I'm asked to prove that $$\cos n\pi=(-1)^n\qquad n\in\mathbb {Z}$$ I'm not sure how to approach the problem, I want to know if there is a different way to use induction

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    $\begingroup$ Do you mean $\cos n\pi = (-1)^n$? $\endgroup$ – JimmyK4542 Sep 16 '14 at 6:32

I'm going to assume that you meant to write the problem as "$\cos n\pi = (-1)^n$".

Hint: For any $x \in \mathbb{R}$, $\cos(x+\pi) = \cos x \cos \pi - \sin x \sin \pi = \cos x \cdot -1 - \sin x \cdot 0 = -\cos x$.

Therefore, $\cos((n+1)\pi) = \cos(n\pi + \pi) = -\cos n\pi$. Can you take it from here?


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