Question

Pre-calculus help!

Rewrite cos4x in terms of cosx

Thanks!

Answer 1

`\bf \textit{Double Angle Identities} \\ \quad \\ sin(2\theta)=2sin(\theta)cos(\theta) \\ \quad \\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ \boxed{2cos^2(\theta)-1} \end{cases} \\ \quad \\\\ tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\ -----------------------------\\\\`


`\bf cos[4x]\iff cos[2(2x)]\implies 2cos^2[2x]-1\iff 2[cos(2x)]^2-1 \\\\\\ 2[\ 2cos^2(x)-1\ ]^2-1\impliedby \textit{now, expanding the binomial} \\\\\\ 2[\ 2^2cos^4(x)-4cos^2(x)+1^2\ ]-1 \\\\\\ 2[\ 4cos^4(x)-4cos^2(x)+1\ ]-1 \\\\\\\ [\ 8cos^4(x)-8cos^2(x)+2\ ]-1 \\\\\\ 8cos^4(x)-8cos^2(x)+2-1\implies 8cos^4(x)-8cos^2(x)+1`

Answer 2

`\cos4x=2\cos^22x-1=2(2\cos^2x-1)^2-1`

using the identity
`\cos2x=\cos^2x-\sin^2x=2\cos^2x-1`. Expanding yields


`\cos4x=8\cos^4x-8\cos^2x+1`