Question

Can someone help me prove this idenity?
cos6x=2cos(squared)3x -1

Answer 1

2cos

Answer 2

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

Answer 3

\[\cos(6x)=\cos(2(3x))=\cos(3x)+\cos(3x)\]\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 \[2\cos^2(3x)-1\]

Answer 4

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