{"id":17105,"date":"2020-04-15T01:19:14","date_gmt":"2020-04-15T01:19:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/?post_type=chapter&#038;p=17105"},"modified":"2020-05-21T06:50:48","modified_gmt":"2020-05-21T06:50:48","slug":"trigonometric-functions-unit-circle-approach","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/chapter\/trigonometric-functions-unit-circle-approach\/","title":{"raw":"Section 4.2: Trigonometric Functions: Unit Circle Approach","rendered":"Section 4.2: Trigonometric Functions: Unit Circle Approach"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Analyze values on the unit circle<\/li>\r\n \t<li style=\"font-weight: 400\">Find function values for the trigonometric functions of special angles.<\/li>\r\n \t<li style=\"font-weight: 400\">Identify the domain and range of sine and cosine functions.<\/li>\r\n \t<li>Evaluate trigonometric values using a calculator.<\/li>\r\n \t<li>Simplify expressions using the even-odd properties and periodic properties<\/li>\r\n<\/ul>\r\n<\/div>\r\nTo define our trigonometric functions, we begin by drawing a unit circle, a circle centered at the origin with radius 1, as shown in Figure 2. The angle (in radians) that [latex]t[\/latex] intercepts forms an arc of length [latex]s[\/latex]. Using the formula [latex]s=rt[\/latex], and knowing that [latex]r=1[\/latex], we see that for a <strong>unit circle<\/strong>, [latex]s=t[\/latex].\r\n\r\nRecall that the <em>x- <\/em>and <em>y-<\/em>axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.\r\n\r\nFor any angle [latex]t[\/latex], we can label the intersection of the terminal side and the unit circle as by its coordinates, [latex]\\left(x,y\\right)[\/latex]. The coordinates [latex]x[\/latex] and [latex]y[\/latex] will be the outputs of the trigonometric functions [latex]f\\left(t\\right)=\\cos t[\/latex] and [latex]f\\left(t\\right)=\\sin t[\/latex], respectively. This means [latex]x=\\cos t[\/latex] and [latex]y=\\sin t[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003528\/CNX_Precalc_Figure_05_02_0022.jpg\" alt=\"Graph of a circle with angle t, radius of 1, and an arc created by the angle with length s. The terminal side of the angle intersects the circle at the point (x,y).\" width=\"487\" height=\"385\" \/> <b>Figure 2.<\/b> Unit circle where the central angle is [latex]t[\/latex] radians[\/caption]\r\n<div class=\"textbox\">\r\n<h3>A General Note: Unit Circle<\/h3>\r\nA <strong>unit circle<\/strong> has a center at [latex]\\left(0,0\\right)[\/latex] and radius [latex]1[\/latex] . In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle [latex]1[\/latex].\r\n\r\nLet [latex]\\left(x,y\\right)[\/latex] be the endpoint on the unit circle of an arc of arc length [latex]s[\/latex]. The [latex]\\left(x,y\\right)[\/latex] coordinates of this point can be described as functions of the angle.\r\n\r\n<\/div>\r\n<h2>Defining Sine and Cosine Functions<\/h2>\r\nNow that we have our unit circle labeled, we can learn how the [latex]\\left(x,y\\right)[\/latex] coordinates relate to the <strong>arc length<\/strong> and <strong>angle<\/strong>. The <strong>sine function<\/strong> relates a real number [latex]t[\/latex] to the <em>y<\/em>-coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle [latex]t[\/latex] equals the <em>y<\/em>-value of the endpoint on the unit circle of an arc of length [latex]t[\/latex]. In Figure 2, the sine is equal to [latex]y[\/latex]. Like all functions, the sine function has an input and an output. Its input is the measure of the angle; its output is the <em>y<\/em>-coordinate of the corresponding point on the unit circle.\r\n\r\nThe <strong>cosine function<\/strong> of an angle [latex]t[\/latex] equals the <em>x<\/em>-value of the endpoint on the unit circle of an arc of length [latex]t[\/latex]. In Figure 3, the cosine is equal to [latex]x[\/latex].<span id=\"fs-id1165137735775\">\r\n<\/span>\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\/27003530\/CNX_Precalc_Figure_05_02_0032.jpg\" alt=\"Illustration of an angle t, with terminal side length equal to 1, and an arc created by angle with length t. The terminal side of the angle intersects the circle at the point (x,y), which is equivalent to (cos t, sin t).\" width=\"487\" height=\"226\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n&nbsp;\r\n\r\nBecause it is understood that sine and cosine are functions, we do not always need to write them with parentheses: [latex]\\sin t[\/latex] is the same as [latex]\\sin \\left(t\\right)[\/latex] and [latex]\\cos t[\/latex]\u00a0is the same as [latex]\\cos \\left(t\\right)[\/latex]. Likewise, [latex]{\\cos }^{2}t[\/latex] is a commonly used shorthand notation for [latex]{\\left(\\cos \\left(t\\right)\\right)}^{2}[\/latex]. Be aware that many calculators and computers do not recognize the shorthand notation. When in doubt, use the extra parentheses when entering calculations into a calculator or computer.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Sine and Cosine Functions<\/h3>\r\nIf [latex]t[\/latex] is a real number and a point [latex]\\left(x,y\\right)[\/latex] on the unit circle corresponds to an angle of [latex]t[\/latex], then\r\n<div style=\"text-align: center\">[latex]\\cos t=x[\/latex]<\/div>\r\n<div style=\"text-align: center\">[latex]\\sin t=y[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a point <em>P<\/em> [latex]\\left(x,y\\right)[\/latex] on the unit circle corresponding to an angle of [latex]t[\/latex], find the sine and cosine.<\/h3>\r\n<ol>\r\n \t<li>The sine of [latex]t[\/latex] is equal to the <em>y<\/em>-coordinate of point [latex]P:\\sin t=y[\/latex].<\/li>\r\n \t<li>The cosine of [latex]t[\/latex] is equal to the <em>x<\/em>-coordinate of point [latex]P: \\text{cos}t=x[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Finding Function Values for Sine and Cosine<\/h3>\r\nPoint [latex]P[\/latex] is a point on the unit circle corresponding to an angle of [latex]t[\/latex], as shown in Figure 4. Find [latex]\\cos \\left(t\\right)[\/latex] and [latex]\\text{sin}\\left(t\\right)[\/latex].<span id=\"fs-id1165137723705\">\r\n<\/span>\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\/27003532\/CNX_Precalc_Figure_05_02_0042.jpg\" alt=\"Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (1\/2, square root of 3 over 2).\" width=\"487\" height=\"385\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"288317\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"288317\"]\r\n\r\nWe know that [latex]\\cos t[\/latex] is the <em>x<\/em>-coordinate of the corresponding point on the unit circle and [latex]\\sin t[\/latex] is the <em>y<\/em>-coordinate of the corresponding point on the unit circle. So:\r\n<p style=\"text-align: center\">[latex]\\begin{align} x&amp;=\\cos t=\\frac{1}{2} \\\\ y&amp;=\\sin t=\\frac{\\sqrt{3}}{2} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nA certain angle [latex]t[\/latex] corresponds to a point on the unit circle at [latex]\\left(-\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2}\\right)[\/latex] as shown in Figure 5. Find [latex]\\cos t[\/latex]\u00a0and [latex]\\sin t[\/latex].<span id=\"fs-id1165137434263\">\r\n<\/span>\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\/27003535\/CNX_Precalc_Figure_05_02_0052.jpg\" alt=\"Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (negative square root of 2 over 2, square root of 2 over 2).\" width=\"487\" height=\"383\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"110315\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"110315\"]\r\n\r\n[latex]\\cos \\left(t\\right)=-\\frac{\\sqrt{2}}{2},\\sin \\left(t\\right)=\\frac{\\sqrt{2}}{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Sines and Cosines of Angles on an Axis<\/h2>\r\nFor quadrantral angles, the corresponding point on the unit circle falls on the <em>x- <\/em>or <em>y<\/em>-axis. In that case, we can easily calculate cosine and sine from the values of [latex]x[\/latex] and [latex]y[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Calculating Sines and Cosines along an Axis<\/h3>\r\nFind [latex]\\cos \\left(90^\\circ \\right)[\/latex] and [latex]\\text{sin}\\left(90^\\circ \\right)[\/latex].\r\n\r\n[reveal-answer q=\"736685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736685\"]\r\n\r\nMoving [latex]90^\\circ [\/latex] counterclockwise around the unit circle from the positive <em>x<\/em>-axis brings us to the top of the circle, where the [latex]\\left(x,y\\right)[\/latex] coordinates are (0, 1), as shown in Figure 6.<span id=\"fs-id1165135641569\">\r\n<\/span>\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\/27003537\/CNX_Precalc_Figure_05_02_0062.jpg\" alt=\"Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (0,1).\" width=\"487\" height=\"383\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n&nbsp;\r\n\r\nUsing our definitions of cosine and sine,\r\n<p style=\"text-align: center\">[latex]\\begin{align}x&amp;=\\cos t=\\cos \\left(90^\\circ \\right)=0\\\\ y&amp;=\\sin t=\\sin \\left(90^\\circ \\right)=1\\end{align}[\/latex]<\/p>\r\nThe cosine of 90\u00b0 is 0; the sine of 90\u00b0 is 1.\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\nFind cosine and sine of the angle [latex]\\pi [\/latex].\r\n\r\n[reveal-answer q=\"479772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"479772\"]\r\n\r\n[latex]\\cos \\left(\\pi \\right)=-1[\/latex], [latex]\\sin \\left(\\pi \\right)=0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Pythagorean Identity<\/h2>\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\/27003540\/CNX_Precalc_Figure_05_02_0072.jpg\" alt=\"Graph of an angle t, with a point (x,y) on the unit circle. And equation showing the equivalence of 1, x^2 + y^2, and cos^2 t + sin^2 t.\" width=\"487\" height=\"210\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\nNow that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is [latex]{x}^{2}+{y}^{2}=1[\/latex]. Because [latex]x=\\cos t[\/latex] and [latex]y=\\sin t[\/latex], we can substitute for [latex]x[\/latex] and [latex]y[\/latex] to get [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]. This equation, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex], is known as the <strong>Pythagorean Identity<\/strong>.<span id=\"fs-id1165137405940\">\r\n<\/span>\r\n\r\nWe can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Pythagorean Identity<\/h3>\r\nThe <strong>Pythagorean Identity<\/strong> states that, for any real number [latex]t[\/latex],\r\n<div style=\"text-align: center\">[latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the sine of some angle [latex]t[\/latex] and its quadrant location, find the cosine of [latex]t[\/latex].<\/h3>\r\n<ol>\r\n \t<li>Substitute the known value of [latex]\\sin \\left(t\\right)[\/latex] into the Pythagorean Identity.<\/li>\r\n \t<li>Solve for [latex]\\cos \\left(t\\right)[\/latex].<\/li>\r\n \t<li>Choose the solution with the appropriate sign for the <em>x<\/em>-values in the quadrant where [latex]t[\/latex] is located.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Finding a Cosine from a Sine or a Sine from a Cosine<\/h3>\r\nIf [latex]\\sin \\left(t\\right)=\\frac{3}{7}[\/latex] and [latex]t[\/latex] is in the second quadrant, find [latex]\\cos \\left(t\\right)[\/latex].\r\n\r\n[reveal-answer q=\"81587\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"81587\"]\r\n\r\nIf we drop a vertical line from the point on the unit circle corresponding to [latex]t[\/latex], we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem.\u00a0<span id=\"fs-id1165135453869\">\r\n<\/span>\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\/27003542\/CNX_Precalc_Figure_05_02_0082.jpg\" alt=\"Graph of a unit circle with an angle that intersects the circle at a point with the y-coordinate equal to 3\/7.\" width=\"487\" height=\"383\" \/> <b>Figure 8<\/b>[\/caption]\r\n\r\n&nbsp;\r\n\r\nSubstituting the known value for sine into the Pythagorean Identity,\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}{\\cos }^{2}\\left(t\\right)+{\\sin }^{2}\\left(t\\right)=1 \\\\ {\\cos }^{2}\\left(t\\right)+\\frac{9}{49}=1 \\\\ {\\cos }^{2}\\left(t\\right)=\\frac{40}{49} \\\\ \\text{cos}\\left(t\\right)=\\pm \\sqrt{\\frac{40}{49}}=\\pm \\frac{\\sqrt{40}}{7}=\\pm \\frac{2\\sqrt{10}}{7} \\end{gathered}[\/latex]<\/p>\r\nBecause the angle is in the second quadrant, we know the <em>x-<\/em>value is a negative real number, so the cosine is also negative. So\r\n[latex]\\text{cos}\\left(t\\right)=-\\frac{2\\sqrt{10}}{7}\\\\[\/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\nIf [latex]\\cos \\left(t\\right)=\\frac{24}{25}[\/latex] and [latex]t[\/latex] is in the fourth quadrant, find [latex]\\sin\\left(t\\right)[\/latex].\r\n\r\n[reveal-answer q=\"656903\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"656903\"]\r\n\r\n[latex]\\sin \\left(t\\right)=-\\frac{7}{25}[\/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]174837[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding Sines and Cosines of Special Angles<\/h2>\r\nWe have already learned some properties of the special angles, such as the conversion from radians to degrees. We can also calculate sines and cosines of the special angles using the <strong>Pythagorean Identity<\/strong> and our knowledge of triangles.\r\n<h3>Finding Sines and Cosines of 45\u00b0 Angles<\/h3>\r\nFirst, we will look at angles of [latex]45^\\circ [\/latex] or [latex]\\frac{\\pi }{4}[\/latex], as shown in Figure 9. A [latex]45^\\circ -45^\\circ -90^\\circ [\/latex] triangle is an isosceles triangle, so the <em>x-<\/em> and <em>y<\/em>-coordinates of the corresponding point on the circle are the same. Because the <em>x-<\/em> and <em>y<\/em>-values are the same, the sine and cosine values will also be equal.<span id=\"fs-id1165134253522\">\r\n<\/span>\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\/27003544\/CNX_Precalc_Figure_05_02_0092.jpg\" alt=\"Graph of 45 degree angle inscribed within a circle with radius of 1. Equivalence between point (x,y) and (x,x) shown.\" width=\"487\" height=\"210\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n&nbsp;\r\n\r\nAt [latex]t=\\frac{\\pi }{4}[\/latex] , which is 45 degrees, the radius of the unit circle bisects the first <strong>quadrantal angle<\/strong>. This means the radius lies along the line [latex]y=x[\/latex]. A unit circle has a radius equal to 1. So, the right triangle formed below the line [latex]y=x[\/latex] has sides [latex]x[\/latex] and [latex]y\\text{ }\\left(y=x\\right)[\/latex], and a radius = 1.<span id=\"fs-id1165137755840\">\r\n<\/span>\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\/27003546\/CNX_Precalc_Figure_05_02_0182.jpg\" alt=\"Graph of circle with pi\/4 angle inscribed and a radius of 1.\" width=\"487\" height=\"366\" \/> <b>Figure 10<\/b>[\/caption]\r\n\r\n&nbsp;\r\n\r\nFrom the Pythagorean Theorem we get\r\n<div style=\"text-align: center\">[latex]{x}^{2}+{y}^{2}=1[\/latex]<\/div>\r\nSubstituting [latex]y=x[\/latex], we get\r\n<div style=\"text-align: center\">[latex]{x}^{2}+{x}^{2}=1[\/latex]<\/div>\r\nCombining like terms we get\r\n<div style=\"text-align: center\">[latex]2{x}^{2}=1[\/latex]<\/div>\r\nAnd solving for [latex]x[\/latex], we get\r\n<div style=\"text-align: center\">[latex]\\begin{gathered}{x}^{2}=\\frac{1}{2}\\\\ x=\\pm \\frac{1}{\\sqrt{2}}\\end{gathered}[\/latex]<\/div>\r\nIn quadrant I, [latex]x=\\frac{1}{\\sqrt{2}}[\/latex].\r\n\r\nAt [latex]t=\\frac{\\pi }{4}[\/latex] or 45 degrees,\r\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\left(x,y\\right)=\\left(x,x\\right)=\\left(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}\\right) \\\\ x=\\frac{1}{\\sqrt{2}},y=\\frac{1}{\\sqrt{2}}\\\\ \\cos t=\\frac{1}{\\sqrt{2}},\\sin t=\\frac{1}{\\sqrt{2}} \\end{gathered}[\/latex]<\/div>\r\nIf we then rationalize the denominators, we get\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos t&amp;=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}} =\\frac{\\sqrt{2}}{2} \\\\ \\sin t&amp;=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2} \\end{align}[\/latex]<\/div>\r\nTherefore, the [latex]\\left(x,y\\right)[\/latex] coordinates of a point on a circle of radius [latex]1[\/latex] at an angle of [latex]45^\\circ [\/latex] are [latex]\\left(\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2}\\right)[\/latex].\r\n<h3>Finding Sines and Cosines of 30\u00b0 and 60\u00b0 Angles<\/h3>\r\nNext, we will find the cosine and sine at an angle of [latex]30^\\circ [\/latex], or [latex]\\frac{\\pi }{6}[\/latex] . First, we will draw a triangle inside a circle with one side at an angle of [latex]30^\\circ [\/latex], and another at an angle of [latex]-30^\\circ [\/latex], as shown in Figure 11. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be [latex]60^\\circ [\/latex], as shown in Figure 12.<span id=\"fs-id1165137846440\">\r\n<\/span>\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\/27003549\/CNX_Precalc_Figure_05_02_0102.jpg\" alt=\"Graph of a circle with 30 degree angle and negative 30 degree angle inscribed to form a trangle.\" width=\"487\" height=\"369\" \/> <b>Figure 11<\/b>[\/caption]\r\n\r\n&nbsp;\r\n<figure id=\"Figure_05_02_011\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"490\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003551\/CNX_Precalc_Figure_05_02_0112.jpg\" alt=\"Image of two 30\/60\/90 triangles back to back. Label for hypoteneuse r and side y.\" width=\"490\" height=\"202\" \/> <b>Figure 12<\/b>[\/caption]<\/figure>\r\nBecause all the angles are equal, the sides are also equal. The vertical line has length [latex]2y[\/latex], and since the sides are all equal, we can also conclude that [latex]r=2y[\/latex] or [latex]y=\\frac{1}{2}r[\/latex]. Since [latex]\\sin t=y[\/latex] ,\r\n<div style=\"text-align: center\">[latex]\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}r[\/latex]<\/div>\r\nAnd since [latex]r=1[\/latex]\u00a0in our <strong>unit circle<\/strong>,\r\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\left(1\\right)=\\frac{1}{2}\\end{align}[\/latex]<\/div>\r\nUsing the Pythagorean Identity, we can find the cosine value.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{cll}{\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)+{\\sin }^{2}\\left(\\frac{\\pi }{6}\\right)=1 \\hfill \\\\ {\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)+{\\left(\\frac{1}{2}\\right)}^{2}=1 \\\\ {\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)=\\frac{3}{4}&amp;&amp; \\text{Use the square root property}. \\\\ \\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\pm \\sqrt{3}}{\\pm \\sqrt{4}}=\\frac{\\sqrt{3}}{2}&amp;&amp; \\text{Since }y\\text{ is positive, choose the positive root}. \\end{array}[\/latex]<\/div>\r\nThe [latex]\\left(x,y\\right)[\/latex] coordinates for the point on a circle of radius [latex]1[\/latex] at an angle of [latex]30^\\circ [\/latex] are [latex]\\left(\\frac{\\sqrt{3}}{2},\\frac{1}{2}\\right)[\/latex].\u00a0At [latex]t=\\frac{\\pi }{3}[\/latex]\u00a0(60\u00b0), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, [latex]BAD[\/latex], as shown in Figure 13 below. Angle [latex]A[\/latex] has measure [latex]60^\\circ [\/latex]. At point [latex]B[\/latex], we draw an angle [latex]ABC[\/latex] with measure of [latex]60^\\circ [\/latex]. We know the angles in a triangle sum to [latex]180^\\circ [\/latex], so the measure of angle [latex]C[\/latex] is also [latex]60^\\circ [\/latex]. Now we have an equilateral triangle. Because each side of the equilateral triangle [latex]ABC[\/latex] is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.\r\n<figure id=\"Figure_05_02_019\" class=\"small\">\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\/27003553\/CNX_Precalc_Figure_05_02_0192.jpg\" alt=\"Graph of circle with an isoceles triangle inscribed.\" width=\"487\" height=\"368\" \/> <b>Figure 13<\/b>[\/caption]<\/figure>\r\nThe measure of angle [latex]ABD[\/latex] is 30\u00b0. So, if double, angle [latex]ABC[\/latex] is 60\u00b0. [latex]BD[\/latex] is the perpendicular bisector of [latex]AC[\/latex], so it cuts [latex]AC[\/latex] in half. This means that [latex]AD[\/latex] is [latex]\\frac{1}{2}[\/latex] the radius, or [latex]\\frac{1}{2}[\/latex]. Notice that [latex]AD[\/latex] is the <em>x<\/em>-coordinate of point [latex]B[\/latex], which is at the intersection of the 60\u00b0 angle and the unit circle. This gives us a triangle [latex]BAD[\/latex] with hypotenuse of 1 and side [latex]x[\/latex] of length [latex]\\frac{1}{2}[\/latex].\r\n\r\nFrom the Pythagorean Theorem, we get\r\n<div style=\"text-align: center\">[latex]{x}^{2}+{y}^{2}=1[\/latex]<\/div>\r\nSubstituting [latex]x=\\frac{1}{2}[\/latex], we get\r\n<div style=\"text-align: center\">[latex]{\\left(\\frac{1}{2}\\right)}^{2}+{y}^{2}=1[\/latex]<\/div>\r\nSolving for [latex]y[\/latex], we get\r\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\frac{1}{4}+{y}^{2}=1\\\\ {y}^{2}=1-\\frac{1}{4}\\\\ {y}^{2}=\\frac{3}{4}\\\\ y=\\pm \\frac{\\sqrt{3}}{2}\\end{gathered}[\/latex]<\/div>\r\nSince [latex]t=\\frac{\\pi }{3}[\/latex] has the terminal side in quadrant I where the <em>y-<\/em>coordinate is positive, we choose [latex]y=\\frac{\\sqrt{3}}{2}\\[\/latex], the positive value.\r\n\r\nAt [latex]t=\\frac{\\pi }{3}[\/latex] (60\u00b0), the [latex]\\left(x,y\\right)[\/latex] coordinates for the point on a circle of radius [latex]1[\/latex] at an angle of [latex]60^\\circ [\/latex] are [latex]\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)\\[\/latex], so we can find the sine and cosine.\r\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\left(x,y\\right)=\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right) \\\\ x=\\frac{1}{2},y=\\frac{\\sqrt{3}}{2}\\\\ \\cos t=\\frac{1}{2},\\sin t=\\frac{\\sqrt{3}}{2} \\end{gathered}[\/latex]<\/div>\r\nWe have now found the cosine and sine values for all of the most commonly encountered angles in the first quadrant of the unit circle. The table below\u00a0summarizes these values.\r\n<table id=\"Table_05_02_01\" summary=\"..\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/> <col \/> <\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong>Angle<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\pi }{6}[\/latex], or 30\u00b0<\/td>\r\n<td>[latex]\\frac{\\pi }{4}[\/latex], or 45\u00b0<\/td>\r\n<td>[latex]\\frac{\\pi }{3}[\/latex], or 60\u00b0<\/td>\r\n<td>[latex]\\frac{\\pi }{2}[\/latex], or 90\u00b0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosine<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Sine<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFigure 14\u00a0shows the common angles in the first quadrant of the unit circle.<span id=\"fs-id1165135181639\">\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"725\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003555\/CNX_Precalc_Figure_05_02_021n2.jpg\" alt=\"Graph of a quarter circle with angles of 0, 30, 45, 60, and 90 degrees inscribed. Equivalence of angles in radians shown. Points along circle are marked.\" width=\"725\" height=\"647\" \/> <b>Figure 14<\/b>[\/caption]\r\n\r\n&nbsp;\r\n<h3>Using a Calculator to Find Sine and Cosine<\/h3>\r\nTo find the cosine and sine of angles other than the <strong>special angles<\/strong>, we turn to a computer or calculator. <strong>Be aware<\/strong>: Most calculators can be set into \"degree\" or \"radian\" mode, which tells the calculator the units for the input value. When we evaluate [latex]\\cos \\left(30\\right)[\/latex] on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an angle in radians, use a graphing calculator to find the cosine.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>If the calculator has degree mode and radian mode, set it to radian mode.<\/li>\r\n \t<li>Press the COS key.<\/li>\r\n \t<li>Enter the radian value of the angle and press the close-parentheses key \")\".<\/li>\r\n \t<li>Press ENTER.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Using a Graphing Calculator to Find Sine and Cosine<\/h3>\r\nEvaluate [latex]\\cos \\left(\\frac{5\\pi }{3}\\right)[\/latex] using a graphing calculator or computer.\r\n\r\n[reveal-answer q=\"949209\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"949209\"]\r\n\r\nEnter the following keystrokes:\r\n\r\nCOS (5 \u00d7 \u03c0 \u00f7 3 ) ENTER\r\n<p style=\"text-align: center\">[latex]\\cos \\left(\\frac{5\\pi }{3}\\right)=0.5[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can find the cosine or sine of an angle in degrees directly on a calculator with degree mode. For calculators or software that use only radian mode, we can find the sign of [latex]20^\\circ [\/latex], for example, by including the conversion factor to radians as part of the input:\r\n<div style=\"text-align: center\">SIN( 20 \u00d7 \u03c0 \u00f7 180 ) ENTER<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\sin \\left(\\frac{\\pi }{3}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"184627\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"184627\"]\r\n\r\napproximately 0.866025403\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/vaG4O6d48mo\r\n<h2>Identifying the Domain and Range of Sine and Cosine Functions<\/h2>\r\nNow that we can find the sine and cosine of an angle, we need to discuss their domains and ranges. What are the domains of the sine and cosine functions? That is, what are the smallest and largest numbers that can be inputs of the functions? Because angles smaller than 0 and angles larger than [latex]2\\pi [\/latex] can still be graphed on the unit circle and have real values of [latex]x,y[\/latex], and [latex]r[\/latex], there is no lower or upper limit to the angles that can be inputs to the sine and cosine functions. The input to the sine and cosine functions is the rotation from the positive <em>x<\/em>-axis, and that may be any real number.\r\n\r\nWhat are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the <strong>unit circle<\/strong>, as shown in Figure 15. The bounds of the <em>x<\/em>-coordinate are [latex]\\left[-1,1\\right][\/latex]. The bounds of the <em>y<\/em>-coordinate are also [latex]\\left[-1,1\\right][\/latex]. Therefore, the range of both the sine and cosine functions is [latex]\\left[-1,1\\right][\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003558\/CNX_Precalc_Figure_05_02_0132.jpg\" alt=\"Graph of unit circle.\" width=\"487\" height=\"369\" \/> <b>Figure 15<\/b>[\/caption]\r\n\r\nWe have discussed finding the sine and cosine for angles in the first quadrant, but what if our angle is in another quadrant? For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Because the sine value is the <em>y<\/em>-coordinate on the unit circle, the other angle with the same sine will share the same <em>y<\/em>-value, but have the opposite <em>x<\/em>-value. Therefore, its cosine value will be the opposite of the first angle\u2019s cosine value.\r\n\r\nLikewise, there will be an angle in the fourth quadrant with the same cosine as the original angle. The angle with the same cosine will share the same <em>x<\/em>-value but will have the opposite <em>y<\/em>-value. Therefore, its sine value will be the opposite of the original angle\u2019s sine value.\r\n\r\nAs shown in Figure 16, angle [latex]\\alpha [\/latex] has the same sine value as angle [latex]t[\/latex]; the cosine values are opposites. Angle [latex]\\beta [\/latex] has the same cosine value as angle [latex]t[\/latex]; the sine values are opposites.\r\n<div style=\"text-align: center\">\r\n\r\n[latex]\\begin{array}{ccc}\\sin \\left(t\\right)=\\sin \\left(\\alpha \\right)\\hfill &amp; \\text{and}\\hfill &amp; \\cos \\left(t\\right)=-\\cos \\left(\\alpha \\right)\\hfill \\\\ \\sin \\left(t\\right)=-\\sin \\left(\\beta \\right)\\hfill &amp; \\text{and}\\hfill &amp; \\cos \\left(t\\right)=\\cos \\left(\\beta \\right)\\hfill \\end{array}[\/latex]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003601\/CNX_Precalc_Figure_05_02_0142.jpg\" alt=\"Graph of two side by side circles. First graph has circle with angle t and angle alpha with radius r. Second graph has circle with angle t and angle beta inscribed with radius r.\" width=\"975\" height=\"369\" \/> <b>Figure 16<\/b>[\/caption]\r\n\r\n<\/div>\r\n<h2>Find exact values of the trigonometric functions secant, cosecant, tangent, and cotangent<\/h2>\r\nTo define the remaining functions, we will once again draw a unit circle with a point [latex]\\left(x,y\\right)[\/latex] corresponding to an angle of [latex]t[\/latex], as shown in Figure 1. As with the sine and cosine, we can use the [latex]\\left(x,y\\right)[\/latex] coordinates to find the other functions.\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\/27003658\/CNX_Precalc_Figure_05_03_0012.jpg\" alt=\"Graph of circle with angle of t inscribed. Point of (x, y) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"198\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\nThe first function we will define is the tangent. The <strong>tangent<\/strong> of an angle is the ratio of the <em>y<\/em>-value to the <em>x<\/em>-value of the corresponding point on the unit circle. In Figure 1, the tangent of angle [latex]t[\/latex] is equal to [latex]\\frac{y}{x},x\\ne 0[\/latex]. Because the <em>y<\/em>-value is equal to the sine of [latex]t[\/latex], and the <em>x<\/em>-value is equal to the cosine of [latex]t[\/latex], the tangent of angle [latex]t[\/latex] can also be defined as [latex]\\frac{\\sin t}{\\cos t},\\cos t\\ne 0[\/latex]. The tangent function is abbreviated as [latex]\\tan[\/latex]. The remaining three functions can all be expressed as reciprocals of functions we have already defined.\r\n<ul>\r\n \t<li>The <strong>secant<\/strong> function is the reciprocal of the cosine function. In Figure 1, the secant of angle [latex]t[\/latex] is equal to [latex]\\frac{1}{\\cos t}=\\frac{1}{x},x\\ne 0[\/latex]. The secant function is abbreviated as [latex]\\sec[\/latex].<\/li>\r\n \t<li>The <strong>cotangent<\/strong> function is the reciprocal of the tangent function. In Figure 1, the cotangent of angle [latex]t[\/latex] is equal to [latex]\\frac{\\cos t}{\\sin t}=\\frac{x}{y},y\\ne 0[\/latex]. The cotangent function is abbreviated as [latex]\\cot[\/latex].<\/li>\r\n \t<li>The <strong>cosecant<\/strong> function is the reciprocal of the sine function. In Figure 1, the cosecant of angle [latex]t[\/latex] is equal to [latex]\\frac{1}{\\sin t}=\\frac{1}{y},y\\ne 0[\/latex]. The cosecant function is abbreviated as [latex]\\csc[\/latex].<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Tangent, Secant, Cosecant, and Cotangent Functions<\/h3>\r\nIf [latex]t[\/latex] is a real number and [latex]\\left(x,y\\right)[\/latex] is a point where the terminal side of an angle of [latex]t[\/latex] radians intercepts the unit circle, then\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\tan t=\\frac{y}{x},x\\ne 0\\\\ \\sec t=\\frac{1}{x},x\\ne 0\\\\ \\csc t=\\frac{1}{y},y\\ne 0\\\\ \\cot t=\\frac{x}{y},y\\ne 0\\end{gathered}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Finding Trigonometric Functions from a Point on the Unit Circle<\/h3>\r\nThe point [latex]\\left(-\\frac{\\sqrt{3}}{2},\\frac{1}{2}\\right)[\/latex] is on the unit circle, as shown in Figure 2. Find [latex]\\sin t,\\cos t,\\tan t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003700\/CNX_Precalc_Figure_05_03_0022.jpg\" alt=\"Graph of circle with angle of t inscribed. Point of (negative square root of 3 over 2, 1\/2) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"216\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"714608\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"714608\"]\r\n\r\nBecause we know the [latex]\\left(x,y\\right)[\/latex] coordinates of the point on the unit circle indicated by angle [latex]t[\/latex], we can use those coordinates to find the six functions:\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin t=y=\\frac{1}{2}\\\\ \\cos t=x=-\\frac{\\sqrt{3}}{2}\\\\ \\tan t=\\frac{y}{x}=\\frac{\\frac{1}{2}}{-\\frac{\\sqrt{3}}{2}}=\\frac{1}{2}\\left(-\\frac{2}{\\sqrt{3}}\\right)=-\\frac{1}{\\sqrt{3}}=-\\frac{\\sqrt{3}}{3}\\\\ \\sec t=\\frac{1}{x}=\\frac{1}{\\frac{-\\frac{\\sqrt{3}}{2}}{}}=-\\frac{2}{\\sqrt{3}}=-\\frac{2\\sqrt{3}}{3}\\\\ \\csc t=\\frac{1}{y}=\\frac{1}{\\frac{1}{2}}=2\\\\ \\cot t=\\frac{x}{y}=\\frac{-\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}=-\\frac{\\sqrt{3}}{2}\\left(\\frac{2}{1}\\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\nThe point [latex]\\left(\\frac{\\sqrt{2}}{2},-\\frac{\\sqrt{2}}{2}\\right)[\/latex] is on the unit circle, as shown in Figure 3. Find [latex]\\sin t,\\cos t,\\tan t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003703\/CNX_Precalc_Figure_05_03_0032.jpg\" alt=\"Graph of circle with angle of t inscribed. Point of (square root of 2 over 2, negative square root of 2 over 2) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"347\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"264592\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"264592\"]\r\n\r\n[latex]\\sin t=-\\frac{\\sqrt{2}}{2},\\cos t=\\frac{\\sqrt{2}}{2},\\tan t=-1,\\sec t=\\sqrt{2},\\csc t=-\\sqrt{2},\\cot t=-1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Finding the Trigonometric Functions of an Angle<\/h3>\r\nFind [latex]\\sin t,\\cos t,\\tan t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex] when [latex]t=\\frac{\\pi }{6}[\/latex].\r\n\r\n[reveal-answer q=\"419551\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"419551\"]\r\n\r\nWe have previously used the properties of equilateral triangles to demonstrate that [latex]\\sin \\frac{\\pi }{6}=\\frac{1}{2}[\/latex] and [latex]\\cos \\frac{\\pi }{6}=\\frac{\\sqrt{3}}{2}[\/latex]. We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values.\r\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\tan \\frac{\\pi }{6}=\\frac{\\sin\\frac{\\pi }{6}}{\\cos \\frac{\\pi }{6}} =\\frac{\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}\\end{gathered}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sec \\frac{\\pi }{6}=\\frac{1}{\\cos \\frac{\\pi }{6}} =\\frac{1}{\\frac{\\sqrt{3}}{2}}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3}\\end{gathered}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\csc \\frac{\\pi }{6}=\\frac{1}{\\sin \\frac{\\pi }{6}}=\\frac{1}{\\frac{1}{2}}=2\\end{gathered}[\/latex]<\/p>\r\n<p style=\"text-align: center\"><span style=\"font-size: 1rem\">[latex]\\begin{gathered}\\cot \\frac{\\pi }{6}=\\frac{\\cos \\frac{\\pi }{6}}{\\sin \\frac{\\pi }{6}} =\\frac{\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}=\\sqrt{3} \\end{gathered}[\/latex]<\/span><\/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 [latex]\\sin t,\\cos t,\\tan t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex] when [latex]t=\\frac{\\pi }{3}[\/latex].\r\n\r\n[reveal-answer q=\"151548\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"151548\"]\r\n\r\n[latex]\\begin{align}&amp;\\sin \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{2}\\\\ &amp;\\cos \\frac{\\pi }{3}=\\frac{1}{2}\\\\ &amp;\\tan \\frac{\\pi }{3}=\\sqrt{3}\\\\ &amp;\\sec \\frac{\\pi }{3}=2\\\\ &amp;\\csc \\frac{\\pi }{3}=\\frac{2\\sqrt{3}}{3}\\\\ &amp;\\cot \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{3}\\end{align}[\/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]173354[\/ohm_question]\r\n\r\n<\/div>\r\nBecause we know the sine and cosine values for the common first-quadrant angles, we can find the other function values for those angles as well by setting [latex]x[\/latex] equal to the cosine and [latex]y[\/latex] equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in the table below.\r\n<table id=\"Table_05_03_01\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Angle<\/strong><\/td>\r\n<td><strong> [latex]0[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{6},\\text{ or }{30}^{\\circ}[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{4},\\text{ or } {45}^{\\circ }[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{3},\\text{ or }{60}^{\\circ }[\/latex] <\/strong><\/td>\r\n<td><strong> [latex]\\frac{\\pi }{2},\\text{ or }{90}^{\\circ }[\/latex] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosine<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Sine<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Tangent<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>Undefined<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Secant<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>2<\/td>\r\n<td>Undefined<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cosecant<\/strong><\/td>\r\n<td>Undefined<\/td>\r\n<td>2<\/td>\r\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Cotangent<\/strong><\/td>\r\n<td>Undefined<\/td>\r\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\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&nbsp;\r\n<h2 style=\"text-align: center\">Section 4.2 Homework Exercises<\/h2>\r\n1. Describe the unit circle.\r\n\r\n2.\u00a0What do the x- and y-coordinates of the points on the unit circle represent?\r\n\r\n3. On an interval of [latex]\\left[0,2\\pi \\right)[\/latex], can the sine and cosine values of a radian measure ever be equal? If so, where?\r\n\r\n4.\u00a0What would you estimate the cosine of [latex]\\pi [\/latex] degrees to be? Explain your reasoning.\r\n\r\n5. For any angle in quadrant II, if you knew the sine of the angle, how could you determine the cosine of the angle?\r\n\r\nFor the following exercises, use the given signs of the trigonometric functions to find the quadrant in which the terminal point determined by [latex]t[\/latex] lies.\r\n\r\n6. [latex]\\text{tan}\\left(t\\right)&gt;0[\/latex] and [latex]\\cos \\left(t\\right)&lt;0[\/latex]\r\n\r\n7. [latex]\\text{tan}\\left(t\\right)&gt;0[\/latex] and [latex]\\cos \\left(t\\right)&gt;0[\/latex]\r\n\r\n8.\u00a0[latex]\\tan \\left(t\\right)&lt;0[\/latex]\u00a0and\u00a0[latex]\\cos \\left(t\\right)&lt;0[\/latex]\r\n\r\n9. [latex]\\tan \\left(t\\right)&lt;0[\/latex]\u00a0and\u00a0[latex]\\cos \\left(t\\right)&gt;0[\/latex]\r\n\r\nFor the following exercises, find the exact value of each trigonometric function.\r\n\r\n10. [latex]\\sin \\frac{\\pi }{2}[\/latex],\u00a0[latex]\\tan \\frac{\\pi }{6}[\/latex]\r\n\r\n11. [latex]\\sin \\frac{\\pi }{3}[\/latex],\u00a0[latex]\\sec \\frac{\\pi }{6}[\/latex]\r\n\r\n12.\u00a0[latex]\\cos \\frac{\\pi }{2}[\/latex],\u00a0[latex]\\csc \\frac{\\pi }{6}[\/latex]\r\n\r\n13. [latex]\\cos \\frac{\\pi }{3}[\/latex],\u00a0[latex]\\csc \\frac{\\pi }{6}[\/latex]\r\n\r\n14.\u00a0[latex]\\sin \\frac{\\pi }{4}[\/latex],\u00a0[latex]\\sec \\frac{\\pi }{6}[\/latex]\r\n\r\n15. [latex]\\cos \\frac{\\pi }{4}[\/latex],\u00a0[latex]\\cot \\frac{\\pi }{6}[\/latex]\r\n\r\n16.\u00a0[latex]\\sin \\frac{\\pi }{6}[\/latex],\u00a0[latex]\\tan \\frac{\\pi }{4}[\/latex]\r\n\r\n17. [latex]\\sin \\pi [\/latex],\u00a0[latex]\\sec \\frac{\\pi }{4}[\/latex]\r\n\r\n18.\u00a0[latex]\\sin \\frac{3\\pi }{2}[\/latex],\u00a0[latex]\\csc \\frac{\\pi }{4}[\/latex]\r\n\r\n19. [latex]\\cos \\pi [\/latex],\u00a0[latex]\\tan \\pi[\/latex]\r\n\r\n20. [latex]\\cos \\frac{\\pi }{6}[\/latex],\u00a0[latex]\\sec \\frac{\\pi }{3}[\/latex]\r\n\r\n21.\u00a0[latex]\\cos 0[\/latex],\u00a0[latex]\\tan 0[\/latex]\r\n\r\nFor the following exercises, find the requested value.\r\n\r\n22. If [latex]\\text{cos}\\left(t\\right)=\\frac{1}{7}[\/latex] and [latex]t[\/latex] is in the 4th quadrant, find [latex]\\text{sin}\\left(t\\right)[\/latex].\r\n\r\n23. If [latex]\\text{cos}\\left(t\\right)=\\frac{2}{9}[\/latex] and [latex]t[\/latex] is in the 1st quadrant, find [latex]\\text{sin}\\left(t\\right)[\/latex].\r\n\r\n24.\u00a0If [latex]\\text{sin}\\left(t\\right)=\\frac{3}{8}[\/latex] and [latex]t[\/latex] is in the 2nd quadrant, find [latex]\\text{cos}\\left(t\\right)[\/latex].\r\n\r\n25. If [latex]\\text{sin}\\left(t\\right)=-\\frac{1}{4}[\/latex] and [latex]t[\/latex] is in the 3rd quadrant, find [latex]\\text{cos}\\left(t\\right)[\/latex].\r\n\r\nFor the following exercises, use the given point on the unit circle to find the value of all six trigonometric functions of [latex]t[\/latex].\r\n\r\n26. [latex]\\left(\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2}\\right)[\/latex]\r\n\r\n27. [latex]\\left(-\\frac{\\sqrt{3}}{2},\\frac{1}{2}\\right)[\/latex]\r\n\r\n28. [latex]\\left(\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex]\r\n\r\n29. [latex]\\left(-\\frac{\\sqrt{2}}{2},-\\frac{\\sqrt{2}}{2}\\right)[\/latex]\r\n\r\n30. [latex]\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)[\/latex]\r\n\r\n31. [latex]\\left(-\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)[\/latex]\r\n\r\n32. [latex]\\left(-\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex]\r\n\r\n33. [latex]\\left(\\frac{\\sqrt{2}}{2},-\\frac{\\sqrt{2}}{2}\\right)[\/latex]\r\n\r\n34. [latex]\\left(1,0\\right)[\/latex]\r\n\r\n35. [latex]\\left(-1,0\\right)[\/latex]\r\n\r\n36. [latex]\\left(0.111,0.994\\right)[\/latex]\r\n\r\n37. [latex]\\left(0.803,-0.596\\right)[\/latex]\r\n\r\nFor the following exercises, use a graphing calculator to evaluate.\r\n\r\n38. [latex]\\sin\\left(\\frac{5\\pi}{9}\\right)[\/latex]\r\n\r\n39. [latex]\\cos\\left(\\frac{5\\pi}{9}\\right)[\/latex]\r\n\r\n40. [latex]\\sin\\left(18^\\circ\\right)[\/latex]\r\n\r\n41. [latex]\\cos\\left(18^\\circ\\right)[\/latex]\r\n\r\n42. [latex]\\sin\\left(135^\\circ\\right)[\/latex]\r\n\r\n43. [latex]\\cos\\left(135^\\circ\\right)[\/latex]\r\n\r\n44. [latex]\\csc\\left(\\frac{5\\pi }{9}\\right)[\/latex]\r\n\r\n45. [latex]\\cot\\left(\\frac{4\\pi }{7}\\right)[\/latex]\r\n\r\n46.\u00a0[latex]\\sec\\left(\\frac{\\pi }{10}\\right)[\/latex]\r\n\r\n47. [latex]\\tan\\left(\\frac{5\\pi }{8}\\right)[\/latex]\r\n\r\n48. [latex]\\cot\\left(140^\\circ\\right)[\/latex]\r\n\r\n49. [latex]\\sec\\left(310^\\circ\\right)[\/latex]\r\n\r\nFor the following exercises, find the exact value.\r\n\r\n50.\u00a0[latex]\\sin\\left(\\frac{7\\pi}{4}\\right)\\cos\\left(\\frac{2\\pi}{3}\\right)[\/latex]\r\n\r\n51.\u00a0[latex]\\sin\\left(\\frac{3\\pi}{4}\\right)\\cos\\left(\\frac{5\\pi}{3}\\right)[\/latex]\r\n\r\n52.\u00a0[latex]\\sin\\left(\\frac{-4\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{2}\\right)[\/latex]\r\n\r\n53.\u00a0[latex]\\sin\\left(\\frac{-9\\pi}{4}\\right)\\cos\\left(\\frac{-\\pi}{6}\\right)[\/latex]\r\n\r\n54.\u00a0[latex]\\sin\\left(\\frac{\\pi}{6}\\right)\\cos\\left(\\frac{-\\pi}{3}\\right)[\/latex]\r\n\r\n55. [latex]\\sin\\left(\\frac{7\\pi}{4}\\right)\\cos\\left(\\frac{-2\\pi}{3}\\right)[\/latex]\r\n\r\n56.\u00a0[latex]\\cos\\left(\\frac{5\\pi}{6}\\right)\\cos\\left(\\frac{2\\pi}{3}\\right)[\/latex]\r\n\r\n57.\u00a0[latex]\\cos\\left(\\frac{-\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{4}\\right)[\/latex]\r\n\r\n58.\u00a0[latex]\\sin\\left(\\frac{-5\\pi}{4}\\right)\\sin\\left(\\frac{11\\pi}{6}\\right)[\/latex]\r\n\r\n59.\u00a0[latex]\\sin\\left(\\pi\\right)\\sin\\left(\\frac{\\pi}{6}\\right)[\/latex]\r\n\r\n60. Use the Even-Odd Properties and Periodic Properties to simplify: [latex]5\\cos\\left(3t+2\\pi\\right)+\\sin(-5t)-\\cos\\left(-3t\\right)[\/latex]\r\n\r\n61. Use the Even-Odd Properties and Periodic Properties to simplify: [latex]2\\cos\\left(6t+2\\pi\\right)+\\sin(-9t)-\\cos\\left(-6t\\right)[\/latex]\r\n\r\n62.\u00a0Determine whether the function [latex]f\\left(x\\right)=2\\sin x\\cos x[\/latex] is even, odd, or neither.\r\n\r\n63. Determine whether the function [latex]f\\left(x\\right)=3{\\sin }^{2}x\\cos x+\\sec x[\/latex] is even, odd, or neither.\r\n\r\n64.\u00a0Determine whether the function [latex]f\\left(x\\right)=\\sin x - 2{\\cos }^{2}x[\/latex] is even, odd, or neither.\r\n\r\n65. Determine whether the function [latex]f\\left(x\\right)={\\csc }^{2}x+\\sec x[\/latex] is even, odd, or neither.\r\n\r\n66. The amount of sunlight in a certain city can be modeled by the function [latex]h=15\\cos \\left(\\frac{1}{600}d\\right)[\/latex], where [latex]h[\/latex] represents the hours of sunlight, and [latex]d[\/latex] is the day of the year. Use the equation to find how many hours of sunlight there are on February 10, the 42<sup>nd<\/sup> day of the year. State the period of the function.\r\n\r\n67. The amount of sunlight in a certain city can be modeled by the function [latex]h=16\\cos \\left(\\frac{1}{500}d\\right)[\/latex], where [latex]h[\/latex]\u00a0represents the hours of sunlight, and [latex]d[\/latex]\u00a0is the day of the year. Use the equation to find how many hours of sunlight there are on September 24, the 267<sup>th<\/sup> day of the year. State the period of the function.\r\n\r\n68.\u00a0The equation [latex]P=20\\sin \\left(2\\pi t\\right)+100[\/latex] models the blood pressure, [latex]P[\/latex], where [latex]t[\/latex] represents time in seconds. (a) Find the blood pressure after 15 seconds. (b) What are the maximum and minimum blood pressures?\r\n\r\n69. The height of a piston, [latex]h[\/latex], in inches, can be modeled by the equation [latex]y=2\\cos x+6[\/latex], where [latex]x[\/latex] represents the crank angle. Find the height of the piston when the crank angle is [latex]55^\\circ [\/latex].\r\n\r\n70.\u00a0The height of a piston, [latex]h[\/latex], in inches, can be modeled by the equation [latex]y=2\\cos x+5[\/latex], where [latex]x[\/latex] represents the crank angle. Find the height of the piston when the crank angle is [latex]55^\\circ [\/latex].","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Analyze values on the unit circle<\/li>\n<li style=\"font-weight: 400\">Find function values for the trigonometric functions of special angles.<\/li>\n<li style=\"font-weight: 400\">Identify the domain and range of sine and cosine functions.<\/li>\n<li>Evaluate trigonometric values using a calculator.<\/li>\n<li>Simplify expressions using the even-odd properties and periodic properties<\/li>\n<\/ul>\n<\/div>\n<p>To define our trigonometric functions, we begin by drawing a unit circle, a circle centered at the origin with radius 1, as shown in Figure 2. The angle (in radians) that [latex]t[\/latex] intercepts forms an arc of length [latex]s[\/latex]. Using the formula [latex]s=rt[\/latex], and knowing that [latex]r=1[\/latex], we see that for a <strong>unit circle<\/strong>, [latex]s=t[\/latex].<\/p>\n<p>Recall that the <em>x- <\/em>and <em>y-<\/em>axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.<\/p>\n<p>For any angle [latex]t[\/latex], we can label the intersection of the terminal side and the unit circle as by its coordinates, [latex]\\left(x,y\\right)[\/latex]. The coordinates [latex]x[\/latex] and [latex]y[\/latex] will be the outputs of the trigonometric functions [latex]f\\left(t\\right)=\\cos t[\/latex] and [latex]f\\left(t\\right)=\\sin t[\/latex], respectively. This means [latex]x=\\cos t[\/latex] and [latex]y=\\sin t[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003528\/CNX_Precalc_Figure_05_02_0022.jpg\" alt=\"Graph of a circle with angle t, radius of 1, and an arc created by the angle with length s. The terminal side of the angle intersects the circle at the point (x,y).\" width=\"487\" height=\"385\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2.<\/b> Unit circle where the central angle is [latex]t[\/latex] radians<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Unit Circle<\/h3>\n<p>A <strong>unit circle<\/strong> has a center at [latex]\\left(0,0\\right)[\/latex] and radius [latex]1[\/latex] . In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle [latex]1[\/latex].<\/p>\n<p>Let [latex]\\left(x,y\\right)[\/latex] be the endpoint on the unit circle of an arc of arc length [latex]s[\/latex]. The [latex]\\left(x,y\\right)[\/latex] coordinates of this point can be described as functions of the angle.<\/p>\n<\/div>\n<h2>Defining Sine and Cosine Functions<\/h2>\n<p>Now that we have our unit circle labeled, we can learn how the [latex]\\left(x,y\\right)[\/latex] coordinates relate to the <strong>arc length<\/strong> and <strong>angle<\/strong>. The <strong>sine function<\/strong> relates a real number [latex]t[\/latex] to the <em>y<\/em>-coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle [latex]t[\/latex] equals the <em>y<\/em>-value of the endpoint on the unit circle of an arc of length [latex]t[\/latex]. In Figure 2, the sine is equal to [latex]y[\/latex]. Like all functions, the sine function has an input and an output. Its input is the measure of the angle; its output is the <em>y<\/em>-coordinate of the corresponding point on the unit circle.<\/p>\n<p>The <strong>cosine function<\/strong> of an angle [latex]t[\/latex] equals the <em>x<\/em>-value of the endpoint on the unit circle of an arc of length [latex]t[\/latex]. In Figure 3, the cosine is equal to [latex]x[\/latex].<span id=\"fs-id1165137735775\"><br \/>\n<\/span><\/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\/27003530\/CNX_Precalc_Figure_05_02_0032.jpg\" alt=\"Illustration of an angle t, with terminal side length equal to 1, and an arc created by angle with length t. The terminal side of the angle intersects the circle at the point (x,y), which is equivalent to (cos t, sin t).\" width=\"487\" height=\"226\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Because it is understood that sine and cosine are functions, we do not always need to write them with parentheses: [latex]\\sin t[\/latex] is the same as [latex]\\sin \\left(t\\right)[\/latex] and [latex]\\cos t[\/latex]\u00a0is the same as [latex]\\cos \\left(t\\right)[\/latex]. Likewise, [latex]{\\cos }^{2}t[\/latex] is a commonly used shorthand notation for [latex]{\\left(\\cos \\left(t\\right)\\right)}^{2}[\/latex]. Be aware that many calculators and computers do not recognize the shorthand notation. When in doubt, use the extra parentheses when entering calculations into a calculator or computer.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Sine and Cosine Functions<\/h3>\n<p>If [latex]t[\/latex] is a real number and a point [latex]\\left(x,y\\right)[\/latex] on the unit circle corresponds to an angle of [latex]t[\/latex], then<\/p>\n<div style=\"text-align: center\">[latex]\\cos t=x[\/latex]<\/div>\n<div style=\"text-align: center\">[latex]\\sin t=y[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a point <em>P<\/em> [latex]\\left(x,y\\right)[\/latex] on the unit circle corresponding to an angle of [latex]t[\/latex], find the sine and cosine.<\/h3>\n<ol>\n<li>The sine of [latex]t[\/latex] is equal to the <em>y<\/em>-coordinate of point [latex]P:\\sin t=y[\/latex].<\/li>\n<li>The cosine of [latex]t[\/latex] is equal to the <em>x<\/em>-coordinate of point [latex]P: \\text{cos}t=x[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Finding Function Values for Sine and Cosine<\/h3>\n<p>Point [latex]P[\/latex] is a point on the unit circle corresponding to an angle of [latex]t[\/latex], as shown in Figure 4. Find [latex]\\cos \\left(t\\right)[\/latex] and [latex]\\text{sin}\\left(t\\right)[\/latex].<span id=\"fs-id1165137723705\"><br \/>\n<\/span><\/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\/27003532\/CNX_Precalc_Figure_05_02_0042.jpg\" alt=\"Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (1\/2, square root of 3 over 2).\" width=\"487\" height=\"385\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q288317\">Show Solution<\/span><\/p>\n<div id=\"q288317\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know that [latex]\\cos t[\/latex] is the <em>x<\/em>-coordinate of the corresponding point on the unit circle and [latex]\\sin t[\/latex] is the <em>y<\/em>-coordinate of the corresponding point on the unit circle. So:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align} x&=\\cos t=\\frac{1}{2} \\\\ y&=\\sin t=\\frac{\\sqrt{3}}{2} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A certain angle [latex]t[\/latex] corresponds to a point on the unit circle at [latex]\\left(-\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2}\\right)[\/latex] as shown in Figure 5. Find [latex]\\cos t[\/latex]\u00a0and [latex]\\sin t[\/latex].<span id=\"fs-id1165137434263\"><br \/>\n<\/span><\/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\/27003535\/CNX_Precalc_Figure_05_02_0052.jpg\" alt=\"Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (negative square root of 2 over 2, square root of 2 over 2).\" width=\"487\" height=\"383\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q110315\">Show Solution<\/span><\/p>\n<div id=\"q110315\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\left(t\\right)=-\\frac{\\sqrt{2}}{2},\\sin \\left(t\\right)=\\frac{\\sqrt{2}}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Sines and Cosines of Angles on an Axis<\/h2>\n<p>For quadrantral angles, the corresponding point on the unit circle falls on the <em>x- <\/em>or <em>y<\/em>-axis. In that case, we can easily calculate cosine and sine from the values of [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 2: Calculating Sines and Cosines along an Axis<\/h3>\n<p>Find [latex]\\cos \\left(90^\\circ \\right)[\/latex] and [latex]\\text{sin}\\left(90^\\circ \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q736685\">Show Solution<\/span><\/p>\n<div id=\"q736685\" class=\"hidden-answer\" style=\"display: none\">\n<p>Moving [latex]90^\\circ[\/latex] counterclockwise around the unit circle from the positive <em>x<\/em>-axis brings us to the top of the circle, where the [latex]\\left(x,y\\right)[\/latex] coordinates are (0, 1), as shown in Figure 6.<span id=\"fs-id1165135641569\"><br \/>\n<\/span><\/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\/27003537\/CNX_Precalc_Figure_05_02_0062.jpg\" alt=\"Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (0,1).\" width=\"487\" height=\"383\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Using our definitions of cosine and sine,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}x&=\\cos t=\\cos \\left(90^\\circ \\right)=0\\\\ y&=\\sin t=\\sin \\left(90^\\circ \\right)=1\\end{align}[\/latex]<\/p>\n<p>The cosine of 90\u00b0 is 0; the sine of 90\u00b0 is 1.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find cosine and sine of the angle [latex]\\pi[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q479772\">Show Solution<\/span><\/p>\n<div id=\"q479772\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\left(\\pi \\right)=-1[\/latex], [latex]\\sin \\left(\\pi \\right)=0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Pythagorean Identity<\/h2>\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\/27003540\/CNX_Precalc_Figure_05_02_0072.jpg\" alt=\"Graph of an angle t, with a point (x,y) on the unit circle. And equation showing the equivalence of 1, x^2 + y^2, and cos^2 t + sin^2 t.\" width=\"487\" height=\"210\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<p>Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is [latex]{x}^{2}+{y}^{2}=1[\/latex]. Because [latex]x=\\cos t[\/latex] and [latex]y=\\sin t[\/latex], we can substitute for [latex]x[\/latex] and [latex]y[\/latex] to get [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]. This equation, [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex], is known as the <strong>Pythagorean Identity<\/strong>.<span id=\"fs-id1165137405940\"><br \/>\n<\/span><\/p>\n<p>We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Pythagorean Identity<\/h3>\n<p>The <strong>Pythagorean Identity<\/strong> states that, for any real number [latex]t[\/latex],<\/p>\n<div style=\"text-align: center\">[latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the sine of some angle [latex]t[\/latex] and its quadrant location, find the cosine of [latex]t[\/latex].<\/h3>\n<ol>\n<li>Substitute the known value of [latex]\\sin \\left(t\\right)[\/latex] into the Pythagorean Identity.<\/li>\n<li>Solve for [latex]\\cos \\left(t\\right)[\/latex].<\/li>\n<li>Choose the solution with the appropriate sign for the <em>x<\/em>-values in the quadrant where [latex]t[\/latex] is located.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Finding a Cosine from a Sine or a Sine from a Cosine<\/h3>\n<p>If [latex]\\sin \\left(t\\right)=\\frac{3}{7}[\/latex] and [latex]t[\/latex] is in the second quadrant, find [latex]\\cos \\left(t\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q81587\">Show Solution<\/span><\/p>\n<div id=\"q81587\" class=\"hidden-answer\" style=\"display: none\">\n<p>If we drop a vertical line from the point on the unit circle corresponding to [latex]t[\/latex], we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem.\u00a0<span id=\"fs-id1165135453869\"><br \/>\n<\/span><\/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\/27003542\/CNX_Precalc_Figure_05_02_0082.jpg\" alt=\"Graph of a unit circle with an angle that intersects the circle at a point with the y-coordinate equal to 3\/7.\" width=\"487\" height=\"383\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Substituting the known value for sine into the Pythagorean Identity,<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}{\\cos }^{2}\\left(t\\right)+{\\sin }^{2}\\left(t\\right)=1 \\\\ {\\cos }^{2}\\left(t\\right)+\\frac{9}{49}=1 \\\\ {\\cos }^{2}\\left(t\\right)=\\frac{40}{49} \\\\ \\text{cos}\\left(t\\right)=\\pm \\sqrt{\\frac{40}{49}}=\\pm \\frac{\\sqrt{40}}{7}=\\pm \\frac{2\\sqrt{10}}{7} \\end{gathered}[\/latex]<\/p>\n<p>Because the angle is in the second quadrant, we know the <em>x-<\/em>value is a negative real number, so the cosine is also negative. So<br \/>\n[latex]\\text{cos}\\left(t\\right)=-\\frac{2\\sqrt{10}}{7}\\\\[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>If [latex]\\cos \\left(t\\right)=\\frac{24}{25}[\/latex] and [latex]t[\/latex] is in the fourth quadrant, find [latex]\\sin\\left(t\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q656903\">Show Solution<\/span><\/p>\n<div id=\"q656903\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sin \\left(t\\right)=-\\frac{7}{25}[\/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=\"ohm174837\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=174837&theme=oea&iframe_resize_id=ohm174837\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding Sines and Cosines of Special Angles<\/h2>\n<p>We have already learned some properties of the special angles, such as the conversion from radians to degrees. We can also calculate sines and cosines of the special angles using the <strong>Pythagorean Identity<\/strong> and our knowledge of triangles.<\/p>\n<h3>Finding Sines and Cosines of 45\u00b0 Angles<\/h3>\n<p>First, we will look at angles of [latex]45^\\circ[\/latex] or [latex]\\frac{\\pi }{4}[\/latex], as shown in Figure 9. A [latex]45^\\circ -45^\\circ -90^\\circ[\/latex] triangle is an isosceles triangle, so the <em>x-<\/em> and <em>y<\/em>-coordinates of the corresponding point on the circle are the same. Because the <em>x-<\/em> and <em>y<\/em>-values are the same, the sine and cosine values will also be equal.<span id=\"fs-id1165134253522\"><br \/>\n<\/span><\/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\/27003544\/CNX_Precalc_Figure_05_02_0092.jpg\" alt=\"Graph of 45 degree angle inscribed within a circle with radius of 1. Equivalence between point (x,y) and (x,x) shown.\" width=\"487\" height=\"210\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>At [latex]t=\\frac{\\pi }{4}[\/latex] , which is 45 degrees, the radius of the unit circle bisects the first <strong>quadrantal angle<\/strong>. This means the radius lies along the line [latex]y=x[\/latex]. A unit circle has a radius equal to 1. So, the right triangle formed below the line [latex]y=x[\/latex] has sides [latex]x[\/latex] and [latex]y\\text{ }\\left(y=x\\right)[\/latex], and a radius = 1.<span id=\"fs-id1165137755840\"><br \/>\n<\/span><\/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\/27003546\/CNX_Precalc_Figure_05_02_0182.jpg\" alt=\"Graph of circle with pi\/4 angle inscribed and a radius of 1.\" width=\"487\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>From the Pythagorean Theorem we get<\/p>\n<div style=\"text-align: center\">[latex]{x}^{2}+{y}^{2}=1[\/latex]<\/div>\n<p>Substituting [latex]y=x[\/latex], we get<\/p>\n<div style=\"text-align: center\">[latex]{x}^{2}+{x}^{2}=1[\/latex]<\/div>\n<p>Combining like terms we get<\/p>\n<div style=\"text-align: center\">[latex]2{x}^{2}=1[\/latex]<\/div>\n<p>And solving for [latex]x[\/latex], we get<\/p>\n<div style=\"text-align: center\">[latex]\\begin{gathered}{x}^{2}=\\frac{1}{2}\\\\ x=\\pm \\frac{1}{\\sqrt{2}}\\end{gathered}[\/latex]<\/div>\n<p>In quadrant I, [latex]x=\\frac{1}{\\sqrt{2}}[\/latex].<\/p>\n<p>At [latex]t=\\frac{\\pi }{4}[\/latex] or 45 degrees,<\/p>\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\left(x,y\\right)=\\left(x,x\\right)=\\left(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}\\right) \\\\ x=\\frac{1}{\\sqrt{2}},y=\\frac{1}{\\sqrt{2}}\\\\ \\cos t=\\frac{1}{\\sqrt{2}},\\sin t=\\frac{1}{\\sqrt{2}} \\end{gathered}[\/latex]<\/div>\n<p>If we then rationalize the denominators, we get<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\cos t&=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}} =\\frac{\\sqrt{2}}{2} \\\\ \\sin t&=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2} \\end{align}[\/latex]<\/div>\n<p>Therefore, the [latex]\\left(x,y\\right)[\/latex] coordinates of a point on a circle of radius [latex]1[\/latex] at an angle of [latex]45^\\circ[\/latex] are [latex]\\left(\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2}\\right)[\/latex].<\/p>\n<h3>Finding Sines and Cosines of 30\u00b0 and 60\u00b0 Angles<\/h3>\n<p>Next, we will find the cosine and sine at an angle of [latex]30^\\circ[\/latex], or [latex]\\frac{\\pi }{6}[\/latex] . First, we will draw a triangle inside a circle with one side at an angle of [latex]30^\\circ[\/latex], and another at an angle of [latex]-30^\\circ[\/latex], as shown in Figure 11. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be [latex]60^\\circ[\/latex], as shown in Figure 12.<span id=\"fs-id1165137846440\"><br \/>\n<\/span><\/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\/27003549\/CNX_Precalc_Figure_05_02_0102.jpg\" alt=\"Graph of a circle with 30 degree angle and negative 30 degree angle inscribed to form a trangle.\" width=\"487\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<figure id=\"Figure_05_02_011\" class=\"small\">\n<div style=\"width: 500px\" 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\/27003551\/CNX_Precalc_Figure_05_02_0112.jpg\" alt=\"Image of two 30\/60\/90 triangles back to back. Label for hypoteneuse r and side y.\" width=\"490\" height=\"202\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12<\/b><\/p>\n<\/div>\n<\/figure>\n<p>Because all the angles are equal, the sides are also equal. The vertical line has length [latex]2y[\/latex], and since the sides are all equal, we can also conclude that [latex]r=2y[\/latex] or [latex]y=\\frac{1}{2}r[\/latex]. Since [latex]\\sin t=y[\/latex] ,<\/p>\n<div style=\"text-align: center\">[latex]\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}r[\/latex]<\/div>\n<p>And since [latex]r=1[\/latex]\u00a0in our <strong>unit circle<\/strong>,<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align}\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\left(1\\right)=\\frac{1}{2}\\end{align}[\/latex]<\/div>\n<p>Using the Pythagorean Identity, we can find the cosine value.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{cll}{\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)+{\\sin }^{2}\\left(\\frac{\\pi }{6}\\right)=1 \\hfill \\\\ {\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)+{\\left(\\frac{1}{2}\\right)}^{2}=1 \\\\ {\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)=\\frac{3}{4}&& \\text{Use the square root property}. \\\\ \\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\pm \\sqrt{3}}{\\pm \\sqrt{4}}=\\frac{\\sqrt{3}}{2}&& \\text{Since }y\\text{ is positive, choose the positive root}. \\end{array}[\/latex]<\/div>\n<p>The [latex]\\left(x,y\\right)[\/latex] coordinates for the point on a circle of radius [latex]1[\/latex] at an angle of [latex]30^\\circ[\/latex] are [latex]\\left(\\frac{\\sqrt{3}}{2},\\frac{1}{2}\\right)[\/latex].\u00a0At [latex]t=\\frac{\\pi }{3}[\/latex]\u00a0(60\u00b0), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, [latex]BAD[\/latex], as shown in Figure 13 below. Angle [latex]A[\/latex] has measure [latex]60^\\circ[\/latex]. At point [latex]B[\/latex], we draw an angle [latex]ABC[\/latex] with measure of [latex]60^\\circ[\/latex]. We know the angles in a triangle sum to [latex]180^\\circ[\/latex], so the measure of angle [latex]C[\/latex] is also [latex]60^\\circ[\/latex]. Now we have an equilateral triangle. Because each side of the equilateral triangle [latex]ABC[\/latex] is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.<\/p>\n<figure id=\"Figure_05_02_019\" class=\"small\">\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\/27003553\/CNX_Precalc_Figure_05_02_0192.jpg\" alt=\"Graph of circle with an isoceles triangle inscribed.\" width=\"487\" height=\"368\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 13<\/b><\/p>\n<\/div>\n<\/figure>\n<p>The measure of angle [latex]ABD[\/latex] is 30\u00b0. So, if double, angle [latex]ABC[\/latex] is 60\u00b0. [latex]BD[\/latex] is the perpendicular bisector of [latex]AC[\/latex], so it cuts [latex]AC[\/latex] in half. This means that [latex]AD[\/latex] is [latex]\\frac{1}{2}[\/latex] the radius, or [latex]\\frac{1}{2}[\/latex]. Notice that [latex]AD[\/latex] is the <em>x<\/em>-coordinate of point [latex]B[\/latex], which is at the intersection of the 60\u00b0 angle and the unit circle. This gives us a triangle [latex]BAD[\/latex] with hypotenuse of 1 and side [latex]x[\/latex] of length [latex]\\frac{1}{2}[\/latex].<\/p>\n<p>From the Pythagorean Theorem, we get<\/p>\n<div style=\"text-align: center\">[latex]{x}^{2}+{y}^{2}=1[\/latex]<\/div>\n<p>Substituting [latex]x=\\frac{1}{2}[\/latex], we get<\/p>\n<div style=\"text-align: center\">[latex]{\\left(\\frac{1}{2}\\right)}^{2}+{y}^{2}=1[\/latex]<\/div>\n<p>Solving for [latex]y[\/latex], we get<\/p>\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\frac{1}{4}+{y}^{2}=1\\\\ {y}^{2}=1-\\frac{1}{4}\\\\ {y}^{2}=\\frac{3}{4}\\\\ y=\\pm \\frac{\\sqrt{3}}{2}\\end{gathered}[\/latex]<\/div>\n<p>Since [latex]t=\\frac{\\pi }{3}[\/latex] has the terminal side in quadrant I where the <em>y-<\/em>coordinate is positive, we choose [latex]y=\\frac{\\sqrt{3}}{2}\\[\/latex], the positive value.<\/p>\n<p>At [latex]t=\\frac{\\pi }{3}[\/latex] (60\u00b0), the [latex]\\left(x,y\\right)[\/latex] coordinates for the point on a circle of radius [latex]1[\/latex] at an angle of [latex]60^\\circ[\/latex] are [latex]\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)\\[\/latex], so we can find the sine and cosine.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{gathered}\\left(x,y\\right)=\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right) \\\\ x=\\frac{1}{2},y=\\frac{\\sqrt{3}}{2}\\\\ \\cos t=\\frac{1}{2},\\sin t=\\frac{\\sqrt{3}}{2} \\end{gathered}[\/latex]<\/div>\n<p>We have now found the cosine and sine values for all of the most commonly encountered angles in the first quadrant of the unit circle. The table below\u00a0summarizes these values.<\/p>\n<table id=\"Table_05_02_01\" summary=\"..\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/> <\/colgroup>\n<tbody>\n<tr>\n<td><strong>Angle<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\pi }{6}[\/latex], or 30\u00b0<\/td>\n<td>[latex]\\frac{\\pi }{4}[\/latex], or 45\u00b0<\/td>\n<td>[latex]\\frac{\\pi }{3}[\/latex], or 60\u00b0<\/td>\n<td>[latex]\\frac{\\pi }{2}[\/latex], or 90\u00b0<\/td>\n<\/tr>\n<tr>\n<td><strong>Cosine<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td><strong>Sine<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Figure 14\u00a0shows the common angles in the first quadrant of the unit circle.<span id=\"fs-id1165135181639\"><br \/>\n<\/span><\/p>\n<div style=\"width: 735px\" 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\/27003555\/CNX_Precalc_Figure_05_02_021n2.jpg\" alt=\"Graph of a quarter circle with angles of 0, 30, 45, 60, and 90 degrees inscribed. Equivalence of angles in radians shown. Points along circle are marked.\" width=\"725\" height=\"647\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14<\/b><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h3>Using a Calculator to Find Sine and Cosine<\/h3>\n<p>To find the cosine and sine of angles other than the <strong>special angles<\/strong>, we turn to a computer or calculator. <strong>Be aware<\/strong>: Most calculators can be set into &#8220;degree&#8221; or &#8220;radian&#8221; mode, which tells the calculator the units for the input value. When we evaluate [latex]\\cos \\left(30\\right)[\/latex] on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an angle in radians, use a graphing calculator to find the cosine.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>If the calculator has degree mode and radian mode, set it to radian mode.<\/li>\n<li>Press the COS key.<\/li>\n<li>Enter the radian value of the angle and press the close-parentheses key &#8220;)&#8221;.<\/li>\n<li>Press ENTER.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Using a Graphing Calculator to Find Sine and Cosine<\/h3>\n<p>Evaluate [latex]\\cos \\left(\\frac{5\\pi }{3}\\right)[\/latex] using a graphing calculator or computer.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q949209\">Show Solution<\/span><\/p>\n<div id=\"q949209\" class=\"hidden-answer\" style=\"display: none\">\n<p>Enter the following keystrokes:<\/p>\n<p>COS (5 \u00d7 \u03c0 \u00f7 3 ) ENTER<\/p>\n<p style=\"text-align: center\">[latex]\\cos \\left(\\frac{5\\pi }{3}\\right)=0.5[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can find the cosine or sine of an angle in degrees directly on a calculator with degree mode. For calculators or software that use only radian mode, we can find the sign of [latex]20^\\circ[\/latex], for example, by including the conversion factor to radians as part of the input:<\/p>\n<div style=\"text-align: center\">SIN( 20 \u00d7 \u03c0 \u00f7 180 ) ENTER<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\sin \\left(\\frac{\\pi }{3}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q184627\">Show Solution<\/span><\/p>\n<div id=\"q184627\" class=\"hidden-answer\" style=\"display: none\">\n<p>approximately 0.866025403<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Introduction to Trigonometric Functions Using Angles\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vaG4O6d48mo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Identifying the Domain and Range of Sine and Cosine Functions<\/h2>\n<p>Now that we can find the sine and cosine of an angle, we need to discuss their domains and ranges. What are the domains of the sine and cosine functions? That is, what are the smallest and largest numbers that can be inputs of the functions? Because angles smaller than 0 and angles larger than [latex]2\\pi[\/latex] can still be graphed on the unit circle and have real values of [latex]x,y[\/latex], and [latex]r[\/latex], there is no lower or upper limit to the angles that can be inputs to the sine and cosine functions. The input to the sine and cosine functions is the rotation from the positive <em>x<\/em>-axis, and that may be any real number.<\/p>\n<p>What are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the <strong>unit circle<\/strong>, as shown in Figure 15. The bounds of the <em>x<\/em>-coordinate are [latex]\\left[-1,1\\right][\/latex]. The bounds of the <em>y<\/em>-coordinate are also [latex]\\left[-1,1\\right][\/latex]. Therefore, the range of both the sine and cosine functions is [latex]\\left[-1,1\\right][\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003558\/CNX_Precalc_Figure_05_02_0132.jpg\" alt=\"Graph of unit circle.\" width=\"487\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 15<\/b><\/p>\n<\/div>\n<p>We have discussed finding the sine and cosine for angles in the first quadrant, but what if our angle is in another quadrant? For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Because the sine value is the <em>y<\/em>-coordinate on the unit circle, the other angle with the same sine will share the same <em>y<\/em>-value, but have the opposite <em>x<\/em>-value. Therefore, its cosine value will be the opposite of the first angle\u2019s cosine value.<\/p>\n<p>Likewise, there will be an angle in the fourth quadrant with the same cosine as the original angle. The angle with the same cosine will share the same <em>x<\/em>-value but will have the opposite <em>y<\/em>-value. Therefore, its sine value will be the opposite of the original angle\u2019s sine value.<\/p>\n<p>As shown in Figure 16, angle [latex]\\alpha[\/latex] has the same sine value as angle [latex]t[\/latex]; the cosine values are opposites. Angle [latex]\\beta[\/latex] has the same cosine value as angle [latex]t[\/latex]; the sine values are opposites.<\/p>\n<div style=\"text-align: center\">\n<p>[latex]\\begin{array}{ccc}\\sin \\left(t\\right)=\\sin \\left(\\alpha \\right)\\hfill & \\text{and}\\hfill & \\cos \\left(t\\right)=-\\cos \\left(\\alpha \\right)\\hfill \\\\ \\sin \\left(t\\right)=-\\sin \\left(\\beta \\right)\\hfill & \\text{and}\\hfill & \\cos \\left(t\\right)=\\cos \\left(\\beta \\right)\\hfill \\end{array}[\/latex]<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003601\/CNX_Precalc_Figure_05_02_0142.jpg\" alt=\"Graph of two side by side circles. First graph has circle with angle t and angle alpha with radius r. Second graph has circle with angle t and angle beta inscribed with radius r.\" width=\"975\" height=\"369\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 16<\/b><\/p>\n<\/div>\n<\/div>\n<h2>Find exact values of the trigonometric functions secant, cosecant, tangent, and cotangent<\/h2>\n<p>To define the remaining functions, we will once again draw a unit circle with a point [latex]\\left(x,y\\right)[\/latex] corresponding to an angle of [latex]t[\/latex], as shown in Figure 1. As with the sine and cosine, we can use the [latex]\\left(x,y\\right)[\/latex] coordinates to find the other functions.<\/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\/27003658\/CNX_Precalc_Figure_05_03_0012.jpg\" alt=\"Graph of circle with angle of t inscribed. Point of (x, y) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"198\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>The first function we will define is the tangent. The <strong>tangent<\/strong> of an angle is the ratio of the <em>y<\/em>-value to the <em>x<\/em>-value of the corresponding point on the unit circle. In Figure 1, the tangent of angle [latex]t[\/latex] is equal to [latex]\\frac{y}{x},x\\ne 0[\/latex]. Because the <em>y<\/em>-value is equal to the sine of [latex]t[\/latex], and the <em>x<\/em>-value is equal to the cosine of [latex]t[\/latex], the tangent of angle [latex]t[\/latex] can also be defined as [latex]\\frac{\\sin t}{\\cos t},\\cos t\\ne 0[\/latex]. The tangent function is abbreviated as [latex]\\tan[\/latex]. The remaining three functions can all be expressed as reciprocals of functions we have already defined.<\/p>\n<ul>\n<li>The <strong>secant<\/strong> function is the reciprocal of the cosine function. In Figure 1, the secant of angle [latex]t[\/latex] is equal to [latex]\\frac{1}{\\cos t}=\\frac{1}{x},x\\ne 0[\/latex]. The secant function is abbreviated as [latex]\\sec[\/latex].<\/li>\n<li>The <strong>cotangent<\/strong> function is the reciprocal of the tangent function. In Figure 1, the cotangent of angle [latex]t[\/latex] is equal to [latex]\\frac{\\cos t}{\\sin t}=\\frac{x}{y},y\\ne 0[\/latex]. The cotangent function is abbreviated as [latex]\\cot[\/latex].<\/li>\n<li>The <strong>cosecant<\/strong> function is the reciprocal of the sine function. In Figure 1, the cosecant of angle [latex]t[\/latex] is equal to [latex]\\frac{1}{\\sin t}=\\frac{1}{y},y\\ne 0[\/latex]. The cosecant function is abbreviated as [latex]\\csc[\/latex].<\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>A General Note: Tangent, Secant, Cosecant, and Cotangent Functions<\/h3>\n<p>If [latex]t[\/latex] is a real number and [latex]\\left(x,y\\right)[\/latex] is a point where the terminal side of an angle of [latex]t[\/latex] radians intercepts the unit circle, then<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\tan t=\\frac{y}{x},x\\ne 0\\\\ \\sec t=\\frac{1}{x},x\\ne 0\\\\ \\csc t=\\frac{1}{y},y\\ne 0\\\\ \\cot t=\\frac{x}{y},y\\ne 0\\end{gathered}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Finding Trigonometric Functions from a Point on the Unit Circle<\/h3>\n<p>The point [latex]\\left(-\\frac{\\sqrt{3}}{2},\\frac{1}{2}\\right)[\/latex] is on the unit circle, as shown in Figure 2. Find [latex]\\sin t,\\cos t,\\tan t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003700\/CNX_Precalc_Figure_05_03_0022.jpg\" alt=\"Graph of circle with angle of t inscribed. Point of (negative square root of 3 over 2, 1\/2) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"216\" \/><\/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=\"q714608\">Show Solution<\/span><\/p>\n<div id=\"q714608\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because we know the [latex]\\left(x,y\\right)[\/latex] coordinates of the point on the unit circle indicated by angle [latex]t[\/latex], we can use those coordinates to find the six functions:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sin t=y=\\frac{1}{2}\\\\ \\cos t=x=-\\frac{\\sqrt{3}}{2}\\\\ \\tan t=\\frac{y}{x}=\\frac{\\frac{1}{2}}{-\\frac{\\sqrt{3}}{2}}=\\frac{1}{2}\\left(-\\frac{2}{\\sqrt{3}}\\right)=-\\frac{1}{\\sqrt{3}}=-\\frac{\\sqrt{3}}{3}\\\\ \\sec t=\\frac{1}{x}=\\frac{1}{\\frac{-\\frac{\\sqrt{3}}{2}}{}}=-\\frac{2}{\\sqrt{3}}=-\\frac{2\\sqrt{3}}{3}\\\\ \\csc t=\\frac{1}{y}=\\frac{1}{\\frac{1}{2}}=2\\\\ \\cot t=\\frac{x}{y}=\\frac{-\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}=-\\frac{\\sqrt{3}}{2}\\left(\\frac{2}{1}\\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>The point [latex]\\left(\\frac{\\sqrt{2}}{2},-\\frac{\\sqrt{2}}{2}\\right)[\/latex] is on the unit circle, as shown in Figure 3. Find [latex]\\sin t,\\cos t,\\tan t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003703\/CNX_Precalc_Figure_05_03_0032.jpg\" alt=\"Graph of circle with angle of t inscribed. Point of (square root of 2 over 2, negative square root of 2 over 2) is at intersection of terminal side of angle and edge of circle.\" width=\"487\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q264592\">Show Solution<\/span><\/p>\n<div id=\"q264592\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sin t=-\\frac{\\sqrt{2}}{2},\\cos t=\\frac{\\sqrt{2}}{2},\\tan t=-1,\\sec t=\\sqrt{2},\\csc t=-\\sqrt{2},\\cot t=-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Finding the Trigonometric Functions of an Angle<\/h3>\n<p>Find [latex]\\sin t,\\cos t,\\tan t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex] when [latex]t=\\frac{\\pi }{6}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q419551\">Show Solution<\/span><\/p>\n<div id=\"q419551\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have previously used the properties of equilateral triangles to demonstrate that [latex]\\sin \\frac{\\pi }{6}=\\frac{1}{2}[\/latex] and [latex]\\cos \\frac{\\pi }{6}=\\frac{\\sqrt{3}}{2}[\/latex]. We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered} \\tan \\frac{\\pi }{6}=\\frac{\\sin\\frac{\\pi }{6}}{\\cos \\frac{\\pi }{6}} =\\frac{\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}=\\frac{1}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}\\end{gathered}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\sec \\frac{\\pi }{6}=\\frac{1}{\\cos \\frac{\\pi }{6}} =\\frac{1}{\\frac{\\sqrt{3}}{2}}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3}\\end{gathered}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{gathered}\\csc \\frac{\\pi }{6}=\\frac{1}{\\sin \\frac{\\pi }{6}}=\\frac{1}{\\frac{1}{2}}=2\\end{gathered}[\/latex]<\/p>\n<p style=\"text-align: center\"><span style=\"font-size: 1rem\">[latex]\\begin{gathered}\\cot \\frac{\\pi }{6}=\\frac{\\cos \\frac{\\pi }{6}}{\\sin \\frac{\\pi }{6}} =\\frac{\\frac{\\sqrt{3}}{2}}{\\frac{1}{2}}=\\sqrt{3} \\end{gathered}[\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find [latex]\\sin t,\\cos t,\\tan t,\\sec t,\\csc t[\/latex], and [latex]\\cot t[\/latex] when [latex]t=\\frac{\\pi }{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q151548\">Show Solution<\/span><\/p>\n<div id=\"q151548\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}&\\sin \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{2}\\\\ &\\cos \\frac{\\pi }{3}=\\frac{1}{2}\\\\ &\\tan \\frac{\\pi }{3}=\\sqrt{3}\\\\ &\\sec \\frac{\\pi }{3}=2\\\\ &\\csc \\frac{\\pi }{3}=\\frac{2\\sqrt{3}}{3}\\\\ &\\cot \\frac{\\pi }{3}=\\frac{\\sqrt{3}}{3}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173354\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173354&theme=oea&iframe_resize_id=ohm173354\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Because we know the sine and cosine values for the common first-quadrant angles, we can find the other function values for those angles as well by setting [latex]x[\/latex] equal to the cosine and [latex]y[\/latex] equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in the table below.<\/p>\n<table id=\"Table_05_03_01\" summary=\"..\">\n<tbody>\n<tr>\n<td><strong>Angle<\/strong><\/td>\n<td><strong> [latex]0[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{6},\\text{ or }{30}^{\\circ}[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{4},\\text{ or } {45}^{\\circ }[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{3},\\text{ or }{60}^{\\circ }[\/latex] <\/strong><\/td>\n<td><strong> [latex]\\frac{\\pi }{2},\\text{ or }{90}^{\\circ }[\/latex] <\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Cosine<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td><strong>Sine<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td><strong>Tangent<\/strong><\/td>\n<td>0<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>Undefined<\/td>\n<\/tr>\n<tr>\n<td><strong>Secant<\/strong><\/td>\n<td>1<\/td>\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>2<\/td>\n<td>Undefined<\/td>\n<\/tr>\n<tr>\n<td><strong>Cosecant<\/strong><\/td>\n<td>Undefined<\/td>\n<td>2<\/td>\n<td>[latex]\\sqrt{2}[\/latex]<\/td>\n<td>[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td><strong>Cotangent<\/strong><\/td>\n<td>Undefined<\/td>\n<td>[latex]\\sqrt{3}[\/latex]<\/td>\n<td>1<\/td>\n<td>[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n<td>0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\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<p>&nbsp;<\/p>\n<h2 style=\"text-align: center\">Section 4.2 Homework Exercises<\/h2>\n<p>1. Describe the unit circle.<\/p>\n<p>2.\u00a0What do the x- and y-coordinates of the points on the unit circle represent?<\/p>\n<p>3. On an interval of [latex]\\left[0,2\\pi \\right)[\/latex], can the sine and cosine values of a radian measure ever be equal? If so, where?<\/p>\n<p>4.\u00a0What would you estimate the cosine of [latex]\\pi[\/latex] degrees to be? Explain your reasoning.<\/p>\n<p>5. For any angle in quadrant II, if you knew the sine of the angle, how could you determine the cosine of the angle?<\/p>\n<p>For the following exercises, use the given signs of the trigonometric functions to find the quadrant in which the terminal point determined by [latex]t[\/latex] lies.<\/p>\n<p>6. [latex]\\text{tan}\\left(t\\right)>0[\/latex] and [latex]\\cos \\left(t\\right)<0[\/latex]\n\n7. [latex]\\text{tan}\\left(t\\right)>0[\/latex] and [latex]\\cos \\left(t\\right)>0[\/latex]<\/p>\n<p>8.\u00a0[latex]\\tan \\left(t\\right)<0[\/latex]\u00a0and\u00a0[latex]\\cos \\left(t\\right)<0[\/latex]\n\n9. [latex]\\tan \\left(t\\right)<0[\/latex]\u00a0and\u00a0[latex]\\cos \\left(t\\right)>0[\/latex]<\/p>\n<p>For the following exercises, find the exact value of each trigonometric function.<\/p>\n<p>10. [latex]\\sin \\frac{\\pi }{2}[\/latex],\u00a0[latex]\\tan \\frac{\\pi }{6}[\/latex]<\/p>\n<p>11. [latex]\\sin \\frac{\\pi }{3}[\/latex],\u00a0[latex]\\sec \\frac{\\pi }{6}[\/latex]<\/p>\n<p>12.\u00a0[latex]\\cos \\frac{\\pi }{2}[\/latex],\u00a0[latex]\\csc \\frac{\\pi }{6}[\/latex]<\/p>\n<p>13. [latex]\\cos \\frac{\\pi }{3}[\/latex],\u00a0[latex]\\csc \\frac{\\pi }{6}[\/latex]<\/p>\n<p>14.\u00a0[latex]\\sin \\frac{\\pi }{4}[\/latex],\u00a0[latex]\\sec \\frac{\\pi }{6}[\/latex]<\/p>\n<p>15. [latex]\\cos \\frac{\\pi }{4}[\/latex],\u00a0[latex]\\cot \\frac{\\pi }{6}[\/latex]<\/p>\n<p>16.\u00a0[latex]\\sin \\frac{\\pi }{6}[\/latex],\u00a0[latex]\\tan \\frac{\\pi }{4}[\/latex]<\/p>\n<p>17. [latex]\\sin \\pi[\/latex],\u00a0[latex]\\sec \\frac{\\pi }{4}[\/latex]<\/p>\n<p>18.\u00a0[latex]\\sin \\frac{3\\pi }{2}[\/latex],\u00a0[latex]\\csc \\frac{\\pi }{4}[\/latex]<\/p>\n<p>19. [latex]\\cos \\pi[\/latex],\u00a0[latex]\\tan \\pi[\/latex]<\/p>\n<p>20. [latex]\\cos \\frac{\\pi }{6}[\/latex],\u00a0[latex]\\sec \\frac{\\pi }{3}[\/latex]<\/p>\n<p>21.\u00a0[latex]\\cos 0[\/latex],\u00a0[latex]\\tan 0[\/latex]<\/p>\n<p>For the following exercises, find the requested value.<\/p>\n<p>22. If [latex]\\text{cos}\\left(t\\right)=\\frac{1}{7}[\/latex] and [latex]t[\/latex] is in the 4th quadrant, find [latex]\\text{sin}\\left(t\\right)[\/latex].<\/p>\n<p>23. If [latex]\\text{cos}\\left(t\\right)=\\frac{2}{9}[\/latex] and [latex]t[\/latex] is in the 1st quadrant, find [latex]\\text{sin}\\left(t\\right)[\/latex].<\/p>\n<p>24.\u00a0If [latex]\\text{sin}\\left(t\\right)=\\frac{3}{8}[\/latex] and [latex]t[\/latex] is in the 2nd quadrant, find [latex]\\text{cos}\\left(t\\right)[\/latex].<\/p>\n<p>25. If [latex]\\text{sin}\\left(t\\right)=-\\frac{1}{4}[\/latex] and [latex]t[\/latex] is in the 3rd quadrant, find [latex]\\text{cos}\\left(t\\right)[\/latex].<\/p>\n<p>For the following exercises, use the given point on the unit circle to find the value of all six trigonometric functions of [latex]t[\/latex].<\/p>\n<p>26. [latex]\\left(\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2}\\right)[\/latex]<\/p>\n<p>27. [latex]\\left(-\\frac{\\sqrt{3}}{2},\\frac{1}{2}\\right)[\/latex]<\/p>\n<p>28. [latex]\\left(\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex]<\/p>\n<p>29. [latex]\\left(-\\frac{\\sqrt{2}}{2},-\\frac{\\sqrt{2}}{2}\\right)[\/latex]<\/p>\n<p>30. [latex]\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)[\/latex]<\/p>\n<p>31. [latex]\\left(-\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)[\/latex]<\/p>\n<p>32. [latex]\\left(-\\frac{1}{2},-\\frac{\\sqrt{3}}{2}\\right)[\/latex]<\/p>\n<p>33. [latex]\\left(\\frac{\\sqrt{2}}{2},-\\frac{\\sqrt{2}}{2}\\right)[\/latex]<\/p>\n<p>34. [latex]\\left(1,0\\right)[\/latex]<\/p>\n<p>35. [latex]\\left(-1,0\\right)[\/latex]<\/p>\n<p>36. [latex]\\left(0.111,0.994\\right)[\/latex]<\/p>\n<p>37. [latex]\\left(0.803,-0.596\\right)[\/latex]<\/p>\n<p>For the following exercises, use a graphing calculator to evaluate.<\/p>\n<p>38. [latex]\\sin\\left(\\frac{5\\pi}{9}\\right)[\/latex]<\/p>\n<p>39. [latex]\\cos\\left(\\frac{5\\pi}{9}\\right)[\/latex]<\/p>\n<p>40. [latex]\\sin\\left(18^\\circ\\right)[\/latex]<\/p>\n<p>41. [latex]\\cos\\left(18^\\circ\\right)[\/latex]<\/p>\n<p>42. [latex]\\sin\\left(135^\\circ\\right)[\/latex]<\/p>\n<p>43. [latex]\\cos\\left(135^\\circ\\right)[\/latex]<\/p>\n<p>44. [latex]\\csc\\left(\\frac{5\\pi }{9}\\right)[\/latex]<\/p>\n<p>45. [latex]\\cot\\left(\\frac{4\\pi }{7}\\right)[\/latex]<\/p>\n<p>46.\u00a0[latex]\\sec\\left(\\frac{\\pi }{10}\\right)[\/latex]<\/p>\n<p>47. [latex]\\tan\\left(\\frac{5\\pi }{8}\\right)[\/latex]<\/p>\n<p>48. [latex]\\cot\\left(140^\\circ\\right)[\/latex]<\/p>\n<p>49. [latex]\\sec\\left(310^\\circ\\right)[\/latex]<\/p>\n<p>For the following exercises, find the exact value.<\/p>\n<p>50.\u00a0[latex]\\sin\\left(\\frac{7\\pi}{4}\\right)\\cos\\left(\\frac{2\\pi}{3}\\right)[\/latex]<\/p>\n<p>51.\u00a0[latex]\\sin\\left(\\frac{3\\pi}{4}\\right)\\cos\\left(\\frac{5\\pi}{3}\\right)[\/latex]<\/p>\n<p>52.\u00a0[latex]\\sin\\left(\\frac{-4\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{2}\\right)[\/latex]<\/p>\n<p>53.\u00a0[latex]\\sin\\left(\\frac{-9\\pi}{4}\\right)\\cos\\left(\\frac{-\\pi}{6}\\right)[\/latex]<\/p>\n<p>54.\u00a0[latex]\\sin\\left(\\frac{\\pi}{6}\\right)\\cos\\left(\\frac{-\\pi}{3}\\right)[\/latex]<\/p>\n<p>55. [latex]\\sin\\left(\\frac{7\\pi}{4}\\right)\\cos\\left(\\frac{-2\\pi}{3}\\right)[\/latex]<\/p>\n<p>56.\u00a0[latex]\\cos\\left(\\frac{5\\pi}{6}\\right)\\cos\\left(\\frac{2\\pi}{3}\\right)[\/latex]<\/p>\n<p>57.\u00a0[latex]\\cos\\left(\\frac{-\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{4}\\right)[\/latex]<\/p>\n<p>58.\u00a0[latex]\\sin\\left(\\frac{-5\\pi}{4}\\right)\\sin\\left(\\frac{11\\pi}{6}\\right)[\/latex]<\/p>\n<p>59.\u00a0[latex]\\sin\\left(\\pi\\right)\\sin\\left(\\frac{\\pi}{6}\\right)[\/latex]<\/p>\n<p>60. Use the Even-Odd Properties and Periodic Properties to simplify: [latex]5\\cos\\left(3t+2\\pi\\right)+\\sin(-5t)-\\cos\\left(-3t\\right)[\/latex]<\/p>\n<p>61. Use the Even-Odd Properties and Periodic Properties to simplify: [latex]2\\cos\\left(6t+2\\pi\\right)+\\sin(-9t)-\\cos\\left(-6t\\right)[\/latex]<\/p>\n<p>62.\u00a0Determine whether the function [latex]f\\left(x\\right)=2\\sin x\\cos x[\/latex] is even, odd, or neither.<\/p>\n<p>63. Determine whether the function [latex]f\\left(x\\right)=3{\\sin }^{2}x\\cos x+\\sec x[\/latex] is even, odd, or neither.<\/p>\n<p>64.\u00a0Determine whether the function [latex]f\\left(x\\right)=\\sin x - 2{\\cos }^{2}x[\/latex] is even, odd, or neither.<\/p>\n<p>65. Determine whether the function [latex]f\\left(x\\right)={\\csc }^{2}x+\\sec x[\/latex] is even, odd, or neither.<\/p>\n<p>66. The amount of sunlight in a certain city can be modeled by the function [latex]h=15\\cos \\left(\\frac{1}{600}d\\right)[\/latex], where [latex]h[\/latex] represents the hours of sunlight, and [latex]d[\/latex] is the day of the year. Use the equation to find how many hours of sunlight there are on February 10, the 42<sup>nd<\/sup> day of the year. State the period of the function.<\/p>\n<p>67. The amount of sunlight in a certain city can be modeled by the function [latex]h=16\\cos \\left(\\frac{1}{500}d\\right)[\/latex], where [latex]h[\/latex]\u00a0represents the hours of sunlight, and [latex]d[\/latex]\u00a0is the day of the year. Use the equation to find how many hours of sunlight there are on September 24, the 267<sup>th<\/sup> day of the year. State the period of the function.<\/p>\n<p>68.\u00a0The equation [latex]P=20\\sin \\left(2\\pi t\\right)+100[\/latex] models the blood pressure, [latex]P[\/latex], where [latex]t[\/latex] represents time in seconds. (a) Find the blood pressure after 15 seconds. (b) What are the maximum and minimum blood pressures?<\/p>\n<p>69. The height of a piston, [latex]h[\/latex], in inches, can be modeled by the equation [latex]y=2\\cos x+6[\/latex], where [latex]x[\/latex] represents the crank angle. Find the height of the piston when the crank angle is [latex]55^\\circ[\/latex].<\/p>\n<p>70.\u00a0The height of a piston, [latex]h[\/latex], in inches, can be modeled by the equation [latex]y=2\\cos x+5[\/latex], where [latex]x[\/latex] represents the crank angle. Find the height of the piston when the crank angle is [latex]55^\\circ[\/latex].<\/p>\n","protected":false},"author":264444,"menu_order":2,"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-17105","chapter","type-chapter","status-publish","hentry"],"part":13934,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/17105","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/users\/264444"}],"version-history":[{"count":44,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/17105\/revisions"}],"predecessor-version":[{"id":17589,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/17105\/revisions\/17589"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/parts\/13934"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/17105\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/media?parent=17105"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=17105"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/contributor?post=17105"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/license?post=17105"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}