## anonymous 5 years ago Can someone help me prove this idenity? cos6x=2cos(squared)3x -1

1. anonymous

2cos

2. anonymous

sry i mean 2cos^2 (3x-1)?

3. anonymous

$\cos(6x)=\cos(2(3x))=\cos(3x)+\cos(3x)$\applying the sum of angles formula for cosine you get $\cos(3x)\cos(3x)-\sin(3x)\sin(3x)=\cos^2(3x)-\sin^2(3x)$ now by the pythagorean identity for sine and consine you get$\cos^2(3x)-(1-\cos^2(3x))$ finally gathering like terms you get $2\cos^2(3x)-1$

4. anonymous

if you can assume the identity for $\cos(2\theta)=2\cos^2(\theta)-1$ its a bit more direct