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
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$
 A: We have
$$
\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}} = 
\frac{\frac{3(\cos^2 \theta + \sin^2 \theta)\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3(\cos^2 \theta + \sin^2 \theta)\cos \theta}{4\cos^3 \theta}} = \\
\frac{\frac{3\cos^2 \theta\sin \theta + 3\sin^3 \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos^3 \theta + 3\sin^2 \theta\cos \theta}{4\cos^3 \theta}} = \\
\frac{\frac{3\cos^2 \theta\sin \theta}{4\cos^3\theta} + \frac{3\sin^3 \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos^3 \theta}{4\cos^3 \theta} -\frac{3\sin^2 \theta\cos \theta}{4\cos^3 \theta}}
$$
and from there, the simplification is straightforward.
A: Use that $$\tan(3x)=\frac{\tan(x)+\tan(2x)}{1-\tan(x)\tan(2x)}$$ and then
$$\tan(x+x)=\frac{2\tan(x)}{1-\tan^2(x)}$$
A: You have achieved the $cos 3x$ and $sin 3x$ answers by matching the Real and Imaginary components to show them in terms of only $cos x$ and $sin x$ I presume, leading on from an earlier question.
It should be a lot easier if you simply do exactly what you've done for the answers before matching the components
A: 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,
$$(1 + i \tan \theta)^3 = 1 + 3i \tan \theta - 3 \tan^2 \theta - i \tan^3 \theta = (1 - 3 \tan^2 \theta) + i (3 \tan \theta - \tan^3 \theta).$$
A: Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice: 
Note that $\frac{\sin\theta}{\cos^3\theta} = \tan\theta\sec^2\theta = \tan\theta(1 + \tan^2\theta) = \tan\theta + \tan^3\theta$. Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice: 
note that $\frac{\sin\theta}{\cos^3\theta} = \tan\theta\sec^2\theta = \tan\theta(1 + \tan^2\theta) = \tan\theta + \tan^3\theta$. That will fix up your numerator nicely. Using $\sec^2 \theta = 1 + \tan^2\theta$ directly (since $\frac{\cos\theta}{\cos^3\theta} = \sec^2\theta$) will similarly fix up the numerator, so what you have so far is equal to 
$$\frac{\frac{3}{4}\tan\theta-\frac{1}{4}\tan^3\theta}{\frac{1}{4} - \frac{3}{4}\tan^2\theta} = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta},$$
as required
