Proof formula for $(\sin(x))^n$ I currently try to proof the following equation:
$$(\sin(x))^n=\sum_{k=0}^{n}{a_k\cos(kx)+b_k\sin(kx)}$$
with $a_0,...,a_k$ and $b_0,...b_k$ being real numbers for each $n$.
I tried to proof this by induction, then I could use the induction hypothesis for replacing $\sin(x)^{n}$ with $\sin(x)\cdot (\sum_{k=0}^{n}{a_k \cos(kx)+b_k\sin(kx)})$
Do you have any idea, how I can get this in the form $\sum_{k=0}^{n+1}{a_k \cos(kx)+b_k\sin(kx)}$?
Thanks
 A: Hint
$$\sin^n(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^n = ?$$
A: One may consider the two formulae
$$\begin{align}\sin(n)\cos(m)&=\frac12(\sin(n+m)+\sin(n-m))\\
\sin(n)\sin(m)&=-\frac12(\cos(n+m)-\cos(n-m))\end{align}$$
which are both a direct consequence of the addition theorems for the sine function and the cosine function respectively. In your case we set $n=x$ and $m=kx$ to get
$$\begin{align}
\sin(x)\cos(kx)&=\frac12(\sin(x+kx)+\sin(x-kx))=\frac12(\sin((k+1)x)-\sin((k-1)x))\\
\sin(x)\sin(kx)&=-\frac12(\cos(x+kx)-\cos(x-kx))=-\frac12(\cos((k+1)x)-\cos((k-1)x))
\end{align}$$
As you can see this produces a term of the type $a_{k+1}\cos((k+1)x)+b_{k+1}\sin((k+1)x)$ which implies a higher upper border and therefore you are done hence this procedure can be applied for any value of $k$ and $n$ respectively.
Concerning the special terms $\sin(-x)$ and $\cos(-x)$ occuring for $k=0$ recall the symmetric properties of the trigonometric functions.

EDIT
Since I realised for myself - by extending my answer within the comments - actually writing down a proof using these given formulae seems to be more difficult than I thought in the first place.
Therefore lets plug in the new terms constructed by using the given formulae. So we get
$$S=\sum_{k=0}^n \left[\frac{a_k}2(\sin((k+1)x)-\sin((k-1)x))-\frac{b_k}2(\cos((k+1)x)-\cos((k-1)x))\right]$$
Hence we are dealing with finite sums we are perfectly fine with rearraging the sums aswell as splitting them up. So lets do a little bit of reshaping
$$\begin{align}
S&=\sum_{k=0}^n \left[\frac{a_k}2(\sin((k+1)x)-\sin((k-1)x))-\frac{b_k}2(\cos((k+1)x)-\cos((k-1)x))\right]\\
&=\sum_{k=0}^n \left[a_k'\sin((k+1)x)-b_k'\cos((k+1)x)+b_k'\cos((k-1)x))-a_k'\sin((k-1)x)\right]\\
&=\sum_{k=0}^n \left[a_k'\sin((k+1)x)-b_k'\cos((k+1)x)\right]+\sum_{k=0}^n \left[b_k'\cos((k-1)x))-a_k'\sin((k-1)x)\right]\\
&=\underbrace{\sum_{k=1}^{n+1}\left[a_{k-1}'\sin(kx)-b_{k-1}'\cos(kx)\right]}_{=S_1}+\sum_{k=0}^n \left[b_k'\cos((k-1)x))-a_k'\sin((k-1)x)\right]\\
&=S_1+b_0'\cos(-x)-a_0'\sin(-x)+\sum_{k=1}^n \left[b_k'\cos((k-1)x))-a_k'\sin((k-1)x)\right]\\
&=S_1+b_0'\cos(x)+a_0'\sin(x)+\underbrace{\sum_{k=0}^{n-1} \left[b_{k+1}'\cos(kx)-a_{k+1}'\sin(kx)\right]}_{=S_2}\\
&=a_0'\sin(x)+b_0'\cos(x)+S_1+S_2
\end{align}$$
Now we can argue there are terms for $k=0$ produced by the sum $S_2$, terms for $k=n+1$ produced by $S_1$, the extra term which matches with the case $k=1$ for either $S_1$ or $S_2$ and terms inbetween with coefficients composed out of these given by $S_1$ and $S_2$. Therefore we can proceed to say that 
$$S=\sum_{k=0}^{n+1}a_k''\cos(kx)+b_k''\sin(kx)$$
for some $a_k''$ and $b_k''$ which can be computed by oberserving the matching constants infront of the sines and cosines for the respective values of $k$ in each sum $S_1$ and $S_2$ and for $k=1$ the remaining term respectively.
