# Showing the identity: $\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$

My knowledge of trigonometry are still insufficient to resolve this problem. Any help would be greatly appreciated. $$\tan \alpha + 2 \tan 2\alpha + 4 \tan 4\alpha = \cot \alpha − 8 \cot 8\alpha$$

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I take it "ctan" means "cotangent"? –  Gerry Myerson Mar 26 '12 at 5:06
Yes, sorry about that. –  ignaces Mar 26 '12 at 5:17
en.wikipedia.org/wiki/… –  pedja Mar 26 '12 at 5:49

Hoping this is not homework: $$\cot x - 8\cot 8x = \cot x -8\cot 8x= \cot x-8\frac{\cot^24x -1}{2\cot 4x}$$ $$= \cot x-4\frac{\cot^24x -1}{\cot 4x}= \cot x -4\cot 4x +4\tan 4x$$

$$= \cot x -4 \frac{\cot^22x -1}{2\cot 2x}+4\tan 4x=\cot x -2\cot 2x + 2\tan 2 x + 4 \tan 4x$$

$$=\cot x -2 \frac{\cot^2x -1}{2\cot x}+2\tan 2x + 4\tan 4x = \cot x - \cot x + \tan x + 2\tan 2x + 4\tan 4x$$ $$=\tan x + 2\tan 2x + 4\tan 4x$$

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+1 very nice   –  draks ... Mar 26 '12 at 13:19

This follows easily from the identity:

$$\cot x - 2 \cot 2x = \tan x$$

Which can easily be seen by using $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$ as follows: $$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin ^2 x}{\sin x \cos x} = 2 \cot 2x$$

We now have

$$\cot \alpha - 2 \cot 2\alpha = \tan \alpha$$ $$2\cot 2\alpha - 4 \cot 4\alpha = 2\tan 2\alpha$$ $$4\cot 2\alpha - 8 \cot 4\alpha = 4\tan 4\alpha$$

Adding gives us the result.

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