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

eyeballfrog
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Neil hawking
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  • You're going to want to give us some context for why you're trying to prove this. And, ideally, what you mean by "from a complex analysis point of view". – eyeballfrog Feb 02 '19 at 19:19
  • Apparently, the terms are the roots of $x^4 - 28 x^3 + 70 x^2 - 28 x + 1$. But that's not complex analysis... – lhf Feb 02 '19 at 19:24
  • You can check this out: https://math.stackexchange.com/questions/309271/proving-csc2-left-frac-pi7-right-csc2-left-frac2-pi7-right?rq=1 – Higurashi Feb 02 '19 at 19:33
  • See https://math.stackexchange.com/questions/951522/trig-sum-tan-21-circ-tan-22-circ-tan2-89-circ https://math.stackexchange.com/questions/175736/evaluate-tan220-circ-tan240-circ-tan280-circ https://math.stackexchange.com/questions/2137609/prove-that-sec2-20-circ-sec2-40-circ-sec2-80-circ-textrm-36 – lab bhattacharjee Feb 05 '19 at 14:10

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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 $S=28$.
Myunghyun Song
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