<s>[INST] Prove the identity for $\tan3\theta$ 
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$

Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\cos 3 \theta}=\frac{3\sin \theta-4\sin^3\theta}{4\cos^3\theta - 3\cos \theta}$$
To then try and get the whole expression in terms of $\tan\theta$ I multiplied top and bottom of the fraction by $(4\cos^3\theta)^{-1}$. This gave me the following but I'm not now sure how to finish it off
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$
 [/INST] A slightly different approach using de Moivre's theorem would be: notice that $1 + i \tan \theta = \sec \theta \cdot e^{i \theta}$.  Therefore, cubing both sides, $(1 + i \tan \theta)^3 = \sec^3 \theta \cdot e^{3 i \theta}$.  Now, if you take the "slope" of this complex number, i.e. the imaginary part divided by the real part, from the right hand side you see the result will be $\tan(3\theta)$.  On the other hand, by binomial expansion,
$$(1 + i \tan \theta)^3 = 1 + 3i \tan \theta - 3 \tan^2 \theta - i \tan^3 \theta = (1 - 3 \tan^2 \theta) + i (3 \tan \theta - \tan^3 \theta).$$
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