Proving $\frac{(2\cos^2 x - 1)^2}{\cos^4 x - \sin^4 x} = 1 - 2 \sin^2x$

Question

Prove that $$\frac{(2\cos^2 x - 1)^2}{\cos^4 x - \sin^4 x} = 1 - 2 \sin^2x$$

Thanks!

Answer 1

Hint: $$2\cos^{2}{(x)} - 1 = \cos{(2x)} = \cos^{2}{(x)} -\sin^{2}{(x)}$$

Can you take it from here?

Answer 2

The denominator is $(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$. Furthermore, you should know something about $\cos^2 x + \sin^2 x$...

Answer 3

1. Factor $\cos^4 x-\sin^4 x=(\cos^2 x)^2-(\sin^2 x)^2$ as a difference of squares.

2. Render $\cos^2 x+\sin^2 x = 1$ to eliminate the cosine terms.

3. Everything falls into place.

Answer 4

In other words, $(\cos 2x)^2/\cos 2x=\cos 2x$, where one of the three famous expressions for $\cos 2x$ has been multiplied by $\cos^2x+\sin^2x=1$ to throw you off the scent. (As @OscarLanzi points out, we could instead think of them as three expressions for $\cos^2x-\sin^2x$ if we didn't know double angle formulae.)