Prove that

$\displaystyle \frac{d^n}{dx^n}(\sin^4x+\cos^4x)=4^{n-1}\cos(4x+\frac{n\pi}{2})$

The first thing to do, obviously, is to find the first derivative, which is

$\displaystyle 4\sin^3x\cos x-4\cos^3x\sin x$

First I tried factoring, then using a couple of trigonometric identities.

$\displaystyle

-4\sin x\cos x(\cos^2x-\sin^2x)=-4\sin x\cos x(\cos(2x))=-4\sin x\cos x(2\cos^2x-1)

$

None of this, however, gets me any closer to $\displaystyle 4^{n-1}\cos(4x+\frac{n\pi}{2})$. I differentiated a second time and got

-4(\cos^4x+\sin^4x-24\sin^2x\cos^2x)

I haven't double-checked my math on this last step but I'm pretty sure I'm right, but, regardless, it's not any closer either! So, I ask you, am i doing something wrong or just trying to follow the wrong steps?