# Prove that $\tan5 \theta = \frac {5\tan \theta -10 \tan ^3 \theta +\tan ^5 \theta} {1-10\tan ^2 \theta +5\tan ^4 \theta}$

As the title suggests, what is required to prove is that $$\tan5 \theta = \frac {5\tan \theta -10 \tan ^3 \theta +\tan ^5 \theta} {1-10\tan ^2 \theta +5\tan ^4 \theta}$$

I was looking back through my old high school tests and came across this monster and have now -as I did then- no idea where to start with it. This was a test on a 'Complex numbers' unit so using complex numbers is required to prove it (which sadly, I'm quite out of practice on). I'd be really interested to see a solution to problem which has been on the back of my mind ever since I put my eyes on it for the first time. I'm assuming the first step should be to open $\tan 5\theta = \frac {\sin 5\theta} {\cos 5\theta}$ but I could be mistaken.

• Try considering the real and imaginary parts on both sides of De Moivre's formula for $n = 5$: $$\cos 5\theta + i \sin 5\theta =(\cos \theta + i\sin\theta)^5$$ and using Binomial expansion for the right hand side. – Yiyuan Lee Aug 31 '15 at 11:42
• @Malcom, See math.stackexchange.com/questions/346368/… – lab bhattacharjee Aug 31 '15 at 13:05

$$\cos5x+i\sin5x=\left(\cos x+i\sin x\right)^{5}=$$$$\cos^{5}x-10\cos^{3}x\sin^{2}x+5\cos x\sin^{4}x+i\left(5\cos^{4}x\sin x-10\cos^{2}x\sin^{3}x+\sin^{5}x\right)$$
• Although equivalent, it's computationally a bit simpler to consider the argument of $(1+i\tan(x))^5$ – robjohn Aug 31 '15 at 12:15
$$\tan (3\theta + 2\theta)= (\tan 3\theta + \tan 2\theta)/(1-\tan 3\theta \tan2\theta )$$