## Calculus: Early Transcendentals 8th Edition

$cos3\theta=4cos^{3}\theta-3cos\theta$
We need to prove the identity $cos3\theta=4cos^{3}\theta-3cos\theta$ Now, $cos3\theta$ can be written as: $cos3\theta=cos(2\theta+\theta)$ Use sum identity for cosine. $cos 3\theta=cos(2\theta+\theta)=cos2\theta cos\theta-sin2\theta sin\theta$ $=(2cos^{2}\theta -1)cos\theta-(2sin\theta cos\theta ) sin\theta$ $=2cos^{3}\theta - cos\theta-2sin^{2}\theta cos\theta$ $=2cos^{3}\theta - cos\theta-2(1-cos^{2}\theta) cos\theta$ $=2cos^{3}\theta - cos\theta-2 cos\theta+2cos^{3}\theta$ Hence, $cos3\theta=4cos^{3}\theta-3cos\theta$