Inverse Trigonometric Functions

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Given \(\sin (\frac{5\pi }{12})\approx 0.96593,\) write a relation involving the inverse sine.

   Откриј одговор.

   Use the relation for the inverse sine. If \(\sin \ y=x,\) then \({\sin }^{-1}x=y\).

   In this problem, \(x=0.96593,\) and \(y=\frac{5\pi }{12}.\)

   \[{\sin }^{-1}(0.96593)\approx \frac{5\pi }{12}\]

2. Given \(\cos (0.5)\approx 0.8776,\) write a relation involving the inverse cosine.

   Откриј одговор.

   \(\text{arccos}(0.8776)\approx 0.5\)

3. Evaluate each of the following.

   1. ⓐ \({\text{sin}}^{-1}(\frac{1}{2})\)
   2. ⓑ \({\text{sin}}^{-1}(-\frac{\sqrt{2}}{2})\)
   3. ⓒ \({\cos }^{-1}(-\frac{\sqrt{3}}{2})\)
   4. ⓓ \({\tan }^{-1}(1)\)
   Откриј одговор.

   1. ⓐ Evaluating \({\sin }^{-1}(\frac{1}{2})\) is the same as determining the angle that would have a sine value of \(\frac{1}{2}.\) In other words, what angle \(x\) would satisfy \(\sin (x)=\frac{1}{2}?\) There are multiple values that would satisfy this relationship, such as \(\frac{\pi }{6}\) and \(\frac{5\pi }{6},\) but we know we need the angle in the interval \([-\frac{\pi }{2},\frac{\pi }{2}],\) so the answer will be \({\sin }^{-1}(\frac{1}{2})=\frac{\pi }{6}.\) Remember that the inverse is a function, so for each input, we will get exactly one output.
   2. ⓑ To evaluate \({\sin }^{-1}(-\frac{\sqrt{2}}{2}),\) 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 \([-\frac{\pi }{2},\frac{\pi }{2}].\) For that, we need the negative angle coterminal with \(\frac{7\pi }{4}:\) \({\text{sin}}^{-1}(-\frac{\sqrt{2}}{2})=-\frac{\pi }{4}.\)
   3. ⓒTo evaluate \({\cos }^{-1}(-\frac{\sqrt{3}}{2}),\) we are looking for an angle in the interval \([0,\pi ]\) with a cosine value of \(-\frac{\sqrt{3}}{2}.\) The angle that satisfies this is \({\cos }^{-1}(-\frac{\sqrt{3}}{2})=\frac{5\pi }{6}.\)
   4. ⓓ Evaluating \({\tan }^{-1}(1),\) we are looking for an angle in the interval \((-\frac{\pi }{2},\frac{\pi }{2})\) with a tangent value of 1. The correct angle is \({\tan }^{-1}(1)=\frac{\pi }{4}.\)

4. Evaluate each of the following.

   1. ⓐ \({\text{sin}}^{-1}(-1)\)
   2. ⓑ \({\tan }^{-1}(-1)\)
   3. ⓒ \({\cos }^{-1}(-1)\)
   4. ⓓ \({\cos }^{-1}(\frac{1}{2})\)
   Откриј одговор.

   1. ⓐ\(-\frac{\pi }{2};\)
   2. ⓑ \(-\frac{\pi }{4};\)
   3. ⓒ \(\pi ;\)
   4. ⓓ \(\frac{\pi }{3}\)

5. Evaluate \({\sin }^{-1}(0.97)\) using a calculator.

   Откриј одговор.

   Because the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using.

   In radian mode, \({\sin }^{-1}(0.97)\approx 1.3252.\) In degree mode, \({\sin }^{-1}(0.97)\approx 75.93^{\circ}.\) Note that in calculus and beyond we will use radians in almost all cases.

6. Evaluate \({\cos }^{-1}(-0.4)\) using a calculator.

   Откриј одговор.

   1.9823 or 113.578°

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

   Откриј одговор.

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

   \[\begin{array}{llll}\cos \ \theta =\frac{9}{12} & \begin{array}{lll} & & \end{array} \\ \theta ={\cos }^{-1}(\frac{9}{12}) & \begin{array}{lll} & & \end{array}\text{Apply definition of the inverse}. \\ \theta \approx 0.7227\text{ or about }41.4096^{\circ} & \begin{array}{lll} & & \end{array}\text{Evaluate}.\end{array}\]

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

   Откриј одговор.

   \({\sin }^{-1}(0.6)=36.87^{\circ}=0.6435\) radians

9. Evaluate the following:

   1. ⓐ \({\sin }^{-1}(\sin (\frac{\pi }{3}))\)
   2. ⓑ \({\sin }^{-1}(\sin (\frac{2\pi }{3}))\)
   3. ⓒ \({\cos }^{-1}(\cos (\frac{2\pi }{3}))\)
   4. ⓓ \({\cos }^{-1}(\cos (-\frac{\pi }{3}))\)
   Откриј одговор.

   1. ⓐ \(\frac{\pi }{3}\text{ is in }[-\frac{\pi }{2},\frac{\pi }{2}],\) so \({\sin }^{-1}(\sin (\frac{\pi }{3}))=\frac{\pi }{3}.\)
   2. ⓑ \(\frac{2\pi }{3}\text{ is not in }[-\frac{\pi }{2},\frac{\pi }{2}],\) but \(\sin (\frac{2\pi }{3})=\sin (\frac{\pi }{3}),\) so \({\sin }^{-1}(\sin (\frac{2\pi }{3}))=\frac{\pi }{3}.\)
   3. ⓒ \(\frac{2\pi }{3}\text{ is in }[0,\pi ],\) so \({\cos }^{-1}(\cos (\frac{2\pi }{3}))=\frac{2\pi }{3}.\)
   4. ⓓ \(-\frac{\pi }{3}\text{ is not in }[0,\pi ],\) but \(\cos (-\frac{\pi }{3})=\cos (\frac{\pi }{3})\) because cosine is an even function. \(\frac{\pi }{3}\text{ is in }[0,\pi ],\) so \({\cos }^{-1}(\cos (-\frac{\pi }{3}))=\frac{\pi }{3}.\)

10. Evaluate \({\tan }^{-1}(\tan (\frac{\pi }{8}))\ \text{and}\ {\tan }^{-1}(\tan (\frac{11\pi }{9})).\)

    Откриј одговор.

    \(\frac{\pi }{8};\frac{2\pi }{9}\)

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

    1. ⓐby direct evaluation.
    2. ⓑ by the method described previously.
    Откриј одговор.

    1. ⓐ Here, we can directly evaluate the inside of the composition. \[\begin{array}{l} \\ \begin{array}{l}\cos (\frac{13\pi }{6})=\cos (\frac{\pi }{6}+2\pi ) \\ =\cos (\frac{\pi }{6}) \\ =\frac{\sqrt{3}}{2}\end{array}\end{array}\]

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

       \[{\sin }^{-1}(\frac{\sqrt{3}}{2})=\frac{\pi }{3}\]
    2. ⓑ We have \(x=\frac{13\pi }{6}\text{,}\ y=\frac{\pi }{6},\) and \[\begin{array}{l}{\sin }^{-1}(\cos (\frac{13\pi }{6}))=\frac{\pi }{2}-\frac{\pi }{6} \\ =\frac{\pi }{3}\ \end{array}\]

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

    Откриј одговор.

    \(\frac{3\pi }{4}\)

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

    Откриј одговор.

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

    \[\begin{array}{llll}{\sin }^{2}\theta +{\cos }^{2}\theta =1 & & & \text{Use our known value for cosine}. \\ {\sin }^{2}\theta +{(\frac{4}{5})}^{2}=1 & & & \text{Solve for sine}. \\ {\sin }^{2}\theta =1-\frac{16}{25} & & & \\ \sin \ \theta =\pm \sqrt{\frac{9}{25}}=\pm \frac{3}{5} & & & \end{array}\]

    Since \(\theta ={\cos }^{-1}(\frac{4}{5})\) is in quadrant I, \(\sin \ \theta\) must be positive, so the solution is \(\frac{3}{5}.\) See .

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

14. Evaluate \(\cos ({\tan }^{-1}(\frac{5}{12})).\)

    Откриј одговор.

    \(\frac{12}{13}\)

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

    Откриј одговор.

    While we could use a similar technique as in , we will demonstrate a different technique here. From the inside, we know there is an angle such that \(\tan \ \theta =\frac{7}{4}.\) We can envision this as the opposite and adjacent sides on a right triangle, as shown in .

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

    \[\begin{array}{l}\begin{array}{l} \\ {4}^{2}+{7}^{2}={\text{hypotenuse}}^{2}\end{array} \\ \text{hypotenuse}=\sqrt{65}\end{array}\]

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

    \[\sin \ \theta =\frac{7}{\sqrt{65}}\]

    This gives us our desired composition.

    \[\begin{array}{l}\sin ({\tan }^{-1}(\frac{7}{4}))=\sin \ \theta \\ =\frac{7}{\sqrt{65}} \\ =\frac{7\sqrt{65}}{65}\end{array}\]

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

    Откриј одговор.

    \(\frac{4\sqrt{2}}{9}\)

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

    Откриј одговор.

    We know there is an angle \(\theta\) such that \(\sin \ \theta =\frac{x}{3}.\)

    \[\begin{array}{ll}{\sin }^{2}\theta +{\cos }^{2}\theta =1 & \text{Use the Pythagorean Theorem}. \\ {(\frac{x}{3})}^{2}+{\cos }^{2}\theta =1 & \text{Solve for cosine}. \\ {\cos }^{2}\theta =1-\frac{{x}^{2}}{9} & \\ \cos \theta =\pm \sqrt{\frac{9-{x}^{2}}{9}}=\pm \frac{\sqrt{9-{x}^{2}}}{3} & \end{array}\]

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

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

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

    Откриј одговор.

    \(\frac{4x}{\sqrt{16{x}^{2}+1}}\)

19. Why do the functions \(f(x)={\sin }^{-1}x\) and \(g(x)={\cos }^{-1}x\) have different ranges?

    Откриј одговор.

    The function \(y=\sin x\) is one-to-one on \([-\frac{\pi }{2},\frac{\pi }{2}];\) thus, this interval is the range of the inverse function of \(y=\sin x,\) \(f(x)={\sin }^{-1}x.\) The function \(y=\cos x\) is one-to-one on \([0,\pi ];\) thus, this interval is the range of the inverse function of \(y=\cos x,f(x)={\cos }^{-1}x.\)

20. Since the functions \(y=\cos \ x\) and \(y={\cos }^{-1}x\) are inverse functions, why is \({\cos }^{-1}(\cos (-\frac{\pi }{6}))\) not equal to \(-\frac{\pi }{6}?\)

21. Explain the meaning of \(\frac{\pi }{6}=\text{arcsin}(0.5).\)

    Откриј одговор.

    \(\frac{\pi }{6}\) is the radian measure of an angle between \(-\frac{\pi }{2}\) and \(\frac{\pi }{2}\) whose sine is 0.5.

22. Most calculators do not have a key to evaluate \({\text{sec}}^{-1}(2).\) Explain how this can be done using the cosine function or the inverse cosine function.

23. Why must the domain of the sine function, \(\sin \ x,\) be restricted to \([-\frac{\pi }{2},\frac{\pi }{2}]\) for the inverse sine function to exist?

    Откриј одговор.

    In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval \([-\frac{\pi }{2},\frac{\pi }{2}]\) so that it is one-to-one and possesses an inverse.

24. Discuss why this statement is incorrect: \(\text{arccos}(\cos \ x)=x\) for all \(x.\)

25. Determine whether the following statement is true or false and explain your answer: \(\text{arccos}(-x)=\pi -\text{arccos}\ x.\)

    Откриј одговор.

    True . The angle, \({\theta }_{1}\) that equals \(\text{arccos}(-x)\), \(x>0\), will be a second quadrant angle with reference angle, \({\theta }_{2}\), where \({\theta }_{2}\) equals \(\text{arccos}x\), \(x>0\). Since \({\theta }_{2}\) is the reference angle for \({\theta }_{1}\), \({\theta }_{2}=\pi -{\theta }_{1}\) and \(\text{arccos}(-x)\) = \(\pi -\text{arccos}x\)-

26. \({\sin }^{-1}(\frac{\sqrt{2}}{2})\)

27. \({\sin }^{-1}(-\frac{1}{2})\)

    Откриј одговор.

    \(-\frac{\pi }{6}\)

28. \({\cos }^{-1}(\frac{1}{2})\)

29. \({\cos }^{-1}(-\frac{\sqrt{2}}{2})\)

    Откриј одговор.

    \(\frac{3\pi }{4}\)

30. \({\tan }^{-1}(1)\)

31. \({\tan }^{-1}(-\sqrt{3})\)

    Откриј одговор.

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

32. \({\tan }^{-1}(-1)\)

33. \({\tan }^{-1}(\sqrt{3})\)

    Откриј одговор.

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

34. \({\tan }^{-1}(\frac{-1}{\sqrt{3}})\)

35. \({\cos }^{-1}(-0.4)\)

    Откриј одговор.

    1.98

36. \(\text{arcsin}(0.23)\)

37. \(\text{arccos}(\frac{3}{5})\)

    Откриј одговор.

    0.93

38. \({\cos }^{-1}(0.8)\)

39. \({\tan }^{-1}(6)\)

    Откриј одговор.

    1.41

40. \({\sin }^{-1}(\cos (\pi ))\)

Покушај сам.

Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Више у Trigonometry

The unit circleTrigonometric equationsTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions