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How do you find the sum: $$\sum_{r=1}^6 \tan^2\left(\frac{r \pi}{n}\right)$$

I managed to solve this question using complex numbers so I thought I'd share the solution. If you know of any better method of solving it, please do share.

wythagoras
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Ashish Gupta
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1 Answers1

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Let $x=\frac{\pi}{7}:$ $$(\cos x + i \sin x)^7 = \cos 7x + i \sin 7x$$

Since $\sin 7x$ = Imaginary part of $(\cos x + i \sin x)^7$

$$\sin7x = \sum_{r=0}^3 \binom{7}{2r+1}(\cos x)^{7-(2r+1)} \cdot(i\sin x)^{2r+1} $$

We also know that $\sin 7x = 0$ $$0 = \binom{7}{1}(\cos x)^{6} (\sin x)^{1} -\binom{7}{3}(\cos x)^{4} (\sin x)^{3}+\binom{7}{5}(\cos x)^{2} (\sin x)^{5}-\binom{7}{7}(\cos x)^{0} (\sin x)^{7}$$

$$0 = 7(\cos x)^{6} (\sin x)^{1} -35(\cos x)^{4} (\sin x)^{3}+21(\cos x)^{2} (\sin x)^{5}-(\cos x)^{0} (\sin x)^{7}$$

$$0 = 7\tan x - 35\tan^3 x + 21\tan^5 x - \tan^7 x$$

$$0 = \tan^6 x - 21\tan^4 x + 35\tan^2 x -7$$

Since the roots of the above equation are $\tan \frac{\pi}{7},\tan \frac{2\pi}{7},\tan \frac{3\pi}{7},\tan \frac{4\pi}{7},\tan \frac{5\pi}{7},\tan \frac{6\pi}{7}$ and $\tan \frac{\pi}{7}$=$-\tan \frac{6\pi}{7}$, on pairing the roots of the equation, we have:

$$\left(\tan^2 x - \tan^2 \frac{\pi}{7}\right)\left(\tan^2 x - \tan^2 \frac{2\pi}{7}\right)\left(\tan^2 x - \tan^2 \frac{3\pi}{7}\right)=0$$

Hence we have an equation in $\tan^2 x$. The given question requires us to find 2 times the sum of the roots, since $$\sum_{r=1}^6 tan^2\left(\frac{r \pi}{n}\right)=2\sum_{r=1}^3 tan^2\left(\frac{r \pi}{n}\right)$$

Thus we find the required sum as: $$2\left(\frac{-b}{a}\right) = 2 \cdot21 =42$$

Ashish Gupta
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