The question is how to prove \begin{eqnarray*} \cos^m(x)= \left\{ \begin{array}{cc} \frac{1}{2^{m-1}} \left( \sum_{r=0}^{2r<m} { m \choose r } \cos (m-2r)x \right) + \frac{1}{2^{m}} {m \choose \frac{m}{2}} & m=2,4,\cdots , \\ \frac{1}{2^{m-1}} \left( \sum_{r=0}^{2r<m} { m \choose r } \cos (m-2r)x \right) & m=1,3,\cdots , \\ \end{array} \right. \end{eqnarray*}
that presented in here.
This is my try: \begin{eqnarray*} \cos^m(x)&=&\left(\frac{e^{ix}+e^{-ix}}{2}\right)^m=2^{-m}\sum_{k=0}^{m}\binom {m}{k} e^{kix} e^{-(m-k)ix}\\ &=& 2^{-m}\sum_{k=0}^{m}\binom {m}{k} e^{kix-mix+kix} \\ &=& 2^{-m}\sum_{k=0}^{m} \binom {m}{k} e^{(2k-m)ix} \\ &=& 2^{-m}\sum_{k=0}^{m} \binom {m}{k} \left\{\cos((2k-m)x)+i\sin((2k-m)x)\right\} \\ &=& 2^{-m}\sum_{k=0}^{m} \binom {m}{k} \left\{\cos((m-2k)x)-i\sin((m-2k)x)\right\} \\ &=& 2^{-m}\sum_{k=0}^{m} \binom {m}{k} \cos((m-2k)x) \end{eqnarray*}
If m is even:
\begin{eqnarray*} \cos^m(x)&=& 2^{-m}\sum_{k=0}^{m} \binom {m}{k} \cos((m-2k)x) \\ &=& 2^{-m}\sum_{k\neq \frac{m}{2}} \binom {m}{k} \cos((m-2k)x)+ 2^{-m}\binom {m}{m/2} \\ &=& 2^{-m}\sum_{2k< m} \binom {m}{k} \cos((m-2k)x)+ 2^{-m}\sum_{2k>m} \binom {m}{k} \cos((m-2k)x)+ 2^{-m}\binom {m}{m/2} \\ &=& 2\times2^{-m}\sum_{2k< m} \binom {m}{k} \cos((m-2k)x)+ 2^{-m}\binom {m}{m/2} \\ &=& \frac{1}{2^{m-1}}\sum_{2k< m} \binom {m}{k} \cos((m-2k)x)+ \frac{1}{2^{m}}\binom {m}{m/2} \end{eqnarray*}
