Evaluate $\sum_{k=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4k+1)\pi}{2^{12}}\sin\frac{(4k+3)\pi}{2^{12}}}$ I need to evaluate
$$\sum_{n=0}^{511}\frac{\sin\frac\pi{2^{11}}}{\sin\frac{(4n+1)\pi}{2^{12}}\sin\frac{(4n+3)\pi}{2^{12}}}$$
Please give me some hint!
The final answer is $2^{10}$.
By CuriousGuest's answer, we need to prove
$$\cot\frac{\pi}{2^{12}}-\cot\frac{2046\pi}{2^{12}}+\cot\frac{2\pi}{2^{12}}-\cot\frac{2047\pi}{2^{12}}=2^{10}$$
any idea?

Hint: note that $$\frac{\pi}{2^{11}}=\frac{(4n+3)\pi}{2^{12}}-\frac{(4n+1)\pi}{2^{12}}.$$
Then use formula for $\sin(\alpha-\beta)$ and you'll get a telescopic sum.
Edit: It turns out to be not exactly telescopic, but something like
$$\cot\frac{\pi}{2^{12}}-\cot\frac{3\pi}{2^{12}}+\cot\frac{5\pi}{2^{12}}-\cot\frac{7\pi}{2^{12}}+\ldots+\cot\frac{2045\pi}{2^{12}}-\cot\frac{2047\pi}{2^{12}}.$$
This one can be handled in the following way: using that $$\cot\frac{(2^{11}-2k-1)\pi}{2^{12}}=\tan\frac{(2k+1)\pi}{2^{12}}$$
and the equality $\cot \alpha-\tan\alpha=2\cot2\alpha$, we can make this sum twice as short, then do the same thing with resulting sum, and so on 10 times, till we get the final result.
