\circ \\right)[\/latex]<\/li>\r\n \t<li>Using the reference angle, find [latex]\\cot \\frac{7\\pi }{4}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"233930\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"233930\"]\r\n<ol>\r\n \t<li>240\u00b0 is located in the third quadrant. The angle it makes with the <em>x<\/em>-axis is 240\u00b0 \u2212 180\u00b0 = 60\u00b0, so the reference angle is 60\u00b0. This tells us that 240\u00b0 has the same tangent values as 60\u00b0, except for the sign. We know that\r\n<div style=\"text-align: center;\">[latex]\\tan \\left(60^\\circ \\right)=\\sqrt{3}[\/latex].<\/div>\r\nSince 240\u00b0 is in the second quadrant, the <em>x<\/em>-coordinate of the point on the circle is positive, so the tangent value is positive.\r\n<div style=\"text-align: center;\">[latex]\\tan \\left(240^\\circ \\right)=\\sqrt{3}[\/latex]<\/div><\/li>\r\n \t<li>[latex]\\frac{7\\pi }{4}[\/latex] is in the fourth quadrant. Its reference angle is [latex]2\\pi-\\frac{7\\pi }{4} =\\frac{\\pi }{4}[\/latex]. The cotangent of [latex]\\frac{\\pi }{4}[\/latex] is [latex]\\frac{2}{\\sqrt{2}}=\\sqrt{2}[\/latex]. In the fourth quadrant, cotangent is negative, so\r\n<div style=\"text-align: center;\">[latex]\\cot \\frac{7\\pi }{4}=-\\sqrt{2}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Using Reference Angles to Find Secant and Cosecant<\/h3>\r\n<ol>\r\n \t<li>Using a reference angle, find the exact value of [latex]\\sec \\left(210^\\circ \\right)[\/latex] and [latex]\\csc\\left(210^\\circ \\right)[\/latex].<\/li>\r\n \t<li>Using the reference angle, find [latex]\\sec \\frac{3\\pi }{4}[\/latex] and [latex]\\csc \\frac{3\\pi }{4}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"233920\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"233920\"]\r\n<ol>\r\n \t<li>150\u00b0 is located in the second quadrant. The angle it makes with the <em>x<\/em>-axis is 210\u00b0 \u2212 180\u00b0 = 30\u00b0, so the reference angle is 30\u00b0. This tells us that 210\u00b0 has the same sine and cosine values as 30\u00b0, except for the sign. We know that\r\n<div style=\"text-align: center;\">[latex]\\sec \\left(30^\\circ \\right)=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3}\\text{ and }\\csc \\left(30^\\circ \\right)=2[\/latex].<\/div>\r\nSince 210\u00b0 is in the second quadrant, the <em>x<\/em>-coordinate of the point on the circle is negative, so the sec value is negative. The <em>y<\/em>-coordinate is negative, so the sine value is also negative.\r\n<div style=\"text-align: center;\">[latex]\\sec \\left(210^\\circ \\right)=-\\frac{2\\sqrt{3}}{3}\\text{ and }\\csc \\left(210^\\circ \\right)=-2[\/latex]<\/div><\/li>\r\n \t<li>[latex]\\frac{3\\pi }{4}[\/latex] is in the second quadrant. Its reference angle is [latex]\\pi-\\frac{3\\pi }{4} =\\frac{\\pi }{4}[\/latex]. The secant and cosecant of [latex]\\frac{\\pi }{4}[\/latex] are both [latex]\\frac{2}{\\sqrt{2}}=\\sqrt{2}[\/latex]. In the second quadrant, [latex]x[\/latex] is negative and [latex]y[\/latex] is positive, so:\r\n<div style=\"text-align: center;\">[latex]\\sec \\frac{3\\pi }{4}=-\\sqrt{2}\\text{ and }\\csc \\frac{3\\pi }{4}=\\sqrt{2}[\/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\na. Use the reference angle of [latex]315^\\circ [\/latex] to find [latex]\\cos \\left(315^\\circ \\right)[\/latex] and [latex]\\tan \\left(315^\\circ \\right)[\/latex].\r\n\r\nb. Use the reference angle of [latex]\\frac{2\\pi }{3}[\/latex] to find [latex]\\sec \\left(\\frac{2\\pi }{3}\\right)[\/latex] and [latex]\\cot \\left(\\frac{2\\pi }{3}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"69984\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"69984\"]\r\n\r\na. [latex]\\text{cos}\\left(315^\\circ \\right)=\\frac{\\sqrt{2}}{2},\\text{tan}\\left(315^\\circ \\right)=-1[\/latex]\r\nb. [latex]\\cos \\left(\\frac{2\\pi }{3}\\right)=-\\frac{1}{2},\\cot \\left(\\frac{2\\pi }{3}\\right)=-\\frac{\\sqrt{3}}{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173155[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Using Reference Angles to Find Trigonometric Functions<\/h3>\r\nUse reference angles to find all six trigonometric functions of [latex]-\\frac{5\\pi }{6}[\/latex].\r\n\r\n[reveal-answer q=\"720177\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"720177\"]\r\n\r\nIn order to find a reference angle for a negative angle, add [latex]2\\pi[\/latex] until the angle becomes positive: [latex]-\\frac{5\\pi}{6}+2\\pi=\\frac{7\\pi}{6}[\/latex]. This angle is in the third quadrant, so the reference angle is: [latex]\\frac{7\\pi}{6}-\\pi=\\frac{\\pi}{6}[\/latex]. Since [latex]\\frac{7\\pi }{6}[\/latex] is in the third quadrant, where both [latex]x[\/latex] and [latex]y[\/latex] are negative, sine, cosecant, cosine, and secant will be negative, while tangent and cotangent will be positive.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(-\\frac{5\\pi }{6}\\right)=-\\frac{\\sqrt{3}}{2} \\\\ \\sin \\left(-\\frac{5\\pi }{6}\\right)=-\\frac{1}{2} \\\\ \\tan\\left(-\\frac{5\\pi }{6}\\right)=\\frac{\\sqrt{3}}{3} \\\\ \\sec\\left(-\\frac{5\\pi }{6}\\right)=-\\frac{2\\sqrt{3}}{3}\\\\ \\csc\\left(-\\frac{5\\pi }{6}\\right)=-2\\\\ \\cot \\left(-\\frac{5\\pi }{6}\\right)=\\sqrt{3} \\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\nUse reference angles to find all six trigonometric functions of [latex]-\\frac{7\\pi }{4}[\/latex].\r\n\r\n[reveal-answer q=\"621482\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"621482\"]\r\n\r\n[latex]\\sin \\left(-\\frac{7\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\cos \\left(-\\frac{7\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\tan \\left(-\\frac{7\\pi }{4}\\right)=1[\/latex],\r\n[latex]\\sec \\left(-\\frac{7\\pi }{4}\\right)=\\sqrt{2},\\csc \\left(-\\frac{7\\pi }{4}\\right)=\\sqrt{2},\\cot \\left(-\\frac{7\\pi }{4}\\right)=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Using Reference Angles to Find Trigonometric Functions<\/h3>\r\nUse reference angles to find all six trigonometric functions of [latex]-585^\\circ[\/latex].\r\n\r\n[reveal-answer q=\"720180\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"720180\"]\r\n\r\nWe need to add [latex]360^\\circ[\/latex] until the angle becomes positive. In this case, we needed to add it twice: [latex]-585^\\circ+360^\\circ+360^\\circ=135^\\circ[\/latex]. This angle is in the second quadrant, so the reference angle is: [latex]180^\\circ-135=45^\\circ[\/latex]. Since [latex]45^\\circ[\/latex] is in the second quadrant, where [latex]x[\/latex] is negative and [latex]y[\/latex] is positive, cosine, secant, tangent, and cotangent will be negative, while sine and cosecant will be positive.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(-585^\\circ\\right)=-\\frac{\\sqrt{2}}{2} \\\\ \\sec \\left(-585^\\circ\\right)=-\\sqrt{2} \\\\ \\tan\\left(-585^\\circ\\right)=-1 \\\\ \\cot\\left(-585^\\circ\\right)=-1\\\\ \\sin\\left(-585^\\circ\\right)=\\frac{\\sqrt{2}}{2}\\\\ \\csc \\left(-585^\\circ\\right)=\\sqrt{2} \\end{gathered}[\/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]100617[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Using the Unit Circle to Find Coordinates<\/h3>\r\nFind the coordinates of the point on the unit circle at an angle of [latex]\\frac{7\\pi }{6}[\/latex].\r\n\r\n[reveal-answer q=\"865133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"865133\"]\r\n\r\nWe know that the angle [latex]\\frac{7\\pi }{6}[\/latex] is in the third quadrant.\r\n\r\nFirst, let\u2019s find the reference angle by measuring the angle to the <em>x<\/em>-axis. To find the reference angle of an angle whose terminal side is in quadrant III, we find the difference of the angle and [latex]\\pi [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\frac{7\\pi }{6}-\\pi =\\frac{\\pi }{6}[\/latex]<\/p>\r\nNext, we find the cosine and sine of the reference angle:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\sqrt{3}}{2} \\\\ \\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\end{gathered}[\/latex]<\/p>\r\nWe must determine the appropriate signs for <em>x<\/em> and <em>y<\/em> in the given quadrant. Because our original angle is in the third quadrant, where both [latex]x[\/latex] and [latex]y[\/latex] are negative, both cosine and sine are negative.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(\\frac{7\\pi }{6}\\right)=-\\frac{\\sqrt{3}}{2} \\\\ \\sin \\left(\\frac{7\\pi }{6}\\right)=-\\frac{1}{2} \\end{gathered}[\/latex]<\/p>\r\nNow we can calculate the [latex]\\left(x,y\\right)[\/latex] coordinates using the identities [latex]x=\\cos \\theta [\/latex] and [latex]y=\\sin \\theta [\/latex].\r\n<p style=\"text-align: center;\">The coordinates of the point are [latex]\\left(-\\frac{\\sqrt{3}}{2},-\\frac{1}{2}\\right)[\/latex] on the unit circle.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the coordinates of the point on the unit circle at an angle of [latex]\\frac{5\\pi }{3}[\/latex].\r\n\r\n[reveal-answer q=\"913342\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"913342\"]\r\n\r\n[latex]\\left(\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Finding 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(\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\csc \\left(\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\left(\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\sec \\left(\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cot \\left(\\theta \\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"477960\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"477960\"]\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 4.\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 4<\/b>[\/caption]\r\n\r\nNow that we know all the sides of the triangle, we can use right triangle trigonometry definitions to find the answers:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\left(\\theta\\right)=\\frac{3}{5} \\\\ \\csc \\left(\\theta\\right)=\\frac{5}{3} \\\\ \\cos\\left(\\theta\\right)=-\\frac{4}{5} \\\\ \\sec\\left(\\theta\\right)=-\\frac{5}{4}\\\\ \\cot\\left(\\theta\\right)=\u2212\\frac{4}{3} \\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\nGiven that [latex]\\tan \\theta =\\frac{5}{12}[\/latex] and [latex]\\theta [\/latex] is in quadrant III, find the following:\r\n<ol>\r\n \t<li>[latex]\\sin \\left(\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\csc \\left(\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\left(\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\sec \\left(\\theta \\right)[\/latex]<\/li>\r\n \t<li>[latex]\\cot \\left(\\theta \\right)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"621480\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"621480\"]\r\n\r\n[latex]\\sin \\left(\\theta\\right)=-\\frac{5}{13},\\csc \\left(\\theta\\right)=-\\frac{13}{5},\\cot \\left(\\theta\\right)=\\frac{12}{5}[\/latex],\r\n[latex]\\cos \\left(\\theta\\right)=-\\frac{12}{13},\\sec \\left(\\theta\\right)=-\\frac{13}{12}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using Even and Odd Trigonometric Functions<\/h2>\r\nTo be able to use our six trigonometric functions freely with both positive and negative angle inputs, we should examine how each function treats a negative input. As it turns out, there is an important difference among the functions in this regard.\r\n\r\nConsider the function [latex]f\\left(x\\right)={x}^{2}[\/latex], shown in Figure 5. The graph of the function is symmetrical about the <em>y<\/em>-axis. All along the curve, any two points with opposite <em>x<\/em>-values have the same function value. This matches the result of calculation: [latex]{\\left(4\\right)}^{2}={\\left(-4\\right)}^{2}[\/latex], [latex]{\\left(-5\\right)}^{2}={\\left(5\\right)}^{2}[\/latex],\u00a0and so on. So [latex]f\\left(x\\right)={x}^{2}[\/latex] is an <strong>even function<\/strong>, a function such that two inputs that are opposites have the same output. That means [latex]f\\left(-x\\right)=f\\left(x\\right)[\/latex].\r\n\r\n<span id=\"fs-id1165137817732\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003707\/CNX_Precalc_Figure_05_03_0052.jpg\" alt=\"Graph of parabola with points (-2, 4) and (2, 4) labeled.\" \/><\/span>\r\n<p style=\"text-align: center;\"><strong>Figure 5.\u00a0<\/strong>The function [latex]f\\left(x\\right)={x}^{2}[\/latex]\u00a0is an even function.<\/p>\r\nNow consider the function [latex]f\\left(x\\right)={x}^{3}[\/latex], shown in Figure 6. The graph is not symmetrical about the <em>y<\/em>-axis. All along the graph, any two points with opposite <em>x<\/em>-values also have opposite <em>y<\/em>-values. So [latex]f\\left(x\\right)={x}^{3}[\/latex] is an <strong>odd function<\/strong>, one such that two inputs that are opposites have outputs that are also opposites. That means [latex]f\\left(-x\\right)=-f\\left(x\\right)[\/latex].\r\n\r\n<span id=\"fs-id1165135545756\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003710\/CNX_Precalc_Figure_05_03_0062.jpg\" alt=\"Graph of function with labels for points (-1, -1) and (1, 1).\" \/><\/span>\r\n<p style=\"text-align: center;\"><strong>Figure 6.\u00a0<\/strong>The function [latex]f\\left(x\\right)={x}^{3}[\/latex]\u00a0is an odd function.<\/p>\r\nWe can test whether a trigonometric function is even or odd by drawing a <strong>unit circle<\/strong> with a positive and a negative angle, as in Figure 7. The sine of the positive angle is [latex]y[\/latex]. The sine of the negative angle is \u2212<em>y<\/em>. The <strong>sine function<\/strong>, then, is an odd function. We can test each of the six trigonometric functions in this fashion. The results are shown in in the table below.\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\/27003712\/CNX_Precalc_Figure_05_03_0072.jpg\" alt=\"Graph of circle with angle of t and -t inscribed. Point of (x, y) is at intersection of terminal side of angle t and edge of circle. Point of (x, -y) is at intersection of terminal side of angle -t and edge of circle.\" width=\"487\" height=\"369\" \/> <b>Figure 7<\/b>[\/caption]\r\n<table id=\"Table_05_03_02\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}\\sin t=y\\hfill \\\\ \\sin \\left(-t\\right)=-y\\hfill \\\\ \\sin t\\ne \\sin \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{l}\\text{cos}t=x\\hfill \\\\ \\cos \\left(-t\\right)=x\\hfill \\\\ \\cos t=\\cos \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{l}\\text{tan}\\left(t\\right)=\\frac{y}{x}\\hfill \\\\ \\tan \\left(-t\\right)=-\\frac{y}{x}\\hfill \\\\ \\tan t\\ne \\tan \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}\\sec t=\\frac{1}{x}\\hfill \\\\ \\sec \\left(-t\\right)=\\frac{1}{x}\\hfill \\\\ \\sec t=\\sec \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{l}\\csc t=\\frac{1}{y}\\hfill \\\\ \\csc \\left(-t\\right)=\\frac{1}{-y}\\hfill \\\\ \\csc t\\ne \\csc \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{l}\\cot t=\\frac{x}{y}\\hfill \\\\ \\cot \\left(-t\\right)=\\frac{x}{-y}\\hfill \\\\ \\cot t\\ne cot\\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Even and Odd Trigonometric Functions<\/h3>\r\nAn <strong>even function<\/strong> is one in which [latex]f\\left(-x\\right)=f\\left(x\\right)[\/latex].\r\n\r\nAn <strong>odd function<\/strong> is one in which [latex]f\\left(-x\\right)=-f\\left(x\\right)[\/latex].\r\n\r\nCosine and secant are even:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(-t\\right)=\\cos t \\\\ \\sec \\left(-t\\right)=\\sec t \\end{gathered}[\/latex]<\/p>\r\nSine, tangent, cosecant, and cotangent are odd:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\left(-t\\right)=-\\sin t \\\\ \\tan \\left(-t\\right)=-\\tan t \\\\ \\csc \\left(-t\\right)=-\\csc t \\\\ \\cot \\left(-t\\right)=-\\cot t \\end{gathered}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Using Even and Odd Properties of Trigonometric Functions<\/h3>\r\nIf the [latex]\\sec t=2[\/latex], what is the [latex]\\sec (-t)[\/latex]?\r\n\r\n[reveal-answer q=\"5363\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"5363\"]\r\n\r\nSecant is an even function. The secant of an angle is the same as the secant of its opposite. So if the secant of angle <em>t<\/em> is 2, the secant of [latex]-t[\/latex] is also 2.\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\nIf the [latex]\\cot t=\\sqrt{3}[\/latex], what is [latex]\\cot (-t)[\/latex]?\r\n\r\n[reveal-answer q=\"840134\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"840134\"]\r\n\r\n[latex]-\\sqrt{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Periodic Properties<\/h2>\r\nIf you add or subtract one revolution [latex]\\left(360^\\circ \\text{ or }2\\pi\\right)[\/latex] to an angle, the result will be the same because going around one full revolution will result in the same place on the unit circle. We will let [latex]k[\/latex] be any integer, and this represents [latex]k[\/latex] revolution in the equations below. These formulas are presented in radians, however they can also be expressed in degrees if we use [latex]360^\\circ k[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Periodic Properties<\/h3>\r\nLet [latex]k[\/latex] be an integer, and [latex]t[\/latex] represent an angle.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin\\left(t\\pm 2\\pi k\\right)=\\sin(t) \\\\ \\cos\\left(t\\pm 2\\pi k\\right)=\\cos(t) \\\\ \\tan\\left(t\\pm 2\\pi k\\right)=\\tan(t) \\\\ \\csc\\left(t\\pm 2\\pi k\\right)=\\csc(t) \\\\ \\sec\\left(t\\pm 2\\pi k\\right)=\\sec(t) \\\\ \\cot\\left(t\\pm 2\\pi k\\right)=\\cot(t)\\end{gathered}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Simplifying using Even-Odd Properties and Periodic Properties<\/h3>\r\nUse the Even-Odd Properties and Periodic Properties to simplify:\r\n[latex]7\\cos\\left(-2t\\right)+4\\sin(-2t)-3\\cos\\left(2t-2\\pi\\right)[\/latex]\r\n\r\n[reveal-answer q=\"5360\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"5360\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} &amp; 7\\cos\\left(2t\\right)-4\\sin\\left(2t\\right)-3\\cos\\left(2t\\right) &amp;&amp; \\text{Apply properties.} \\\\ &amp; 4\\cos\\left(2t\\right)-4\\sin\\left(2t\\right) &amp;&amp; \\text{Add like terms.}\\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table id=\"eip-id1165134284283\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Cosine<\/td>\r\n<td>[latex]\\cos t=x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Sine<\/td>\r\n<td>[latex]\\sin t=y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Pythagorean Identity<\/td>\r\n<td>[latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<section id=\"fs-id1165137692629\" class=\"key-concepts\">\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165137772132\">\r\n \t<li>The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.<\/li>\r\n \t<li>The signs of the sine and cosine are determined from the <em>x<\/em>- and <em>y<\/em>-values in the quadrant of the original angle.<\/li>\r\n \t<li>An angle\u2019s reference angle is the size angle, [latex]t[\/latex], formed by the terminal side of the angle [latex]t[\/latex] and the horizontal axis.<\/li>\r\n \t<li>Reference angles can be used to find the sine and cosine of the original angle.<\/li>\r\n \t<li>Reference angles can also be used to find the coordinates of a point on a circle.<\/li>\r\n<\/ul>\r\n<\/section>&nbsp;\r\n<h2 style=\"text-align: center;\">Section 6.3 Homework Exercises<\/h2>\r\n1. Discuss the difference between a coterminal angle and a reference angle.\r\n\r\n2.\u00a0Explain how the cosine of an angle in the second quadrant differs from the cosine of its reference angle in the unit circle.\r\n\r\n3. Explain how the sine of an angle in the second quadrant differs from the sine of its reference angle in the unit circle.\r\n\r\n4. What is the purpose of a reference angle?\r\n\r\nFor the following exercises, state the reference angle for the given angle.\r\n\r\n5. [latex]240^\\circ [\/latex]\r\n\r\n6.\u00a0[latex]-170^\\circ [\/latex]\r\n\r\n7. [latex]460^\\circ [\/latex]\r\n\r\n8.\u00a0[latex]-675^\\circ [\/latex]\r\n\r\n9. [latex]135^\\circ [\/latex]\r\n\r\n10.\u00a0[latex]\\frac{5\\pi }{4}[\/latex]\r\n\r\n11. [latex]\\frac{2\\pi }{3}[\/latex]\r\n\r\n12.\u00a0[latex]\\frac{17\\pi }{6}[\/latex]\r\n\r\n13. [latex]-\\frac{17\\pi }{3}[\/latex]\r\n\r\n14.\u00a0[latex]-\\frac{7\\pi }{4}[\/latex]\r\n\r\n15. [latex]-\\frac{\\pi }{8}[\/latex]\r\n\r\nFor the following exercises, find the reference angle, the quadrant of the terminal side, and the sine, cosine of each angle.\r\n\r\n16. [latex]225^\\circ [\/latex]\r\n\r\n17. [latex]300^\\circ [\/latex]\r\n\r\n18.\u00a0[latex]315^\\circ [\/latex]\r\n\r\n19. [latex]135^\\circ [\/latex]\r\n\r\n20.\u00a0[latex]570^\\circ [\/latex]\r\n\r\n21. [latex]480^\\circ [\/latex]\r\n\r\n22.\u00a0[latex]-120^\\circ [\/latex]\r\n\r\n23. [latex]-210^\\circ [\/latex]\r\n\r\n24.\u00a0[latex]\\frac{5\\pi }{4}[\/latex]\r\n\r\n25. [latex]\\frac{7\\pi }{6}[\/latex]\r\n\r\n26.\u00a0[latex]\\frac{5\\pi }{3}[\/latex]\r\n\r\n27. [latex]\\frac{3\\pi }{4}[\/latex]\r\n\r\n28.\u00a0[latex]\\frac{4\\pi }{3}[\/latex]\r\n\r\n29. [latex]\\frac{2\\pi }{3}[\/latex]\r\n\r\n30.\u00a0[latex]\\frac{-19\\pi }{6}[\/latex]\r\n\r\n31. [latex]\\frac{-9\\pi }{4}[\/latex]\r\n\r\nFor the following exercises, find the reference angle, the quadrant of the terminal side, and the exact value of the trigonometric function.\r\n\r\n32. [latex]\\tan \\frac{5\\pi }{6}[\/latex]\r\n\r\n33. [latex]\\sec \\frac{7\\pi }{6}[\/latex]\r\n\r\n34.\u00a0[latex]\\csc \\frac{11\\pi }{6}[\/latex]\r\n\r\n35. [latex]\\cot \\frac{13\\pi }{6}[\/latex]\r\n\r\n36.\u00a0[latex]\\tan \\frac{15\\pi }{4}[\/latex]\r\n\r\n37. [latex]\\sec \\frac{3\\pi }{4}[\/latex]\r\n\r\n38.\u00a0[latex]\\csc \\frac{5\\pi }{4}[\/latex]\r\n\r\n39. [latex]\\cot \\frac{11\\pi }{4}[\/latex]\r\n\r\n40.\u00a0[latex]\\tan \\left(-\\frac{4\\pi }{3}\\right)[\/latex]\r\n\r\n41. [latex]\\sec \\left(-\\frac{2\\pi }{3}\\right)[\/latex]\r\n\r\n42.\u00a0[latex]\\csc \\left(-\\frac{10\\pi }{3}\\right)[\/latex]\r\n\r\n43. [latex]\\cot \\left(-\\frac{7\\pi }{3}\\right)[\/latex]\r\n\r\n44.\u00a0[latex]\\tan 225^\\circ [\/latex]\r\n\r\n45. [latex]\\sec 300^\\circ [\/latex]\r\n\r\n46.\u00a0[latex]\\csc 510^\\circ [\/latex]\r\n\r\n47. [latex]\\cot 600^\\circ [\/latex]\r\n\r\n48.\u00a0[latex]\\tan \\left(-30^\\circ\\right) [\/latex]\r\n\r\n49. [latex]\\sec \\left(-210^\\circ\\right) [\/latex]\r\n\r\n50.\u00a0[latex]\\csc \\left(-510^\\circ\\right) [\/latex]\r\n\r\n51. [latex]\\cot \\left(-405^\\circ\\right) [\/latex]\r\n\r\nIn the following exercises, use a right triangle to find the exact value.\r\n\r\n52.\u00a0If [latex]\\text{sin}t=\\frac{3}{4}[\/latex], and [latex]t[\/latex] is in quadrant II, find [latex]\\cos t,\\sec t,\\csc t,\\tan t,\\cot t[\/latex].\r\n\r\n53. If [latex]\\text{cos}t=-\\frac{1}{3}[\/latex], and [latex]t[\/latex] is in quadrant III, find [latex]\\sin t,\\sec t,\\csc t,\\tan t,\\cot t[\/latex].\r\n\r\n54.\u00a0If [latex]\\tan t=\\frac{12}{5}[\/latex], and [latex]0\\le t&lt;\\frac{\\pi }{2}[\/latex], find [latex]\\sin t,\\cos t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex].\r\n\r\n55. If [latex]\\sin t=\\frac{\\sqrt{3}}{2}[\/latex] and [latex]\\cos t=\\frac{1}{2}[\/latex], find [latex]\\sec t,\\csc t,\\tan t[\/latex], and [latex]\\cot t[\/latex].\r\n\r\nFor the following exercises, find the exact value using reference angles.\r\n\r\n56.\u00a0[latex]\\sin\\left(\\frac{11\\pi}{3}\\right)\\cos\\left(\\frac{-5\\pi}{6}\\right)[\/latex]\r\n\r\n57.\u00a0[latex]\\sin\\left(\\frac{3\\pi}{4}\\right)\\cos\\left(\\frac{5\\pi}{3}\\right)[\/latex]\r\n\r\n58.\u00a0[latex]\\sin\\left(\\frac{-4\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{2}\\right)[\/latex]\r\n\r\n59.\u00a0[latex]\\sin\\left(\\frac{-9\\pi}{4}\\right)\\cos\\left(\\frac{-\\pi}{6}\\right)[\/latex]\r\n\r\n60.\u00a0[latex]\\sin\\left(\\frac{\\pi}{6}\\right)\\cos\\left(\\frac{-\\pi}{3}\\right)[\/latex]\r\n\r\n61. [latex]\\sin\\left(\\frac{7\\pi}{4}\\right)\\cos\\left(\\frac{-2\\pi}{3}\\right)[\/latex]\r\n\r\n62.\u00a0[latex]\\cos\\left(\\frac{5\\pi}{6}\\right)\\cos\\left(\\frac{2\\pi}{3}\\right)[\/latex]\r\n\r\n63.\u00a0[latex]\\cos\\left(\\frac{-\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{4}\\right)[\/latex]\r\n\r\n64.\u00a0[latex]\\sin\\left(\\frac{-5\\pi}{4}\\right)\\sin\\left(\\frac{11\\pi}{6}\\right)[\/latex]\r\n\r\n65.\u00a0[latex]\\sin\\left(\\pi\\right)\\sin\\left(\\frac{\\pi}{6}\\right)[\/latex]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"font-weight: 400;\">Use reference angles to evaluate trigonometric functions.<\/li>\n<li>Use even-odd properties to find the exact values of the trigonometric functions.<\/li>\n<li>Use periodic properties to find the exact values of the trigonometric functions.<\/li>\n<\/ul>\n<\/div>\n<h2>Reference Angles<\/h2>\n<p>An angle\u2019s <strong>reference angle<\/strong> is the measure of the smallest, positive, acute angle [latex]t[\/latex] formed by the terminal side of the angle [latex]t[\/latex] and the horizontal axis. Thus positive reference angles have terminal sides that lie in the first quadrant and can be used as models for angles in other quadrants. See Figure 1 for examples of reference angles for angles in different quadrants.<span id=\"fs-id1165137542464\">\u00a0<\/span><\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/923\/2015\/04\/25180227\/CNX_Precalc_Figure_05_01_0194.jpg\" alt=\"Four side by side graphs. First graph shows an angle of t in quadrant 1 in it's normal position. Second graph shows an angle of t in quadrant 2 due to a rotation of pi minus t. Third graph shows an angle of t in quadrant 3 due to a rotation of t minus pi. Fourth graph shows an angle of t in quadrant 4 due to a rotation of two pi minus t.\" width=\"975\" height=\"331\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A GENERAL NOTE: REFERENCE ANGLES<\/h3>\n<p>An angle\u2019s <strong>reference angle<\/strong> is the size of the smallest acute angle, [latex]{t}^{\\prime }[\/latex], formed by the terminal side of the angle [latex]t[\/latex]\u00a0and the horizontal axis.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an angle between [latex]0[\/latex] and [latex]2\\pi[\/latex], find its reference angle.<\/h3>\n<ol>\n<li>An angle in the first quadrant is its own reference angle.<\/li>\n<li>For an angle in the second or third quadrant, the reference angle is [latex]|\\pi -t|[\/latex] or [latex]|180^\\circ \\mathrm{-t}|[\/latex].<\/li>\n<li>For an angle in the fourth quadrant, the reference angle is [latex]2\\pi -t[\/latex] or [latex]360^\\circ \\mathrm{-t}[\/latex].<\/li>\n<li>If an angle is less than [latex]0[\/latex] or greater than [latex]2\\pi[\/latex], add or subtract [latex]2\\pi[\/latex] as many times as needed to find an equivalent angle between [latex]0[\/latex] and [latex]2\\pi[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Finding a Reference Angle<\/h3>\n<p>Find the reference angle of [latex]225^\\circ[\/latex] as 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\/27003606\/CNX_Precalc_Figure_05_02_0162.jpg\" alt=\"Graph of circle with 225 degree angle inscribed.\" width=\"487\" height=\"383\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q770468\">Show Solution<\/span><\/p>\n<div id=\"q770468\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because [latex]225^\\circ[\/latex] is in the third quadrant, the reference angle is<\/p>\n<p style=\"text-align: center;\">[latex]|\\left(180^\\circ -225^\\circ \\right)|=|-45^\\circ |=45^\\circ[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the reference angle of [latex]\\frac{5\\pi }{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q227547\">Show Solution<\/span><\/p>\n<div id=\"q227547\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi }{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Using Reference Angles<\/span><\/p>\n<p>Reference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find [latex]\\left(x,y\\right)[\/latex] coordinates for those angles. We will use the <strong>reference angle<\/strong> of the angle of rotation combined with the quadrant in which the terminal side of the angle lies.\u00a0 We can find the exact trig value of any angle in any quadrant if we apply the trig function to the reference angle.\u00a0 The sign depends on the quadrant of the original angle.<\/p>\n<p>The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by <em>x<\/em>&#8211; and <em>y<\/em>-values in the original quadrant. Figure 3\u00a0shows which functions are positive in which quadrant.<\/p>\n<p>To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase &#8220;All Students Take Calculus&#8221; Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is &#8220;<strong>A<\/strong>,&#8221; <strong><u>a<\/u><\/strong>ll of the six trigonometric functions are positive. In quadrant II, &#8220;<strong>S<\/strong>tudents,&#8221; only <strong><u>s<\/u><\/strong>ine and its reciprocal function, cosecant, are positive. In quadrant III, &#8220;<strong>T<\/strong>ake,&#8221; only <strong><u>t<\/u><\/strong>angent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, &#8220;<strong>C<\/strong>alculus&#8221; only <strong><u>c<\/u><\/strong>osine and its reciprocal function, secant, are positive.<\/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\/27003705\/CNX_Precalc_Figure_05_03_0042.jpg\" alt=\"Graph of circle with each quadrant labeled. Under quadrant 1, labels fro sin t, cos t, tan t, sec t, csc t, and cot t. Under quadrant 2, labels for sin t and csc t. Under quadrant 3, labels for tan t and cot t. Under quadrant 4, labels for cos t, sec t.\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Find the trigonometric value for any angle<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Measure the angle between the terminal side of the given angle and the horizontal axis. This is the reference angle.<\/li>\n<li>Apply the trig function to the reference angle.<\/li>\n<li>Apply the appropriate sign using the table above.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Using Reference Angles to Find Sine and Cosine<\/h3>\n<ol>\n<li>Using a reference angle, find the exact value of [latex]\\cos \\left(150^\\circ \\right)[\/latex] and [latex]\\text{sin}\\left(150^\\circ \\right)[\/latex].<\/li>\n<li>Using the reference angle, find [latex]\\cos \\frac{5\\pi }{4}[\/latex] and [latex]\\sin \\frac{5\\pi }{4}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q233925\">Show Solution<\/span><\/p>\n<div id=\"q233925\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>150\u00b0 is located in the second quadrant. The angle it makes with the <em>x<\/em>-axis is 180\u00b0 \u2212 150\u00b0 = 30\u00b0, so the reference angle is 30\u00b0. This tells us that 150\u00b0 has the same sine and cosine values as 30\u00b0, except for the sign. We know that\n<div style=\"text-align: center;\">[latex]\\cos \\left(30^\\circ \\right)=\\frac{\\sqrt{3}}{2}\\text{ and }\\sin \\left(30^\\circ \\right)=\\frac{1}{2}[\/latex].<\/div>\n<p>Since 150\u00b0 is in the second quadrant, the <em>x<\/em>-coordinate of the point on the circle is negative, so the cosine value is negative. The <em>y<\/em>-coordinate is positive, so the sine value is positive.<\/p>\n<div style=\"text-align: center;\">[latex]\\cos \\left(150^\\circ \\right)=-\\frac{\\sqrt{3}}{2}\\text{ and }\\sin \\left(150^\\circ \\right)=\\frac{1}{2}[\/latex]<\/div>\n<\/li>\n<li>[latex]\\frac{5\\pi }{4}[\/latex] is in the third quadrant. Its reference angle is [latex]\\frac{5\\pi }{4}-\\pi =\\frac{\\pi }{4}[\/latex]. The cosine and sine of [latex]\\frac{\\pi }{4}[\/latex] are both [latex]\\frac{\\sqrt{2}}{2}[\/latex]. In the third quadrant, both [latex]x[\/latex] and [latex]y[\/latex] are negative, so:\n<div style=\"text-align: center;\">[latex]\\cos \\frac{5\\pi }{4}=-\\frac{\\sqrt{2}}{2}\\text{ and }\\sin \\frac{5\\pi }{4}=-\\frac{\\sqrt{2}}{2}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Using Reference Angles to Find Tangent and Cotangent<\/h3>\n<ol>\n<li>Using a reference angle, find the exact value of [latex]\\tan \\left(240^\\circ \\right)[\/latex]<\/li>\n<li>Using the reference angle, find [latex]\\cot \\frac{7\\pi }{4}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q233930\">Show Solution<\/span><\/p>\n<div id=\"q233930\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>240\u00b0 is located in the third quadrant. The angle it makes with the <em>x<\/em>-axis is 240\u00b0 \u2212 180\u00b0 = 60\u00b0, so the reference angle is 60\u00b0. This tells us that 240\u00b0 has the same tangent values as 60\u00b0, except for the sign. We know that\n<div style=\"text-align: center;\">[latex]\\tan \\left(60^\\circ \\right)=\\sqrt{3}[\/latex].<\/div>\n<p>Since 240\u00b0 is in the second quadrant, the <em>x<\/em>-coordinate of the point on the circle is positive, so the tangent value is positive.<\/p>\n<div style=\"text-align: center;\">[latex]\\tan \\left(240^\\circ \\right)=\\sqrt{3}[\/latex]<\/div>\n<\/li>\n<li>[latex]\\frac{7\\pi }{4}[\/latex] is in the fourth quadrant. Its reference angle is [latex]2\\pi-\\frac{7\\pi }{4} =\\frac{\\pi }{4}[\/latex]. The cotangent of [latex]\\frac{\\pi }{4}[\/latex] is [latex]\\frac{2}{\\sqrt{2}}=\\sqrt{2}[\/latex]. In the fourth quadrant, cotangent is negative, so\n<div style=\"text-align: center;\">[latex]\\cot \\frac{7\\pi }{4}=-\\sqrt{2}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Using Reference Angles to Find Secant and Cosecant<\/h3>\n<ol>\n<li>Using a reference angle, find the exact value of [latex]\\sec \\left(210^\\circ \\right)[\/latex] and [latex]\\csc\\left(210^\\circ \\right)[\/latex].<\/li>\n<li>Using the reference angle, find [latex]\\sec \\frac{3\\pi }{4}[\/latex] and [latex]\\csc \\frac{3\\pi }{4}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q233920\">Show Solution<\/span><\/p>\n<div id=\"q233920\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>150\u00b0 is located in the second quadrant. The angle it makes with the <em>x<\/em>-axis is 210\u00b0 \u2212 180\u00b0 = 30\u00b0, so the reference angle is 30\u00b0. This tells us that 210\u00b0 has the same sine and cosine values as 30\u00b0, except for the sign. We know that\n<div style=\"text-align: center;\">[latex]\\sec \\left(30^\\circ \\right)=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3}\\text{ and }\\csc \\left(30^\\circ \\right)=2[\/latex].<\/div>\n<p>Since 210\u00b0 is in the second quadrant, the <em>x<\/em>-coordinate of the point on the circle is negative, so the sec value is negative. The <em>y<\/em>-coordinate is negative, so the sine value is also negative.<\/p>\n<div style=\"text-align: center;\">[latex]\\sec \\left(210^\\circ \\right)=-\\frac{2\\sqrt{3}}{3}\\text{ and }\\csc \\left(210^\\circ \\right)=-2[\/latex]<\/div>\n<\/li>\n<li>[latex]\\frac{3\\pi }{4}[\/latex] is in the second quadrant. Its reference angle is [latex]\\pi-\\frac{3\\pi }{4} =\\frac{\\pi }{4}[\/latex]. The secant and cosecant of [latex]\\frac{\\pi }{4}[\/latex] are both [latex]\\frac{2}{\\sqrt{2}}=\\sqrt{2}[\/latex]. In the second quadrant, [latex]x[\/latex] is negative and [latex]y[\/latex] is positive, so:\n<div style=\"text-align: center;\">[latex]\\sec \\frac{3\\pi }{4}=-\\sqrt{2}\\text{ and }\\csc \\frac{3\\pi }{4}=\\sqrt{2}[\/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>a. Use the reference angle of [latex]315^\\circ[\/latex] to find [latex]\\cos \\left(315^\\circ \\right)[\/latex] and [latex]\\tan \\left(315^\\circ \\right)[\/latex].<\/p>\n<p>b. Use the reference angle of [latex]\\frac{2\\pi }{3}[\/latex] to find [latex]\\sec \\left(\\frac{2\\pi }{3}\\right)[\/latex] and [latex]\\cot \\left(\\frac{2\\pi }{3}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q69984\">Show Solution<\/span><\/p>\n<div id=\"q69984\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. [latex]\\text{cos}\\left(315^\\circ \\right)=\\frac{\\sqrt{2}}{2},\\text{tan}\\left(315^\\circ \\right)=-1[\/latex]<br \/>\nb. [latex]\\cos \\left(\\frac{2\\pi }{3}\\right)=-\\frac{1}{2},\\cot \\left(\\frac{2\\pi }{3}\\right)=-\\frac{\\sqrt{3}}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173155\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173155&theme=oea&iframe_resize_id=ohm173155\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Using Reference Angles to Find Trigonometric Functions<\/h3>\n<p>Use reference angles to find all six trigonometric functions of [latex]-\\frac{5\\pi }{6}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q720177\">Show Solution<\/span><\/p>\n<div id=\"q720177\" class=\"hidden-answer\" style=\"display: none\">\n<p>In order to find a reference angle for a negative angle, add [latex]2\\pi[\/latex] until the angle becomes positive: [latex]-\\frac{5\\pi}{6}+2\\pi=\\frac{7\\pi}{6}[\/latex]. This angle is in the third quadrant, so the reference angle is: [latex]\\frac{7\\pi}{6}-\\pi=\\frac{\\pi}{6}[\/latex]. Since [latex]\\frac{7\\pi }{6}[\/latex] is in the third quadrant, where both [latex]x[\/latex] and [latex]y[\/latex] are negative, sine, cosecant, cosine, and secant will be negative, while tangent and cotangent will be positive.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(-\\frac{5\\pi }{6}\\right)=-\\frac{\\sqrt{3}}{2} \\\\ \\sin \\left(-\\frac{5\\pi }{6}\\right)=-\\frac{1}{2} \\\\ \\tan\\left(-\\frac{5\\pi }{6}\\right)=\\frac{\\sqrt{3}}{3} \\\\ \\sec\\left(-\\frac{5\\pi }{6}\\right)=-\\frac{2\\sqrt{3}}{3}\\\\ \\csc\\left(-\\frac{5\\pi }{6}\\right)=-2\\\\ \\cot \\left(-\\frac{5\\pi }{6}\\right)=\\sqrt{3} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Use reference angles to find all six trigonometric functions of [latex]-\\frac{7\\pi }{4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q621482\">Show Solution<\/span><\/p>\n<div id=\"q621482\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sin \\left(-\\frac{7\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\cos \\left(-\\frac{7\\pi }{4}\\right)=\\frac{\\sqrt{2}}{2},\\tan \\left(-\\frac{7\\pi }{4}\\right)=1[\/latex],<br \/>\n[latex]\\sec \\left(-\\frac{7\\pi }{4}\\right)=\\sqrt{2},\\csc \\left(-\\frac{7\\pi }{4}\\right)=\\sqrt{2},\\cot \\left(-\\frac{7\\pi }{4}\\right)=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Using Reference Angles to Find Trigonometric Functions<\/h3>\n<p>Use reference angles to find all six trigonometric functions of [latex]-585^\\circ[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q720180\">Show Solution<\/span><\/p>\n<div id=\"q720180\" class=\"hidden-answer\" style=\"display: none\">\n<p>We need to add [latex]360^\\circ[\/latex] until the angle becomes positive. In this case, we needed to add it twice: [latex]-585^\\circ+360^\\circ+360^\\circ=135^\\circ[\/latex]. This angle is in the second quadrant, so the reference angle is: [latex]180^\\circ-135=45^\\circ[\/latex]. Since [latex]45^\\circ[\/latex] is in the second quadrant, where [latex]x[\/latex] is negative and [latex]y[\/latex] is positive, cosine, secant, tangent, and cotangent will be negative, while sine and cosecant will be positive.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(-585^\\circ\\right)=-\\frac{\\sqrt{2}}{2} \\\\ \\sec \\left(-585^\\circ\\right)=-\\sqrt{2} \\\\ \\tan\\left(-585^\\circ\\right)=-1 \\\\ \\cot\\left(-585^\\circ\\right)=-1\\\\ \\sin\\left(-585^\\circ\\right)=\\frac{\\sqrt{2}}{2}\\\\ \\csc \\left(-585^\\circ\\right)=\\sqrt{2} \\end{gathered}[\/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=\"ohm100617\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100617&theme=oea&iframe_resize_id=ohm100617\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Using the Unit Circle to Find Coordinates<\/h3>\n<p>Find the coordinates of the point on the unit circle at an angle of [latex]\\frac{7\\pi }{6}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q865133\">Show Solution<\/span><\/p>\n<div id=\"q865133\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know that the angle [latex]\\frac{7\\pi }{6}[\/latex] is in the third quadrant.<\/p>\n<p>First, let\u2019s find the reference angle by measuring the angle to the <em>x<\/em>-axis. To find the reference angle of an angle whose terminal side is in quadrant III, we find the difference of the angle and [latex]\\pi[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{7\\pi }{6}-\\pi =\\frac{\\pi }{6}[\/latex]<\/p>\n<p>Next, we find the cosine and sine of the reference angle:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\sqrt{3}}{2} \\\\ \\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\end{gathered}[\/latex]<\/p>\n<p>We must determine the appropriate signs for <em>x<\/em> and <em>y<\/em> in the given quadrant. Because our original angle is in the third quadrant, where both [latex]x[\/latex] and [latex]y[\/latex] are negative, both cosine and sine are negative.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(\\frac{7\\pi }{6}\\right)=-\\frac{\\sqrt{3}}{2} \\\\ \\sin \\left(\\frac{7\\pi }{6}\\right)=-\\frac{1}{2} \\end{gathered}[\/latex]<\/p>\n<p>Now we can calculate the [latex]\\left(x,y\\right)[\/latex] coordinates using the identities [latex]x=\\cos \\theta[\/latex] and [latex]y=\\sin \\theta[\/latex].<\/p>\n<p style=\"text-align: center;\">The coordinates of the point are [latex]\\left(-\\frac{\\sqrt{3}}{2},-\\frac{1}{2}\\right)[\/latex] on the unit circle.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the coordinates of the point on the unit circle at an angle of [latex]\\frac{5\\pi }{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q913342\">Show Solution<\/span><\/p>\n<div id=\"q913342\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Finding 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(\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\csc \\left(\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\cos \\left(\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\sec \\left(\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\cot \\left(\\theta \\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q477960\">Show Solution<\/span><\/p>\n<div id=\"q477960\" 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 4.<\/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 4<\/b><\/p>\n<\/div>\n<p>Now that we know all the sides of the triangle, we can use right triangle trigonometry definitions to find the answers:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\left(\\theta\\right)=\\frac{3}{5} \\\\ \\csc \\left(\\theta\\right)=\\frac{5}{3} \\\\ \\cos\\left(\\theta\\right)=-\\frac{4}{5} \\\\ \\sec\\left(\\theta\\right)=-\\frac{5}{4}\\\\ \\cot\\left(\\theta\\right)=\u2212\\frac{4}{3} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given that [latex]\\tan \\theta =\\frac{5}{12}[\/latex] and [latex]\\theta[\/latex] is in quadrant III, find the following:<\/p>\n<ol>\n<li>[latex]\\sin \\left(\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\csc \\left(\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\cos \\left(\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\sec \\left(\\theta \\right)[\/latex]<\/li>\n<li>[latex]\\cot \\left(\\theta \\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q621480\">Show Solution<\/span><\/p>\n<div id=\"q621480\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sin \\left(\\theta\\right)=-\\frac{5}{13},\\csc \\left(\\theta\\right)=-\\frac{13}{5},\\cot \\left(\\theta\\right)=\\frac{12}{5}[\/latex],<br \/>\n[latex]\\cos \\left(\\theta\\right)=-\\frac{12}{13},\\sec \\left(\\theta\\right)=-\\frac{13}{12}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using Even and Odd Trigonometric Functions<\/h2>\n<p>To be able to use our six trigonometric functions freely with both positive and negative angle inputs, we should examine how each function treats a negative input. As it turns out, there is an important difference among the functions in this regard.<\/p>\n<p>Consider the function [latex]f\\left(x\\right)={x}^{2}[\/latex], shown in Figure 5. The graph of the function is symmetrical about the <em>y<\/em>-axis. All along the curve, any two points with opposite <em>x<\/em>-values have the same function value. This matches the result of calculation: [latex]{\\left(4\\right)}^{2}={\\left(-4\\right)}^{2}[\/latex], [latex]{\\left(-5\\right)}^{2}={\\left(5\\right)}^{2}[\/latex],\u00a0and so on. So [latex]f\\left(x\\right)={x}^{2}[\/latex] is an <strong>even function<\/strong>, a function such that two inputs that are opposites have the same output. That means [latex]f\\left(-x\\right)=f\\left(x\\right)[\/latex].<\/p>\n<p><span id=\"fs-id1165137817732\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003707\/CNX_Precalc_Figure_05_03_0052.jpg\" alt=\"Graph of parabola with points (-2, 4) and (2, 4) labeled.\" \/><\/span><\/p>\n<p style=\"text-align: center;\"><strong>Figure 5.\u00a0<\/strong>The function [latex]f\\left(x\\right)={x}^{2}[\/latex]\u00a0is an even function.<\/p>\n<p>Now consider the function [latex]f\\left(x\\right)={x}^{3}[\/latex], shown in Figure 6. The graph is not symmetrical about the <em>y<\/em>-axis. All along the graph, any two points with opposite <em>x<\/em>-values also have opposite <em>y<\/em>-values. So [latex]f\\left(x\\right)={x}^{3}[\/latex] is an <strong>odd function<\/strong>, one such that two inputs that are opposites have outputs that are also opposites. That means [latex]f\\left(-x\\right)=-f\\left(x\\right)[\/latex].<\/p>\n<p><span id=\"fs-id1165135545756\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003710\/CNX_Precalc_Figure_05_03_0062.jpg\" alt=\"Graph of function with labels for points (-1, -1) and (1, 1).\" \/><\/span><\/p>\n<p style=\"text-align: center;\"><strong>Figure 6.\u00a0<\/strong>The function [latex]f\\left(x\\right)={x}^{3}[\/latex]\u00a0is an odd function.<\/p>\n<p>We can test whether a trigonometric function is even or odd by drawing a <strong>unit circle<\/strong> with a positive and a negative angle, as in Figure 7. The sine of the positive angle is [latex]y[\/latex]. The sine of the negative angle is \u2212<em>y<\/em>. The <strong>sine function<\/strong>, then, is an odd function. We can test each of the six trigonometric functions in this fashion. The results are shown in in the table below.<\/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\/27003712\/CNX_Precalc_Figure_05_03_0072.jpg\" alt=\"Graph of circle with angle of t and -t inscribed. Point of (x, y) is at intersection of terminal side of angle t and edge of circle. Point of (x, -y) is at intersection of terminal side of angle -t and edge of circle.\" width=\"487\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<table id=\"Table_05_03_02\" summary=\"..\">\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{l}\\sin t=y\\hfill \\\\ \\sin \\left(-t\\right)=-y\\hfill \\\\ \\sin t\\ne \\sin \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\n<td>[latex]\\begin{array}{l}\\text{cos}t=x\\hfill \\\\ \\cos \\left(-t\\right)=x\\hfill \\\\ \\cos t=\\cos \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\n<td>[latex]\\begin{array}{l}\\text{tan}\\left(t\\right)=\\frac{y}{x}\\hfill \\\\ \\tan \\left(-t\\right)=-\\frac{y}{x}\\hfill \\\\ \\tan t\\ne \\tan \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}\\sec t=\\frac{1}{x}\\hfill \\\\ \\sec \\left(-t\\right)=\\frac{1}{x}\\hfill \\\\ \\sec t=\\sec \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\n<td>[latex]\\begin{array}{l}\\csc t=\\frac{1}{y}\\hfill \\\\ \\csc \\left(-t\\right)=\\frac{1}{-y}\\hfill \\\\ \\csc t\\ne \\csc \\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\n<td>[latex]\\begin{array}{l}\\cot t=\\frac{x}{y}\\hfill \\\\ \\cot \\left(-t\\right)=\\frac{x}{-y}\\hfill \\\\ \\cot t\\ne cot\\left(-t\\right)\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox\">\n<h3>A General Note: Even and Odd Trigonometric Functions<\/h3>\n<p>An <strong>even function<\/strong> is one in which [latex]f\\left(-x\\right)=f\\left(x\\right)[\/latex].<\/p>\n<p>An <strong>odd function<\/strong> is one in which [latex]f\\left(-x\\right)=-f\\left(x\\right)[\/latex].<\/p>\n<p>Cosine and secant are even:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(-t\\right)=\\cos t \\\\ \\sec \\left(-t\\right)=\\sec t \\end{gathered}[\/latex]<\/p>\n<p>Sine, tangent, cosecant, and cotangent are odd:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\left(-t\\right)=-\\sin t \\\\ \\tan \\left(-t\\right)=-\\tan t \\\\ \\csc \\left(-t\\right)=-\\csc t \\\\ \\cot \\left(-t\\right)=-\\cot t \\end{gathered}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Using Even and Odd Properties of Trigonometric Functions<\/h3>\n<p>If the [latex]\\sec t=2[\/latex], what is the [latex]\\sec (-t)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q5363\">Show Solution<\/span><\/p>\n<div id=\"q5363\" class=\"hidden-answer\" style=\"display: none\">\n<p>Secant is an even function. The secant of an angle is the same as the secant of its opposite. So if the secant of angle <em>t<\/em> is 2, the secant of [latex]-t[\/latex] is also 2.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>If the [latex]\\cot t=\\sqrt{3}[\/latex], what is [latex]\\cot (-t)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q840134\">Show Solution<\/span><\/p>\n<div id=\"q840134\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\sqrt{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Periodic Properties<\/h2>\n<p>If you add or subtract one revolution [latex]\\left(360^\\circ \\text{ or }2\\pi\\right)[\/latex] to an angle, the result will be the same because going around one full revolution will result in the same place on the unit circle. We will let [latex]k[\/latex] be any integer, and this represents [latex]k[\/latex] revolution in the equations below. These formulas are presented in radians, however they can also be expressed in degrees if we use [latex]360^\\circ k[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Periodic Properties<\/h3>\n<p>Let [latex]k[\/latex] be an integer, and [latex]t[\/latex] represent an angle.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin\\left(t\\pm 2\\pi k\\right)=\\sin(t) \\\\ \\cos\\left(t\\pm 2\\pi k\\right)=\\cos(t) \\\\ \\tan\\left(t\\pm 2\\pi k\\right)=\\tan(t) \\\\ \\csc\\left(t\\pm 2\\pi k\\right)=\\csc(t) \\\\ \\sec\\left(t\\pm 2\\pi k\\right)=\\sec(t) \\\\ \\cot\\left(t\\pm 2\\pi k\\right)=\\cot(t)\\end{gathered}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Simplifying using Even-Odd Properties and Periodic Properties<\/h3>\n<p>Use the Even-Odd Properties and Periodic Properties to simplify:<br \/>\n[latex]7\\cos\\left(-2t\\right)+4\\sin(-2t)-3\\cos\\left(2t-2\\pi\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q5360\">Show Solution<\/span><\/p>\n<div id=\"q5360\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{gathered} & 7\\cos\\left(2t\\right)-4\\sin\\left(2t\\right)-3\\cos\\left(2t\\right) && \\text{Apply properties.} \\\\ & 4\\cos\\left(2t\\right)-4\\sin\\left(2t\\right) && \\text{Add like terms.}\\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Equations<\/h2>\n<table id=\"eip-id1165134284283\" summary=\"..\">\n<tbody>\n<tr>\n<td>Cosine<\/td>\n<td>[latex]\\cos t=x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Sine<\/td>\n<td>[latex]\\sin t=y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Pythagorean Identity<\/td>\n<td>[latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<section id=\"fs-id1165137692629\" class=\"key-concepts\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137772132\">\n<li>The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.<\/li>\n<li>The signs of the sine and cosine are determined from the <em>x<\/em>&#8211; and <em>y<\/em>-values in the quadrant of the original angle.<\/li>\n<li>An angle\u2019s reference angle is the size angle, [latex]t[\/latex], formed by the terminal side of the angle [latex]t[\/latex] and the horizontal axis.<\/li>\n<li>Reference angles can be used to find the sine and cosine of the original angle.<\/li>\n<li>Reference angles can also be used to find the coordinates of a point on a circle.<\/li>\n<\/ul>\n<\/section>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: center;\">Section 6.3 Homework Exercises<\/h2>\n<p>1. Discuss the difference between a coterminal angle and a reference angle.<\/p>\n<p>2.\u00a0Explain how the cosine of an angle in the second quadrant differs from the cosine of its reference angle in the unit circle.<\/p>\n<p>3. Explain how the sine of an angle in the second quadrant differs from the sine of its reference angle in the unit circle.<\/p>\n<p>4. What is the purpose of a reference angle?<\/p>\n<p>For the following exercises, state the reference angle for the given angle.<\/p>\n<p>5. [latex]240^\\circ[\/latex]<\/p>\n<p>6.\u00a0[latex]-170^\\circ[\/latex]<\/p>\n<p>7. [latex]460^\\circ[\/latex]<\/p>\n<p>8.\u00a0[latex]-675^\\circ[\/latex]<\/p>\n<p>9. [latex]135^\\circ[\/latex]<\/p>\n<p>10.\u00a0[latex]\\frac{5\\pi }{4}[\/latex]<\/p>\n<p>11. [latex]\\frac{2\\pi }{3}[\/latex]<\/p>\n<p>12.\u00a0[latex]\\frac{17\\pi }{6}[\/latex]<\/p>\n<p>13. [latex]-\\frac{17\\pi }{3}[\/latex]<\/p>\n<p>14.\u00a0[latex]-\\frac{7\\pi }{4}[\/latex]<\/p>\n<p>15. [latex]-\\frac{\\pi }{8}[\/latex]<\/p>\n<p>For the following exercises, find the reference angle, the quadrant of the terminal side, and the sine, cosine of each angle.<\/p>\n<p>16. [latex]225^\\circ[\/latex]<\/p>\n<p>17. [latex]300^\\circ[\/latex]<\/p>\n<p>18.\u00a0[latex]315^\\circ[\/latex]<\/p>\n<p>19. [latex]135^\\circ[\/latex]<\/p>\n<p>20.\u00a0[latex]570^\\circ[\/latex]<\/p>\n<p>21. [latex]480^\\circ[\/latex]<\/p>\n<p>22.\u00a0[latex]-120^\\circ[\/latex]<\/p>\n<p>23. [latex]-210^\\circ[\/latex]<\/p>\n<p>24.\u00a0[latex]\\frac{5\\pi }{4}[\/latex]<\/p>\n<p>25. [latex]\\frac{7\\pi }{6}[\/latex]<\/p>\n<p>26.\u00a0[latex]\\frac{5\\pi }{3}[\/latex]<\/p>\n<p>27. [latex]\\frac{3\\pi }{4}[\/latex]<\/p>\n<p>28.\u00a0[latex]\\frac{4\\pi }{3}[\/latex]<\/p>\n<p>29. [latex]\\frac{2\\pi }{3}[\/latex]<\/p>\n<p>30.\u00a0[latex]\\frac{-19\\pi }{6}[\/latex]<\/p>\n<p>31. [latex]\\frac{-9\\pi }{4}[\/latex]<\/p>\n<p>For the following exercises, find the reference angle, the quadrant of the terminal side, and the exact value of the trigonometric function.<\/p>\n<p>32. [latex]\\tan \\frac{5\\pi }{6}[\/latex]<\/p>\n<p>33. [latex]\\sec \\frac{7\\pi }{6}[\/latex]<\/p>\n<p>34.\u00a0[latex]\\csc \\frac{11\\pi }{6}[\/latex]<\/p>\n<p>35. [latex]\\cot \\frac{13\\pi }{6}[\/latex]<\/p>\n<p>36.\u00a0[latex]\\tan \\frac{15\\pi }{4}[\/latex]<\/p>\n<p>37. [latex]\\sec \\frac{3\\pi }{4}[\/latex]<\/p>\n<p>38.\u00a0[latex]\\csc \\frac{5\\pi }{4}[\/latex]<\/p>\n<p>39. [latex]\\cot \\frac{11\\pi }{4}[\/latex]<\/p>\n<p>40.\u00a0[latex]\\tan \\left(-\\frac{4\\pi }{3}\\right)[\/latex]<\/p>\n<p>41. [latex]\\sec \\left(-\\frac{2\\pi }{3}\\right)[\/latex]<\/p>\n<p>42.\u00a0[latex]\\csc \\left(-\\frac{10\\pi }{3}\\right)[\/latex]<\/p>\n<p>43. [latex]\\cot \\left(-\\frac{7\\pi }{3}\\right)[\/latex]<\/p>\n<p>44.\u00a0[latex]\\tan 225^\\circ[\/latex]<\/p>\n<p>45. [latex]\\sec 300^\\circ[\/latex]<\/p>\n<p>46.\u00a0[latex]\\csc 510^\\circ[\/latex]<\/p>\n<p>47. [latex]\\cot 600^\\circ[\/latex]<\/p>\n<p>48.\u00a0[latex]\\tan \\left(-30^\\circ\\right)[\/latex]<\/p>\n<p>49. [latex]\\sec \\left(-210^\\circ\\right)[\/latex]<\/p>\n<p>50.\u00a0[latex]\\csc \\left(-510^\\circ\\right)[\/latex]<\/p>\n<p>51. [latex]\\cot \\left(-405^\\circ\\right)[\/latex]<\/p>\n<p>In the following exercises, use a right triangle to find the exact value.<\/p>\n<p>52.\u00a0If [latex]\\text{sin}t=\\frac{3}{4}[\/latex], and [latex]t[\/latex] is in quadrant II, find [latex]\\cos t,\\sec t,\\csc t,\\tan t,\\cot t[\/latex].<\/p>\n<p>53. If [latex]\\text{cos}t=-\\frac{1}{3}[\/latex], and [latex]t[\/latex] is in quadrant III, find [latex]\\sin t,\\sec t,\\csc t,\\tan t,\\cot t[\/latex].<\/p>\n<p>54.\u00a0If [latex]\\tan t=\\frac{12}{5}[\/latex], and [latex]0\\le t<\\frac{\\pi }{2}[\/latex], find [latex]\\sin t,\\cos t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex].\n\n55. If [latex]\\sin t=\\frac{\\sqrt{3}}{2}[\/latex] and [latex]\\cos t=\\frac{1}{2}[\/latex], find [latex]\\sec t,\\csc t,\\tan t[\/latex], and [latex]\\cot t[\/latex].\n\nFor the following exercises, find the exact value using reference angles.\n\n56.\u00a0[latex]\\sin\\left(\\frac{11\\pi}{3}\\right)\\cos\\left(\\frac{-5\\pi}{6}\\right)[\/latex]\n\n57.\u00a0[latex]\\sin\\left(\\frac{3\\pi}{4}\\right)\\cos\\left(\\frac{5\\pi}{3}\\right)[\/latex]\n\n58.\u00a0[latex]\\sin\\left(\\frac{-4\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{2}\\right)[\/latex]\n\n59.\u00a0[latex]\\sin\\left(\\frac{-9\\pi}{4}\\right)\\cos\\left(\\frac{-\\pi}{6}\\right)[\/latex]\n\n60.\u00a0[latex]\\sin\\left(\\frac{\\pi}{6}\\right)\\cos\\left(\\frac{-\\pi}{3}\\right)[\/latex]\n\n61. [latex]\\sin\\left(\\frac{7\\pi}{4}\\right)\\cos\\left(\\frac{-2\\pi}{3}\\right)[\/latex]\n\n62.\u00a0[latex]\\cos\\left(\\frac{5\\pi}{6}\\right)\\cos\\left(\\frac{2\\pi}{3}\\right)[\/latex]\n\n63.\u00a0[latex]\\cos\\left(\\frac{-\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{4}\\right)[\/latex]\n\n64.\u00a0[latex]\\sin\\left(\\frac{-5\\pi}{4}\\right)\\sin\\left(\\frac{11\\pi}{6}\\right)[\/latex]\n\n65.\u00a0[latex]\\sin\\left(\\pi\\right)\\sin\\left(\\frac{\\pi}{6}\\right)[\/latex]\n<\/p>\n","protected":false},"author":264444,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-17059","chapter","type-chapter","status-publish","hentry"],"part":13934,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17059","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":65,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17059\/revisions"}],"predecessor-version":[{"id":18074,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17059\/revisions\/18074"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/parts\/13934"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapters\/17059\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/media?parent=17059"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/pressbooks\/v2\/chapter-type?post=17059"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/contributor?post=17059"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculusv2\/wp-json\/wp\/v2\/license?post=17059"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}