Hi, got a problem that I need some help with.
How can I express cos 4x in cos x?
Help appreciated.
$\displaystyle \cos 4x=\cos(3x+x)=\cos3x\cdot\cos x - \sin 3x\cdot \sin x$
$\displaystyle =(4\cos^3x-3\cos x)\cos x-(3\sin x-4\sin^3 x)\cdot \sin x$
$\displaystyle =4\cos^4x-3\cos^2x-3\sin^2x+4\sin^4x$
$\displaystyle =4\cos^4x+4\sin^4x-3$
$\displaystyle =4\cos^4x+4(1-\cos^2 x)^2-3$
$\displaystyle =4\cos^4x+4(1+\cos^4x-2\cos^2x)-3$
$\displaystyle =4\cos^4x+4+4\cos^4x-8\cos^2x-3$
$\displaystyle =8\cos^4x-8\cos^2x+1$