I currently try to prove 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
Hint
$$\sin^n(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^n = ?$$
This is an example to illustrate Hamam Abdallah's method. Let's compute $\sin^6(x)$. \begin{align*} \sin^6(x)&=\frac{1}{(2i)^6}(e^{ix}-e^{-ix})^6\\ &=\frac{-1}{2^6}(e^{6ix}-6e^{(5-1)ix}+15e^{(4-2)ix}-20e^{(3-3)ix}+15e^{(2-4)ix}-6e^{(1-5)ix}+e^{-6ix})\\ &=-\frac{1}{2^6}((e^{6ix}+e^{-6ix})-6(e^{4ix}+e^{-4ix})+15(e^{2ix}+e^{-2ix})-20)\\ &=-\frac{1}{2^6}\left(2\cos(6x)-12\cos(4x)+30\cos(2x)-20\right) \end{align*} In general you can derive formulas for $\sin^n(x)$ based on the parity of $n$ modulo $4$. Symmetry among the terms in the binomial expansion will cause terms to group together to form sines or cosines.
Edit (11/29/18): Having nothing better to do on a rainy day, I worked out the general formulae. Enjoy:
\begin{align*} \sin^{2m}(x)&=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2m}\\ &=\frac{(-1)^m}{2^{2m}}\sum_{k=0}^{2m}(-1)^k\binom{2m}{k}e^{ikx-i(2m-k)x}\\ &=\frac{(-1)^m}{2^{2m}}\sum_{k=0}^{2m}(-1)^k\binom{2m}{k}e^{i(2k-2m)x}\\ &=\frac{(-1)^m}{2^{2m}}\left[\sum_{k=0}^{m-1} (-1)^k\binom{2m}{k}e^{i(2k-2m)x} + (-1)^m\binom{2m}{m}+\sum_{k=m+1}^{2m}(-1)^k \binom{2m}{k}e^{i(2k-2m)x}\right]\\ &=\frac{(-1)^m}{2^{2m}}\left[\sum_{u=1}^{m} (-1)^{m-u}\binom{2m}{m-u}e^{-i(2u)x} +(-1)^m \binom{2m}{m}+\sum_{u=1}^{m} (-1)^{m+u}\binom{2m}{m+u}e^{i(2u)x}\right]\\ &=\frac{(-1)^m}{2^{2m}}\left[(-1)^m \binom{2m}{m}+\sum_{u=1}^{m} (-1)^{m-u}\binom{2m}{m-u}(e^{-i(2u)x}+e^{i(2u)x})\right]\\ &=\frac{1}{2^{2m}}\left[ \binom{2m}{m}+\sum_{u=1}^{m}(-1)^{u} \binom{2m}{m-u}2\cos(2ux)\right]\\ \end{align*} \begin{align*} \sin^{2m+1}(x)&=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2m+1}\\ &=\frac{(-1)^m}{2^{2m+1}i}\sum_{k=0}^{2m+1}(-1)^{k+1}\binom{2m+1}{k}e^{ikx-i(2m+1-k)x}\\ &=\frac{(-1)^m}{2^{2m+1}i}\sum_{k=0}^{2m+1}(-1)^{k+1}\binom{2m+1}{k}e^{i(2k-2m-1)x}\\ &=\frac{(-1)^m}{2^{2m+1}i}\left[\sum_{k=0}^{m}(-1)^{k+1}\binom{2m+1}{k}e^{i(2k-2m-1)x}+\sum_{k=m+1}^{2m+1}(-1)^{k+1}\binom{2m+1}{k}e^{i(2k-2m-1)x}\right]\\ &=\frac{(-1)^m}{2^{2m+1}i}\left[\sum_{u=0}^{m}(-1)^{m-u+1}\binom{2m+1}{m-u}e^{-i(2u+1)x}+\sum_{u=0}^{m}(-1)^{u+m}\binom{2m+1}{u+m+1}e^{i(2u+1)x}\right]\\ &=\frac{(-1)^m}{2^{2m}}\sum_{u=0}^{m}(-1)^{m-u}\binom{2m+1}{m-u}\left(\frac{e^{i(2u+1)x}-e^{-i(2u+1)x}}{2i}\right)\\ &=\frac{1}{2^{2m}}\sum_{u=0}^{m}(-1)^{u}\binom{2m+1}{m-u}\sin((2u+1)x) \end{align*}
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.
