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Verify the triple angle formula

$$\tan(3x) = \frac{3 \tan(x) − \tan^3(x)}{1 − 3 \tan^2(x)}$$

I have tried simplifying the right side by the following

$$\tan(3x) = \frac{\tan(x)(3 − \tan^2(x))}{1 − 3 \tan^2(x)}$$

but then I am getting stuck trying to verify the equation

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    $\begingroup$ I suggest you try to simplify the LHS with the formula $$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$ because, as long as your problem remains in the form $f(3x)\stackrel?=g(x)$, you hare not moving too far from the starting point. $\endgroup$ – user228113 Oct 9 '16 at 15:27
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$$\tan 3x =\tan(x+2x)= \frac{\tan x + \tan 2x}{1 − \tan x \tan 2x}=$$ $$=\frac{\tan x + \frac{2\tan x}{1-\tan^2x}}{1 − \tan x \frac{2\tan x}{1-\tan^2x}}=\frac{3\tan x- \tan ^3x}{1-3\tan^2x}$$

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You can try.

$$\tan (3x) = \frac{\sin 3x}{\cos 3x} = \frac{\sin x \cos 2x + \cos x \sin 2x}{\cos x \cos 2x - \sin x\sin 2x} = \frac{\sin x(\cos^2 x- \sin^2x) + 2\sin x\cos^2 x}{\cos x(2\cos^2 x -1) - 2\sin^2 x\cos x} = \frac{\tan x(1-\tan^2 x) + 2\tan x}{2 - \frac{1}{\cos^2 x} - 2\tan^2 x} = \frac{3\tan x - \tan^3x}{1-3\tan^2x}$$

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If you are allowed to use complex number, there is a simple way to derive the triple angle formula. For real $x$ with $\cos x, \cos 3x \ne 0$, if one expand $e^{i3x}$ in two different ways, one get

$$\cos 3x( 1 + i\tan 3x) =\cos 3x + i\sin 3x\\ = e^{i3x} = (e^{ix})^3\\ = (\cos x + i\sin x)^3 = \cos^3x ( 1 + i\tan x)^3$$ If one compare the ratio of real and imaginary part of both sides, one immediately obtain $$\tan x = \frac{\Im (1 + i\tan x)^3 }{\Re (1+i\tan x)^3} = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$$

You can use the same trick to expand $\tan n x$ for other integer $n > 3$.

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