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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.

Malcolm
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  • 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 http://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series – lab bhattacharjee Aug 31 '15 at 13:05

2 Answers2

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Hint:

$$\cos5x+i\sin5x=\left(\cos x+i\sin x\right)^{5}=$$$$\cos^{5}x-10\cos^{3}x\sin^{2}x+5\cos x\sin^{4}x+i\left(5\cos^{4}x\sin x-10\cos^{2}x\sin^{3}x+\sin^{5}x\right)$$

(De Moivre)

drhab
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  • 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
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$$\tan (3\theta + 2\theta)= (\tan 3\theta + \tan 2\theta)/(1-\tan 3\theta \tan2\theta )$$

Put the value Just expand the terms and u will get the answer

Harambe
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Ankit Sahu
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    Welcome to Math SE! Please take a look at MathJax for information about formatting your answer. As it stands it is very difficult to read. – DMcMor May 20 '17 at 19:50