Prove that $2^{2m}\sin^{2m} x = \sum\limits_{k=0}^{m-1} (-1)^{m+k} C^{2m}_k \cos (2m-2k)x + C^{2m}_m$ So I've been doing some complex analysis in uni and I don't see how to approach this:
I have to prove that $2^{2m}\sin^{2m} x = \sum\limits_{k=0}^{m-1} (-1)^{m+k} C^{2m}_k \cos (2m-2k)x + C^{2m}_m$.
So far I tried to represent $\sin^{2m}x = (\frac{z-1/z}{2i})^{2m}$, but I don't know what to do next. Any help is appreciated. Thanks
 A: Use Euler's formula
$$ \sin^{2m} x= \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2m}$$
$$\Longrightarrow (2i)^{2m}\sin^{2m} x= (-1)^{m}2^{2m}\sin^{2m}= (e^{ix}-e^{-i})^{2m} \tag{1}$$
Expand the right hand side with the binomial theorem
$$(e^{ix}-e^{-ix})^{2m} = \sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j} e^{ix(2m-j)}e^{-ijx} = \sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j} e^{ix(2m-2j)} \tag{2} $$
Now, using Euler's formula again
$$e^{ix(2m-2j)} = \cos(x(2m-2j))+i\sin(x(2m-2j)) \tag{3}$$
From (1),(2),(3):
$$(-1)^{m}2^{2m}\sin^{2m}x = \sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\left[\cos(x(2m-2j))+i\sin(x(2m-2j)) \right] = \underbrace{\sum_{j=0}^{m-1}\binom{2m}{j}(-1)^{j}\left[\cos(x(2m-2j))+i\sin(x(2m-2j)) \right]}_{A} +\underbrace{\binom{2m}{m}}_{B} +\underbrace{\sum_{j=m+1}^{2m}\binom{2m}{j}(-1)^{j}\left[\cos(x(2m-2j))+i\sin(x(2m-2j)) \right]}_{C} $$
Note that in $C$:
$$\binom{2m}{j} = \binom{2m}{2m-j}$$
Hence
$$ C = \sum_{j=m+1}^{2m}\binom{2m}{j}(-1)^{j}\left[\cos(x(2m-2j))+i\sin(x(2m-2j)) \right] = \sum_{j=m+1}^{2m}\binom{2m}{2m-j}(-1)^{j}\left[\cos(x(2m-2j))+i\sin(x(2m-2j)) \right]$$
Do the following chang of variable $k = 2m-j$
$$ C = \sum_{k=0}^{m-1}\binom{2m}{k}(-1)^{k}\left[\cos(-x(2m-2k))+i\sin(-x(2m-2k)) \right] $$
Therefore
$$(-1)^{m}2^{2m}\sin^{2m} x = \underbrace{\sum_{j=0}^{m-1}\binom{2m}{j}(-1)^{j}\left[\cos(x(2m-2j))+i\sin(x(2m-2j)) \right]}_{A} +\underbrace{\binom{2m}{m}}_{B} + \underbrace{\sum_{k=0}^{m-1}\binom{2m}{k}(-1)^{k}\left[\cos(-x(2m-2k))+i\sin(-x(2m-2k)) \right]}_{C} =  \sum_{j=0}^{m-1}\binom{2m}{j} (-1)^{j} \left[\cos(x(2m-2j))+  \cos(-x(2m-2j))\right] + i\sum_{j=0}^{m-1}\binom{2m}{j} (-1)^{j}\left[\sin(x(2m-2k)) + \sin(-x(2m-2k))\right]+ \binom{2m}{m}$$
Note that given that $\sin(x)$ is odd and $\cos(x)$ is even:
$$ \left[\cos(x(2m-2j))+  \cos(-x(2m-2j))\right] = 2 \cos(x(2m-2j))$$
$$ \left[\sin(x(2m-2k)) + \sin(-x(2m-2k))\right] = 0$$
Therefore
$$(-1)^{m}2^{2m}\sin^{2m}x = 2\sum_{j=0}^{m-1}\binom{2m}{j} (-1)^{j} \cos(x(2m-2j))+ \binom{2m}{m}$$
Hence
$$\boxed{\sin^{2m}x = \frac{1}{2^{m-1}}\sum_{j=0}^{m-1}\binom{2m}{j} (-1)^{j+m} \cos(x(2m-2j))+ \binom{2m}{m}} $$
A: Yes, go ahead
$$2^{2m}\cos^{2m} x=(e^{ix}+e^{-ix})^{2m}=\sum_{k=0}^{2m} {2m \choose k}~ e^{-ikx} ~e^{i(2m-k)x}$$
$$2^{2m}\cos^{2m} x=\sum_{k=0}^{m-1} {2m \choose k} ~e^{i(2m-2k)x}+{2m \choose m}+\sum_{k=m+1}^{2m}
 {2m \choose k} ~e^{i(2m-2k)x}.$$
In the third term on the RHS let $k=2m-j$ , then
$$2^{2m}\cos^{2m} x=\sum_{k=0}^{m-1} {2m \choose k} ~e^{i(2m-2k)x}+{2m \choose m}+\sum_{j=0}^{m-1}
 {2m \choose 2m-j} ~e^{-i(2m-2j)x}.$$
$j$ being dummy replace it with $k$ to get
$$2^{2m}\cos^{2m} x=\sum_{k=0}^{m-1} {2m \choose k} ~[e^{i(2m-2k)x}+e^{-i(2m-2k)}]+{2m \choose m}.$$
$$\implies 2^{2m}\cos^{2m} x=\sum_{k=0}^{m-1} {2m \choose k} ~ 2\cos (2m-2k)x+{2m \choose m}.$$
Finally replace $x$ by $x-\pi/2$, to get
$$\implies 2^{2m}\sin^{2m} x=\sum_{k=0}^{m-1} (-1)^{m+k} {2m \choose k} ~ 2\cos (2m-2k)x+{2m \choose m}.$$
Note that $(-1)^{m-k}=(-1)^{m+k}$
