## Solutions to Try Its

1. $\arccos(0.8776)\approx0.5$

2. a. $−\frac{\pi}{2}$; b. $−\frac{\pi}{4}$ c. $\pi$ d. $\frac{\pi}{3}$

3. 1.9823 or 113.578°

4. $\sin^{−1}(0.6)=36.87^{\circ}=0.6435$ radians

5. $\frac{\pi}{8}\text{; }\frac{2\pi}{9}$

6. $\frac{3\pi}{4}$

7. $\frac{12}{13}$

8. $\frac{4\sqrt{2}}{9}$

9. $\frac{4x}{\sqrt{16x^{2}+1}}$

## Solutions to Odd-Numbered Exercises

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

3. $\frac{\pi}{6}$ is the radian measure of an angle between $−\frac{\pi}{2}$ and $\frac{\pi}{2}$ whose sine is 0.5.

5. 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 $\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right]$ so that it is one-to-one and possesses an inverse.

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

9. $−\frac{\pi}{6}$

11. $\frac{3\pi}{4}$

13. $−\frac{\pi}{3}$

15. $\frac{\pi}{3}$

17. 1.98

19. 0.93

21. 1.41

25. 0

27. 0.71

29. −0.71

31. $−\frac{\pi}{4}$

33. 0.8

35. $\frac{5}{13}$

37. $\frac{x−1}{\sqrt{−x^{2}+2x}}$

39. $\frac{\sqrt{x^{2}−1}}{x}$

41. $\frac{x+0.5}{\sqrt{−x^{2}−x+\frac{3}{4}}}$

43. $\frac{\sqrt{2x+1}}{x+1}$

45. $\frac{\sqrt{2x+1}}{x+1}$

47. t

49. domain [−1,1]; range [0,π]

51. approximately $x=0.00$