Encuentre $\tan(π/16)+\tan(5π/16)+\tan(9π/16)+\tan(13π/16)$

Question

Tuve que calcular $\tan(/16)+\tan(5/16)+\tan(9/16)+\tan(13/16)$

Intenté escribir $/16=x$ y luego escribir la suma como $\tan x+\tan(9x-4x)+\tan9x+\tan(9x+4x)$ y luego simplificando con $\tan(x+y)$ y $\tan(x-y)$ pero no se simplificó a nada, puede alguien por favor dar una pista sobre cómo debo abordar esto.

Answer 1

$$a=\tan(\frac{\pi}{16})+\tan(\frac{5\pi}{16})+\tan(\frac{9\pi}{16})+\tan(\frac{13\pi}{16})\\= \tan(\frac{\pi}{16})+\tan(\frac{5\pi}{16})+\tan(\frac{1\pi}{16}+\frac{\pi}{2})+\tan(\frac{5\pi}{16}+\frac{\pi}{2})\\=\\ \tan(\frac{\pi}{16})-\cot(\frac{\pi}{16})+\tan(\frac{5\pi}{16})-\cot(\frac{5\pi}{16})\\=?$$ ahora permítanme señalar que $\color{red} {2\cot(2x)=\cot x- \tan x}\tag{1}$ así que $$a=-(-\tan(\frac{\pi}{16})+\cot(\frac{\pi}{16})-\tan(\frac{5\pi}{16})+\cot(\frac{5\pi}{16}))\\= -(2\cot(\frac{2\pi}{16})+2\cot(\frac{10\pi}{16}))\\=$$ ¿puede seguir? $$a=-2(\cot(\frac{\pi}{8})+\cot(\frac{5\pi}{8}))\\=-2(\cot(\frac{\pi}{8})+\cot(\frac{\pi}{8}+\frac{\pi}{2}))\\\text{from formula (1)}\\ -2(\cot(\frac{\pi}{8})-\tan(\frac{\pi}{8}))\\=-2(2\cot(2(\frac{\pi}{8}))\\=-4$$

Answer 2

Una pista: $$\tan\left(x+\pi\right) = \tan(x).$$ Si $x=\frac{3\pi}{16}$ y $y=\frac{2\pi}{16}$ entonces $x+y = ?$ y $x-y=?$ .