{"id":135,"date":"2021-02-03T21:52:26","date_gmt":"2021-02-03T21:52:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=135"},"modified":"2022-03-11T21:39:02","modified_gmt":"2022-03-11T21:39:02","slug":"basic-trigonometric-functions-and-identities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/basic-trigonometric-functions-and-identities\/","title":{"raw":"Basic Trigonometric Functions and Identities","rendered":"Basic Trigonometric Functions and Identities"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Convert angle measures between degrees and radians<\/li>\r\n \t<li>Recognize the triangular and circular definitions of the basic trigonometric functions<\/li>\r\n \t<li>Write the basic trigonometric identities<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Radian Measure<\/h2>\r\n<p id=\"fs-id1170572296912\">To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, <strong>radians<\/strong> are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle [latex]\\theta[\/latex], let [latex]s[\/latex] be the length of the corresponding arc on the unit circle (Figure 1). We say the angle corresponding to the arc of length 1 has radian measure 1.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202419\/CNX_Calc_Figure_01_03_001.jpg\" alt=\"An image of a circle. At the exact center of the circle there is a point. From this point, there is one line segment that extends horizontally to the right a point on the edge of the circle and another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. These line segments have a length of 1 unit. The curved segment on the edge of the circle that connects the two points at the end of the line segments is labeled \u201cs\u201d. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label \u201ctheta = s radians\u201d.\" width=\"325\" height=\"258\" \/> Figure 1. The radian measure of an angle [latex]\\theta [\/latex] is the arc length [latex]s[\/latex] of the associated arc on the unit circle.[\/caption]\r\n<p id=\"fs-id1170572151010\">Since an angle of 360\u00b0 corresponds to the circumference of a circle, or an arc of length [latex]2\\pi [\/latex], we conclude that an angle with a degree measure of 360\u00b0 has a radian measure of [latex]2\\pi [\/latex]. Similarly, we see that 180\u00b0 is equivalent to [latex]\\pi [\/latex] radians. The table below shows the relationship between common degree and radian values.<\/p>\r\n\r\n<table id=\"fs-id1170572205698\" summary=\"A table with 9 rows and 2 columns. The first column is labeled \u201cDegrees\u201d and has the values \u201c0; 30; 45; 60; 90; 120; 135; 150; 180\u201d. The second column is labeled \u201cRadians\u201d and the values are \u201c0; (pi\/6); (pi\/4); (pi\/3); (pi\/2); (2pi\/3); (3pi\/4); (5pi\/6); pi\u201d.\"><caption>Table 1. Common Angles Expressed in Degrees and Radians<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Degrees<\/th>\r\n<th>Radians<\/th>\r\n<th>Degrees<\/th>\r\n<th>Radians<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<td>120<\/td>\r\n<td>[latex]2\\pi\/3[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>30<\/td>\r\n<td>[latex]\\pi\/6[\/latex]<\/td>\r\n<td>135<\/td>\r\n<td>[latex]3\\pi\/4[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>45<\/td>\r\n<td>[latex]\\pi\/4[\/latex]<\/td>\r\n<td>150<\/td>\r\n<td>[latex]5\\pi\/6[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>60<\/td>\r\n<td>[latex]\\pi\/3[\/latex]<\/td>\r\n<td>180<\/td>\r\n<td>[latex]\\pi [\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>90<\/td>\r\n<td>[latex]\\pi\/2[\/latex]<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div>Though the common radian and degree equivalents in the table above are worth memorizing, if you can't remember them, don't forget that we have a conversion formula!<\/div>\r\n<div><\/div>\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: radian and degree conversions<\/h3>\r\n<h2>Converting between Radians and Degrees<\/h2>\r\nBecause degrees and radians both measure angles, we need to be able to convert between them. We can easily do so using a proportion.\r\n<div style=\"text-align: center;\">[latex]\\frac{\\theta }{180}=\\frac{{\\theta }^{R}}{\\pi }[\/latex]<\/div>\r\nThis proportion shows that the measure of angle [latex]\\theta [\/latex] in degrees divided by 180 equals the measure of angle [latex]\\theta [\/latex] in radians divided by [latex]\\pi . [\/latex] Or, phrased another way, degrees is to 180 as radians is to [latex]\\pi [\/latex].\r\n<div style=\"text-align: center;\">[latex]\\frac{\\text{Degrees}}{180}=\\frac{\\text{Radians}}{\\pi }[\/latex]<span class=\"fontsize-ensurer reset-size5 size5\" style=\"font-size: 1rem; text-align: initial;\"><span class=\"\">\u200b<\/span><\/span><span style=\"font-size: 1rem; text-align: initial;\">\u200b<\/span><\/div>\r\n<div>\r\n<h2>Converting between Radians and Degrees<\/h2>\r\nTo convert between degrees and radians, use the proportion\r\n<div style=\"text-align: center;\">[latex]\\frac{\\theta }{180}=\\frac{{\\theta }^{R}}{\\pi }[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572139592\" class=\"textbook exercises\">\r\n<h3>Example: Converting between Radians and Degrees<\/h3>\r\n<ol id=\"fs-id1170572204405\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Express 225\u00b0 using radians.<\/li>\r\n \t<li>Express [latex]\\dfrac{5\\pi}{3}[\/latex] rad using degrees.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572224739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572224739\"]\r\n<p id=\"fs-id1170572224739\">Use the fact that 180\u00b0 is equivalent to [latex]\\pi [\/latex] radians as a conversion factor: [latex]1=\\dfrac{\\pi \\, \\text{rad}}{180^{\\circ}}=\\dfrac{180^{\\circ}}{\\pi \\, \\text{rad}}[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170572240548\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]225^{\\circ}=225^{\\circ}\u00b7\\dfrac{\\pi }{180^{\\circ}}=\\dfrac{5\\pi }{4}\\text{ rad}[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{5\\pi }{3}\\text{ rad}[\/latex]= [latex]\\dfrac{5\\pi }{3}\u00b7\\dfrac{180^{\\circ}}{\\pi }=300^{\\circ}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572212258\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572137475\">Express 210\u00b0 using radians. Express [latex]\\dfrac{11\\pi}{6}[\/latex] rad using degrees.<\/p>\r\n[reveal-answer q=\"668822\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"668822\"]\r\n<div id=\"fs-id1165041814727\" class=\"commentary\">\r\n<p id=\"fs-id1165041962818\">[latex]\\pi [\/latex] radians is equal to [latex]180^{\\circ}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n[reveal-answer q=\"fs-id1170572482409\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572482409\"]\r\n<p id=\"fs-id1170572482409\">[latex]\\dfrac{7\\pi}{6}[\/latex] rad; 330\u00b0<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]196627[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>The Six Basic Trigonometric Functions<\/h2>\r\n<p id=\"fs-id1170572175201\">Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle\u2014not only on a unit circle\u2014or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.<\/p>\r\n<p id=\"fs-id1170572094630\">To define the trigonometric functions, first consider the unit circle centered at the origin and a point [latex]P=(x,y)[\/latex] on the unit circle. Let [latex]\\theta [\/latex] be an angle with an initial side that lies along the positive [latex]x[\/latex]-axis and with a terminal side that is the line segment [latex]OP[\/latex]. An angle in this position is said to be in <em>standard position<\/em> (Figure 2). We can then define the values of the six trigonometric functions for [latex]\\theta [\/latex] in terms of the coordinates [latex]x[\/latex] and [latex]y[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"292\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202422\/CNX_Calc_Figure_01_03_002.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled \u201cP = (x, y)\u201d. These line segments have a length of 1 unit. From the point \u201cP\u201d, there is a dotted vertical line that extends downwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label \u201ctheta\u201d.\" width=\"292\" height=\"297\" \/> Figure 2. The angle [latex]\\theta [\/latex] is in standard position. The values of the trigonometric functions for [latex]\\theta [\/latex] are defined in terms of the coordinates [latex]x[\/latex] and [latex]y[\/latex].[\/caption]\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572111833\">Let [latex]P=(x,y)[\/latex] be a point on the unit circle centered at the origin [latex]O[\/latex]. Let [latex]\\theta [\/latex] be an angle with an initial side along the positive [latex]x[\/latex]-axis and a terminal side given by the line segment [latex]OP[\/latex]. The <strong>trigonometric functions<\/strong> are then defined as<\/p>\r\n\r\n<div id=\"fs-id1170572111492\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{cccc}\\sin \\theta =y &amp; &amp; &amp; \\csc \\theta =\\large{\\frac{1}{y}} \\normalsize \\\\ \\cos \\theta =x &amp; &amp; &amp; \\sec \\theta =\\large{\\frac{1}{x}} \\normalsize \\\\ \\tan \\theta =\\large{\\frac{y}{x}} \\normalsize &amp; &amp; &amp; \\cot \\theta =\\large{\\frac{x}{y}} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572228639\">If [latex]x=0[\/latex], then [latex]\\sec \\theta [\/latex] and [latex]\\tan \\theta [\/latex] are undefined. If [latex]y=0[\/latex], then [latex]\\cot \\theta [\/latex] and [latex]\\csc \\theta [\/latex] are undefined.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572151609\">We can see that for a point [latex]P=(x,y)[\/latex] on a circle of radius [latex]r[\/latex] with a corresponding angle [latex]\\theta[\/latex], the coordinates [latex]x[\/latex] and [latex]y[\/latex] satisfy<\/p>\r\n\r\n<div id=\"fs-id1170572205569\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c} \\cos \\theta =\\large{\\frac{x}{r}} \\\\ x=r \\cos \\theta \\\\ \\\\ \\sin \\theta =\\large{\\frac{y}{r}} \\\\ y=r \\sin \\theta \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572454076\">The values of the other trigonometric functions can be expressed in terms of [latex]x, \\, y[\/latex], and [latex]r[\/latex] (Figure 3).<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"466\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202426\/CNX_Calc_Figure_01_03_003.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one blue line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another blue line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled \u201cP = (x, y)\u201d. These line segments have a length of \u201cr\u201d units. Between these line segments within the circle is the label \u201ctheta\u201d, representing the angle between the segments. From the point \u201cP\u201d, there is a blue vertical line that extends downwards until it hits the x axis and thus hits the horizontal line segment, at a point labeled \u201cx\u201d. At the intersection horizontal line segment and vertical line segment at the point x, there is a right triangle symbol. From the point \u201cP\u201d, there is a dotted horizontal line segment that extends left until it hits the y axis at a point labeled \u201cy\u201d.\" width=\"466\" height=\"297\" \/> Figure 3. For a point [latex]P=(x,y)[\/latex] on a circle of radius [latex]r[\/latex], the coordinates [latex]x[\/latex] and [latex]y[\/latex] satisfy [latex]x=r \\cos \\theta [\/latex] and [latex]y=r \\sin \\theta[\/latex].[\/caption]\r\n<p id=\"fs-id1170572229687\">The table below shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of [latex]\\sin \\theta [\/latex] and [latex]\\cos \\theta[\/latex].<\/p>\r\n\r\n<table id=\"fs-id1170572167612\" summary=\"A table with 5 rows and 3 columns. The first column is labeled \u201ctheta\u201d and has the values \u201c0; (pi\/6); (pi\/4); (pi\/3); (pi\/2)\u201d. The second column is labeled \u201csin(theta)\u201d and the values are \u201c0; (1\/2); ((square root of 2)\/2); ((square root of 3)\/2); 1\u201d. The third column is labeled \u201ccos(theta)\u201d and the values are \u201c0; ((square root of 3)\/2); ((square root of 2)\/2); (1\/2); 0\u201d.\"><caption>Values of [latex]\\sin \\theta [\/latex] and [latex]\\cos \\theta [\/latex] at Major Angles [latex]\\theta[\/latex] in the First Quadrant<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]\\theta [\/latex]<\/th>\r\n<th>[latex]\\sin \\theta [\/latex]<\/th>\r\n<th>[latex]\\cos \\theta [\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\large{\\frac{\\pi}{6}}[\/latex]<\/td>\r\n<td>[latex]\\large{\\frac{1}{2}}[\/latex]<\/td>\r\n<td>[latex]\\large{\\frac{\\sqrt{3}}{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\large{\\frac{\\pi}{4}}[\/latex]<\/td>\r\n<td>[latex]\\large{\\frac{\\sqrt{2}}{2}}[\/latex]<\/td>\r\n<td>[latex]\\large{\\frac{\\sqrt{2}}{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\large{\\frac{\\pi}{3}}[\/latex]<\/td>\r\n<td>[latex]\\large{\\frac{\\sqrt{3}}{2}}[\/latex]<\/td>\r\n<td>[latex]\\large{\\frac{1}{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\large{\\frac{\\pi}{2}}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1170572205827\" class=\"textbook exercises\">\r\n<h3>Example: Evaluating Trigonometric Functions<\/h3>\r\n<p id=\"fs-id1170572247984\">Evaluate each of the following expressions.<\/p>\r\n\r\n<ol id=\"fs-id1170572107237\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\sin \\Big(\\large\\frac{2\\pi}{3}\\Big)[\/latex]<\/li>\r\n \t<li>[latex]\\cos \\Big(-\\large\\frac{5\\pi}{6}\\Big)[\/latex]<\/li>\r\n \t<li>[latex]\\tan \\Big(\\large\\frac{15\\pi}{4}\\Big)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572086904\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572086904\"]\r\n<ol id=\"fs-id1170572086904\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>On the unit circle, the angle [latex]\\theta =\\large\\frac{2\\pi}{3}[\/latex] corresponds to the point [latex]\\Big(-\\large\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\Big)[\/latex]. Therefore, [latex]\\sin \\Big(\\large\\frac{2\\pi}{3}\\Big) \\normalsize = y = \\large\\frac{\\sqrt{3}}{2}[\/latex].\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202429\/CNX_Calc_Figure_01_03_004.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the left to another point on the edge of the circle. This point is labeled \u201c(-(1\/2), ((square root of 3)\/2))\u201d. These line segments have a length of 1 unit. From the point \u201c(-(1\/2), ((square root of 3)\/2))\u201d, there is a vertical line that extends downwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise until it hits the diagonal line segment. This arrow has the label \u201ctheta = (2 pi)\/3\u201d.\" width=\"325\" height=\"300\" \/> Figure 4. Unit circle depiction of [latex]\\theta[\/latex][\/caption]<\/li>\r\n \t<li>An angle [latex]\\theta =-\\large\\frac{5\\pi}{6}[\/latex] corresponds to a revolution in the negative direction, as shown. Therefore, [latex]\\cos \\Big(-\\large\\frac{5\\pi}{6}\\Big)[\/latex] [latex]=x=-\\large\\frac{\\sqrt{3}}{2}[\/latex].\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202433\/CNX_Calc_Figure_01_03_005.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the left to another point on the edge of the circle. This point is labeled \u201c(-((square root of 3)\/2)), -(1\/2))\u201d. These line segments have a length of 1 unit. From the point \u201c(-((square root of 3)\/2)), -(1\/2))\u201d, there is a vertical line that extends upwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels clockwise until it hits the diagonal line segment. This arrow has the label \u201ctheta = -(5 pi)\/6\u201d.\" width=\"325\" height=\"300\" \/> Figure 5. Unit circle depiction of [latex]\\theta[\/latex][\/caption]<\/li>\r\n \t<li>An angle [latex]\\theta =\\large\\frac{15\\pi}{4} \\normalsize = 2\\pi +\\large\\frac{7\\pi}{4}[\/latex]. Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of [latex]\\large\\frac{7\\pi}{4}[\/latex] corresponds to the point [latex]\\Big(\\large\\frac{\\sqrt{2}}{2}, -\\frac{\\sqrt{2}}{2}\\Big)[\/latex], we can conclude that [latex]\\tan \\Big(\\large\\frac{15\\pi}{4}\\Big)\\normalsize =\\large\\frac{y}{x}[\/latex] [latex]=-1[\/latex].\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202436\/CNX_Calc_Figure_01_03_006.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the right to another point on the edge of the circle. This point is labeled \u201c(((square root of 2)\/2), -((square root of 2)\/2))\u201d. These line segments have a length of 1 unit. From the point \u201c(((square root of 2)\/2), -((square root of 2)\/2))\u201d, there is a vertical line that extends upwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise. The arrow makes one full rotation around the circle and then keeps traveling until it hits the diagonal line segment. This arrow has the label \u201ctheta = (15 pi)\/4\u201d.\" width=\"325\" height=\"300\" \/> Figure 6. Unit circle depiction of [latex]\\theta[\/latex][\/caption]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572295456\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572168320\">Evaluate [latex]\\cos \\left(\\dfrac{3\\pi}{4}\\right)[\/latex] and [latex]\\sin \\left(\\dfrac{\u2212\\pi}{6}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"565722\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"565722\"]\r\n<p id=\"fs-id1165041836203\">Look at angles on the unit circle.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572222562\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572222562\"]\r\n<p id=\"fs-id1170572222562\">[latex]\\cos \\left(\\dfrac{3\\pi}{4}\\right)=\\dfrac{\u2212\\sqrt{2}}{2}; \\, \\sin\\left(\\dfrac{\u2212\\pi}{6}\\right)=\\dfrac{-1}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]203731[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572546942\">As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let [latex]\\theta [\/latex] be one of the acute angles. Let [latex]A[\/latex] be the length of the adjacent leg, [latex]O[\/latex] be the length of the opposite leg, and [latex]H[\/latex] be the length of the hypotenuse. By inscribing the triangle into a circle of radius [latex]H[\/latex], as shown in Figure 7, we see that [latex]A, \\, H[\/latex], and [latex]O[\/latex] satisfy the following relationships with [latex]\\theta[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1170572169799\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc}\\sin \\theta =\\large \\frac{O}{H} &amp; &amp; &amp; \\normalsize \\csc \\theta =\\large \\frac{H}{O} \\\\ \\cos \\theta =\\large \\frac{A}{H} &amp; &amp; &amp; \\sec \\theta =\\large \\frac{H}{A} \\\\ \\tan \\theta =\\large \\frac{O}{A} &amp; &amp; &amp; \\cot \\theta =\\large \\frac{A}{O} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div><\/div>\r\n<div>[caption id=\"\" align=\"aligncenter\" width=\"313\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202440\/CNX_Calc_Figure_01_03_007.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment with length labeled \u201cH\u201d that extends diagonally upwards and to the right to another point on the edge of the circle. From the point, there is vertical line with a length labeled \u201cO\u201d that extends downwards until it hits the x axis and thus the horizontal line segment at a point with a right triangle symbol. The distance from this point to the center of the circle is labeled \u201cA\u201d. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label \u201ctheta\u201d.\" width=\"313\" height=\"297\" \/> Figure 7. By inscribing a right triangle in a circle, we can express the ratios of the side lengths in terms of the trigonometric functions evaluated at [latex]\\theta[\/latex].[\/caption]<\/div>\r\n<div id=\"fs-id1170572203718\" class=\"textbook exercises\">\r\n<h3>Example: Constructing a Wooden Ramp<\/h3>\r\n<p id=\"fs-id1170572544701\">A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is 4 ft from the ground and the angle between the ground and the ramp is to be [latex]10^{\\circ}[\/latex], how long does the ramp need to be?<\/p>\r\n[reveal-answer q=\"fs-id1170572224156\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572224156\"]\r\n<p id=\"fs-id1170572224156\">Let [latex]x[\/latex] denote the length of the ramp. In the following image, we see that [latex]x[\/latex] needs to satisfy the equation [latex]\\sin(10^{\\circ})=\\dfrac{4}{x}[\/latex]. Solving this equation for [latex]x[\/latex], we see that [latex]x=\\frac{4}{ \\sin(10^{\\circ})} \\approx 23.035[\/latex] ft.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202443\/CNX_Calc_Figure_01_03_008.jpg\" alt=\"An image of a ramp and a staircase. The ramp starts at a point and increases diagonally upwards and to the right at an angle of 10 degrees for x feet. At the end of the ramp, which is 4 feet off the ground, a staircase descends downwards and to the right.\" width=\"487\" height=\"62\" \/> Figure 8. Sketch of the ramp and staircase.[\/caption]\r\n\r\n<span id=\"fs-id1170572286595\"><\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Constructing a Wooden Ramp[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/N1YoLqhe_pw?controls=0&amp;start=583&amp;end=703&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/TrigonometricFunctions583to703_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.3 Trigonometric Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572453725\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572480688\">A house painter wants to lean a 20-ft ladder against a house. If the angle between the base of the ladder and the ground is to be [latex]60^{\\circ}[\/latex], how far from the house should she place the base of the ladder?<\/p>\r\n[reveal-answer q=\"478821\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"478821\"]\r\n<p id=\"fs-id1165041814643\">Draw a right triangle with hypotenuse 20 ft.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572450935\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572450935\"]\r\n<p id=\"fs-id1170572450935\">10 ft<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Trigonometric Identities<\/h2>\r\n<p id=\"fs-id1170572229415\">A <strong>trigonometric identity<\/strong> is an equation involving trigonometric functions that is true for all angles [latex]\\theta [\/latex] for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Trigonometric Identities<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572208589\"><strong>Reciprocal identities<\/strong><\/p>\r\n\r\n<div id=\"fs-id1170572167703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc}\\tan \\theta =\\large \\frac{\\sin \\theta}{\\cos \\theta} &amp; &amp; &amp; \\cot \\theta =\\large \\frac{\\cos \\theta}{\\sin \\theta} \\\\ \\csc \\theta =\\large \\frac{1}{\\sin \\theta} &amp; &amp; &amp; \\sec \\theta =\\large \\frac{1}{\\cos \\theta} \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572140422\"><strong>Pythagorean identities<\/strong><\/p>\r\n\r\n<div id=\"fs-id1170572455704\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin^2 \\theta +\\cos^2 \\theta =1\\phantom{\\rule{2em}{0ex}}1+\\tan^2 \\theta =\\sec^2 \\theta \\phantom{\\rule{2em}{0ex}}1+\\cot^2 \\theta =\\csc^2 \\theta [\/latex]<\/div>\r\n<div><\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572480920\"><strong>Addition and subtraction formulas<\/strong><\/p>\r\n\r\n<div id=\"fs-id1170572216911\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin(\\alpha \\pm \\beta)=\\sin \\alpha \\cos \\beta \\pm \\cos \\alpha \\sin \\beta [\/latex]<\/div>\r\n<div id=\"fs-id1170572098934\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos(\\alpha \\pm \\beta)=\\cos \\alpha \\cos \\beta \\mp \\sin \\alpha \\sin \\beta [\/latex]<\/div>\r\n<div><\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572240527\"><strong>Double-angle formulas<\/strong><\/p>\r\n\r\n<div id=\"fs-id1170572215732\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin(2\\theta)=2\\sin \\theta \\cos \\theta [\/latex]<\/div>\r\n<div id=\"fs-id1170572130013\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos(2\\theta)=2\\cos^2 \\theta -1=1-2\\sin^2 \\theta =\\cos^2 \\theta -\\sin^2 \\theta [\/latex]<\/div>\r\n<\/div>\r\n<div>Remember, solving a trigonometric equation is not very different from solving an algebraic equation.<\/div>\r\n<div><\/div>\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall:\u00a0 Given a trigonometric equation, solve using algebra.<\/h3>\r\n<ul>\r\n \t<li>Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.<\/li>\r\n \t<li>Substitute the trigonometric expression with a single variable, such as [latex]x[\/latex] or [latex]u[\/latex].<\/li>\r\n \t<li>Solve the equation the same way an algebraic equation would be solved.<\/li>\r\n \t<li>Substitute the trigonometric expression back in for the variable in the resulting expressions.<\/li>\r\n \t<li>Solve for the angle.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div style=\"text-align: center;\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]15978[\/ohm_question]\r\n\r\n<\/div>\r\nAdditionally, trigonometric equations will have an infinite number of solutions.\u00a0If we need to find all possible solutions, then we must add [latex]2\\pi k[\/latex], where [latex]k[\/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\\pi :[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\sin \\theta =\\sin \\left(\\theta \\pm 2k\\pi \\right)[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572455784\" class=\"textbook exercises\">\r\n<h3>Example: Solving Trigonometric Equations<\/h3>\r\n<p id=\"fs-id1170572232837\">For each of the following equations, use a trigonometric identity to find all solutions.<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]1+\\cos(2\\theta)=\\cos \\theta [\/latex]<\/li>\r\n \t<li>[latex]\\sin(2\\theta)=\\tan \\theta [\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"501288\"]Show Solution[\/reveal-answer][hidden-answer a=\"501288\"]a. Using the double-angle formula for [latex]\\cos(2\\theta)[\/latex], we see that [latex]\\theta [\/latex] is a solution of\r\n<p style=\"text-align: center;\">[latex]1+\\cos(2\\theta)=\\cos \\theta [\/latex]<\/p>\r\nif and only if\r\n<p style=\"text-align: center;\">[latex]1+2\\cos^2 \\theta -1=\\cos \\theta[\/latex],<\/p>\r\nwhich is true if and only if\r\n<p style=\"text-align: center;\">[latex]2\\cos^2 \\theta -\\cos \\theta =0[\/latex].<\/p>\r\nTo solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by [latex]\\cos \\theta[\/latex]. The problem with dividing by [latex]\\cos \\theta[\/latex] is that it is possible that [latex]\\cos \\theta[\/latex] is zero. In fact, if we did divide both sides of the equation by [latex]\\cos \\theta[\/latex], we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that [latex]\\theta [\/latex] is a solution of this equation if and only if\r\n<p style=\"text-align: center;\">[latex]\\cos \\theta (2\\cos \\theta -1)=0[\/latex]<\/p>\r\nSince [latex]\\cos \\theta =0[\/latex] when\r\n<p style=\"text-align: center;\">[latex]\\theta =\\large \\frac{\\pi }{2}, \\, \\frac{\\pi}{2} \\normalsize \\pm \\pi, \\, \\large \\frac{\\pi}{2} \\normalsize \\pm 2\\pi, \\cdots[\/latex],<\/p>\r\nand [latex]\\cos \\theta =1\/2[\/latex] when\r\n<p style=\"text-align: center;\">[latex]\\theta =\\large \\frac{\\pi}{3}, \\, \\frac{\\pi}{3} \\normalsize \\pm 2\\pi, \\cdots[\/latex], or [latex]\\theta =-\\large \\frac{\\pi}{3}, \\, -\\frac{\\pi}{3} \\normalsize \\pm 2\\pi, \\cdots[\/latex],<\/p>\r\nwe conclude that the set of solutions to this equation is\r\n<p style=\"text-align: center;\">[latex]\\theta =\\large \\frac{\\pi}{2} \\normalsize +n\\pi, \\, \\theta =\\large \\frac{\\pi}{3} \\normalsize +2n\\pi[\/latex], and [latex]\\theta =-\\large \\frac{\\pi}{3}\\normalsize +2n\\pi, \\, n=0, \\pm 1, \\pm 2,\\cdots[\/latex]<\/p>\r\nb. Using the double-angle formula for [latex]\\sin(2\\theta)[\/latex] and the reciprocal identity for [latex]\\tan(\\theta)[\/latex], the equation can be written as\r\n<p style=\"text-align: center;\">[latex]2\\sin \\theta \\cos \\theta =\\large \\frac{\\sin\\theta}{\\cos \\theta}[\/latex]<\/p>\r\nTo solve this equation, we multiply both sides by [latex]\\cos \\theta [\/latex] to eliminate the denominator, and say that if [latex]\\theta [\/latex] satisfies this equation, then [latex]\\theta [\/latex] satisfies the equation\r\n<p style=\"text-align: center;\">[latex]2\\sin \\theta \\cos^2 \\theta -\\sin \\theta =0[\/latex]<\/p>\r\nHowever, we need to be a little careful here. Even if [latex]\\theta [\/latex] satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by [latex]\\cos \\theta[\/latex]. However, if [latex]\\cos \\theta =0[\/latex], we cannot divide both sides of the equation by [latex]\\cos \\theta[\/latex]. Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor [latex]\\sin \\theta [\/latex] out of both terms on the left-hand side instead of dividing both sides of the equation by [latex]\\sin \\theta[\/latex]. Factoring the left-hand side of the equation, we can rewrite this equation as\r\n<p style=\"text-align: center;\">[latex]\\sin \\theta (2\\cos^2 \\theta -1)=0[\/latex]<\/p>\r\nTherefore, the solutions are given by the angles [latex]\\theta [\/latex] such that [latex]\\sin \\theta =0[\/latex] or [latex]\\cos^2 \\theta =1\/2[\/latex]. The solutions of the first equation are [latex]\\theta =0, \\pm \\pi, \\pm 2\\pi, \\cdots[\/latex]. The solutions of the second equation are [latex]\\theta =\\pi \/4, \\, (\\pi\/4) \\pm (\\pi\/2), \\, (\\pi\/4) \\pm \\pi, \\cdots[\/latex]. After checking for extraneous solutions, the set of solutions to the equation is\r\n<p style=\"text-align: center;\">[latex]\\theta =n\\pi[\/latex] and [latex]\\theta =\\large \\frac{\\pi}{4}+\\frac{n\\pi}{2}, \\, \\normalsize n=0, \\pm 1, \\pm 2, \\cdots[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Solving Trigonometric Equations[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6239436&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=p56ltnfoFuA&amp;video_target=tpm-plugin-9gl9o8ol-p56ltnfoFuA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.3.3_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.3.3\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572204063\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572482544\">Find all solutions to the equation [latex]\\cos(2\\theta)=\\sin \\theta[\/latex].<\/p>\r\n[reveal-answer q=\"772243\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"772243\"]\r\n<p id=\"fs-id1165042005368\">Use the double-angle formula for cosine.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1170572229679\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572229679\"]\r\n<p id=\"fs-id1170572229679\">[latex]\\theta =\\large \\frac{3\\pi}{2} \\normalsize +2n\\pi, \\, \\large \\frac{\\pi}{6} \\normalsize +2n\\pi, \\, \\large \\frac{5\\pi}{6} \\normalsize +2n\\pi[\/latex] for [latex]n=0, \\pm 1, \\pm 2, \\cdots[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTrigonometric identities are not only useful for solving trigonometric equations, they can also help us verify identities.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Given a trigonometric identity, verify that it is true.<strong>\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id2191946\">\r\n \t<li>Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.<\/li>\r\n \t<li>Look for opportunities to factor expressions, square a binomial, or add fractions.<\/li>\r\n \t<li>Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.<\/li>\r\n \t<li>If these steps do not yield the desired result, try converting all terms to sines and cosines.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170572111814\" class=\"textbook exercises\">\r\n<h3>Example: Proving a Trigonometric Identity<\/h3>\r\n<p id=\"fs-id1170572550142\">Prove the trigonometric identity [latex]1+\\tan^2 \\theta =\\sec^2 \\theta[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572224130\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572224130\"]\r\n<p id=\"fs-id1170572224130\">We start with the identity<\/p>\r\n\r\n<div id=\"fs-id1170572167191\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin^2 \\theta +\\cos^2 \\theta =1[\/latex]<\/div>\r\n<p id=\"fs-id1170572140245\">Dividing both sides of this equation by [latex]\\cos^2 \\theta[\/latex], we obtain<\/p>\r\n\r\n<div id=\"fs-id1170572481849\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{\\sin^2 \\theta}{\\cos^2 \\theta}+1=\\frac{1}{\\cos^2 \\theta}[\/latex]<\/div>\r\nSince [latex]\\sin \\theta \/ \\cos \\theta =\\tan \\theta[\/latex] and [latex]1 \/ \\cos \\theta =\\sec \\theta[\/latex], we conclude that\r\n<div id=\"fs-id1170572481963\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\tan^2 \\theta +1=\\sec^2 \\theta[\/latex].[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572239833\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572226907\">Prove the trigonometric identity [latex]1+\\cot^2 \\theta =\\csc^2 \\theta[\/latex].<\/p>\r\n[reveal-answer q=\"265389\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"265389\"]\r\n<p id=\"fs-id1165041809092\">Divide both sides of the identity [latex]\\sin^2 \\theta + \\cos^2 \\theta =1[\/latex] by [latex]\\sin^2 \\theta[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Convert angle measures between degrees and radians<\/li>\n<li>Recognize the triangular and circular definitions of the basic trigonometric functions<\/li>\n<li>Write the basic trigonometric identities<\/li>\n<\/ul>\n<\/div>\n<h2>Radian Measure<\/h2>\n<p id=\"fs-id1170572296912\">To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, <strong>radians<\/strong> are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle [latex]\\theta[\/latex], let [latex]s[\/latex] be the length of the corresponding arc on the unit circle (Figure 1). We say the angle corresponding to the arc of length 1 has radian measure 1.<\/p>\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202419\/CNX_Calc_Figure_01_03_001.jpg\" alt=\"An image of a circle. At the exact center of the circle there is a point. From this point, there is one line segment that extends horizontally to the right a point on the edge of the circle and another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. These line segments have a length of 1 unit. The curved segment on the edge of the circle that connects the two points at the end of the line segments is labeled \u201cs\u201d. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label \u201ctheta = s radians\u201d.\" width=\"325\" height=\"258\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The radian measure of an angle [latex]\\theta [\/latex] is the arc length [latex]s[\/latex] of the associated arc on the unit circle.<\/p>\n<\/div>\n<p id=\"fs-id1170572151010\">Since an angle of 360\u00b0 corresponds to the circumference of a circle, or an arc of length [latex]2\\pi[\/latex], we conclude that an angle with a degree measure of 360\u00b0 has a radian measure of [latex]2\\pi[\/latex]. Similarly, we see that 180\u00b0 is equivalent to [latex]\\pi[\/latex] radians. The table below shows the relationship between common degree and radian values.<\/p>\n<table id=\"fs-id1170572205698\" summary=\"A table with 9 rows and 2 columns. The first column is labeled \u201cDegrees\u201d and has the values \u201c0; 30; 45; 60; 90; 120; 135; 150; 180\u201d. The second column is labeled \u201cRadians\u201d and the values are \u201c0; (pi\/6); (pi\/4); (pi\/3); (pi\/2); (2pi\/3); (3pi\/4); (5pi\/6); pi\u201d.\">\n<caption>Table 1. Common Angles Expressed in Degrees and Radians<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>Degrees<\/th>\n<th>Radians<\/th>\n<th>Degrees<\/th>\n<th>Radians<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>0<\/td>\n<td>0<\/td>\n<td>120<\/td>\n<td>[latex]2\\pi\/3[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>30<\/td>\n<td>[latex]\\pi\/6[\/latex]<\/td>\n<td>135<\/td>\n<td>[latex]3\\pi\/4[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>45<\/td>\n<td>[latex]\\pi\/4[\/latex]<\/td>\n<td>150<\/td>\n<td>[latex]5\\pi\/6[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>60<\/td>\n<td>[latex]\\pi\/3[\/latex]<\/td>\n<td>180<\/td>\n<td>[latex]\\pi[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>90<\/td>\n<td>[latex]\\pi\/2[\/latex]<\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div>Though the common radian and degree equivalents in the table above are worth memorizing, if you can&#8217;t remember them, don&#8217;t forget that we have a conversion formula!<\/div>\n<div><\/div>\n<div>\n<div class=\"textbox examples\">\n<h3>Recall: radian and degree conversions<\/h3>\n<h2>Converting between Radians and Degrees<\/h2>\n<p>Because degrees and radians both measure angles, we need to be able to convert between them. We can easily do so using a proportion.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\theta }{180}=\\frac{{\\theta }^{R}}{\\pi }[\/latex]<\/div>\n<p>This proportion shows that the measure of angle [latex]\\theta[\/latex] in degrees divided by 180 equals the measure of angle [latex]\\theta[\/latex] in radians divided by [latex]\\pi .[\/latex] Or, phrased another way, degrees is to 180 as radians is to [latex]\\pi[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\text{Degrees}}{180}=\\frac{\\text{Radians}}{\\pi }[\/latex]<span class=\"fontsize-ensurer reset-size5 size5\" style=\"font-size: 1rem; text-align: initial;\"><span class=\"\">\u200b<\/span><\/span><span style=\"font-size: 1rem; text-align: initial;\">\u200b<\/span><\/div>\n<div>\n<h2>Converting between Radians and Degrees<\/h2>\n<p>To convert between degrees and radians, use the proportion<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{\\theta }{180}=\\frac{{\\theta }^{R}}{\\pi }[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572139592\" class=\"textbook exercises\">\n<h3>Example: Converting between Radians and Degrees<\/h3>\n<ol id=\"fs-id1170572204405\" style=\"list-style-type: lower-alpha;\">\n<li>Express 225\u00b0 using radians.<\/li>\n<li>Express [latex]\\dfrac{5\\pi}{3}[\/latex] rad using degrees.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572224739\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572224739\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572224739\">Use the fact that 180\u00b0 is equivalent to [latex]\\pi[\/latex] radians as a conversion factor: [latex]1=\\dfrac{\\pi \\, \\text{rad}}{180^{\\circ}}=\\dfrac{180^{\\circ}}{\\pi \\, \\text{rad}}[\/latex].<\/p>\n<ol id=\"fs-id1170572240548\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]225^{\\circ}=225^{\\circ}\u00b7\\dfrac{\\pi }{180^{\\circ}}=\\dfrac{5\\pi }{4}\\text{ rad}[\/latex]<\/li>\n<li>[latex]\\dfrac{5\\pi }{3}\\text{ rad}[\/latex]= [latex]\\dfrac{5\\pi }{3}\u00b7\\dfrac{180^{\\circ}}{\\pi }=300^{\\circ}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572212258\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572137475\">Express 210\u00b0 using radians. Express [latex]\\dfrac{11\\pi}{6}[\/latex] rad using degrees.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q668822\">Hint<\/span><\/p>\n<div id=\"q668822\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041814727\" class=\"commentary\">\n<p id=\"fs-id1165041962818\">[latex]\\pi[\/latex] radians is equal to [latex]180^{\\circ}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572482409\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572482409\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572482409\">[latex]\\dfrac{7\\pi}{6}[\/latex] rad; 330\u00b0<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm196627\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=196627&theme=oea&iframe_resize_id=ohm196627&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>The Six Basic Trigonometric Functions<\/h2>\n<p id=\"fs-id1170572175201\">Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle\u2014not only on a unit circle\u2014or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.<\/p>\n<p id=\"fs-id1170572094630\">To define the trigonometric functions, first consider the unit circle centered at the origin and a point [latex]P=(x,y)[\/latex] on the unit circle. Let [latex]\\theta[\/latex] be an angle with an initial side that lies along the positive [latex]x[\/latex]-axis and with a terminal side that is the line segment [latex]OP[\/latex]. An angle in this position is said to be in <em>standard position<\/em> (Figure 2). We can then define the values of the six trigonometric functions for [latex]\\theta[\/latex] in terms of the coordinates [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<div style=\"width: 302px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202422\/CNX_Calc_Figure_01_03_002.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled \u201cP = (x, y)\u201d. These line segments have a length of 1 unit. From the point \u201cP\u201d, there is a dotted vertical line that extends downwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label \u201ctheta\u201d.\" width=\"292\" height=\"297\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The angle [latex]\\theta [\/latex] is in standard position. The values of the trigonometric functions for [latex]\\theta [\/latex] are defined in terms of the coordinates [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170572111833\">Let [latex]P=(x,y)[\/latex] be a point on the unit circle centered at the origin [latex]O[\/latex]. Let [latex]\\theta[\/latex] be an angle with an initial side along the positive [latex]x[\/latex]-axis and a terminal side given by the line segment [latex]OP[\/latex]. The <strong>trigonometric functions<\/strong> are then defined as<\/p>\n<div id=\"fs-id1170572111492\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{cccc}\\sin \\theta =y & & & \\csc \\theta =\\large{\\frac{1}{y}} \\normalsize \\\\ \\cos \\theta =x & & & \\sec \\theta =\\large{\\frac{1}{x}} \\normalsize \\\\ \\tan \\theta =\\large{\\frac{y}{x}} \\normalsize & & & \\cot \\theta =\\large{\\frac{x}{y}} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572228639\">If [latex]x=0[\/latex], then [latex]\\sec \\theta[\/latex] and [latex]\\tan \\theta[\/latex] are undefined. If [latex]y=0[\/latex], then [latex]\\cot \\theta[\/latex] and [latex]\\csc \\theta[\/latex] are undefined.<\/p>\n<\/div>\n<p id=\"fs-id1170572151609\">We can see that for a point [latex]P=(x,y)[\/latex] on a circle of radius [latex]r[\/latex] with a corresponding angle [latex]\\theta[\/latex], the coordinates [latex]x[\/latex] and [latex]y[\/latex] satisfy<\/p>\n<div id=\"fs-id1170572205569\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c} \\cos \\theta =\\large{\\frac{x}{r}} \\\\ x=r \\cos \\theta \\\\ \\\\ \\sin \\theta =\\large{\\frac{y}{r}} \\\\ y=r \\sin \\theta \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572454076\">The values of the other trigonometric functions can be expressed in terms of [latex]x, \\, y[\/latex], and [latex]r[\/latex] (Figure 3).<\/p>\n<div style=\"width: 476px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202426\/CNX_Calc_Figure_01_03_003.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one blue line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another blue line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled \u201cP = (x, y)\u201d. These line segments have a length of \u201cr\u201d units. Between these line segments within the circle is the label \u201ctheta\u201d, representing the angle between the segments. From the point \u201cP\u201d, there is a blue vertical line that extends downwards until it hits the x axis and thus hits the horizontal line segment, at a point labeled \u201cx\u201d. At the intersection horizontal line segment and vertical line segment at the point x, there is a right triangle symbol. From the point \u201cP\u201d, there is a dotted horizontal line segment that extends left until it hits the y axis at a point labeled \u201cy\u201d.\" width=\"466\" height=\"297\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. For a point [latex]P=(x,y)[\/latex] on a circle of radius [latex]r[\/latex], the coordinates [latex]x[\/latex] and [latex]y[\/latex] satisfy [latex]x=r \\cos \\theta [\/latex] and [latex]y=r \\sin \\theta[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170572229687\">The table below shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex].<\/p>\n<table id=\"fs-id1170572167612\" summary=\"A table with 5 rows and 3 columns. The first column is labeled \u201ctheta\u201d and has the values \u201c0; (pi\/6); (pi\/4); (pi\/3); (pi\/2)\u201d. The second column is labeled \u201csin(theta)\u201d and the values are \u201c0; (1\/2); ((square root of 2)\/2); ((square root of 3)\/2); 1\u201d. The third column is labeled \u201ccos(theta)\u201d and the values are \u201c0; ((square root of 3)\/2); ((square root of 2)\/2); (1\/2); 0\u201d.\">\n<caption>Values of [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex] at Major Angles [latex]\\theta[\/latex] in the First Quadrant<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>[latex]\\theta[\/latex]<\/th>\n<th>[latex]\\sin \\theta[\/latex]<\/th>\n<th>[latex]\\cos \\theta[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>0<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\large{\\frac{\\pi}{6}}[\/latex]<\/td>\n<td>[latex]\\large{\\frac{1}{2}}[\/latex]<\/td>\n<td>[latex]\\large{\\frac{\\sqrt{3}}{2}}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\large{\\frac{\\pi}{4}}[\/latex]<\/td>\n<td>[latex]\\large{\\frac{\\sqrt{2}}{2}}[\/latex]<\/td>\n<td>[latex]\\large{\\frac{\\sqrt{2}}{2}}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\large{\\frac{\\pi}{3}}[\/latex]<\/td>\n<td>[latex]\\large{\\frac{\\sqrt{3}}{2}}[\/latex]<\/td>\n<td>[latex]\\large{\\frac{1}{2}}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\large{\\frac{\\pi}{2}}[\/latex]<\/td>\n<td>1<\/td>\n<td>0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1170572205827\" class=\"textbook exercises\">\n<h3>Example: Evaluating Trigonometric Functions<\/h3>\n<p id=\"fs-id1170572247984\">Evaluate each of the following expressions.<\/p>\n<ol id=\"fs-id1170572107237\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\sin \\Big(\\large\\frac{2\\pi}{3}\\Big)[\/latex]<\/li>\n<li>[latex]\\cos \\Big(-\\large\\frac{5\\pi}{6}\\Big)[\/latex]<\/li>\n<li>[latex]\\tan \\Big(\\large\\frac{15\\pi}{4}\\Big)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572086904\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572086904\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572086904\" style=\"list-style-type: lower-alpha;\">\n<li>On the unit circle, the angle [latex]\\theta =\\large\\frac{2\\pi}{3}[\/latex] corresponds to the point [latex]\\Big(-\\large\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\Big)[\/latex]. Therefore, [latex]\\sin \\Big(\\large\\frac{2\\pi}{3}\\Big) \\normalsize = y = \\large\\frac{\\sqrt{3}}{2}[\/latex].\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202429\/CNX_Calc_Figure_01_03_004.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the left to another point on the edge of the circle. This point is labeled \u201c(-(1\/2), ((square root of 3)\/2))\u201d. These line segments have a length of 1 unit. From the point \u201c(-(1\/2), ((square root of 3)\/2))\u201d, there is a vertical line that extends downwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise until it hits the diagonal line segment. This arrow has the label \u201ctheta = (2 pi)\/3\u201d.\" width=\"325\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Unit circle depiction of [latex]\\theta[\/latex]<\/p>\n<\/div>\n<\/li>\n<li>An angle [latex]\\theta =-\\large\\frac{5\\pi}{6}[\/latex] corresponds to a revolution in the negative direction, as shown. Therefore, [latex]\\cos \\Big(-\\large\\frac{5\\pi}{6}\\Big)[\/latex] [latex]=x=-\\large\\frac{\\sqrt{3}}{2}[\/latex].\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202433\/CNX_Calc_Figure_01_03_005.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the left to another point on the edge of the circle. This point is labeled \u201c(-((square root of 3)\/2)), -(1\/2))\u201d. These line segments have a length of 1 unit. From the point \u201c(-((square root of 3)\/2)), -(1\/2))\u201d, there is a vertical line that extends upwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels clockwise until it hits the diagonal line segment. This arrow has the label \u201ctheta = -(5 pi)\/6\u201d.\" width=\"325\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. Unit circle depiction of [latex]\\theta[\/latex]<\/p>\n<\/div>\n<\/li>\n<li>An angle [latex]\\theta =\\large\\frac{15\\pi}{4} \\normalsize = 2\\pi +\\large\\frac{7\\pi}{4}[\/latex]. Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of [latex]\\large\\frac{7\\pi}{4}[\/latex] corresponds to the point [latex]\\Big(\\large\\frac{\\sqrt{2}}{2}, -\\frac{\\sqrt{2}}{2}\\Big)[\/latex], we can conclude that [latex]\\tan \\Big(\\large\\frac{15\\pi}{4}\\Big)\\normalsize =\\large\\frac{y}{x}[\/latex] [latex]=-1[\/latex].\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202436\/CNX_Calc_Figure_01_03_006.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the right to another point on the edge of the circle. This point is labeled \u201c(((square root of 2)\/2), -((square root of 2)\/2))\u201d. These line segments have a length of 1 unit. From the point \u201c(((square root of 2)\/2), -((square root of 2)\/2))\u201d, there is a vertical line that extends upwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise. The arrow makes one full rotation around the circle and then keeps traveling until it hits the diagonal line segment. This arrow has the label \u201ctheta = (15 pi)\/4\u201d.\" width=\"325\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Unit circle depiction of [latex]\\theta[\/latex]<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572295456\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572168320\">Evaluate [latex]\\cos \\left(\\dfrac{3\\pi}{4}\\right)[\/latex] and [latex]\\sin \\left(\\dfrac{\u2212\\pi}{6}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q565722\">Hint<\/span><\/p>\n<div id=\"q565722\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165041836203\">Look at angles on the unit circle.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572222562\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572222562\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572222562\">[latex]\\cos \\left(\\dfrac{3\\pi}{4}\\right)=\\dfrac{\u2212\\sqrt{2}}{2}; \\, \\sin\\left(\\dfrac{\u2212\\pi}{6}\\right)=\\dfrac{-1}{2}[\/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=\"ohm203731\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=203731&theme=oea&iframe_resize_id=ohm203731&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1170572546942\">As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let [latex]\\theta[\/latex] be one of the acute angles. Let [latex]A[\/latex] be the length of the adjacent leg, [latex]O[\/latex] be the length of the opposite leg, and [latex]H[\/latex] be the length of the hypotenuse. By inscribing the triangle into a circle of radius [latex]H[\/latex], as shown in Figure 7, we see that [latex]A, \\, H[\/latex], and [latex]O[\/latex] satisfy the following relationships with [latex]\\theta[\/latex]:<\/p>\n<div id=\"fs-id1170572169799\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc}\\sin \\theta =\\large \\frac{O}{H} & & & \\normalsize \\csc \\theta =\\large \\frac{H}{O} \\\\ \\cos \\theta =\\large \\frac{A}{H} & & & \\sec \\theta =\\large \\frac{H}{A} \\\\ \\tan \\theta =\\large \\frac{O}{A} & & & \\cot \\theta =\\large \\frac{A}{O} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div><\/div>\n<div>\n<div style=\"width: 323px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202440\/CNX_Calc_Figure_01_03_007.jpg\" alt=\"An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment with length labeled \u201cH\u201d that extends diagonally upwards and to the right to another point on the edge of the circle. From the point, there is vertical line with a length labeled \u201cO\u201d that extends downwards until it hits the x axis and thus the horizontal line segment at a point with a right triangle symbol. The distance from this point to the center of the circle is labeled \u201cA\u201d. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label \u201ctheta\u201d.\" width=\"313\" height=\"297\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. By inscribing a right triangle in a circle, we can express the ratios of the side lengths in terms of the trigonometric functions evaluated at [latex]\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572203718\" class=\"textbook exercises\">\n<h3>Example: Constructing a Wooden Ramp<\/h3>\n<p id=\"fs-id1170572544701\">A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is 4 ft from the ground and the angle between the ground and the ramp is to be [latex]10^{\\circ}[\/latex], how long does the ramp need to be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572224156\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572224156\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572224156\">Let [latex]x[\/latex] denote the length of the ramp. In the following image, we see that [latex]x[\/latex] needs to satisfy the equation [latex]\\sin(10^{\\circ})=\\dfrac{4}{x}[\/latex]. Solving this equation for [latex]x[\/latex], we see that [latex]x=\\frac{4}{ \\sin(10^{\\circ})} \\approx 23.035[\/latex] ft.<\/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\/2332\/2018\/01\/11202443\/CNX_Calc_Figure_01_03_008.jpg\" alt=\"An image of a ramp and a staircase. The ramp starts at a point and increases diagonally upwards and to the right at an angle of 10 degrees for x feet. At the end of the ramp, which is 4 feet off the ground, a staircase descends downwards and to the right.\" width=\"487\" height=\"62\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. Sketch of the ramp and staircase.<\/p>\n<\/div>\n<p><span id=\"fs-id1170572286595\"><\/span><\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Constructing a Wooden Ramp<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/N1YoLqhe_pw?controls=0&amp;start=583&amp;end=703&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/TrigonometricFunctions583to703_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.3 Trigonometric Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572453725\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572480688\">A house painter wants to lean a 20-ft ladder against a house. If the angle between the base of the ladder and the ground is to be [latex]60^{\\circ}[\/latex], how far from the house should she place the base of the ladder?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q478821\">Hint<\/span><\/p>\n<div id=\"q478821\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165041814643\">Draw a right triangle with hypotenuse 20 ft.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572450935\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572450935\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572450935\">10 ft<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Trigonometric Identities<\/h2>\n<p id=\"fs-id1170572229415\">A <strong>trigonometric identity<\/strong> is an equation involving trigonometric functions that is true for all angles [latex]\\theta[\/latex] for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Trigonometric Identities<\/h3>\n<hr \/>\n<p id=\"fs-id1170572208589\"><strong>Reciprocal identities<\/strong><\/p>\n<div id=\"fs-id1170572167703\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc}\\tan \\theta =\\large \\frac{\\sin \\theta}{\\cos \\theta} & & & \\cot \\theta =\\large \\frac{\\cos \\theta}{\\sin \\theta} \\\\ \\csc \\theta =\\large \\frac{1}{\\sin \\theta} & & & \\sec \\theta =\\large \\frac{1}{\\cos \\theta} \\end{array}[\/latex]<\/div>\n<div><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572140422\"><strong>Pythagorean identities<\/strong><\/p>\n<div id=\"fs-id1170572455704\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin^2 \\theta +\\cos^2 \\theta =1\\phantom{\\rule{2em}{0ex}}1+\\tan^2 \\theta =\\sec^2 \\theta \\phantom{\\rule{2em}{0ex}}1+\\cot^2 \\theta =\\csc^2 \\theta[\/latex]<\/div>\n<div><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572480920\"><strong>Addition and subtraction formulas<\/strong><\/p>\n<div id=\"fs-id1170572216911\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin(\\alpha \\pm \\beta)=\\sin \\alpha \\cos \\beta \\pm \\cos \\alpha \\sin \\beta[\/latex]<\/div>\n<div id=\"fs-id1170572098934\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos(\\alpha \\pm \\beta)=\\cos \\alpha \\cos \\beta \\mp \\sin \\alpha \\sin \\beta[\/latex]<\/div>\n<div><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572240527\"><strong>Double-angle formulas<\/strong><\/p>\n<div id=\"fs-id1170572215732\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin(2\\theta)=2\\sin \\theta \\cos \\theta[\/latex]<\/div>\n<div id=\"fs-id1170572130013\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos(2\\theta)=2\\cos^2 \\theta -1=1-2\\sin^2 \\theta =\\cos^2 \\theta -\\sin^2 \\theta[\/latex]<\/div>\n<\/div>\n<div>Remember, solving a trigonometric equation is not very different from solving an algebraic equation.<\/div>\n<div><\/div>\n<div>\n<div class=\"textbox examples\">\n<h3>Recall:\u00a0 Given a trigonometric equation, solve using algebra.<\/h3>\n<ul>\n<li>Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.<\/li>\n<li>Substitute the trigonometric expression with a single variable, such as [latex]x[\/latex] or [latex]u[\/latex].<\/li>\n<li>Solve the equation the same way an algebraic equation would be solved.<\/li>\n<li>Substitute the trigonometric expression back in for the variable in the resulting expressions.<\/li>\n<li>Solve for the angle.<\/li>\n<\/ul>\n<\/div>\n<div style=\"text-align: center;\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm15978\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15978&theme=oea&iframe_resize_id=ohm15978&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Additionally, trigonometric equations will have an infinite number of solutions.\u00a0If we need to find all possible solutions, then we must add [latex]2\\pi k[\/latex], where [latex]k[\/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\\pi :[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\sin \\theta =\\sin \\left(\\theta \\pm 2k\\pi \\right)[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572455784\" class=\"textbook exercises\">\n<h3>Example: Solving Trigonometric Equations<\/h3>\n<p id=\"fs-id1170572232837\">For each of the following equations, use a trigonometric identity to find all solutions.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]1+\\cos(2\\theta)=\\cos \\theta[\/latex]<\/li>\n<li>[latex]\\sin(2\\theta)=\\tan \\theta[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q501288\">Show Solution<\/span><\/p>\n<div id=\"q501288\" class=\"hidden-answer\" style=\"display: none\">a. Using the double-angle formula for [latex]\\cos(2\\theta)[\/latex], we see that [latex]\\theta[\/latex] is a solution of<\/p>\n<p style=\"text-align: center;\">[latex]1+\\cos(2\\theta)=\\cos \\theta[\/latex]<\/p>\n<p>if and only if<\/p>\n<p style=\"text-align: center;\">[latex]1+2\\cos^2 \\theta -1=\\cos \\theta[\/latex],<\/p>\n<p>which is true if and only if<\/p>\n<p style=\"text-align: center;\">[latex]2\\cos^2 \\theta -\\cos \\theta =0[\/latex].<\/p>\n<p>To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by [latex]\\cos \\theta[\/latex]. The problem with dividing by [latex]\\cos \\theta[\/latex] is that it is possible that [latex]\\cos \\theta[\/latex] is zero. In fact, if we did divide both sides of the equation by [latex]\\cos \\theta[\/latex], we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that [latex]\\theta[\/latex] is a solution of this equation if and only if<\/p>\n<p style=\"text-align: center;\">[latex]\\cos \\theta (2\\cos \\theta -1)=0[\/latex]<\/p>\n<p>Since [latex]\\cos \\theta =0[\/latex] when<\/p>\n<p style=\"text-align: center;\">[latex]\\theta =\\large \\frac{\\pi }{2}, \\, \\frac{\\pi}{2} \\normalsize \\pm \\pi, \\, \\large \\frac{\\pi}{2} \\normalsize \\pm 2\\pi, \\cdots[\/latex],<\/p>\n<p>and [latex]\\cos \\theta =1\/2[\/latex] when<\/p>\n<p style=\"text-align: center;\">[latex]\\theta =\\large \\frac{\\pi}{3}, \\, \\frac{\\pi}{3} \\normalsize \\pm 2\\pi, \\cdots[\/latex], or [latex]\\theta =-\\large \\frac{\\pi}{3}, \\, -\\frac{\\pi}{3} \\normalsize \\pm 2\\pi, \\cdots[\/latex],<\/p>\n<p>we conclude that the set of solutions to this equation is<\/p>\n<p style=\"text-align: center;\">[latex]\\theta =\\large \\frac{\\pi}{2} \\normalsize +n\\pi, \\, \\theta =\\large \\frac{\\pi}{3} \\normalsize +2n\\pi[\/latex], and [latex]\\theta =-\\large \\frac{\\pi}{3}\\normalsize +2n\\pi, \\, n=0, \\pm 1, \\pm 2,\\cdots[\/latex]<\/p>\n<p>b. Using the double-angle formula for [latex]\\sin(2\\theta)[\/latex] and the reciprocal identity for [latex]\\tan(\\theta)[\/latex], the equation can be written as<\/p>\n<p style=\"text-align: center;\">[latex]2\\sin \\theta \\cos \\theta =\\large \\frac{\\sin\\theta}{\\cos \\theta}[\/latex]<\/p>\n<p>To solve this equation, we multiply both sides by [latex]\\cos \\theta[\/latex] to eliminate the denominator, and say that if [latex]\\theta[\/latex] satisfies this equation, then [latex]\\theta[\/latex] satisfies the equation<\/p>\n<p style=\"text-align: center;\">[latex]2\\sin \\theta \\cos^2 \\theta -\\sin \\theta =0[\/latex]<\/p>\n<p>However, we need to be a little careful here. Even if [latex]\\theta[\/latex] satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by [latex]\\cos \\theta[\/latex]. However, if [latex]\\cos \\theta =0[\/latex], we cannot divide both sides of the equation by [latex]\\cos \\theta[\/latex]. Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor [latex]\\sin \\theta[\/latex] out of both terms on the left-hand side instead of dividing both sides of the equation by [latex]\\sin \\theta[\/latex]. Factoring the left-hand side of the equation, we can rewrite this equation as<\/p>\n<p style=\"text-align: center;\">[latex]\\sin \\theta (2\\cos^2 \\theta -1)=0[\/latex]<\/p>\n<p>Therefore, the solutions are given by the angles [latex]\\theta[\/latex] such that [latex]\\sin \\theta =0[\/latex] or [latex]\\cos^2 \\theta =1\/2[\/latex]. The solutions of the first equation are [latex]\\theta =0, \\pm \\pi, \\pm 2\\pi, \\cdots[\/latex]. The solutions of the second equation are [latex]\\theta =\\pi \/4, \\, (\\pi\/4) \\pm (\\pi\/2), \\, (\\pi\/4) \\pm \\pi, \\cdots[\/latex]. After checking for extraneous solutions, the set of solutions to the equation is<\/p>\n<p style=\"text-align: center;\">[latex]\\theta =n\\pi[\/latex] and [latex]\\theta =\\large \\frac{\\pi}{4}+\\frac{n\\pi}{2}, \\, \\normalsize n=0, \\pm 1, \\pm 2, \\cdots[\/latex].<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Solving Trigonometric Equations<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6239436&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=p56ltnfoFuA&amp;video_target=tpm-plugin-9gl9o8ol-p56ltnfoFuA\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.3.3_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.3.3&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572204063\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572482544\">Find all solutions to the equation [latex]\\cos(2\\theta)=\\sin \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q772243\">Hint<\/span><\/p>\n<div id=\"q772243\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042005368\">Use the double-angle formula for cosine.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572229679\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572229679\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572229679\">[latex]\\theta =\\large \\frac{3\\pi}{2} \\normalsize +2n\\pi, \\, \\large \\frac{\\pi}{6} \\normalsize +2n\\pi, \\, \\large \\frac{5\\pi}{6} \\normalsize +2n\\pi[\/latex] for [latex]n=0, \\pm 1, \\pm 2, \\cdots[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Trigonometric identities are not only useful for solving trigonometric equations, they can also help us verify identities.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Given a trigonometric identity, verify that it is true.<strong><br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id2191946\">\n<li>Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.<\/li>\n<li>Look for opportunities to factor expressions, square a binomial, or add fractions.<\/li>\n<li>Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.<\/li>\n<li>If these steps do not yield the desired result, try converting all terms to sines and cosines.<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170572111814\" class=\"textbook exercises\">\n<h3>Example: Proving a Trigonometric Identity<\/h3>\n<p id=\"fs-id1170572550142\">Prove the trigonometric identity [latex]1+\\tan^2 \\theta =\\sec^2 \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572224130\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572224130\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572224130\">We start with the identity<\/p>\n<div id=\"fs-id1170572167191\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\sin^2 \\theta +\\cos^2 \\theta =1[\/latex]<\/div>\n<p id=\"fs-id1170572140245\">Dividing both sides of this equation by [latex]\\cos^2 \\theta[\/latex], we obtain<\/p>\n<div id=\"fs-id1170572481849\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{\\sin^2 \\theta}{\\cos^2 \\theta}+1=\\frac{1}{\\cos^2 \\theta}[\/latex]<\/div>\n<p>Since [latex]\\sin \\theta \/ \\cos \\theta =\\tan \\theta[\/latex] and [latex]1 \/ \\cos \\theta =\\sec \\theta[\/latex], we conclude that<\/p>\n<div id=\"fs-id1170572481963\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\tan^2 \\theta +1=\\sec^2 \\theta[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572239833\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572226907\">Prove the trigonometric identity [latex]1+\\cot^2 \\theta =\\csc^2 \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265389\">Hint<\/span><\/p>\n<div id=\"q265389\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165041809092\">Divide both sides of the identity [latex]\\sin^2 \\theta + \\cos^2 \\theta =1[\/latex] by [latex]\\sin^2 \\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-135\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>1.3 Trigonometric Functions. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>1.3.3. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"1.3 Trigonometric Functions\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"1.3.3\",\"author\":\"Ryan 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