# The “sum” of all fear

I have come across an identity problem:

Prove $\sin(\theta+k\pi)=-1^{k}\sin(\theta)$.

I can get to the point where $\sin(\theta)\cos({k}\pi)+\cos(\theta)\sin(k\pi)$ and I assume evaluating $\sin(k\pi)$ to where $\sin(\pi)$ equals zero would eliminate $\sin(k)$ and $\cos(\theta)$. However, $\cos(k\pi)$ evaluating to $-1^k$ is beyond me. I understand that $\cos(\pi)$ is $-1$, the part where $k$ becomes the exponent is tripping me up at the moment. Any thoughts or explanations would be greatly appreciated!

What is the value of $cos(k\pi)$ when $k$ is 0, 1, 2,...

What is the value of $(-1)^k$ when $k$ is 0, 1, 2,...

You will see that only two possible values are involved.

Then look at the parity of $k$.

Hint:

The correct formula is $$\sin(\theta+k\pi)=(-1)^{k}\sin\theta$$ ( note the parenthesis)

where $(-1)^{k}$ is $+1$ for $k$ is even and $-1$ if $k$ is odd. I think that you can see how this is linked to $\cos (k \pi)$.

Sine is periodic with period $2\pi$. Hence $\sin k\pi=0$ follows by induction on $k$ using $k=0$ and $k=1$ as base cases.

Cosine is periodic with period $2\pi$. Hence $\cos k\pi=(-1)^k$ follows by induction on $k$ using $k=0$ and $k=1$ as base cases.

Actualy, the target equation follows directly from the $2\pi$ periodicity of $\sin$ and $\sin(\theta+\pi)=-\sin\theta$.