{"id":13786,"date":"2018-06-14T23:50:27","date_gmt":"2018-06-14T23:50:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/finding-function-values-for-the-sine-and-cosine\/"},"modified":"2021-11-22T21:44:35","modified_gmt":"2021-11-22T21:44:35","slug":"unit-circle","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/unit-circle\/","title":{"raw":"Unit circle","rendered":"Unit circle"},"content":{"raw":"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 unit circle<\/strong>, [latex]s=t[\/latex].\r\n\r\nRecall that the x- <\/em>and 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\"]\"Graph Figure 2.<\/b> Unit circle where the central angle is [latex]t[\/latex] radians[\/caption]\r\n
\r\n

A General Note: Unit Circle<\/h3>\r\nA 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

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 arc length<\/strong> and angle<\/strong>. The sine function<\/strong> relates a real number [latex]t[\/latex] to the 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 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 y<\/em>-coordinate of the corresponding point on the unit circle.\r\n\r\nThe cosine function<\/strong> of an angle [latex]t[\/latex] equals the 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].\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Illustration Figure 3<\/b>[\/caption]\r\n\r\n \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
\r\n

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
[latex]\\cos t=x[\/latex]<\/div>\r\n
[latex]\\sin t=y[\/latex]<\/div>\r\n<\/div>\r\n
\r\n

How To: Given a point 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
    \r\n \t
  1. The sine of [latex]t[\/latex] is equal to the y<\/em>-coordinate of point [latex]P:\\sin t=y[\/latex].<\/li>\r\n \t
  2. The cosine of [latex]t[\/latex] is equal to the x<\/em>-coordinate of point [latex]P: \\text{cos}t=x[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n

    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].\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 4<\/b>[\/caption]\r\n\r\n \r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nWe know that [latex]\\cos t[\/latex] is the x<\/em>-coordinate of the corresponding point on the unit circle and [latex]\\sin t[\/latex] is the y<\/em>-coordinate of the corresponding point on the unit circle. So:\r\n
    [latex]\\begin{array}{l}\\begin{array}{l}\\\\ x=\\cos t=\\frac{1}{2}\\end{array}\\hfill \\\\ y=\\sin t=\\frac{\\sqrt{3}}{2}\\hfill \\end{array}\\\\[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Try It 1<\/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].\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 5<\/b>[\/caption]\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

    Finding Sines and Cosines of Angles on an Axis<\/h2>\r\nFor quadrantral angles, the corresponding point on the unit circle falls on the x- <\/em>or 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
    \r\n

    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<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nMoving [latex]90^\\circ [\/latex] counterclockwise around the unit circle from the positive 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.\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 6<\/b>[\/caption]\r\n\r\n \r\n\r\nUsing our definitions of cosine and sine,\r\n
    [latex]\\begin{array}{l}x=\\cos t=\\cos \\left(90^\\circ \\right)=0\\\\ y=\\sin t=\\sin \\left(90^\\circ \\right)=1\\end{array}\\\\[\/latex]<\/div>\r\nThe cosine of 90\u00b0 is 0; the sine of 90\u00b0 is 1.\r\n\r\n<\/div>\r\n
    \r\n

    Try It 2<\/h3>\r\nFind cosine and sine of the angle [latex]\\pi [\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

    The Pythagorean Identity<\/h2>\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph 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 Pythagorean Identity<\/strong>.\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
    \r\n

    A General Note: Pythagorean Identity<\/h3>\r\nThe Pythagorean Identity<\/strong> states that, for any real number [latex]t[\/latex],\r\n
    [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    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
      \r\n \t
    1. Substitute the known value of [latex]\\sin \\left(t\\right)[\/latex] into the Pythagorean Identity.<\/li>\r\n \t
    2. Solve for [latex]\\cos \\left(t\\right)[\/latex].<\/li>\r\n \t
    3. Choose the solution with the appropriate sign for the x<\/em>-values in the quadrant where [latex]t[\/latex] is located.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
      \r\n

      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<\/div>\r\n
      \r\n

      Solution<\/h3>\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\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 8<\/b>[\/caption]\r\n\r\n \r\n\r\nSubstituting the known value for sine into the Pythagorean Identity,\r\n
      [latex]\\begin{array}{l}{\\cos }^{2}\\left(t\\right)+{\\sin }^{2}\\left(t\\right)=1\\hfill \\\\ {\\cos }^{2}\\left(t\\right)+\\frac{9}{49}=1\\hfill \\\\ {\\cos }^{2}\\left(t\\right)=\\frac{40}{49}\\hfill \\\\ \\text{cos}\\left(t\\right)=\\pm \\sqrt{\\frac{40}{49}}=\\pm \\frac{\\sqrt{40}}{7}=\\pm \\frac{2\\sqrt{10}}{7}\\hfill \\end{array}\\\\[\/latex]<\/div>\r\nBecause the angle is in the second quadrant, we know the 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<\/div>\r\n
      \r\n

      Try It 3<\/h3>\r\nIf [latex]\\cos \\left(t\\right)=\\frac{24}{25}\\\\[\/latex] and [latex]t[\/latex] is in the fourth quadrant, find [latex]\\text{sin}\\left(t\\right)\\\\[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

      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 Pythagorean Identity<\/strong> and our knowledge of triangles.\r\n

      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 x-<\/em> and y<\/em>-coordinates of the corresponding point on the circle are the same. Because the x-<\/em> and y<\/em>-values are the same, the sine and cosine values will also be equal.\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 9<\/b>[\/caption]\r\n\r\n \r\n\r\nAt [latex]t=\\frac{\\pi }{4}[\/latex] , which is 45 degrees, the radius of the unit circle bisects the first 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.\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 10<\/b>[\/caption]\r\n\r\n \r\n\r\nFrom the Pythagorean Theorem we get\r\n
      [latex]{x}^{2}+{y}^{2}=1[\/latex]<\/div>\r\nSubstituting [latex]y=x[\/latex], we get\r\n
      [latex]{x}^{2}+{x}^{2}=1[\/latex]<\/div>\r\nCombining like terms we get\r\n
      [latex]2{x}^{2}=1[\/latex]<\/div>\r\nAnd solving for [latex]x[\/latex], we get\r\n
      [latex]\\begin{array}{c}{x}^{2}=\\frac{1}{2}\\\\ \\text{ }x=\\pm \\frac{1}{\\sqrt{2}}\\end{array}\\\\[\/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
      [latex]\\begin{array}{l}\\left(x,y\\right)=\\left(x,x\\right)=\\left(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}\\right)\\hfill \\\\ x=\\frac{1}{\\sqrt{2}},y=\\frac{1}{\\sqrt{2}}\\hfill \\\\ \\cos t=\\frac{1}{\\sqrt{2}},\\sin t=\\frac{1}{\\sqrt{2}}\\hfill \\end{array}\\\\[\/latex]<\/div>\r\nIf we then rationalize the denominators, we get\r\n
      [latex]\\begin{array}{l}\\cos t=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}}\\hfill \\\\ =\\frac{\\sqrt{2}}{2}\\hfill \\\\ \\sin t=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}}\\hfill \\\\ =\\frac{\\sqrt{2}}{2}\\hfill \\end{array}\\\\[\/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

      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.\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 11<\/b>[\/caption]\r\n\r\n \r\n
      \r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"490\"]\"Image 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
      [latex]\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}r\\\\[\/latex]<\/div>\r\nAnd since [latex]r=1[\/latex]\u00a0in our unit circle<\/strong>,\r\n
      [latex]\\begin{array}{l}\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\left(1\\right)\\hfill \\\\ \\text{ }=\\frac{1}{2}\\hfill \\end{array}\\\\[\/latex]<\/div>\r\nUsing the Pythagorean Identity, we can find the cosine value.\r\n
      [latex]\\begin{array}{ll}{\\cos }^{2}\\frac{\\pi }{6}+{\\sin }^{2}\\left(\\frac{\\pi }{6}\\right)=1\\hfill & \\hfill \\\\ \\text{ }{\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)+{\\left(\\frac{1}{2}\\right)}^{2}=1\\hfill & \\hfill \\\\ \\text{ }{\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)=\\frac{3}{4}\\hfill & \\text{Use the square root property}.\\hfill \\\\ \\text{ }\\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\pm \\sqrt{3}}{\\pm \\sqrt{4}}=\\frac{\\sqrt{3}}{2}\\hfill & \\text{Since }y\\text{ is positive, choose the positive root}.\\hfill \\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 [link]<\/a>. 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
      \r\n<\/span><\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph 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 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
      [latex]{x}^{2}+{y}^{2}=1[\/latex]<\/div>\r\nSubstituting [latex]x=\\frac{1}{2}[\/latex], we get\r\n
      [latex]{\\left(\\frac{1}{2}\\right)}^{2}+{y}^{2}=1[\/latex]<\/div>\r\nSolving for [latex]y[\/latex], we get\r\n
      [latex]\\begin{array}{l}\\frac{1}{4}+{y}^{2}=1\\\\ \\text{ }{y}^{2}=1-\\frac{1}{4}\\\\ \\text{ }{y}^{2}=\\frac{3}{4}\\\\ \\text{ }y=\\pm \\frac{\\sqrt{3}}{2}\\end{array}\\\\[\/latex]<\/div>\r\nSince [latex]t=\\frac{\\pi }{3}[\/latex] has the terminal side in quadrant I where the 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
      [latex]\\begin{array}{l}\\left(x,y\\right)=\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ x=\\frac{1}{2},y=\\frac{\\sqrt{3}}{2}\\hfill \\\\ \\cos t=\\frac{1}{2},\\sin t=\\frac{\\sqrt{3}}{2}\\hfill \\end{array}\\\\[\/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<\/colgroup>\r\n\r\n\r\n\r\n\r\n
      Angle<\/strong><\/td>\r\n0<\/td>\r\n[latex]\\frac{\\pi }{6}\\\\[\/latex], or 30<\/td>\r\n[latex]\\frac{\\pi }{4}\\\\[\/latex], or 45\u00b0<\/td>\r\n[latex]\\frac{\\pi }{3}\\\\[\/latex], or 60\u00b0<\/td>\r\n[latex]\\frac{\\pi }{2}\\\\[\/latex], or 90\u00b0<\/td>\r\n<\/tr>\r\n
      Cosine<\/strong><\/td>\r\n1<\/td>\r\n[latex]\\frac{\\sqrt{3}}{2}\\\\[\/latex]<\/td>\r\n[latex]\\frac{\\sqrt{2}}{2}\\\\[\/latex]<\/td>\r\n[latex]\\frac{1}{2}\\\\[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
      Sine<\/strong><\/td>\r\n0<\/td>\r\n[latex]\\frac{1}{2}\\\\[\/latex]<\/td>\r\n[latex]\\frac{\\sqrt{2}}{2}\\\\[\/latex]<\/td>\r\n[latex]\\frac{\\sqrt{3}}{2}\\\\[\/latex]<\/td>\r\n1<\/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.\r\n<\/span>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"725\"]\"Graph Figure 14<\/b>[\/caption]\r\n\r\n \r\n

      Using a Calculator to Find Sine and Cosine<\/h3>\r\nTo find the cosine and sine of angles other than the special angles<\/strong>, we turn to a computer or calculator. 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
      \r\n

      How To: Given an angle in radians, use a graphing calculator to find the cosine.\r\n<\/strong><\/h3>\r\n
        \r\n \t
      1. If the calculator has degree mode and radian mode, set it to radian mode.<\/li>\r\n \t
      2. Press the COS key.<\/li>\r\n \t
      3. Enter the radian value of the angle and press the close-parentheses key \")\".<\/li>\r\n \t
      4. Press ENTER.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        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<\/div>\r\n
        \r\n

        Solution<\/h3>\r\nEnter the following keystrokes:\r\n\r\nCOS (5 \u00d7 \u03c0 \u00f7 3 ) ENTER\r\n
        [latex]\\cos \\left(\\frac{5\\pi }{3}\\right)=0.5\\\\[\/latex]<\/div>\r\n<\/div>\r\n
        \r\n

        Analysis of the Solution<\/h3>\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
        SIN( 20 \u00d7 \u03c0 \u00f7 180 ) ENTER<\/div>\r\n<\/div>\r\n
        \r\n

        Try It 4<\/h3>\r\nEvaluate [latex]\\sin \\left(\\frac{\\pi }{3}\\right)\\\\[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/vaG4O6d48mo\r\n

        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 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 unit circle<\/strong>, as shown in Figure 15. The bounds of the x<\/em>-coordinate are [latex]\\left[-1,1\\right][\/latex]. The bounds of the 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\"]\"Graph Figure 15<\/b>[\/caption]","rendered":"

        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 unit circle<\/strong>, [latex]s=t[\/latex].<\/p>\n

        Recall that the x- <\/em>and 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

        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

        \"Graph<\/p>\n

        Figure 2.<\/b> Unit circle where the central angle is [latex]t[\/latex] radians<\/p>\n<\/div>\n

        \n

        A General Note: Unit Circle<\/h3>\n

        A 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

        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

        Defining Sine and Cosine Functions<\/h2>\n

        Now that we have our unit circle labeled, we can learn how the [latex]\\left(x,y\\right)[\/latex] coordinates relate to the arc length<\/strong> and angle<\/strong>. The sine function<\/strong> relates a real number [latex]t[\/latex] to the 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 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 y<\/em>-coordinate of the corresponding point on the unit circle.<\/p>\n

        The cosine function<\/strong> of an angle [latex]t[\/latex] equals the 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].
        \n<\/span><\/p>\n

        \"Illustration<\/p>\n

        Figure 3<\/b><\/p>\n<\/div>\n

         <\/p>\n

        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

        \n

        A General Note: Sine and Cosine Functions<\/h3>\n

        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

        [latex]\\cos t=x[\/latex]<\/div>\n
        [latex]\\sin t=y[\/latex]<\/div>\n<\/div>\n
        \n

        How To: Given a point 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
          \n
        1. The sine of [latex]t[\/latex] is equal to the y<\/em>-coordinate of point [latex]P:\\sin t=y[\/latex].<\/li>\n
        2. The cosine of [latex]t[\/latex] is equal to the x<\/em>-coordinate of point [latex]P: \\text{cos}t=x[\/latex].<\/li>\n<\/ol>\n<\/div>\n
          \n

          Example 1: Finding Function Values for Sine and Cosine<\/h3>\n

          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].
          \n<\/span><\/p>\n

          \"Graph<\/p>\n

          Figure 4<\/b><\/p>\n<\/div>\n

           <\/p>\n<\/div>\n

          \n

          Solution<\/h3>\n

          We know that [latex]\\cos t[\/latex] is the x<\/em>-coordinate of the corresponding point on the unit circle and [latex]\\sin t[\/latex] is the y<\/em>-coordinate of the corresponding point on the unit circle. So:<\/p>\n

          [latex]\\begin{array}{l}\\begin{array}{l}\\\\ x=\\cos t=\\frac{1}{2}\\end{array}\\hfill \\\\ y=\\sin t=\\frac{\\sqrt{3}}{2}\\hfill \\end{array}\\\\[\/latex]<\/div>\n<\/div>\n
          \n

          Try It 1<\/h3>\n

          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].
          \n<\/span><\/p>\n

          \"Graph<\/p>\n

          Figure 5<\/b><\/p>\n<\/div>\n

          Solution<\/a><\/p>\n<\/div>\n

          Finding Sines and Cosines of Angles on an Axis<\/h2>\n

          For quadrantral angles, the corresponding point on the unit circle falls on the x- <\/em>or 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

          \n

          Example 2: Calculating Sines and Cosines along an Axis<\/h3>\n

          Find [latex]\\cos \\left(90^\\circ \\right)\\\\[\/latex] and [latex]\\text{sin}\\left(90^\\circ \\right)\\\\[\/latex].<\/p>\n<\/div>\n

          \n

          Solution<\/h3>\n

          Moving [latex]90^\\circ[\/latex] counterclockwise around the unit circle from the positive 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.
          \n<\/span><\/p>\n

          \"Graph<\/p>\n

          Figure 6<\/b><\/p>\n<\/div>\n

           <\/p>\n

          Using our definitions of cosine and sine,<\/p>\n

          [latex]\\begin{array}{l}x=\\cos t=\\cos \\left(90^\\circ \\right)=0\\\\ y=\\sin t=\\sin \\left(90^\\circ \\right)=1\\end{array}\\\\[\/latex]<\/div>\n

          The cosine of 90\u00b0 is 0; the sine of 90\u00b0 is 1.<\/p>\n<\/div>\n

          \n

          Try It 2<\/h3>\n

          Find cosine and sine of the angle [latex]\\pi[\/latex].<\/p>\n

          Solution<\/a><\/p>\n<\/div>\n

          The Pythagorean Identity<\/h2>\n
          \"Graph<\/p>\n

          Figure 7<\/b><\/p>\n<\/div>\n

          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 Pythagorean Identity<\/strong>.
          \n<\/span><\/p>\n

          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

          \n

          A General Note: Pythagorean Identity<\/h3>\n

          The Pythagorean Identity<\/strong> states that, for any real number [latex]t[\/latex],<\/p>\n

          [latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/div>\n<\/div>\n
          \n

          How To: Given the sine of some angle [latex]t[\/latex] and its quadrant location, find the cosine of [latex]t[\/latex].<\/h3>\n
            \n
          1. Substitute the known value of [latex]\\sin \\left(t\\right)[\/latex] into the Pythagorean Identity.<\/li>\n
          2. Solve for [latex]\\cos \\left(t\\right)[\/latex].<\/li>\n
          3. Choose the solution with the appropriate sign for the x<\/em>-values in the quadrant where [latex]t[\/latex] is located.<\/li>\n<\/ol>\n<\/div>\n
            \n

            Example 3: Finding a Cosine from a Sine or a Sine from a Cosine<\/h3>\n

            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>\n

            \n

            Solution<\/h3>\n

            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
            \n<\/span><\/p>\n

            \"Graph<\/p>\n

            Figure 8<\/b><\/p>\n<\/div>\n

             <\/p>\n

            Substituting the known value for sine into the Pythagorean Identity,<\/p>\n

            [latex]\\begin{array}{l}{\\cos }^{2}\\left(t\\right)+{\\sin }^{2}\\left(t\\right)=1\\hfill \\\\ {\\cos }^{2}\\left(t\\right)+\\frac{9}{49}=1\\hfill \\\\ {\\cos }^{2}\\left(t\\right)=\\frac{40}{49}\\hfill \\\\ \\text{cos}\\left(t\\right)=\\pm \\sqrt{\\frac{40}{49}}=\\pm \\frac{\\sqrt{40}}{7}=\\pm \\frac{2\\sqrt{10}}{7}\\hfill \\end{array}\\\\[\/latex]<\/div>\n

            Because the angle is in the second quadrant, we know the x-<\/em>value is a negative real number, so the cosine is also negative. So
            \n[latex]\\text{cos}\\left(t\\right)=-\\frac{2\\sqrt{10}}{7}\\\\[\/latex]<\/p>\n<\/div>\n

            \n

            Try It 3<\/h3>\n

            If [latex]\\cos \\left(t\\right)=\\frac{24}{25}\\\\[\/latex] and [latex]t[\/latex] is in the fourth quadrant, find [latex]\\text{sin}\\left(t\\right)\\\\[\/latex].<\/p>\n

            Solution<\/a><\/p>\n<\/div>\n

            Finding Sines and Cosines of Special Angles<\/h2>\n

            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 Pythagorean Identity<\/strong> and our knowledge of triangles.<\/p>\n

            Finding Sines and Cosines of 45\u00b0 Angles<\/h3>\n

            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 x-<\/em> and y<\/em>-coordinates of the corresponding point on the circle are the same. Because the x-<\/em> and y<\/em>-values are the same, the sine and cosine values will also be equal.
            \n<\/span><\/p>\n

            \"Graph<\/p>\n

            Figure 9<\/b><\/p>\n<\/div>\n

             <\/p>\n

            At [latex]t=\\frac{\\pi }{4}[\/latex] , which is 45 degrees, the radius of the unit circle bisects the first 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.
            \n<\/span><\/p>\n

            \"Graph<\/p>\n

            Figure 10<\/b><\/p>\n<\/div>\n

             <\/p>\n

            From the Pythagorean Theorem we get<\/p>\n

            [latex]{x}^{2}+{y}^{2}=1[\/latex]<\/div>\n

            Substituting [latex]y=x[\/latex], we get<\/p>\n

            [latex]{x}^{2}+{x}^{2}=1[\/latex]<\/div>\n

            Combining like terms we get<\/p>\n

            [latex]2{x}^{2}=1[\/latex]<\/div>\n

            And solving for [latex]x[\/latex], we get<\/p>\n

            [latex]\\begin{array}{c}{x}^{2}=\\frac{1}{2}\\\\ \\text{ }x=\\pm \\frac{1}{\\sqrt{2}}\\end{array}\\\\[\/latex]<\/div>\n

            In quadrant I, [latex]x=\\frac{1}{\\sqrt{2}}\\\\[\/latex].<\/p>\n

            At [latex]t=\\frac{\\pi }{4}[\/latex] or 45 degrees,<\/p>\n

            [latex]\\begin{array}{l}\\left(x,y\\right)=\\left(x,x\\right)=\\left(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}\\right)\\hfill \\\\ x=\\frac{1}{\\sqrt{2}},y=\\frac{1}{\\sqrt{2}}\\hfill \\\\ \\cos t=\\frac{1}{\\sqrt{2}},\\sin t=\\frac{1}{\\sqrt{2}}\\hfill \\end{array}\\\\[\/latex]<\/div>\n

            If we then rationalize the denominators, we get<\/p>\n

            [latex]\\begin{array}{l}\\cos t=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}}\\hfill \\\\ =\\frac{\\sqrt{2}}{2}\\hfill \\\\ \\sin t=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}}\\hfill \\\\ =\\frac{\\sqrt{2}}{2}\\hfill \\end{array}\\\\[\/latex]<\/div>\n

            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

            Finding Sines and Cosines of 30\u00b0 and 60\u00b0 Angles<\/h3>\n

            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.
            \n<\/span><\/p>\n

            \"Graph<\/p>\n

            Figure 11<\/b><\/p>\n<\/div>\n

             <\/p>\n

            \n
            \"Image<\/p>\n

            Figure 12<\/b><\/p>\n<\/div>\n<\/figure>\n

            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

            [latex]\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}r\\\\[\/latex]<\/div>\n

            And since [latex]r=1[\/latex]\u00a0in our unit circle<\/strong>,<\/p>\n

            [latex]\\begin{array}{l}\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\left(1\\right)\\hfill \\\\ \\text{ }=\\frac{1}{2}\\hfill \\end{array}\\\\[\/latex]<\/div>\n

            Using the Pythagorean Identity, we can find the cosine value.<\/p>\n

            [latex]\\begin{array}{ll}{\\cos }^{2}\\frac{\\pi }{6}+{\\sin }^{2}\\left(\\frac{\\pi }{6}\\right)=1\\hfill & \\hfill \\\\ \\text{ }{\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)+{\\left(\\frac{1}{2}\\right)}^{2}=1\\hfill & \\hfill \\\\ \\text{ }{\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)=\\frac{3}{4}\\hfill & \\text{Use the square root property}.\\hfill \\\\ \\text{ }\\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\pm \\sqrt{3}}{\\pm \\sqrt{4}}=\\frac{\\sqrt{3}}{2}\\hfill & \\text{Since }y\\text{ is positive, choose the positive root}.\\hfill \\end{array}\\\\[\/latex]<\/div>\n

            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 [link]<\/a>. 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


            \n<\/span><\/span><\/p>\n
            \"Graph<\/p>\n

            Figure 13<\/b><\/p>\n<\/div>\n<\/figure>\n

            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 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

            From the Pythagorean Theorem, we get<\/p>\n

            [latex]{x}^{2}+{y}^{2}=1[\/latex]<\/div>\n

            Substituting [latex]x=\\frac{1}{2}[\/latex], we get<\/p>\n

            [latex]{\\left(\\frac{1}{2}\\right)}^{2}+{y}^{2}=1[\/latex]<\/div>\n

            Solving for [latex]y[\/latex], we get<\/p>\n

            [latex]\\begin{array}{l}\\frac{1}{4}+{y}^{2}=1\\\\ \\text{ }{y}^{2}=1-\\frac{1}{4}\\\\ \\text{ }{y}^{2}=\\frac{3}{4}\\\\ \\text{ }y=\\pm \\frac{\\sqrt{3}}{2}\\end{array}\\\\[\/latex]<\/div>\n

            Since [latex]t=\\frac{\\pi }{3}[\/latex] has the terminal side in quadrant I where the y-<\/em>coordinate is positive, we choose [latex]y=\\frac{\\sqrt{3}}{2}\\\\[\/latex], the positive value.<\/p>\n

            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

            [latex]\\begin{array}{l}\\left(x,y\\right)=\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ x=\\frac{1}{2},y=\\frac{\\sqrt{3}}{2}\\hfill \\\\ \\cos t=\\frac{1}{2},\\sin t=\\frac{\\sqrt{3}}{2}\\hfill \\end{array}\\\\[\/latex]<\/div>\n

            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\n\n\n\n\n\n\n<\/colgroup>\n\n\n\n\n
            Angle<\/strong><\/td>\n0<\/td>\n[latex]\\frac{\\pi }{6}\\\\[\/latex], or 30<\/td>\n[latex]\\frac{\\pi }{4}\\\\[\/latex], or 45\u00b0<\/td>\n[latex]\\frac{\\pi }{3}\\\\[\/latex], or 60\u00b0<\/td>\n[latex]\\frac{\\pi }{2}\\\\[\/latex], or 90\u00b0<\/td>\n<\/tr>\n
            Cosine<\/strong><\/td>\n1<\/td>\n[latex]\\frac{\\sqrt{3}}{2}\\\\[\/latex]<\/td>\n[latex]\\frac{\\sqrt{2}}{2}\\\\[\/latex]<\/td>\n[latex]\\frac{1}{2}\\\\[\/latex]<\/td>\n0<\/td>\n<\/tr>\n
            Sine<\/strong><\/td>\n0<\/td>\n[latex]\\frac{1}{2}\\\\[\/latex]<\/td>\n[latex]\\frac{\\sqrt{2}}{2}\\\\[\/latex]<\/td>\n[latex]\\frac{\\sqrt{3}}{2}\\\\[\/latex]<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            Figure 14\u00a0shows the common angles in the first quadrant of the unit circle.
            \n<\/span><\/p>\n

            \"Graph<\/p>\n

            Figure 14<\/b><\/p>\n<\/div>\n

             <\/p>\n

            Using a Calculator to Find Sine and Cosine<\/h3>\n

            To find the cosine and sine of angles other than the special angles<\/strong>, we turn to a computer or calculator. 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.<\/p>\n

            \n

            How To: Given an angle in radians, use a graphing calculator to find the cosine.
            \n<\/strong><\/h3>\n
              \n
            1. If the calculator has degree mode and radian mode, set it to radian mode.<\/li>\n
            2. Press the COS key.<\/li>\n
            3. Enter the radian value of the angle and press the close-parentheses key “)”.<\/li>\n
            4. Press ENTER.<\/li>\n<\/ol>\n<\/div>\n
              \n

              Example 4: Using a Graphing Calculator to Find Sine and Cosine<\/h3>\n

              Evaluate [latex]\\cos \\left(\\frac{5\\pi }{3}\\right)\\\\[\/latex] using a graphing calculator or computer.<\/p>\n<\/div>\n

              \n

              Solution<\/h3>\n

              Enter the following keystrokes:<\/p>\n

              COS (5 \u00d7 \u03c0 \u00f7 3 ) ENTER<\/p>\n

              [latex]\\cos \\left(\\frac{5\\pi }{3}\\right)=0.5\\\\[\/latex]<\/div>\n<\/div>\n
              \n

              Analysis of the Solution<\/h3>\n

              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

              SIN( 20 \u00d7 \u03c0 \u00f7 180 ) ENTER<\/div>\n<\/div>\n
              \n

              Try It 4<\/h3>\n

              Evaluate [latex]\\sin \\left(\\frac{\\pi }{3}\\right)\\\\[\/latex].<\/p>\n

              Solution<\/a><\/p>\n<\/div>\n