## Precalculus: Mathematics for Calculus, 7th Edition

a. $\cos 68^\circ$ b. $\cos 10\theta$
a. Use the formula $\cos 2x=\cos^2 x-\sin^2 x$ with $x=34^\circ$. Then $\cos^2 34^\circ-\sin^2 34^\circ=\cos(2*34^\circ)=\cos 68^\circ$. a. Use the formula $\cos 2x=\cos^2 x-\sin^2 x$ with $x=5\theta$. Then $\cos^2 5\theta-\sin^2 5\theta=\cos(2*5\theta)=\cos 10\theta$.