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 :<.
Best Answer
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:
$$ \tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta} $$