Prove $\frac{1}{\sin{2\alpha}}+\frac{1}{\sin{2^2\alpha}}+\cdots+\frac{1}{\sin{2^n\alpha}}=\frac{1}{\tan{\alpha}}-\frac{1}{\tan{2^n\alpha}}$

Question

Can someone help me with the problem? Let $\alpha \in \mathbb{R} \backslash \{2^{-m}k\pi; k\in\mathbb{Z},m\in\mathbb{N} \}$. Prove that the following equation is true for $\forall n\in N$: $$\frac{1}{\sin{2\alpha}}+\frac{1}{\sin{2^2\alpha}}+\cdots+\frac{1}{\sin{2^n\alpha}}=\frac{1}{\tan{\alpha}}-\frac{1}{\tan{2^n\alpha}}$$

Answer 1

This can be shown by Induction ( I will leave you to show the base case).

Observe that \begin{eqnarray*} \frac{1}{\sin (2^{n+1} \alpha)} &= & \frac{1}{ 2 \sin (2^{n} \alpha) \cos (2^{n} \alpha)} \\ &= & \frac{1}{ 2 \cos^2 (2^{n} \alpha) \tan (2^{n} \alpha)} \\ \end{eqnarray*} Now add this term on \begin{eqnarray*} -\frac{1}{\tan (2^{n} \alpha)} +\frac{1}{\sin (2^{n+1} \alpha)} &= & -\frac{1}{\tan (2^{n} \alpha)} \left( \frac{1}{ 2 \cos^2 (2^{n} \alpha) } -1\right) \\ &= & \frac{1 -2 \cos^2 (2^{n} \alpha) }{ 2 \cos^2 (2^{n} \alpha) \sin (2^{n} \alpha)} \\ &= & -\frac{1 }{ \tan (2^{n+1} \alpha)}. \\ \end{eqnarray*}