Using De Moivre to show $\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$ 
Use the De Moivre Theorem to show that
$$\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$

I got this question on my exam today and had no idea how to do it.
There must be a connection to complex numbers since its the topic its connected to. I only learnt how to use the De Moivre theorem with complex numbers in polar form so I was completely lost here.
Sorry if its a dumb question :<.

So you begin by writing
$$
\tan 6\theta \equiv {\sin 6\theta \over \cos 6\theta} 
$$
From there you apply De Moivre's Theorem which states:
$$
(\cos \theta +\iota\sin\theta)^n \equiv (\cos n\theta + \iota\sin n\theta)
$$
In this case we have $n = 6$, so:
$$
(\cos \theta +\iota\sin\theta)^6 \equiv (\cos 6\theta + \iota\sin 6\theta)
$$
Expanding the left side with the binomial theorem gives:
$$
\cos6\theta + \iota \sin6\theta = C^6 + 6 i C^5 S - 15 C^4 S^2 - 20 i C^3 S^3 + 15 C^2 S^4 + 6 i C S^5 - S^6 
$$
Where $C = \cos\theta$ & $S = \sin\theta$
Separating the real and imaginary parts, we get:
$$
\cos6\theta = C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6 \\
\sin6\theta = 6  C^5 S - 20  C^3 S^3 + 6  C S^5
$$
From here
$$
\tan6\theta = {6  C^5 S - 20  C^3 S^3 + 6  C S^5 \over C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6}
$$
Dividing both the numerator and the denominator by $\cos^6\theta$ or $C^6$
$$
\tan6\theta = {{[6  C^5 S - 20  C^3 S^3 + 6  C S^5]\over C^6} \over {[C^6 - 15 C^4 S^2 + 15 C^2 S^4 - S^6]\over C^6}}
$$
Which simplifies to:
$$
\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}
$$
