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!
 A: As we know in a Fourier series, when $f(x)$ is odd then the series reduces to
$$f(x)=\sum_{n=0}^\infty b_n\sin nx$$
here $\sin x$ is an odd function and for odd $N$, $\sin^Nx$ is odd as well. This means
$$\sin^Nx=\sum_{n=0}^\infty b_n\sin nx$$
for $n>N>0$ we have $b_n=0$ because
\begin{align}
2\pi b_n
&= \int_{-\pi}^{\pi} \sin^{N}x\sin nx\,dx \\
&= {\bf Im}\int_{-\pi}^{\pi} \sin^{N}xe^{inx}\,dx \\
&= {\bf Im}\int_{-\pi}^{\pi} \left(\dfrac{e^{ix}-e^{-ix}}{2i}\right)^{N}e^{inx}\,dx \\
&= {\bf Im}\left(\dfrac{1}{2i}\right)^{N}\int_{-\pi}^{\pi} \sum_{k=0}^N{N \choose k}(-1)^{N-k}\left(e^{ix}\right)^{k}\left(e^{-ix}\right)^{N-k}\left(e^{inx}\right)\,dx \\
&= {\bf Im}\left(\dfrac{1}{2i}\right)^{N}\sum_{k=0}^N {N \choose k}(-1)^{N-k}\int_{-\pi}^{\pi} e^{i(2k-N+n)x}\,dx \\
&= {\bf Im}\left(\dfrac{1}{2i}\right)^{N}\sum_{k=0}^N {N \choose k}(-1)^{N-k}0\hspace{2cm};\hspace{2cm}2k-N+n>0\\
&= 0
\end{align}
A: 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?
