# How do you find the exact value of sin^-1 (sqrt2/2)?

${\sin}^{-} 1 \left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$
Let ${\sin}^{-} 1 \left(\frac{\sqrt{2}}{2}\right) = \theta$ , then $\sin \theta = \frac{\sqrt{2}}{2}$. We know $\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \therefore \theta = \frac{\pi}{4} \mathmr{and} {\sin}^{-} 1 \left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$ [Ans]