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Show that $$\tan x +\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}-2\cot(2x), \quad (n=0,1,\ldots). $$ I tried to prove this using induction but with no result. I'm not sure how to begin to solve this.

Michael Hardy
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Gjekaks
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    That seems unlikely, since you can cancel $\frac{1}{2^n}\tan\frac{x}{2^n}$ from both sides, and you have a constant equal to a sum which varies as $n$ varies. – Thomas Andrews May 09 '16 at 17:20
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    Also, as $x\to 0$ the left hand side goes to zero, while the right side goes to infinity. So I'm having a hard time seeing how this can be fixed easily. – Thomas Andrews May 09 '16 at 17:22
  • Nah, even with a product on the left side, the value tends to zero as $n\infty$. @zyx – Thomas Andrews May 09 '16 at 17:43
  • Indeed, when you subtract the $1/2^n$-term from both sides, you get an apparent positive equaling an apparent negative. What about the case $n=0$? – Lubin May 09 '16 at 18:20
  • See also : http://math.stackexchange.com/questions/1743881/value-of-tana – lab bhattacharjee May 10 '16 at 05:03
  • Also : http://math.stackexchange.com/questions/389850/need-help-with-calculating-this-sum-sum-n-0-infty-frac12n-tan-frac1 – lab bhattacharjee May 10 '16 at 11:06
  • This is the finite form of question 389850 if tan() is replaced by cot( ) in the first term on the right side of the equation. – zyx May 10 '16 at 19:34

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Hint:

$$\tan \beta - \cot \beta= -2\cot 2 \beta$$ Then $$\tan x +\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}-\color{red}{2\cot(2x)} $$

$$\color{red}{\tan x} +\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}+\color{red}{\tan x - \cot x} $$

$$\frac{1}{2}\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}\color{red}{ - \cot x} $$

$$\tan \frac{x}{2}+\dots + \frac{1}{2^n}\tan \frac{x}{2^n} = \frac{1}{2^n}\tan \frac{x}{2^n}\color{red}{ - 2\cot x} $$ ...

Roman83
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    Maybe that solves the real problem, but as the problem is stated now I can't see how that helps. – mickep May 09 '16 at 17:24