evaluate the following expressions. your answer must be an exact angle in radians and in the interval -pi/2…

Explanation:

Step1: Recall inverse - sine values

We know that $\sin^{-1}(x)$ gives an angle $\theta$ such that $\sin(\theta)=x$ and $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$. For $\sin^{-1}(-\frac{\sqrt{3}}{2})$, we know that $\sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$, so $\sin^{-1}(-\frac{\sqrt{3}}{2})=-\frac{\pi}{3}$.

Step2: For $\sin^{-1}(-\frac{\sqrt{2}}{2})$

Since $\sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$, then $\sin^{-1}(-\frac{\sqrt{2}}{2}) = -\frac{\pi}{4}$.

Step3: For $\sin^{-1}(0)$

We know that $\sin(0) = 0$, so $\sin^{-1}(0)=0$.

Step4: For $\sin(\cos^{-1}(\frac{\sqrt{2}}{2}))$

Let $\theta=\cos^{-1}(\frac{\sqrt{2}}{2})$, then $\cos\theta=\frac{\sqrt{2}}{2}$ and $\theta\in[0,\pi]$. So $\theta = \frac{\pi}{4}$, and $\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$.

Step5: For $\tan(\cos^{-1}(1))$

Let $\alpha=\cos^{-1}(1)$, then $\cos\alpha = 1$ and $\alpha\in[0,\pi]$. So $\alpha = 0$, and $\tan(0)=0$.

Step6: Convert 80 degrees to radians

Use the conversion formula $x$ (in radians)=$\frac{\pi}{180}\times x$ (in degrees). So $80\times\frac{\pi}{180}=\frac{4\pi}{9}$.

Step7: Convert $\frac{25\pi}{18}$ to degrees

Use the conversion formula $x$ (in degrees)=$\frac{180}{\pi}\times x$ (in radians). So $\frac{180}{\pi}\times\frac{25\pi}{18}=250$.

Answer:

(a) $-\frac{\pi}{3}$ (b) $-\frac{\pi}{4}$ (c) $0$ (d) $\frac{\sqrt{2}}{2}$ (e) $0$ (f) $\frac{4\pi}{9}$ (g) $250$