# How do you calculate the #cos[2 sin^ -1 (1/2)]#?

##### 1 Answer

Feb 26, 2016

#### Explanation:

First, focus on just

#sin(pi/6)=1/2#

Thus, since

#sin^-1(1/2)=pi/6#

Substitute this into the original expression:

#cos[2sin^-1(1/2)]=cos(2*pi/6)=cos(pi/3)=1/2#