<s>[INST] Prove that $\frac{d^n}{dx^n} (\sin^4 x + \cos^4 x) = 4^{n-1}\cos (4x + \frac{n\pi}{2})$ 
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*Question
Prove that $\frac{d^n}{dx^n} (\sin^4 x + \cos^4 x) = 4^{n-1}\cos (4x + \frac{n\pi}{2})$



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*My attempt


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*First calculate $\frac{d}{dx} (\sin^4 x + \cos^4 x)$, that is, 
$$\frac{d}{dx} (\sin^4 x + \cos^4 x) =4\sin^3 x \cos x - 4\cos^3 x \sin x $$
$$= 4\sin x \cos x(\sin^2 x - \cos^2 x)$$
$$\tag {n=1} = -2\sin 2x \cos 2x = - \sin 4x$$

*Now, using the value of $n=1$, I calculate the derivatives for a few more values of $n$:
$$\tag {n=2} -4\cos 4x$$
$$\tag {n=3} 4^2\sin 4x$$
$$\tag {n=4} 4^3\cos 4x$$
$$\tag {n=2} -4^4\sin 4x$$


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*From this I observe the consistency of the $4^{n-1}$ factor.

*Now I will expand $\cos (4x + \frac{n\pi}{2})$, which results in $$\cos 4x \cos \frac{n\pi}{2} - \sin 4x \sin \frac{n\pi}{2}$$

*From what I know of of $\sin$ and $\cos$ functions, this fits the pattern I observe in calculating the derivatives of $\sin^4 x + \cos^4 x$



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*My concerns
I feel uncomfortable putting so much weight on observing a pattern, instead I feel I should be able to put it down in terms of mathematics. 
One more thing, I don't feel comfortable with 

From what I know of of $\sin$ and $\cos$ functions, this fits the pattern I observe in calculating the derivatives of $\sin^4 x + \cos^4 x$.

, there must be a better way to relate $$\frac{d^{n-1}}{dx^{n-1}} (-\sin 4x) $$ with $$\cos (4x + \frac{n\pi}{2})$$

I would appreciate any hints, suggestions, and alternative approaches.
Keep in mind that this should be approached with the tools of elementary introductory derivatives and trigonometry.
 [/INST] Hint: Note that if 
$$\cos (4x + \frac{n\pi}{2})=\cos 4x \cos \frac{n\pi}{2} - \sin 4x \sin \frac{n\pi}{2}$$
If $n$ is odd, then $n=4k-3$ or $n=4k-1$, $\forall k\in\mathbb{N}$, then $\cos\frac{n\pi}{2}=0$: $$\sin{\frac{(4k-3)\pi}{2}}=1$$ and
$$\sin{\frac{(4k-1)\pi}{2}}=-1$$
Now If $n$ is even, then $n=4k$ or $n=4k-2$, $\forall k\in\mathbb{N}$ then $\sin\frac{n\pi}{2}=0$:
$$\cos{\frac{(4k)\pi}{2}}=\cos{(2k\pi)}=1$$
and
$$\cos{\frac{(4k-2)\pi}{2}}=\cos{[(2k-1)\pi]}=-1$$
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