# Solve $\tanα+2\tan2α+4\tan4α+8\tan8α+16\tanα=\cotα$ for $\alpha$

My knowledge of trigonometry are still insufficient to resolve this problem. Any help would be greatly appreciated.

Solving for $\alpha$:

$$\tanα+2\tan2α+4\tan4α+8\tan8α+16\tanα=\cotα$$

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What are your thoughts? – Ataraxia Jun 16 '13 at 17:43
It would be better if you show some of your work – user67258 Jun 16 '13 at 17:44
Is the equation right? $\tan\alpha+\dots+16\tan\alpha=$ – Américo Tavares Jun 16 '13 at 17:48
@RAsa, can you please the last part? Should not it be $8\tan8\alpha+16\tan16\alpha?$ – lab bhattacharjee Jun 16 '13 at 17:49
– vadim123 Jun 16 '13 at 17:54

As $\cos2A=\cos^2A-\sin^2A, \sin2A+2\sin A\cos A,$

$$\cot A-\tan A=\frac{\cos^2A-\sin^2A}{\cos A\sin A}=2\cot 2A$$

$$2\tan2\alpha+4\tan4\alpha+8\tan8\alpha+16\tan\alpha=\cot\alpha-\tan\alpha=2\cot2\alpha$$

$$\implies 4\tan4\alpha+8\tan8\alpha+16\tan\alpha=2(\cot2\alpha-\tan2\alpha)=2\cdot2\cot4\alpha$$

$$\implies 8\tan8\alpha+16\tan\alpha=4(\cdot2\cot4\alpha-\tan4\alpha)=4\cdot(2\cot8\alpha)$$

$$\implies 16\tan\alpha=8\cdot(\cot8\alpha-\tan8\alpha)=8\cdot2\cot16\alpha$$

$$\implies \cot16\alpha=\tan\alpha=\cot\left(\frac\pi2-\alpha\right)$$

$$\implies 16\alpha=n\pi+\frac\pi2-\alpha$$ where $n$ is any integer

$$\implies \alpha=\frac{(2n+1)\pi}{2\cdot17}$$

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@vadim123, please have a look into the edited version. Have you noticed how the multiple angle ratios of $\cot$ get cancelled? – lab bhattacharjee Jun 16 '13 at 18:09

While there may be a more slick way to solve a problem like this, you can repeatedly apply $$\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$$ until you have a rational equation in $\tan(\alpha)$. I haven't checked how unwieldly the resulting rational equation gets, since there is some uncertainty if the equation has the intended coefficients.

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tanα+2tan2α+4tan4α+8tan8α+16tanα=cotα – Rasa Jun 16 '13 at 17:56
@Rasa This is not the same as the equation in your title, when you look at the "degree 8" term. It also seems odd that you have left out a coefficient of 16 as in $16\tan(16\alpha)$, rather than $16\tan\alpha$, since the way you have it this term can be combined with the first to give $17\tan\alpha$. – alex.jordan Jun 16 '13 at 18:03

Iterating the identity $\cot(\alpha)-\tan(\alpha)=2\cot(2\alpha)$, we get $$\cot(\alpha)-\tan(\alpha)-2\tan(2\alpha)-4\tan(4\alpha)-8\tan(8\alpha)=16\cot(16\alpha)\tag{1}$$ Therefore, using $(1)$, the equation in question becomes $$\cot(16\alpha)=\tan(\alpha)\tag{2}$$ which, due to the periodicity of $\tan$, is equivalent to $$n\pi+\frac\pi2-16\alpha=\alpha\tag{3}$$ That is, $$\alpha=\frac\pi{34}(2n+1)\tag{4}$$

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