Answer
$-\dfrac{ \pi}{8}$
Work Step by Step
Value of certain composite functions formula:
$f^{-1}(f(x)= \sin^{-1}{(\sin{x})} = x \hspace{25pt} -\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2}$
Note that $\dfrac{9 \pi}{8} \text{ isn't in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$.
Recall that:
$\sin{\theta} = \sin{(\pi - \theta)}$
Thus,
$\sin{\left(\dfrac{9 \pi}{8} \right)} = \sin{\left(\pi - \dfrac{9 \pi}{8} \right) }= \sin{\left(-\dfrac{\pi}{8} \right)}$
Since $-\dfrac{\pi}{8} \text{ is in the interval } \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$, then
$\sin^{-1} {\left[\sin{\left(-\dfrac{ \pi}{8} \right)} \right]} = -\dfrac{ \pi}{8}$
Therefore,
$\sin^{-1} {\left[\sin{\left(\dfrac{ 9\pi}{8} \right)} \right]} = \boxed{-\dfrac{ \pi}{8}}$
