Proving that $\cos\left(\frac{\pi}{2}+a\pi\right)=0$ when a is an integer

Question

So, I'm having a bit of trouble proving this. I tried using the sum formula, but the problem with that is that I end up having to prove that $\sin \left(a\pi\right)$ is equal to $0$. This is kind of counter-productive as it's essentially what I'm trying to do in the question! Is there a better way of proving it by avoiding this?

I do get a $0\times\cos(k\pi)$ during the proof, but I don't need to do anything with it as it becomes $0$ anyway. However, along with that I get a $1\times\sin(k\pi)$, which is why I'm asking this question.

Answer 1

Hint: using the fact that $\cos(\pi)=-1$ and $\sin(\pi)=0$, $$ \begin{align} \cos(x+\pi) &=\cos(x)\cos(\pi)-\sin(x)\sin(\pi)\\ &=-\cos(x) \end{align} $$ now use induction and the fact that $\cos\left(\frac\pi2\right)=0$.

Answer 2

$sin (n\pi)=0$ for $n\in I $ thats all now use $\sin (x)=\frac {e^{ix}-e^{-ix}}{2i} $ to complete the proof