From Complex Analysis point of view

Question

How to prove the following equation: $$\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\tan^2\left(\frac{5\pi}{16}\right)+\tan^2\left(\frac{7\pi}{16}\right)=28$$

Answer 1

It can also be calculated by applying residue theorem to the function $$f(z) = \tan (8z)\tan^2(z) $$ and the box contour $\{\Re (z)=0 \}\cup\{\Re (z)=\pi \}\cup \{t+\pm\infty i:0\le t\le \pi\}$. Then,

• $\int_{\Re(z)=0}f(z)dz$ and $\int_{\Re(z)=\pi}f(z)dz$ are cancelled.
  
• $\lim_{y\to \pm\infty}\int_{0}^\pi f(t+iy)dt =\int_0^\pi (\pm i)^3dt =\mp\pi i.$
  
• Sum of residues at $z=\frac{2k+1}{16}\pi$, $k=0,1,\ldots, 7$ is equal to $-\frac{S}{4}$ where $S$ is the given sum.
• Residue at $z=\frac{\pi}{2}$ is equal to $$ \lim_{z\to \frac\pi 2} \frac d {dz}\left(z-\frac{\pi}2 \right)^2f(z) = 8. $$ Combining them, by residue theorem we have $$ 2\pi i=2\pi i\left(-\frac S 4+8\right) $$which gives $S=28$.