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

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

Thanks!

• Please show your work and where you faced problems – Vizag Apr 20 '19 at 17:16
• Are you able to solve now? Or are you still facing problems? – Vizag Apr 20 '19 at 18:34

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

Can you take it from here?

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.)

• You don't need the double angle formulas at all. They throw the scent off in this problem. – Oscar Lanzi Apr 20 '19 at 17:22
• @OscarLanzi Thanks; edited. – J.G. Apr 20 '19 at 17:32

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

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.