{"id":1926,"date":"2025-07-30T21:28:52","date_gmt":"2025-07-30T21:28:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1926"},"modified":"2025-08-13T16:52:55","modified_gmt":"2025-08-13T16:52:55","slug":"inverse-trigonometric-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/inverse-trigonometric-functions-learn-it-3\/","title":{"raw":"Inverse Trigonometric Functions: Learn It 3","rendered":"Inverse Trigonometric Functions: Learn It 3"},"content":{"raw":"

Finding Exact Values of Composite Functions with Inverse Trigonometric Functions<\/h2>\r\nThere are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let f<\/em>(x<\/em>) and g<\/em>(x<\/em>) be two different trigonometric functions belonging to the set {sin(x<\/em>), cos(x<\/em>), tan(x<\/em>)} and let [latex]f^{\u22121}(y)[\/latex] and [latex]g^{\u22121}(y)[\/latex] be their inverses.\r\n

Evaluating Compositions of the Form [latex]f\\left(f^{\u22121}(y)\\right)[\/latex] and [latex]f^{\u22121}(f(x))[\/latex]<\/h3>\r\nFor any trigonometric function, [latex]f(f^{\u22121}(y))=y[\/latex] for all y<\/em> in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f<\/em> was defined to be identical to the domain of [latex]f^{\u22121}[\/latex]. However, we have to be a little more careful with expressions of the form [latex]f^{\u22121}(f(x))[\/latex].\r\n\r\n
\r\n

compositions of a trigonometric function and its inverse<\/h3>\r\n

[latex]\\begin{align} &\\sin(\\sin^{\u22121}x)=x\\text{ for }\u22121\\leq x\\leq1\\\\ &\\cos(\\cos^{\u22121}x)=x\\text{ for }\u22121\\leq x\\leq1 \\\\ &\\tan(\\tan^{\u22121}x)=x\\text{ for }\u2212\\infty\\text{ < }x\\text{ < }\\infty \\end{align}[\/latex]<\/p>\r\n

[latex]\\begin{align} \\hfill &\\sin^{\u22121}(\\sin x)=x\\text{ only for }\u2212\\frac{\\pi}{2} \\leq x \\leq \\frac{\\pi}{2} \\hfill \\\\ &\\cos^{\u22121}(\\cos x)=x\\text{ only for }0\\leq x\\leq\\pi \\hfill \\\\ &\\tan^{\u22121}(\\tan x)=x\\text{ only for }\u2212\\frac{\\pi}{2}\\text{ < }x\\text{ < }\\frac{\\pi}{2} \\end{align}[\/latex]<\/p>\r\n\r\n<\/section>

Is it correct that [latex]\\sin^{\u22121}(\\sin x)=x[\/latex]?<\/strong><\/strong> \r\n\r\nNo. This equation is correct if x belongs to the restricted domain [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], but sine is defined for all real input values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex]. The situation is similar for cosine and tangent and their inverses. For example, [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{3\\pi}{4}\\right)\\right)=\\frac{\\pi}{4}[\/latex].<\/section>
How To: <\/strong>Given an expression of the form [latex]f^{\u22121}(f(\\theta))[\/latex] where [latex]f(\\theta)=\\sin\\theta\\text{, }\\cos\\theta\\text{, or }\\tan\\theta[\/latex], evaluate.<\/strong>\r\n
    \r\n \t
  1. If \u03b8 is in the restricted domain of f<\/em>,\u00a0then [latex]f^{\u22121}(f(\\theta))=\\theta[\/latex].<\/li>\r\n \t
  2. If not, then find an angle \u03d5 within the restricted domain of f<\/em> such that [latex]f(\\phi)=f(\\theta)[\/latex]. Then [latex]f^{\u22121}(f(\\theta))=\\phi[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
    Evaluate the following:\r\n
      \r\n \t
    1. [latex]\\sin^{\u22121}(\\sin(\\frac{\\pi}{3}))[\/latex]<\/li>\r\n \t
    2. [latex]\\sin^{\u22121}(\\sin(\\frac{2\\pi}{3}))[\/latex]<\/li>\r\n \t
    3. [latex]\\cos^{\u22121}(\\cos(\\frac{2\\pi}{3}))[\/latex]<\/li>\r\n \t
    4. [latex]\\cos^{\u22121}(\\cos(\u2212\\frac{\\pi}{3}))[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"611200\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"611200\"]\r\n
        \r\n \t
      1. [latex]\\frac{\\pi}{3}[\/latex] is in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], so [latex]\\sin^{\u22121}(\\sin(\\frac{\\pi}{3}))=\\frac{\\pi}{3}[\/latex].<\/li>\r\n \t
      2. [latex]\\frac{2\\pi}{3}[\/latex] is not in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], but [latex]\\sin\\left(\\frac{2\\pi}{3}\\right)=\\sin\\left(\\frac{\\pi}{3}\\right)[\/latex], so [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{2\\pi}{3}\\right)\\right)=\\frac{\\pi}{3}[\/latex].<\/li>\r\n \t
      3. [latex]\\frac{2\\pi}{3}[\/latex] is in [0,\u03c0], so [latex]\\cos^{\u22121}\\left(\\cos\\left(\\frac{2\\pi}{3}\\right)\\right)=\\frac{2\\pi}{3}[\/latex].<\/li>\r\n \t
      4. [latex]\u2212\\frac{\\pi}{3}[\/latex] is not in [0,\u03c0], but [latex]\\cos(\u2212\\frac{\\pi}{3})=\\cos\\left(\\frac{\\pi}{3}\\right)[\/latex] because cosine is an even function.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
        \r\n
        \r\n\r\nEvaluate [latex]\\tan^{\u22121}\\left(\\tan\\left(\\frac{\\pi}{8}\\right)\\right)[\/latex] and [latex]\\tan^{\u22121}\\left(\\tan\\left(\\frac{11\\pi}{9}\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"356884\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"356884\"]\r\n\r\n[latex]\\frac{\\pi}{8}\\text{; }\\frac{2\\pi}{9}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n

        Evaluating Compositions of the Form\u00a0[latex]f^{\u22121}(g(x))[\/latex]<\/h2>\r\nNow that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form [latex]f^{\u22121}(g(x))[\/latex]. For special values of x<\/em>, we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is \u03b8, making the other [latex]\\frac{\\pi}{2}\u2212\\theta[\/latex]. Consider the sine and cosine of each angle of the right triangle.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"An Right triangle illustrating the cofunction relationships[\/caption]\r\n\r\nBecause [latex]\\cos\\theta=\\frac{b}{c}=\\sin\\left(\\frac{\\pi}{2}\u2212\\theta\\right)[\/latex], we have [latex]\\sin^{\u22121}(\\cos\\theta)=\\frac{\\pi}{2}\u2212\\theta\\text{ if }0\\leq\\theta\\leq\\pi[\/latex]. If \u03b8 is not in this domain, then we need to find another angle that has the same cosine as \u03b8 and does belong to the restricted domain; we then subtract this angle from [latex]\\frac{\\pi}{2}[\/latex]. Similarly, [latex]\\sin\\theta=\\frac{a}{c}=\\cos\\left(\\frac{\\pi}{2}\u2212\\theta\\right)[\/latex], so [latex]\\cos^{\u22121}(\\sin\\theta)=\\frac{\\pi}{2}\u2212\\theta\\text{ if }\u2212\\frac{\\pi}{2}\\leq\\theta\\leq\\frac{\\pi}{2}[\/latex]. These are just the function-cofunction relationships presented in another way.\r\n\r\n
        How To:\u00a0Given functions of the form [latex]\\sin^{\u22121}(\\cos x)\\text{ and }\\cos^{\u22121}(\\sin x)[\/latex], evaluate them.<\/strong>\r\n
          \r\n \t
        1. If x<\/em>\u00a0is\u00a0in\u00a0[0,\u03c0], then [latex]\\sin^{\u22121}(\\cos x)=\\frac{\\pi}{2}\u2212x[\/latex].<\/li>\r\n \t
        2. If x<\/em>\u00a0is\u00a0not\u00a0in\u00a0[0,\u03c0], then find another angle y<\/em>\u00a0in\u00a0[0,\u03c0] such that [latex]\\cos y=\\cos x[\/latex].\r\n
          \r\n
          [latex]\\sin^{\u22121}(\\cos x)=\\frac{\\pi}{2}\u2212y[\/latex]<\/div>\r\n<\/div><\/li>\r\n \t
        3. If x<\/em>\u00a0is\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], then [latex]\\cos^{\u22121}(\\sin x)=\\frac{\\pi}{2}\u2212x[\/latex].<\/li>\r\n \t
        4. If x<\/em>\u00a0is\u00a0not\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], then find another angle y<\/em>\u00a0in [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex] such that [latex]\\sin y=\\sin x[\/latex].\r\n
          \r\n
          [latex]\\cos^{\u22121}(\\sin x)=\\frac{\\pi}{2}\u2212y[\/latex]<\/div>\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/section>
          Evaluate [latex]\\sin^{\u22121}(\\cos(\\frac{13\\pi}{6}))[\/latex]\r\n
            \r\n \t
          1. by direct evaluation.<\/li>\r\n \t
          2. by the method described previously.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"651517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"651517\"]\r\n
              \r\n \t
            1. Here, we can directly evaluate the inside of the composition.\r\n
              [latex]\\begin{align}\\cos\\left(\\frac{13\\pi}{6}\\right)&=\\cos\\left(\\frac{\\pi}{6}+2\\pi\\right) \\\\ &=\\cos\\left(\\frac{\\pi}{6}\\right) \\\\ &=\\frac{\\sqrt{3}}{2} \\end{align}[\/latex]<\/div>\r\nNow, we can evaluate the inverse function as we did earlier.\r\n
              [latex]\\sin^{\u22121}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\pi}{3}[\/latex]<\/div><\/li>\r\n \t
            2. We have [latex]x=\\frac{13\\pi}{6}[\/latex], [latex]y=\\frac{\\pi}{6}[\/latex], and\r\n
              [latex]\\begin{align}\\sin^{\u22121}\\left(\\cos\\left(\\frac{13\\pi}{6}\\right)\\right)=\\frac{\\pi}{2}\u2212\\frac{\\pi}{6} =\\frac{\\pi}{3} \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
              Evaluate [latex]\\cos^{\u22121}(\\sin(\u2212\\frac{11\\pi}{4}))[\/latex].[reveal-answer q=\"325594\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"325594\"][latex]\\frac{3\\pi}{4}[\/latex][\/hidden-answer]<\/section>
              [ohm_question hide_question_numbers=1]129738[\/ohm_question]<\/section>\r\n

              Evaluating Compositions of the Form [latex]f(g^{\u22121}(x))[\/latex]<\/h2>\r\nTo evaluate compositions of the form [latex]f(g^{\u22121}(x))[\/latex], where f<\/em> and g<\/em> are any two of the functions sine, cosine, or tangent and x<\/em> is any input in the domain of [latex]g\u22121[\/latex], we have exact formulas, such as [latex]\\sin\\left({\\cos}^{\u22121}x\\right)=\\sqrt{1\u2212{x}^{2}}[\/latex]. When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras\u2019s relation between the lengths of the sides. We can use the Pythagorean identity, [latex]\\sin^{2}x+cos^{2}x=1[\/latex], to solve for one when given the other. We can also use the inverse trigonometric functions<\/strong> to find compositions involving algebraic expressions.\r\n\r\n
              Find an exact value for [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)[\/latex].[reveal-answer q=\"530604\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530604\"]Beginning with the inside, we can say there is some angle such that [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex], which means [latex]\\cos\\theta=\\frac{4}{5}[\/latex], and we are looking for [latex]\\sin\\theta[\/latex]. We can use the Pythagorean identity to do this.\r\n

              [latex]\\begin{align} &\\sin^{2}\\theta+\\cos^{2}\\theta=1 && \\text{Use our known value for cosine.} \\\\ &\\sin^{2}\\theta+\\left(\\frac{4}{5}\\right)^{2}=1 && \\text{Solve for sine.} \\\\ &\\sin^{2}\\theta=1\u2212\\frac{16}{25} \\\\ &\\sin\\theta=\\pm\\sqrt{\\frac{9}{25}}=\\pm\\frac{3}{5} \\end{align}[\/latex]<\/p>\r\nSince [latex]\\theta=\\cos^{\u22121}(\\frac{4}{5})[\/latex] is in quadrant I, [latex]\\sin{\\theta}[\/latex] must be positive, so the solution is [latex]\\frac{3}{5}[\/latex].\r\n

              \"An<\/p>\r\n

              \u00a0Right triangle illustrating that if [latex]\\cos\\theta=\\frac{4}{5}[\/latex], then [latex]\\sin\\theta=\\frac{3}{5}[\/latex]<\/p>\r\nWe know that the inverse cosine always gives an angle on the interval [0,\u00a0\u03c0], so we know that the sine of that angle must be positive; therefore [latex]\\sin\\left(\\cos^{\u22121}\\left(\\frac{4}{5}\\right)\\right)=\\sin\\theta=\\frac{3}{5}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

              Evaluate [latex]\\cos(\\tan^{\u22121}(\\frac{5}{12}))[\/latex].[reveal-answer q=\"754416\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"754416\"][latex]\\frac{12}{13}[\/latex][\/hidden-answer]<\/section>
              \r\n

              Example 8: Evaluating the Composition of a Sine with an Inverse Tangent<\/h3>\r\nFind an exact value for [latex]\\sin\\left(\\tan^{\u22121}\\left(\\frac{7}{4}\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"488532\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"488532\"]\r\n\r\nWhile we could use a similar technique as in Example 6, we will demonstrate a different technique here. From the inside, we know there is an angle such that [latex]\\tan\\theta=\\frac{7}{4}[\/latex]. We can envision this as the opposite and adjacent sides on a right triangle.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"An A right triangle with two sides known[\/caption]\r\n\r\nUsing the Pythagorean Theorem, we can find the hypotenuse of this triangle.\r\n

              [latex]\\begin{gathered}4^{2}+7^{2}=\\text{hypotenuse}^{2} \\\\ \\text{hypotenuse}=\\sqrt{65} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

              \r\n
              \r\n\r\nEvaluate [latex]\\cos(\\sin^{\u22121}(\\frac{7}{9}))[\/latex].\r\n\r\n[reveal-answer q=\"930711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"930711\"]\r\n\r\n[latex]\\frac{4\\sqrt{2}}{9}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
              [ohm_question hide_question_numbers=1]129751[\/ohm_question]<\/section>
              Find a simplified expression for [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{x}{3}\\right)\\right)[\/latex] for [latex]\u22123\\leq x\\leq3[\/latex].[reveal-answer q=\"764834\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"764834\"]We know there is an angle\u00a0\u03b8 such that [latex]\\sin\\theta=\\frac{x}{3}\\\\[\/latex]\r\n

              [latex]\\begin{align}&\\sin^{2}\\theta+\\cos^{2}\\theta=1 7&& \\text{Use the Pythagorean Theorem.} \\\\ &\\left(\\frac{x}{3}\\right)^{2}+\\cos^{2}+\\cos^2\\theta=1 && \\text{Solve for cosine.} \\\\ &\\cos^{2}\\theta=1\u2212\\frac{x^{2}}{9} \\\\ &\\cos\\theta=\\pm\\sqrt{\\frac{9\u2212x^{2}}{9}}=\\pm\\frac{\\sqrt{9\u2212x^{2}}}{3} \\end{align}[\/latex]<\/p>\r\nBecause we know that the inverse sine must give an angle on the interval [latex]\\left[\u2212\\frac{\\pi}{2}\\text{, }\\frac{\\pi}{2}\\right][\/latex], we can deduce that the cosine of that angle must be positive.\r\n

              [latex]\\cos\\left(\\sin^{\u22121}\\left(\\frac{x}{3}\\right)\\right)=\\frac{\\sqrt{9\u2212x^{2}}}{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

              Find a simplified expression for [latex]\\sin\\left(\\tan^{\u22121}\\left(4x\\right)\\right)\\\\[\/latex] for [latex]\u2212\\frac{1}{4}\\leq x \\leq\\frac{1}{4}[\/latex].[reveal-answer q=\"979016\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979016\"][latex]\\frac{4x}{\\sqrt{16x^{2}+1}}[\/latex][\/hidden-answer]<\/section>
              [ohm_question hide_question_numbers=1]129755[\/ohm_question]<\/section>","rendered":"

              Finding Exact Values of Composite Functions with Inverse Trigonometric Functions<\/h2>\n

              There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let f<\/em>(x<\/em>) and g<\/em>(x<\/em>) be two different trigonometric functions belonging to the set {sin(x<\/em>), cos(x<\/em>), tan(x<\/em>)} and let [latex]f^{\u22121}(y)[\/latex] and [latex]g^{\u22121}(y)[\/latex] be their inverses.<\/p>\n

              Evaluating Compositions of the Form [latex]f\\left(f^{\u22121}(y)\\right)[\/latex] and [latex]f^{\u22121}(f(x))[\/latex]<\/h3>\n

              For any trigonometric function, [latex]f(f^{\u22121}(y))=y[\/latex] for all y<\/em> in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f<\/em> was defined to be identical to the domain of [latex]f^{\u22121}[\/latex]. However, we have to be a little more careful with expressions of the form [latex]f^{\u22121}(f(x))[\/latex].<\/p>\n

              \n

              compositions of a trigonometric function and its inverse<\/h3>\n

              [latex]\\begin{align} &\\sin(\\sin^{\u22121}x)=x\\text{ for }\u22121\\leq x\\leq1\\\\ &\\cos(\\cos^{\u22121}x)=x\\text{ for }\u22121\\leq x\\leq1 \\\\ &\\tan(\\tan^{\u22121}x)=x\\text{ for }\u2212\\infty\\text{ < }x\\text{ < }\\infty \\end{align}[\/latex]<\/p>\n

              [latex]\\begin{align} \\hfill &\\sin^{\u22121}(\\sin x)=x\\text{ only for }\u2212\\frac{\\pi}{2} \\leq x \\leq \\frac{\\pi}{2} \\hfill \\\\ &\\cos^{\u22121}(\\cos x)=x\\text{ only for }0\\leq x\\leq\\pi \\hfill \\\\ &\\tan^{\u22121}(\\tan x)=x\\text{ only for }\u2212\\frac{\\pi}{2}\\text{ < }x\\text{ < }\\frac{\\pi}{2} \\end{align}[\/latex]<\/p>\n<\/section>\n

              Is it correct that [latex]\\sin^{\u22121}(\\sin x)=x[\/latex]?<\/strong><\/strong> <\/p>\n

              No. This equation is correct if x belongs to the restricted domain [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex], but sine is defined for all real input values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in [latex]\\left[\u2212\\frac{\\pi}{2},\\frac{\\pi}{2}\\right][\/latex]. The situation is similar for cosine and tangent and their inverses. For example, [latex]\\sin^{\u22121}\\left(\\sin\\left(\\frac{3\\pi}{4}\\right)\\right)=\\frac{\\pi}{4}[\/latex].<\/section>\n

              How To: <\/strong>Given an expression of the form [latex]f^{\u22121}(f(\\theta))[\/latex] where [latex]f(\\theta)=\\sin\\theta\\text{, }\\cos\\theta\\text{, or }\\tan\\theta[\/latex], evaluate.<\/strong><\/p>\n
                \n
              1. If \u03b8 is in the restricted domain of f<\/em>,\u00a0then [latex]f^{\u22121}(f(\\theta))=\\theta[\/latex].<\/li>\n
              2. If not, then find an angle \u03d5 within the restricted domain of f<\/em> such that [latex]f(\\phi)=f(\\theta)[\/latex]. Then [latex]f^{\u22121}(f(\\theta))=\\phi[\/latex].<\/li>\n<\/ol>\n<\/section>\n
                Evaluate the following:<\/p>\n
                  \n
                1. [latex]\\sin^{\u22121}(\\sin(\\frac{\\pi}{3}))[\/latex]<\/li>\n
                2. [latex]\\sin^{\u22121}(\\sin(\\frac{2\\pi}{3}))[\/latex]<\/li>\n
                3. [latex]\\cos^{\u22121}(\\cos(\\frac{2\\pi}{3}))[\/latex]<\/li>\n
                4. [latex]\\cos^{\u22121}(\\cos(\u2212\\frac{\\pi}{3}))[\/latex]<\/li>\n<\/ol>\n