If $N$ is odd, show $\sin^N x$ can be written as a finite sum of the form $\sum _{k=1} ^{N} a_k \sin(kx)$? I am reading a Fourier series book, I got this exercise from the book but have no clue how to prove it. Would you mind kindly giving me some hints?
Thanks!

To show this via induction:


*

*Base case --- $\sin^1(x) = \sin(x)$, so we're done.

*Assume that $\sin^N(x) = \sum_{k = 1}^N a_k\sin(kx)$ is true.
Now, consider $\sin^{N+2}(x)$.
This is:
$$\sin^2(x)\sum_{k = 1}^N a_k\sin(kx)$$
We have that:
$$\sin^2(x) = \frac{1}{2}(1-\cos(2x))$$
So, this sum is:
$$\sum_{k = 1}^N \frac{a_k}{2}\sin(kx)-\frac{a_k}{2}\sin(kx)\cos(2x)$$
From here, we have that:
$$\sin(a)\cos(b) = \frac{1}{2}(\sin(a+b)+\sin(a-b))$$
Do you see how we can finish the proof from here?
