{"id":1798,"date":"2025-07-28T19:44:25","date_gmt":"2025-07-28T19:44:25","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1798"},"modified":"2026-03-24T06:39:43","modified_gmt":"2026-03-24T06:39:43","slug":"sine-and-cosine-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/sine-and-cosine-functions-learn-it-3\/","title":{"raw":"Sine and Cosine Functions: Learn It 3","rendered":"Sine and Cosine Functions: Learn It 3"},"content":{"raw":"

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

Finding Sines and Cosines of 45\u00b0 Angles<\/h3>\r\n\"Graph\r\n\r\nFirst, we will look at angles of [latex]45^\\circ [\/latex] or [latex]\\frac{\\pi }{4}[\/latex]. 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.<\/span>\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\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.\r\n<\/span>\"Graph \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{gathered}{x}^{2}=\\frac{1}{2}\\\\ x=\\pm \\frac{1}{\\sqrt{2}}\\end{gathered}[\/latex]<\/div>\r\nIn quadrant I, [latex]x=\\frac{1}{\\sqrt{2}}[\/latex].\r\n\r\nAt [latex]t=\\frac{\\pi }{4}[\/latex] or 45 degrees,\r\n
[latex]\\begin{gathered}\\left(x,y\\right)=\\left(x,x\\right)=\\left(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}\\right) \\\\ x=\\frac{1}{\\sqrt{2}},y=\\frac{1}{\\sqrt{2}}\\\\ \\cos t=\\frac{1}{\\sqrt{2}},\\sin t=\\frac{1}{\\sqrt{2}} \\end{gathered}[\/latex]<\/div>\r\nIf we then rationalize the denominators, we get\r\n
[latex]\\begin{align}\\cos t&=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}} =\\frac{\\sqrt{2}}{2} \\\\ \\sin t&=\\frac{1}{\\sqrt{2}}\\frac{\\sqrt{2}}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2} \\end{align}[\/latex]<\/div>\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\r\n<\/section>\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]. 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].\r\n<\/span>\r\n\r\n\"Graph\r\n\r\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] ,\"Image\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{align}\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\left(1\\right)=\\frac{1}{2}\\end{align}[\/latex]<\/div>\r\nUsing the Pythagorean Identity, we can find the cosine value.\r\n
[latex]\\begin{array}{cll}{\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)+{\\sin }^{2}\\left(\\frac{\\pi }{6}\\right)=1 \\hfill \\\\ {\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)+{\\left(\\frac{1}{2}\\right)}^{2}=1 \\\\ {\\cos }^{2}\\left(\\frac{\\pi }{6}\\right)=\\frac{3}{4}&& \\text{Use the square root property}. \\\\ \\cos \\left(\\frac{\\pi }{6}\\right)=\\frac{\\pm \\sqrt{3}}{\\pm \\sqrt{4}}=\\frac{\\sqrt{3}}{2}&& \\text{Since }y\\text{ is positive, choose the positive root}. \\end{array}[\/latex]<\/div>\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]. At [latex]t=\\frac{\\pi }{3}[\/latex] (60\u00b0), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, [latex]BAD[\/latex]. 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\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{gathered}\\frac{1}{4}+{y}^{2}=1\\\\ {y}^{2}=1-\\frac{1}{4}\\\\ {y}^{2}=\\frac{3}{4}\\\\ y=\\pm \\frac{\\sqrt{3}}{2}\\end{gathered}[\/latex]<\/div>\r\nSince [latex]t=\\frac{\\pi }{3}[\/latex] has the terminal side in quadrant I where the 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{gathered}\\left(x,y\\right)=\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right) \\\\ x=\\frac{1}{2},y=\\frac{\\sqrt{3}}{2}\\\\ \\cos t=\\frac{1}{2},\\sin t=\\frac{\\sqrt{3}}{2} \\end{gathered}[\/latex]<\/div>\r\n<\/section>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.\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\u00b0<\/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\n
The common angles in the first quadrant of the unit circle.\r\n<\/span>\"Graph<\/section>\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>\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<\/section>
    Evaluate [latex]\\cos \\left(\\frac{5\\pi }{3}\\right)[\/latex] using a graphing calculator or computer.[reveal-answer q=\"949209\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"949209\"]Enter 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]<\/p>\r\n\r\n

    Analysis of the Solution<\/h4>\r\nWe can find the cosine or sine of an angle in degrees directly on a calculator with degree mode. For calculators or software that use only radian mode, we can find the sign of [latex]20^\\circ [\/latex], for example, by including the conversion factor to radians as part of the input:\r\n
    SIN( 20 \u00d7 \u03c0 \u00f7 180 ) ENTER<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
    Evaluate [latex]\\sin \\left(\\frac{\\pi }{3}\\right)[\/latex].[reveal-answer q=\"184627\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"184627\"]approximately [latex]0.866025403[\/latex][\/hidden-answer]<\/section>","rendered":"

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

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

    \"Graph<\/p>\n

    First, we will look at angles of [latex]45^\\circ[\/latex] or [latex]\\frac{\\pi }{4}[\/latex]. 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.<\/span><\/p>\n

     <\/p>\n

     <\/p>\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>\"Graph <\/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{gathered}{x}^{2}=\\frac{1}{2}\\\\ x=\\pm \\frac{1}{\\sqrt{2}}\\end{gathered}[\/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{gathered}\\left(x,y\\right)=\\left(x,x\\right)=\\left(\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}\\right) \\\\ x=\\frac{1}{\\sqrt{2}},y=\\frac{1}{\\sqrt{2}}\\\\ \\cos t=\\frac{1}{\\sqrt{2}},\\sin t=\\frac{1}{\\sqrt{2}} \\end{gathered}[\/latex]<\/div>\n

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

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

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

    \"Graph<\/p>\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] ,\"Image<\/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{align}\\sin \\left(\\frac{\\pi }{6}\\right)=\\frac{1}{2}\\left(1\\right)=\\frac{1}{2}\\end{align}[\/latex]<\/div>\n

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

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

    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]. At [latex]t=\\frac{\\pi }{3}[\/latex] (60\u00b0), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, [latex]BAD[\/latex]. 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

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

    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{gathered}\\left(x,y\\right)=\\left(\\frac{1}{2},\\frac{\\sqrt{3}}{2}\\right) \\\\ x=\\frac{1}{2},y=\\frac{\\sqrt{3}}{2}\\\\ \\cos t=\\frac{1}{2},\\sin t=\\frac{\\sqrt{3}}{2} \\end{gathered}[\/latex]<\/div>\n<\/section>\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\u00b0<\/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
    The common angles in the first quadrant of the unit circle.
    \n<\/span>\"Graph<\/section>\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

    How To: Given an angle in radians, use a graphing calculator to find the cosine.
    \n<\/strong><\/p>\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<\/section>\n
      Evaluate [latex]\\cos \\left(\\frac{5\\pi }{3}\\right)[\/latex] using a graphing calculator or computer.<\/p>\n