Binomial expansion question of complex sine How to show that the binomial expansion of $$\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2n+1}$$ equals
$$\sin ^{2n+1}(x)=\sum_{k=0}^{n}(-1)^{k+n}{{2n+1}\choose{k}}\frac{\sin ((2n-2k+1))x}{2^{2n}}$$
Please help! Sorry for asking such a dumb question!
 A: I think the upper bound of the summation should be $2n +1$ not $n$, and it should be $2^{2n + 1}$ not $2^{2n}$. You want to use the binomial theorem
$$(x + y)^n = \sum_{k = 0}^n {n \choose k } x^k y^{n - k}$$
This tells us
$$\left( e^{ix} - e^{-ix} \right)^{2n + 1} = \sum_{k = 0}^{2n + 1} {2n + 1 \choose k} e^{i(2n + 1 - k)x} \left(-e^{-ix} \right)^k = \sum_{k = 0}^{2n + 1} { 2n + 1 \choose k} (-1)^ke^{(2n - 2k + 1)ix} \\
= \sum_{k = 0}^{2n + 1} { 2n + 1 \choose k} (-1)^k \cos{ \left((2n -2k+1)x\right) } + i \sum_{k = 0}^{2n + 1} { 2n + 1 \choose k} (-1)^k \sin{ \left((2n -2k+1)x\right) }$$
See that both sums are real. This means that the sum with the sines is the imaginary part of $\left( e^{ix} - e^{-ix} \right)^{2n + 1}$. Then because the left hand side of
$$
\sin^{2n + 1} x = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^{2n + 1} = \frac{(-1)^n}{2^{2n + 1} i} \left( e^{ix} - e^{-ix} \right)^{2n + 1}
$$
is purely real, we only want the imaginary part of $\left(e^{ix} - e^{-ix} \right)^{2n + 1}$ so that the right hand side is purely real (this also tells us that the sum with the cosines must be 0). Substituting the sum with the sines gives the answer.
$$
\sin^{2n + 1} x = \frac{1}{2^{2n + 1}}\sum_{k = 0}^{2n + 1} (-1)^{n + k} { 2n + 1 \choose k }  \sin{ \left((2n -2k+1)x\right) }
$$
Edit: It seems that the upper bound being $n$ and it being $2^{2n}$ is actually correct, as the expression above simplifies because the terms in the sum are equal when $k = n + i$ and $k = n - (i - 1)$
We will split the sum as:
$$\sum_{k = 0}^{2n + 1} = \sum_{k = 0}^n + \sum_{k = n + 1}^{2n + 1}$$
We will manipulate the second sum $\sum_{k = n + 1}^{2n + 1}$ as follows by substituting $k = j + (n + 1)$ so that the sum starts at $j = 0$ and ends at $j = n$
$$
\sum_{k = n + 1}^{2n + 1} = \sum_{j = 0}^{n} (-1)^{n + j + (n + 1)} { 2n + 1 \choose j + n + 1 }  \sin{ \left((2n - 2(j + n + 1) +1)x\right) } \\
= \sum_{j = 0}^{n} (-1)^{j + 1} { 2n + 1 \choose j + n + 1 }  \sin{ \left(-2j - 1)x\right) }
$$
And now we substitute $j = n - k$ so the sum starts at $k = n$ and ends at $k = 0$. That is, we are changing the order of summation to be backwards so we have
$$
\sum_{j = 0}^{n} (-1)^{j + 1} { 2n + 1 \choose j + n + 1 }  \sin{ \left(\left(-2j - 1\right)x\right) } = \sum_{k = 0}^n (-1)^{n - k + 1} { 2n + 1 \choose 2n - k + 1 }  \sin{ \left( (-2n + 2k - 1)x\right) } \\
= \sum_{k = 0}^n (-1)^{n + k} { 2n + 1 \choose k }  \sin{ \left( (2n - 2k + 1)x\right) }
$$
where in the last step we used the oddness of $\sin$ and ${ n \choose k} = { n \choose n - k}$ for any positive integers $k \le n$. This is the same as the $\sum_{k = 0}^n$ sum from before so we have
$$
\sum_{k = 0}^{2n + 1} = 2 \sum_{k = 0}^n
$$
The two in the front will cancel reduce the $2^{2n + 1}$ to $2^{2n}$ so we'll have
$$
\sin^{2n + 1} x = \frac{1}{2^{2n}}\sum_{k = 1}^{n} (-1)^{n + k} { 2n + 1 \choose k }  \sin{ \left((2n -2k+1)x\right) }
$$
