# How do you find the exact value of cos^-1(sin(pi/6))?

$\frac{\pi}{3}$
In the first quadrant, $\sin \left(\frac{\pi}{6}\right) = \cos \left(\frac{\pi}{2} - \frac{\pi}{6}\right) = \cos \left(\frac{\pi}{3}\right)$
The given expression = ${\cos}^{- 1} \cos \left(\frac{\pi}{3}\right) = \frac{\pi}{3}$.
Note: For general values also, $\sin a = \cos \left(\frac{\pi}{2} - a\right)$.