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Usually, proofs by mathematical induction are not that difficult (at least the ones I encountered so far in my mathematical journey), but I am stuck with this one... Spent at least 2-3 hours on it. I just can't see it...

$$ \sum_{i=0}^n \frac{1}{2^i}\tan\left(\frac{x}{2^i}\right)=\frac{1}{2^n}\cot\left(\frac{x}{2^n}\right) - 2\cot(2x) , (n = 1,2...)$$

What I tried:

For $n=1$, I expressed the LHS in terms of $\sin$ and $\cos$, but that didn't work... For me, at least... The expression got really complicated, which is a sign that I, probably, did something wrong.

Then I tried getting that $2\cot(2x)$ from RHS to LHS, but that didn't work also, because the expressions get really complicated...

Please, give me a detailed answer, and thank you for your time.

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    $\begingroup$ Hint: if you can show $LHS(n+1) - LHS(n) = RHS(n+1) - RHS(n)$ and $LHS(0) = RHS(0)$, then the induction becomes easy. Then, simplifying $LHS(n+1) - LHS(n)$ and $RHS(n+1) - RHS(n)$ will help you focus on the exact trigonometric identity you need to prove. $\endgroup$ Jun 22 '17 at 20:42
  • $\begingroup$ Note that in general to prove $\sum_{i=0}^n a_n = s_n$, you can just prove $s_0 = a_0$ (for the base case) and that $s_n - s_{n-1} = a_n$ (for the induction step). $\endgroup$ Jun 22 '17 at 20:42
  • $\begingroup$ @JairTaylor The problem is, I can't prove the base case. I just can't. Spent at least 2-3 hours on it. $\endgroup$
    – Shocky2
    Jun 22 '17 at 20:49
  • $\begingroup$ @DanielSchepler Wow. Awesome concept, I never looked at it that way. It can really bring some elegance to my, usually bulky & ugly, solutions. $\endgroup$
    – Shocky2
    Jun 22 '17 at 20:50
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\begin{align} S_{n-1}+T_{n} &=\frac{1}{2^{n-1}}\cot \left(\frac{x}{2^{n-1}}\right)-2\cot (2x)+\frac{1}{2^{n}}\tan \left(\frac{x}{2^{n}}\right)\\ &\text{let} \, \frac{x}{2^{n}}=y\\ &=\frac{1}{2^{n}}\left( 2\cot (2y)+\tan (y)\right)-2\cot (2x)\\ &=\frac{1}{2^{n}}\left( \frac{2(\cos (2y))}{\sin (2y)}+\tan (y)\right)-2\cot (2x)\\ &=\frac{1}{2^{n}}\left( \frac{2(\cos ^2 (y)-\sin^2 (y))}{2\sin (y)\cos (y)}+\tan (y)\right)-2\cot (2x)\\ &=\frac{1}{2^{n}}\left( \cot(y)-\tan (y)+\tan (y)\right)-2\cot (2x)\\ &=\frac{1}{2^{n}}\left(\cot \left(\frac{1}{2^{n}}\right)\right)-2\cot (2x)\\ &=S_n\\ \end{align}

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  • $\begingroup$ Thank you for sharing your knowledge. Truly, a beautiful proof. $\endgroup$
    – Shocky2
    Jun 22 '17 at 21:19
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** just a hint**

It is equivalent to prove that

$$\cot (x/2^{n+1})-2\cot (x/2^n)=$$

$$\tan (x/2^{n+1})$$

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  • $\begingroup$ To do that, notice that $\cot(x/2^n) = \cot(2\cdot x/2^{n+1})$. $\endgroup$
    – md2perpe
    Jun 22 '17 at 20:57

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