How do I prove $32\sin^2\theta*\cos^4\theta=2+\cos(2\theta)-2\cos(4\theta)-\cos(6\theta)?$

Question

Here's a problem in my pre-cal homework.

$32\sin^2\theta*\cos^4\theta=2+\cos(2\theta)-2\cos(4\theta)-\cos(6\theta)$

Anyone have an idea how to solve this? Thanks very much.

Answer 1

Basically, with such questions you would like to show that LHS = RHS by either putting LHS in the form of RHS or vice-versa.

Note that

$1 = \sin^2\theta + \cos^2\theta$

$\sin(2\theta) = 2\sin\theta\cos\theta$

$\cos(2\theta) = \cos^2\theta - \sin^2\theta$

$\cos(4\theta) = \cos^22\theta - \sin^22\theta \\ \qquad = \cos^2\theta - \sin^2\theta - 2\sin\theta\cos\theta$

$\cos(6\theta) = \cos^23\theta - \sin^23\theta \\ \qquad = \cos\theta(\cos^2\theta - \sin^2\theta) - \sin\theta(2\sin\theta\cos\theta) \\ \qquad = \cos^3\theta - sin^2\theta\cos\theta - 2\sin\theta\cos\theta$

Answer 2

Expand the right hand side using, for any $t,$ $$ \cos 2t = 2 \cos^2 t - 1, $$ $$ \cos 3t = 4 \cos^3 t - 3 \cos t $$

Using $t= 2 \theta$ in the first and second identities gives expansions for $\cos 4 \theta$ and $\cos 6 \theta$

Any polynomial in $\cos \theta , \sin \theta$ can be written as a finite Fourier series, as here. Be liberal in using $$ \sin^2 \theta = 1 - \cos^2 \theta $$