$\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$

Question

Encuentra el valor de la expresión $\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$

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He identificado que $\frac{\pi}{16}+\frac{13\pi}{16}=\frac{5\pi}{16}+\frac{9\pi}{16}=\frac{14\pi}{16}$
$\tan(\frac{\pi}{16}+\frac{13\pi}{16})=\tan(\frac{5\pi}{16}+\frac{9\pi}{16})$
$\frac{\tan\frac{\pi}{16}+\tan\frac{13\pi}{16}}{1-\tan\frac{\pi}{16}\tan\frac{13\pi}{16}}=\frac{\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}}{1-\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}}$
Pero estoy atascado aquí. Por favor, ayúdame. Gracias.

Answer 1

dejar $\frac{\pi}{16}=A$ entonces $tan13A=-tan3A$ y $tan9A=-tan7A$

así que

$$S=tanA-tan3A+tan5A-tan7A=\frac{-sin2A}{cosAcos3A}+\frac{-sin2A}{cos5Acos7A}$$

$$S=-2sin2A\left(\frac{1}{2cosAcos3A}+\frac{1}{2cos5Acos7A}\right)$$

$$S=-2sin2A\left(\frac{1}{cos4A+cos2A}+\frac{1}{cos12A+cos2A}\right)$$

$$S=-2sin2A\left(\frac{1}{\frac{1}{\sqrt{2}}+cos2A}+\frac{1}{cos2A-\frac{1}{\sqrt{2}}}\right)$$

$$S=\frac{-4sin2Acos2A}{cos^22A-\frac{1}{2}}=\frac{-8sin2Acos2A}{2cos^22A-1}=\frac{-4sin4A}{cos4A}=-4$$

Answer 2

Como $\tan4\left(\dfrac\pi4+x\right)=\tan(\pi+4x)=\tan4x,$

Si $\tan4x=\tan4A,4x=n\pi+4A\implies x=\dfrac{n\pi}4+A$ donde $n=0,1,2,3$

como $$\tan4x=\dfrac{4\tan x-\binom41\tan^3x}{1-\binom42\tan^2x+\tan^4x}$$

$$\dfrac{4\tan x-\binom41\tan^3x}{1-\binom42\tan^2x+\tan^4x}=\tan4A$$

$$\iff\tan4A\tan^4x+4\tan^3x-6\tan4A\tan^2x-4\tan x+\tan4A=0$$

$$\implies\sum_{r=0}^1\tan\left(r\dfrac\pi4+x\right)=-\dfrac4{\tan4A}$$

Aquí $4A=\dfrac\pi4$