Find the value $\sum_{n=a}^b\frac1{\sin (2^{n+3})}$

Question

Find the value of: $$\sum_{n=0}^{10}\frac1{\sin (2^{n+3})}$$

I'm stuck on this problem, can someone please help?

Answer 1

Note that

$$\begin{align}\cot(2^k)-\cot(2^{k+1})&=\frac{\sin (2^{k+1})\cos( 2^{k})-\sin (2^{k})\cos(2^{k+1})}{\sin (2^k)\sin( 2^{k+1})}\\ \\ &=\frac{\sin( 2^{k+1}- 2^{k})}{\sin( 2^k)\sin (2^{k+1})}\\ \\ &=\frac{1}{\sin (2^{k+1})}\end{align}$$

Therefore,

$$\sum_{k=a}^{b}\frac{1}{\sin 2^{k+1}}=\cot(2^a)-\cot(2^{b+1})$$