## anonymous one year ago simplify given that all angles are between 0 and pi Cos(2arccos sqrt2/2)

1. Mertsj

$\cos^{-1} \frac{\sqrt{2}}{2}=\frac{\pi}{4}, \frac{7\pi}{4}$

2. Mertsj

When you multiply those angles by 2 you get: $\frac{\pi}{2},\frac{7\pi}{2}$

3. Mertsj

$\cos \frac{\pi}{2}=0; \cos \frac{7\pi}{2}=0$