Prove that for every natural number $n$ and for every real numbers $x\neq \dfrac {k\pi}{2^t }$ $(t=0,1,....n;)$ (where $k$ is any integer) $$\dfrac {1}{\sin 2x}+\dfrac {1}{\sin 4x}+.....+\dfrac {1}{\sin 2^n x}=\cot x-\cot 2^nx$$
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I offer a proof that doesn't use the induction. If you don't understand it ,I can edit it clearly. – Daisy Apr 01 '17 at 09:20
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@Daisy, please u may edit – Aryabhatta Apr 01 '17 at 09:24
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See also : http://math.stackexchange.com/questions/1591220/frac1-sin-8-circ-frac1-sin-16-circ-frac1-sin-4096-circ – lab bhattacharjee Apr 01 '17 at 11:12
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Here's a proof by induction
This is true for base case $n=1$ Assume true for $n$ ...So $$\cot x-\cot2^{n+1}x=\cot x-\cot(2(2^{n}x))=\cot x-\frac12((\cot(2^nx)-\tan(2^nx))$$$$= \cot x-\frac12\cot(2^nx)+\frac12\tan(2^nx)=\cot x-\cot(2^nx)+\frac12(\cot(2^nx)+\tan(2^nx))$$ Note that
$$\frac12(\cot(2^nx)+\tan(2^nx))=\frac{1}{\sin(2^{n+1}x)}$$ where I used $\cot x +\tan x=\frac{1}{\sin x\cos x}$ and $\cot2x=\frac12(\cot x-\tan x)$
So..can you complete now?
LM2357
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Since
\begin{align} \frac{1}{\sin 2x}=\frac{2{\cos}^2x-\cos2x}{2\sin x\cos x}=\cot x-\cot 2x, \end{align} then, if we regard $x$ as $2x$ \begin{align} \frac{1}{\sin 4x}=\frac{2{\cos}^22x-\cos 4x}{2\sin 2x \cos 2x}=\cot 2x-\cot 4x. \end{align} Similarly, repeat the above step until $n$, we have \begin{align} LHS=\cot x-\cot 2x+\cot 2x-\cot 4x+...+\cot 2^{n-1}x-\cot 2^nx=\cot x-\cot 2^nx=RHS. \end{align}
