Prove that: $\;$ $\sin^{-1} \left[\dfrac{\sin x + \cos x}{\sqrt{2}}\right] = \dfrac{\pi}{4} + x$, $\;\;\;$ $- \dfrac{\pi}{4} < x < \dfrac{\pi}{4}$
$\begin{aligned}
\sin^{-1} \left[\dfrac{\sin x + \cos x}{\sqrt{2}}\right] & = \sin^{-1} \left[\dfrac{1}{\sqrt{2}} \times \sin x + \dfrac{1}{\sqrt{2}} \times \cos x\right] \\\\
& = \sin^{-1} \left[\cos \left(\dfrac{\pi}{4}\right) \sin x + \sin \left(\dfrac{\pi}{4}\right) \cos x\right] \\\\
& = \sin^{-1} \left[\sin \left(\dfrac{\pi}{4} + x\right)\right] \\\\
& = \dfrac{\pi}{4} + x
\end{aligned}$
Hence proved.