Question: User: 10/21/26 HW angle 9. Sin^-1(sin 5\pi/4) sin^-1 225° -180 sin 45° sin^-1(\sqrt{2}/2) = -\sqrt{2}/2 45\cdot\pi/180 = \pi/4 ratio 10. sin[cos^-1(-1/2)] 120° 180° - 60° = 120° sin(120) sin 60 = \sqrt{3}/2 angle 11. Cos^-1(cos(-2\pi/3)) between 0 and 180 60° cos^-120° 180 - 120 = 60 cos^-1(-1/2) = 120\cdot\pi/180 = 2\pi/3 top-left: The number line / quadrant sketch for problem 9 is in the left-middle area; it shows the angle in Quadrant II and the final answer \pi/4 boxed near the lower-right of that sketch. top-right: The coordinate/quadrant sketch for problem 10 is in the upper-right area; it shows an angle of 120° and the final value \sqrt{3}/2 boxed to the right. bottom-left: The coordinate/quadrant sketch for problem 11 is in the lower-left area; it shows an angle of 60°/120° reasoning and the final answer 2\pi/3 boxed near the lower-right of that sketch.Are these correct?

1. **Problem 9: Evaluate $\sin^{-1}(\sin \frac{5\pi}{4})$.** 2. The angle $\frac{5\pi}{4}$ radians equals $225^\circ$, which lies in the third quadrant. 3. The function $\sin^{-1}(x)$ (arcsin) returns values in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (Quadrants I and IV). 4. Since $\sin(\frac{5\pi}{4}) = \sin(225^\circ) = -\frac{\sqrt{2}}{2}$, we need to find an angle $\theta$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin \theta = -\frac{\sqrt{2}}{2}$. 5. That angle is $\theta = -\frac{\pi}{4}$ (or $-45^\circ$), because $\sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$. 6. Therefore, $\sin^{-1}(\sin \frac{5\pi}{4}) = -\frac{\pi}{4}$. --- 1. **Problem 10: Evaluate $\sin\left(\cos^{-1}\left(-\frac{1}{2}\right)\right)$.** 2. Let $\alpha = \cos^{-1}\left(-\frac{1}{2}\right)$. Since $\cos \alpha = -\frac{1}{2}$ and $\alpha \in [0, \pi]$, $\alpha = \frac{2\pi}{3}$ (or $120^\circ$). 3. Using the Pythagorean identity: $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. 4. Therefore, $\sin\left(\cos^{-1}\left(-\frac{1}{2}\right)\right) = \frac{\sqrt{3}}{2}$. --- 1. **Problem 11: Evaluate $\cos^{-1}(\cos(-\frac{2\pi}{3}))$.** 2. The angle $-\frac{2\pi}{3}$ is $-120^\circ$, which is outside the principal range of $\cos^{-1}$, which is $[0, \pi]$. 3. Since cosine is even, $\cos(-\frac{2\pi}{3}) = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$. 4. The angle in $[0, \pi]$ with cosine $-\frac{1}{2}$ is $\frac{2\pi}{3}$ (or $120^\circ$). 5. Therefore, $\cos^{-1}(\cos(-\frac{2\pi}{3})) = \frac{2\pi}{3}$. --- **Summary:** - Problem 9 answer: $-\frac{\pi}{4}$ (not $\frac{\pi}{4}$). - Problem 10 answer: $\frac{\sqrt{3}}{2}$ (correct). - Problem 11 answer: $\frac{2\pi}{3}$ (correct).