Question

$\sum_{R=1}^{7}\tan^{2}\frac{R\pi}{16}$

Answer 1

35

Answer 2

The sum $\sum_{R=1}^{7}\tan^{2}\frac{R\pi}{16}$ is the sum of the roots of the polynomial in $y=t^2$ derived from setting the numerator of $\tan(16x)$ to zero, where $t=\tan x$. The polynomial is $P(y) = \binom{16}{1} - \binom{16}{3}y + \dots - \binom{16}{15}y^7$. The roots are $\tan^2(\frac{k\pi}{16})$ for $k=1, \dots, 7$. By Vieta's formulas, the sum of these roots is $-\frac{\text{coeff of } y^6}{\text{coeff of } y^7} = -\frac{\binom{16}{13}}{-\binom{16}{15}} = \frac{\binom{16}{3}}{\binom{16}{1}} = \frac{560}{16} = 35$.