# Math Help - Calculus

1. ## Calculus

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?

2. ## Re: Calculus

Some brackets to avoid ambiguity would be nice... Is it \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 \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*}?

3. ## Re: Calculus

Hi it is the first one....

4. ## Re: Calculus

Hi it is \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!

6. ## Re: Calculus

Thanks man that's brilliant!
Originally Posted by ibdutt

7. ## Re: Calculus

Originally Posted by ibdutt
Another idea. Prove the following by an easy induction

$\prod _{k=1}^n \text{cos}\left(\frac{x}{2^k}\right)=\frac{1}{2^n} \sin x \csc \left(\frac{x}{2^n}\right)$

Then take ln of both sides and differentiate

8. ## Re: Calculus

I am sorry could you elaborate further? I don't get the eqn... Thanks!

Originally Posted by Idea
Another idea. Prove the following by an easy induction

$\prod _{k=1}^n \text{cos}\left(\frac{x}{2^k}\right)=\frac{1}{2^n} \sin x \csc \left(\frac{x}{2^n}\right)$

Then take ln of both sides and differentiate

9. ## Re: Calculus

The left side is a product

$\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{2^2}\right)\cos \left(\frac{x}{2^3}\right)\text{...}\text{..}\cos \left(\frac{x}{2^n}\right)=\frac{1}{2^n} \frac{\sin x}{\sin \left(x\left/2^n\right.\right)}$

10. ## Re: Calculus

Ummgh how did you arrive at this equation? I am really lost.

11. ## Re: Calculus

You have to prove this formula by induction

for n=1

$\cos \left(\frac{x}{2}\right)=\frac{1}{2}\frac{\sin x}{\sin (x/2)}$