Prove this equation with calculus knowledge

[aerograce]

I feel it quite difficult to prove this equation:
[itex]\frac{1}{2}[/itex]tan([itex]\frac{1}{2}[/itex]x)+[itex]\frac{1}{2^2}[/itex]tan([itex]\frac{1}{2^2}[/itex]x)+...+[itex]\frac{1}{2^n}[/itex]tan([itex]\frac{1}{2^n}[/itex]x)=[itex]\frac{1}{2^n}[/itex]cot([itex]\frac{1}{2^n}[/itex]x)-cotx

Can you help me with it?

 

[lurflurf]

What have you tried?
$$\text{Let $0<x<\pi/2$, and n a positive integer} \\\text{prove that}\\
\sum_{k=1}^n \tan(x/2^k)/2^k=\cot(x/2^n)/2^n-cot(x)$$
hint
$$\tan(x/2^k)/2^k=\cot(x/2^k)/2^k-\cot(x/2^{k-1})/2^{k-1}$$

 

[aerograce]

So it has nothing to do with calculus or derivative?

 

[lurflurf]

That is a usual problem in finite calculus which deals with sums and differences. It does not seem particularly related to infinitesimal calculus which deals with derivatives and integrals. Though the two are closely related.

 

[aerograce]

Aww.. But I didn't use any finite calculus knowledge with your hint. I just proved your hint and then I found that I need nothing more.

 

[lurflurf]

A basic result in finite calculus is
$$\sum_{k=1}^n (a_k-a_{k-1})=a_n-a_0$$
Which my hint was an example of. The tricky part is figuring out how to write the terms of the sum as a difference.
There are some examples here.
https://en.wikipedia.org/wiki/Indefinite_sum