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 :<.
 A: 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:
$$
\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}
$$
A: From $$(\cos{\theta}+i\sin{\theta})^6=\cos{6\theta}+i\sin{6\theta},$$ by using binomial theorem $$(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^k b^{n-k},$$ you can get $\cos{6\theta}$ as $Re\{(\cos{\theta}+i\sin{\theta})^6\}$ and $\sin{6\theta}$ as $Im\{(\cos{\theta}+i\sin{\theta})^6\}$. Now, $$\tan{6\theta}=\frac{\sin{6\theta}}{\cos{6\theta}}.$$
A: By De Moivre's theorem we have
$$\cos(6\theta)+i\sin(6\theta)=(\cos(\theta)+i\sin(\theta))^6$$
$$=-\sin^6(\theta) + \cos^6(\theta) + 6 i \sin(\theta) \cos^5(\theta) - 15 \sin^2(\theta) \cos^4(\theta) - 20 i \sin^3(\theta) \cos^3(\theta) + 15 \sin^4(\theta) \cos^2(\theta) + 6 i \sin^5(\theta) \cos(\theta)$$ by the Binomial expansion. Then comparing the real and imaginary parts we have:
$$\cos(6\theta)=-\sin^6(\theta) + \cos^6(\theta)- 15 \sin^2(\theta) \cos^4(\theta) + 15 \sin^4(\theta) \cos^2(\theta)$$
$$\sin(6\theta)= 6  \sin(\theta) \cos^5(\theta) - 20 \sin^3(\theta) \cos^3(\theta) + 6  \sin^5(\theta) \cos(\theta)$$
So we have $$\tan(5\theta)=\frac{\sin(6\theta)}{\cos(6\theta)}$$
$$=\frac{6  \sin(\theta) \cos^5(\theta) - 20 \sin^3(\theta) \cos^3(\theta) + 6  \sin^5(\theta) \cos(\theta)}{-\sin^6(\theta) + \cos^6(\theta)- 15 \sin^2(\theta) \cos^4(\theta) + 15 \sin^4(\theta) \cos^2(\theta)}$$
$$=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$
after dividing through by $\cos^6(\theta)$
