# How do you find the exact value of Cos[sin^-1(1/3) - arcsin(1/2)]?

$\frac{1}{6} + \sqrt{\frac{2}{3}}$
First we will consider that ${\sin}^{- 1} \equiv \arcsin$. We will need the identity
$\cos \left(a - b\right) = \cos \left(a\right) \cos \left(b\right) + \sin \left(a\right) \sin \left(b\right)$ and also that
$\cos \left(\arcsin \left(x\right)\right) = \sqrt{1 - {x}^{2}}$. With all this in mind
$\cos \left(\arcsin \left(\frac{1}{3}\right) - \arcsin \left(\frac{1}{2}\right)\right) = \frac{1}{6} + \sqrt{\frac{2}{3}}$