Answer
$arcsin[sin(\frac{9\pi}{4})]=\frac{\pi}{4}$
Work Step by Step
$\frac{9\pi}{4}$ does not lie in the range of $arcsin~x$: ($-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}$). But, we know that the period of $sin~x=2\pi$ and that $\frac{9\pi}{4}=2\pi+\frac{\pi}{4}$. So:
$sin(\frac{9\pi}{4})=sin(\frac{\pi}{4})$. Now, $\frac{\pi}{4}$ lies in the range of $arcsin~x$.
$arcsin[sin(\frac{9\pi}{4})]=\frac{\pi}{4}$
