Sum $\sum_{r=1}^n \cos(2.(\frac {3^rx}{3})).\csc (3^rx)$

Question

Prove that $$\sum_{r=1}^n \cos(2.(\frac {3^rx}{3})).\csc (3^rx) = \frac{1}{2\sin x}-\frac{1}{2\sin (3^nx)}$$

My attempt: $$\Sigma \frac{\cos(\frac {2.3^rx}{3})}{\sin(3^rx)} = \Sigma \frac{1-2\sin^2(\frac {3^rx}{3})}{\sin(3^rx)}$$

I am trying to convert it to a form $f(r+1)-f(r)$ where $f(r)$ represents a general term in the series. But I am unable to do so.

Answer 1

\begin{align} \frac{1-2\sin^2(3^{r-1}x)}{\sin(3^rx)} &= \frac{1-2\sin^2(3^{r-1}x)}{\sin(3\cdot3^{r-1}x)} \\ &= \frac{1-2\sin^2(3^{r-1}x)}{3\sin(3^{r-1}x)-4\sin^3(3^{r-1}x)} \\ &= \frac{(3-4\sin^2(3^{r-1}x))-1}{2\sin(3^{r-1}x)\left(3-4\sin^2(3^{r-1}x)\right)} \\ &= \frac{1}{2\sin(3^{r-1}x)}-\frac{1}{2\sin(3\cdot3^{r-1}x)} \\ &= f(r-1)-f(r) \end{align}