Hi this appeared on my differentiation homework. for 0<x< π/2
Prove 1/2 tan (x/2) + 1/ (2^2) tan (x/ (2^2)) + ..... 1/ (2^n) tan (x/ (2^n)) = 1/ (2^n) cot (x/ (2^n)) - cot x.
Anyone knows how to solve it?
Some brackets to avoid ambiguity would be nice... Is it $\displaystyle \displaystyle \begin{align*} \frac{1}{2}\tan{\left( \frac{x}{2} \right) } + \frac{1}{2^2}\tan{ \left( \frac{x}{2^2} \right) } + \dots + \frac{1}{2^n}\tan{ \left( \frac{x}{2^n} \right) } = \frac{1}{2^n}\cot{ \left( \frac{x}{2^n} \right) } - \cot{(x)} \end{align*}$ or $\displaystyle \displaystyle \begin{align*} \frac{1}{2\tan{ \left( \frac{x}{2} \right) } } + \frac{1}{2^2 \tan{ \left( \frac{x}{2^2} \right) } } + \dots + \frac{1}{2^n\tan{ \left( \frac{x}{2^n} \right) } } = \frac{1}{2^n\cot{ \left( \frac{x}{2^n} \right) } } - \cot{(x)} \end{align*}$?
Hi it is $\displaystyle \displaystyle \begin{align*} \frac{1}{2}\tan{\left( \frac{x}{2} \right) } + \frac{1}{2^2}\tan{ \left( \frac{x}{2^2} \right) } + \dots + \frac{1}{2^n}\tan{ \left( \frac{x}{2^n} \right) } = \frac{1}{2^n}\cot{ \left( \frac{x}{2^n} \right) } - \cot{(x)} \end{align*}$ ... Thanks!
The left side is a product
$\displaystyle \cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{2^2}\right)\cos \left(\frac{x}{2^3}\right)\text{...}\text{..}\cos \left(\frac{x}{2^n}\right)=\frac{1}{2^n} \frac{\sin x}{\sin \left(x\left/2^n\right.\right)}$