Prove $$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$ where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.
My try. If $n=2$, we have
$$\begin{align} \frac{1}{\sin 2x} + \frac{ 1}{\sin 4 x} &= \frac{1}{\sin 2x} + \frac{1}{2 \sin 2x \cos 2x } \\[6pt] &= \frac{ 2 \cos 2x + 1 }{2 \sin 2x \cos 2x} \\[6pt] &= \frac{2 \cos^2 x - 2 \sin^2 x + \cos^2 x + \sin^2 x}{2 \sin 2x \cos 2x} \\[6pt] &= \frac{3 \cos^2 x - \sin^2 x}{2 \sin 2x \cos 2x} \end{align}$$
but here I got stuck. I am on the right track? My goal is to ultimately use induction.
$\endgroup$4 Answers
$\begingroup$Skipping over the base case ($n=1$) for the moment, let's do the inductive step: If we take the equation
$${1\over\sin2x}+{1\over\sin4x}+\cdots+{1\over\sin2^nx}=\cot x-\cot2^nx$$
and replace $x$ with $2x$, we get
$${1\over\sin4x}+{1\over\sin8x}+\cdots+{1\over\sin2^{n+1}x}=\cot2x-\cot2^{n+1}x$$
Adding $1/\sin2x$ to both sides now gives
$${1\over\sin2x}+{1\over\sin4x}+\cdots+{1\over\sin2^{n+1}x}={1\over\sin2x}+\cot2x-\cot2^{n+1}x$$
so, to complete the inductive step, it suffices to prove
$${1\over\sin2x}+\cot2x=\cot x$$
But this is the base case, just rewritten with the $\cot2x$ moved to the left hand side! So let's prove it:
$$\begin{align} {1\over\sin2x}+\cot2x&={1+\cos2x\over\sin2x}\\ &={1+(2\cos^2x-1)\over2\sin x\cos x}\\ &={\cos x\over\sin x}\\ &=\cot x \end{align}$$
$\endgroup$ $\begingroup$Without Induction
We can werite series as $$\sum^{n}_{r=1}\frac{1}{\sin 2^{r}x} = \sum^{n}_{r=1}\frac{\sin(2^{r}x-2^{r-1}x)}{\sin 2^{r-1}x\cdot \sin 2^{r}x} = \sum^{n}_{r=1}(\cot 2^{r-1}x-\cot 2^{r}x )$$
Now use Telescopic Sum
$\endgroup$ $\begingroup$Let $S_n$ be the sum $\sum_{k=1}^n\frac{1}{\sin(2^kx)}$. we want prove by induction that
$\forall n\geq1 \;\; S_n=cot(x)-cot(2^nx)$.
the formula is true for $n=1$.
let $n\geq1 : S_n true$.
we must prove that
$S_{n+1}-S_n=cot(2^nx)-cot(2^{n+1})$.
$cot(2^nx)-cot(2^{n+1}x)=$
$\frac{\cos(2^nx)}{\sin(2^nx)}-\frac{\cos(2^{n+1}x)}{2\sin(2^nx)\cos(2^nx)}=$
$\frac{1}{\sin(2^{n+1})}=S_{n+1}-S_n$.
using $\sin(2X)=2\sin(X)\cos(X)$
and
$\cos(2X)=2\cos^2(X)-1$.
thus
$\color{green}{S_{n+1}=S_n+cot(2^nx)-cot(2^{n+1}) =cot(x)-cot(2^{n+1})}$.
$\endgroup$ 1 $\begingroup$$$\csc2A+\cot2A=\dfrac{1+\cos2A}{\sin2A}=\cot A$$
$$\iff\csc2A=\cot A-\cot 2A$$
Put $2A=2x,4x,8x,\cdots,2^nx$ and add
See also: Telescoping series
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