My textbook without any reasoning, writes the simplified form of $\sum_{r=0}^{100}{100\choose r}\sin(2rx)$ as $2^{101}\cos^{100}x\sin(100x)$ and $\sum_{r=0}^{100}{100\choose r}\cos(2rx)$ as $2^{101}\cos^{100}x\cos(100x)$.
I have tried writing out the binomial expansion to $(\cos(2rx)+i\sin(2rx))^{n}$, but not sure of that would help. Any hints on how to prove this result, or whether any citation to a known such identity would be appreciated. Thanks.
Best Answer
Do it the other way around - let $z=e^{2ix}$, then $$\sin 2rx=\frac1{2i}(z^r-z^{-r})$$ Now the $z^r$ give one binomial sum, and the $z^{-r}$ give another