Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 23

$-\dfrac{\pi}{3}$

Note that $\cos{\left(-\dfrac{7\pi}{6}\right)}=-\dfrac{\sqrt3}{2}$. Thus $\sin^{-1}\left(\cos{\left(-\dfrac{7\pi}{6}\right)}\right)=\sin^{-1}{\left(-\dfrac{\sqrt3}{2}\right)}=-\dfrac{\pi}{3}$, because $\sin{\left(-\dfrac{\pi}{3}\right)}=-\dfrac{\sqrt3}{2}$ and $-\dfrac{\pi}{3}$ is in the range of $\sin^{-1}{x}$, which is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$