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Mathematics LibreTexts

6.3: Inverse Trigonometric Functions

  • Page ID
    13866
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    In previous sections, we have evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that for a one-to-one function, if \(f(a)=b\), then an inverse function would satisfy \(f^{-1} (b)=a\).

    You probably are already recognizing an issue – that the sine, cosine, and tangent functions are not one-to-one functions. To define an inverse of these functions, we will need to restrict the domain of these functions to yield a new function that is one-to-one. We choose a domain for each function that includes the angle zero.

    屏幕快照 2019-07-09 上午4.16.33.png

    On these restricted domains, we can define the inverse sine, inverse cosine, and inverse tangent functions.

     

    INVERSE SINE, COSINE, AND TANGENT FUNCTIONS

    For angles in the interval \(\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]\), if \(\sin \left(\theta \right)=a\), then \(\sin ^{-1} \left(a\right)=\theta\)

    For angles in the interval \(\left[0,\pi \right]\), if \(\cos \left(\theta \right)=a\), then \(\cos ^{-1} \left(a\right)=\theta\)

    For angles in the interval \(\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)\), if \(\tan \left(\theta \right)=a\), then \(\tan ^{-1} \left(a\right)=\theta\)

     

    \(\sin ^{-1} \left(x\right)\) has domain [-1, 1] and range \(\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]\)

    \(\cos ^{-1} \left(x\right)\) has domain [-1, 1] and range \(\left[0,\pi \right]\)

    \(\tan ^{-1} \left(x\right)\) has domain of all real numbers and range \(\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)\)

     

    The \(\sin ^{-1} \left(x\right)\) is sometimes called the arcsine function, and notated \(\arcsin \left(a\right)\).

    The \(\cos ^{-1} \left(x\right)\) is sometimes called the arccosine function, and notated \(\arccos \left(a\right)\).

    The \(\tan ^{-1} \left(x\right)\) is sometimes called the arctangent function, and notated \(\arctan \left(a\right)\).

    The graphs of the inverse functions are shown here:

    屏幕快照 2019-07-09 上午4.18.43.png

    Notice that the output of each of these inverse functions is an \(angle\).

     

    EXAMPLE 1

    Evaluate

    a) \(\sin ^{-1} \left(\frac{1}{2} \right)\)

    b) \(\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)\)

    c) \(\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)\)

    d) \(\tan ^{-1} \left(1\right)\)

    Solution

    a) Evaluating \(\sin ^{-1} \left(\frac{1}{2} \right)\) is the same as asking what angle would have a sine value of \(\frac{1}{2}\). In other words, what angle \(\theta\) would satisfy \(\sin \left(\theta \right)=\frac{1}{2}\)?

    There are multiple angles that would satisfy this relationship, such as \(\frac{\pi }{6}\) and \(\frac{5\pi }{6}\) , but we know we need the angle in the range of \(\sin ^{-1} \left(x\right)\), the interval \(\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]\), so the answer will be \(\sin ^{-1} \left(\frac{1}{2} \right)=\frac{\pi }{6}\).

    Remember that the inverse is a function so for each input, we will get exactly one output.

    b) Evaluating \(\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)\), we know that \(\frac{5\pi }{4}\) and \(\frac{7\pi }{4}\) both have a sine value of \(-\frac{\sqrt{2} }{2}\), but neither is in the interval \(\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]\). For that, we need the negative angle coterminal with \(\frac{7\pi }{4}\). \(\sin ^{-1} \left(-\frac{\sqrt{2} }{2} \right)=-\frac{\pi }{4}\).

    c) Evaluating \(\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)\), we are looking for an angle in the interval \(\left[0,\pi \right]\) with a cosine value of \(-\frac{\sqrt{3} }{2}\). The angle that satisfies this is \(\cos ^{-1} \left(-\frac{\sqrt{3} }{2} \right)=\frac{5\pi }{6}\).

    d) Evaluating \(\tan ^{-1} \left(1\right)\), we are looking for an angle in the interval \(\left(-\frac{\pi }{2} ,\frac{\pi }{2} \right)\) with a tangent value of 1. The correct angle is \(\tan ^{-1} \left(1\right)=\frac{\pi }{4}\).

     

    TRY IT NOW

    Evaluate

    1. \(\sin ^{-1} \left(-1\right)\)
    2. \(\tan ^{-1} \left(-1\right)\)
    3. \(\cos ^{-1} \left(-1\right)\)
    4. \(\cos ^{-1} \left(\frac{1}{2} \right)\)
    Answer

    a) \(-\frac{\pi }{2}\)

    b) \(-\frac{\pi }{4}\)

    c) \(\pi\)

    d) \(\frac{\pi }{3}\)

     

    EXAMPLE 2

    Evaluate \(\sin ^{-1} \left(0.97\right)\) using your calculator.

    Solution

    Since the output of the inverse function is an angle, your calculator will give you a degree value if in degree mode, and a radian value if in radian mode.

    In radian mode, \(\sin ^{-1} (0.97) \approx 1.3252\) In degree mode, \(\sin ^{-1} \left(0.97\right)\approx 75.93{}^\circ\)

     

    TRY IT NOW

    Evaluate \(\cos ^{-1} \left(-0.4\right)\) using your calculator.

    Answer

    1.9823 or 113.578\(\mathrm{{}^\circ}\)

     

    In Section 5.5, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trig functions, we can solve for the angles of a right triangle given two sides.

     

    EXAMPLE 3

    Solve the triangle for the angle \(\theta\).

    屏幕快照 2019-07-09 上午4.22.15.png

    Solution

    Since we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

    \(\cos \left(\theta \right)=\frac{9}{12}\) Using the definition of the inverse,

    \(\theta =\cos ^{-1} \left(\frac{9}{12} \right)\) Evaluating

    \(\theta \approx 0.7227\), or about 41.4096\(\mathrm{{}^\circ}\)

     

    There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can find exact values for the resulting expressions

     

    EXAMPLE 4

    Evaluate \(\sin ^{-1} \left(\cos \left(\frac{13\pi }{6} \right)\right)\).

    Solution

    a) Here, we can directly evaluate the inside of the composition.

    \(\cos \left(\frac{13\pi }{6} \right)=\frac{\sqrt{3} }{2}\)

    Now, we can evaluate the inverse function as we did earlier.

    \(\sin ^{-1} \left(\frac{\sqrt{3} }{2} \right)=\frac{\pi }{3}\)

     

    TRY IT NOW

    Evaluate \(\cos ^{-1} \left(\sin \left(-\frac{11\pi }{4} \right)\right)\).

    Answer

    \(\sin \left(-\frac{11\pi }{4} \right)=-\frac{\sqrt{2} }{2} . \cos ^{-1} \left(-\frac{\sqrt{2} }{2} \right)=\frac{3\pi }{4} \)

     

    EXAMPLE 5

    Find an exact value for \(\sin \left(\cos ^{-1} \left(\frac{4}{5} \right)\right)\).

    Solution

    Beginning with the inside, we can say there is some angle so \(\theta =\cos ^{-1} \left(\frac{4}{5} \right)\), which means \(\cos \left(\theta \right)=\frac{4}{5}\), and we are looking for \(\sin \left(\theta \right)\). We can use the Pythagorean identity to do this.

    \(\sin ^{2} \left(\theta \right)+\cos ^{2} \left(\theta \right)=1\)                 Using our known value for cosine
    \(\sin ^{2} \left(\theta \right)+\left(\frac{4}{5} \right)^{2} =1\)                 Solving for sine
    \(\sin ^{2} \left(\theta \right)=1-\frac{16}{25}\)
    \(\sin \left(\theta \right)=\pm \sqrt{\frac{9}{25} } =\pm \frac{3}{5}\)

    Since we know that the inverse cosine always gives an angle on the interval \(\left[0,\pi \right]\), we know that the sine of that angle must be positive, so \(\sin \left(\cos ^{-1} \left(\frac{4}{5} \right)\right)=\sin (\theta )=\frac{3}{5}\)

     

    EXAMPLE 6

    Find an exact value for \(\sin \left(\tan ^{-1} \left(\frac{7}{4} \right)\right)\).

    Solution

    While we could use a similar te屏幕快照 2019-07-09 上午4.28.27.pngchnique as in the last example, we will demonstrate a different technique here. From the inside, we know there is an angle so \(\tan \left(\theta \right)=\frac{7}{4}\). We can envision this as the opposite and adjacent sides on a right triangle.

    Using the Pythagorean Theorem, we can find the hypotenuse of this triangle:

    \(4^{2} +7^{2} =hypotenuse ^{2}\) 
    \(hypotenuse=\sqrt{65}\)

    Now, we can represent the sine of the angle as opposite side divided by hypotenuse.

    \(\sin \left(\theta \right)=\frac{7}{\sqrt{65} }\)

    This gives us our desired composition

    \(\sin \left(\tan ^{-1} \left(\frac{7}{4} \right)\right)=\sin (\theta )=\frac{7}{\sqrt{65} } .\)

     

    TRY IT NOW

    Evaluate \(\cos \left(\sin ^{-1} \left(\frac{7}{9} \right)\right)\).

    Answer

    Let \(\theta =\sin ^{-1} \left(\frac{7}{9} \right)\) so \(\sin (\theta )=\frac{7}{9}\). .

    Using Pythagorean Identity, \(\sin ^{2} \theta +\cos ^{2} \theta =1\), so \(\left(\frac{7}{9} \right)^{2} +\cos ^{2} \theta =1\).

    Solving, \(\cos \left(\sin ^{-1} \left(\frac{7}{9} \right)\right)=\cos \left(\theta \right)=\frac{4\sqrt{2} }{9}\) .

     

    We can also find compositions involving algebraic expressions

     

    EXAMPLE 7

    Find a simplified expression for \(\cos \left(\sin ^{-1} \left(\frac{x}{3} \right)\right)\), for \(-3\le x\le 3\).

    Solution

    We know there is an angle \(\theta\) so that \(\sin \left(\theta \right)=\frac{x}{3}\). Using the Pythagorean Theorem,

    \(\sin ^{2} \left(\theta \right)+\cos ^{2} \left(\theta \right)=1\)             Using our known expression for sine
    \(\left(\frac{x}{3} \right)^{2} +\cos ^{2} \left(\theta \right)=1\)             Solving for cosine
    \(\cos ^{2} \left(\theta \right)=1-\frac{x^{2} }{9}\) 
    \(\cos \left(\theta \right)=\pm \sqrt{\frac{9-x^{2} }{9} } =\pm \frac{\sqrt{9-x^{2} } }{3}\) 

    Since we know that the inverse sine must give an angle on the interval \(\left[-\frac{\pi }{2} ,\frac{\pi }{2} \right]\), we can deduce that the cosine of that angle must be positive. This gives us

    \(\cos \left(\sin ^{-1} \left(\frac{x}{3} \right)\right)=\frac{\sqrt{9-x^{2} } }{3}\)

     

     

    TRY IT NOW

    Find a simplified expression for \(\sin \left(\tan ^{-1} \left(4x\right)\right)\), for \(-\frac{1}{4} \le x\le \frac{1}{4}\).

    Answer

    Let \(\theta =\tan ^{-1} \left(4x\right)\), so \(\tan (\theta )=4x\). We can represent this on a triangle as \(\tan (\theta )=\frac{4x}{1}\).屏幕快照 2019-07-09 上午4.31.32.png

    The hypotenuse of the triangle would be \(\sqrt{\left(4x\right)^{2} +1}\). \(\sin \left(\tan ^{-1} \left(4x\right)\right)=\sin (\theta )=\frac{4x}{\sqrt{16x^{2} +1} }\)

    Important Topics of This Section

    • Inverse trig functions: arcsine, arccosine and arctangent
    • Domain restrictions
    • Evaluating inverses using unit circle values and the calculator
    • Simplifying numerical expressions involving the inverse trig functions
    • Simplifying algebraic expressions involving the inverse trig functions