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?
 A: 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.
A: Using the partial fractional decomposition, the sum can be restated as
$$\huge {\sum_{k=0}^{511} \cot{\frac{4k+1}{2^{12}}\pi} - \cot{\frac{4k+3}{2^{12}}\pi}}$$
Summing in reverse order gives
$$\huge {\sum_{k=0}^{511} \tan{\frac{4k+3}{2^{12}}\pi} - \tan{\frac{4k+1}{2^{12}}\pi}}$$
Using the identity $\cot{\theta} - \tan{\theta} \equiv 2\cot{2\theta}$
Gives us
$$\huge {\sum_{k=0}^{511} \cot{\frac{4k+1}{2^{11}}\pi} - \cot{\frac{4k+3}{2^{11}}\pi}}$$
Repeat this process until the denominator is equal to $4$.
The observant reader will realise that another application results in a vanishing sum, and therefore, is invalid.
$$\huge {\sum_{k=0}^{511} \cot{\left(k+\frac{1}{4}\right)\pi} - \cot{\left(k+\frac{3}{4}\right)\pi}}$$
$$\huge {\sum_{k=0}^{511} 1-(-1)} = 512 \times 2 = 1024$$
