1. **Problem:** Find $\sin^{-1}(\frac{1}{2})$. Step 1: Recall that $\sin^{-1}(x)$ is the angle whose sine is $x$. Step 2: Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, we have $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$. 2. **Problem:** Find $\cos^{-1}(0)$. Step 1: $\cos^{-1}(x)$ is the angle whose cosine is $x$. Step 2: $\cos(\frac{\pi}{2}) = 0$, so $\cos^{-1}(0) = \frac{\pi}{2}$. 3. **Problem:** Find $\tan^{-1}(1)$. Step 1: $\tan^{-1}(x)$ is the angle whose tangent is $x$. Step 2: $\tan(\frac{\pi}{4}) = 1$, so $\tan^{-1}(1) = \frac{\pi}{4}$. 4. **Problem:** Find $\sin^{-1}(-\frac{\sqrt{3}}{2})$. Step 1: $\sin^{-1}(x)$ returns values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Step 2: $\sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$, so $\sin^{-1}(-\frac{\sqrt{3}}{2}) = -\frac{\pi}{3}$. 5. **Problem:** Find $\cos^{-1}(-1)$. Step 1: $\cos^{-1}(x)$ returns values in $[0, \pi]$. Step 2: $\cos(\pi) = -1$, so $\cos^{-1}(-1) = \pi$. 6. **Problem:** Find $\tan^{-1}(0)$. Step 1: $\tan^{-1}(0)$ is the angle whose tangent is 0. Step 2: $\tan(0) = 0$, so $\tan^{-1}(0) = 0$. 7. **Problem:** Find $\sin(\cos^{-1}(\frac{1}{2}))$. Step 1: Let $\theta = \cos^{-1}(\frac{1}{2})$, so $\cos \theta = \frac{1}{2}$. Step 2: Using $\sin^2 \theta + \cos^2 \theta = 1$, we get $\sin \theta = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$. Step 3: Since $\theta \in [0, \pi]$, $\sin \theta \geq 0$, so $\sin(\cos^{-1}(\frac{1}{2})) = \frac{\sqrt{3}}{2}$. 8. **Problem:** Find $\cos(\sin^{-1}(\frac{3}{5}))$. Step 1: Let $\alpha = \sin^{-1}(\frac{3}{5})$, so $\sin \alpha = \frac{3}{5}$. Step 2: Using $\sin^2 \alpha + \cos^2 \alpha = 1$, $\cos \alpha = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$. Step 3: Since $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $\cos \alpha \geq 0$, so $\cos(\sin^{-1}(\frac{3}{5})) = \frac{4}{5}$. 9. **Problem:** Find $\tan(\cos^{-1}(\frac{4}{5}))$. Step 1: Let $\beta = \cos^{-1}(\frac{4}{5})$, so $\cos \beta = \frac{4}{5}$. Step 2: Using $\sin^2 \beta + \cos^2 \beta = 1$, $\sin \beta = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$. Step 3: $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{3/5}{4/5} = \frac{3}{4}$. 10. **Problem:** Find $\sin^{-1}(\sin(\frac{7\pi}{6}))$. Step 1: $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$. Step 2: $\sin^{-1}(x)$ returns values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Step 3: Since $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$, we have $\sin^{-1}(\sin(\frac{7\pi}{6})) = -\frac{\pi}{6}$.