TITLE: Prove that $\frac{d^n}{dx^n} \cos(ax) = a^n \cos (ax + \frac{n\pi}{2})$
QUESTION [5 upvotes]: Question
Prove that $\frac{d^n}{dx^n} \cos(ax) =  a^n \cos (ax + \frac{n\pi}{2})$



I could calculate the first few derivatives of $\cos(ax)$ and consequently observe the pattern that unfolds, which in turn I can then correlate with $a^n \cos (ax + \frac{n\pi}{2})$.
However, I'm curious for a proof that does not rely on so many words, rather something more concrete(could I put it that way?).
I would include some of my attempts, but it's all rather trivial.

REPLY [2 votes]: A bit different approach: from the Euler representation $\cos x=\frac{\operatorname{e}^{ix}+\operatorname{e}^{-ix}}{2}$ we have
$$
\frac{\operatorname{d^n}}{\operatorname{d}x^n}\cos(ax)=\frac{(ia)^n\operatorname{e}^{iax}+(-ia)^n\operatorname{e}^{-iax}}{2}=a^n\cos\left(a+n\frac{\pi}{2}\right)
$$
since $(\pm i)^n=\operatorname{e}^{\pm in\frac{\pi}{2}}$.