Demostrar que $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$

Question

Si $\alpha + \beta + \gamma = \pi $ $\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1$

Demostrar que:

$1 + \cos\alpha + \cos\beta + \cos\gamma = 0$.

No tengo idea de cómo ir sobre esto.

Por favor, ayudar.

Answer 1

Como $\displaystyle \frac{-\alpha+\beta+\gamma}4=\frac{-\alpha+\pi-\alpha}4=\frac\pi4-\frac\alpha2$

$\displaystyle \tan\frac{-\alpha+\beta+\gamma}4=\tan\left(\frac\pi4-\frac\alpha2\right)=\frac{1-\tan\frac\alpha2}{1+\tan\frac\alpha2}=\frac{\cos\frac\alpha2-\sin\frac\alpha2}{\cos\frac\alpha2+\sin\frac\alpha2}$

$\displaystyle\implies \tan^2\left(\frac{-\alpha+\beta+\gamma}4\right)=\left(\frac{\cos\frac\alpha2-\sin\frac\alpha2}{\cos\frac\alpha2+\sin\frac\alpha2}\right)^2=\frac{1-\sin\alpha}{1+\sin\alpha}$

$\displaystyle\implies \prod\left(\frac{1-\sin\alpha}{1+\sin\alpha}\right)=1$

$\displaystyle\implies\sum\sin\alpha+\prod\sin\alpha=0 $

Ahora establezca $\displaystyle\sin\alpha=2\sin\frac{\alpha}2\cos\frac{\alpha}2$ $\displaystyle\prod\sin\alpha$

y utilizar este para $\sum\sin\alpha$

La cancelación $\displaystyle\prod\cos\frac{\alpha}2,$ (asumiendo $\displaystyle\prod\cos\frac{\alpha}2\ne0$ )

llegamos $\displaystyle4\prod \sin\frac{\alpha}2=-2$

Por último, el uso de este, $\displaystyle\sum \cos A=1+4\prod\sin\frac{\alpha}2$

Answer 2

Un poco de Generalización :

Deje $\alpha+\beta+\gamma=4C$

y $\displaystyle\tan\left(\frac{-\alpha + \beta + \gamma}4\right)\tan\left(\frac{\alpha - \beta + \gamma}4\right)\tan\left(\frac{\alpha + \beta - \gamma}4\right) = \cot C$

Como $\displaystyle \frac{-\alpha+\beta+\gamma}4=\frac{-\alpha+4C-\alpha}4=C-\frac\alpha2,$

$\displaystyle \tan\frac{-\alpha+\beta+\gamma}4=\tan\left(C-\frac\alpha2\right)=\frac{\sin\left(C-\frac\alpha2\right)}{\cos\left(C-\frac\alpha2\right)}$

Así, el problema se reduce a $\displaystyle\frac{\sin\left(C-\frac\alpha2\right)\sin\left(C-\frac\beta2\right)\sin\left(C-\frac\gamma2\right)}{\cos\left(C-\frac\alpha2\right)\cos\left(C-\frac\beta2\right)\cos\left(C-\frac\gamma2\right)}=\cot C$

$\displaystyle\implies\frac{\sin\left(C-\frac\alpha2\right)\sin\left(C-\frac\beta2\right)}{\cos\left(C-\frac\alpha2\right)\cos\left(C-\frac\beta2\right)} =\frac{\cos\left(C-\frac\gamma2\right)\cos C}{\sin\left(C-\frac\gamma2\right)\pecado C}$

La aplicación de $\displaystyle2\sin A\sin B,2\cos A\cos B$ fórmula,

$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2-\cos\frac{4C-\beta-\alpha}2}{\cos\frac{\beta-\alpha}2+\cos\frac{4C-\beta-\alpha}2} =\frac{\cos\left(-\frac\gamma2\right)+\cos\frac{4C-\gamma}2}{\cos\left(-\frac\gamma2\right)-\cos\frac{4C-\gamma}2}$

Ahora como $\displaystyle\alpha+\beta+\gamma=4C,$ esto se convierte en

$\displaystyle\implies\frac{\cos\frac{\beta-\alpha}2-\cos\frac{\gamma}2}{\cos\frac{\beta-\alpha}2+\cos\frac{\gamma}2} =\frac{\cos\frac\gamma2+\cos\frac{\alpha+\beta}2}{\cos\frac\gamma2-\cos\frac{\alpha+\beta}2}$

La aplicación de Componendo y de la fundación " dividendo,

$\displaystyle\implica\frac{\cos\frac{\beta\alpha}2}{-\cos\frac{\gamma}2} =\frac{\cos\frac{\alpha+\beta}2}{\cos\frac{\alpha+\beta}2}$

$\displaystyle\implies\cos\frac{\beta-\alpha}2\cos\frac{\beta+\alpha}2=-\cos^2\frac{\gamma}2 $

La aplicación de $2\cos A\cos B,\cos2x=2\cos^2x-1,$ $\displaystyle\frac{\cos\alpha+\cos\beta}2=-\frac{1+\cos\gamma}2$

Aquí $\displaystyle C=\frac\pi4\implies \alpha+\beta+\gamma=4C=\pi$

Si $\displaystyle C=-\frac\pi4\implies \alpha+\beta+\gamma=4C=-\pi\equiv\pi\pmod{2\pi}$

Por eso, $\displaystyle\tan\left(\frac{-\alpha + \beta + \gamma}4\right)\tan\left(\frac{\alpha - \beta + \gamma}4\right)\tan\left(\frac{\alpha + \beta - \gamma}4\right)=-1$ $\alpha+\beta+\gamma=-\pi$ o $\pi$ va a satisfacer la necesaria identidad

Answer 3

Vamos a introducir las notaciones de taquigrafía $$ \begin{align} a &= \frac{-\alpha+\beta+\gamma}{4} = \frac{\pi}{4} - \frac{\alpha}{2},\\ b &= \frac{\alpha-\beta+\gamma}{4} = \frac{\pi}{4} - \frac{\beta}{2},\\ c &= \frac{\alpha+\beta-\gamma}{4} = \frac{\pi}{4} - \frac{\gamma}{2}, \end{align} $$ entonces $$ a+b = \frac{\gamma}{2},\quad a+c = \frac{\beta}{2},\quad b+c = \frac{\alpha}{2}, $$ $$ a+b+c = \frac{\pi}{4}, $$ $$ \sin\pecado b \sen c = \cos\cos b\cos c.\la etiqueta{1} $$ Ahora $$ \cos(a+b+c) = \cos(a+b)\cos(c) - \sin(a+b)\sen c= \frac{\sqrt{2}}{2}, $$ así que $$ \cos(a+b)\cos(c) - \sin\pecado c\cos b - \pecado b\pecado c\cos a = \frac{\sqrt{2}}{2} $$ agregar $(2\cos a\cos b\cos c)$ a ambos lados, y obtenemos $$ \cos(a+b)\cos(c) + \cos(a+c)\cos b+ \cos(b+c)\cos(a)= \frac{\sqrt{2}}{2} + 2\cos a\cos b\cos c, $$ lo que se reduce a $$ \begin{multline} \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\beta}{2}\right) +\\ \cos\left(\frac{\gamma}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\gamma}{2}\right) = \frac{\sqrt{2}}{2} + 2\cos a\cos b\cos c. \tag{2} \end{multline} $$ Por otro lado, $$ \sin(a+b+c) = \sin(a+b)\cos c + \cos(a+b)\sen c = \frac{\sqrt{2}}{2}, $$ de modo que, $$ \sin(a+b)\cos c + \cos\cos b\sen c = \frac{\sqrt{2}}{2} + \sin\pecado b\pecado c.\la etiqueta{3} $$ Del mismo modo, si tomamos $\sin(a+b+c) = \sin((a+c)+b)$, obtenemos $$ \sin(a+c)\cos b + \cos\cos c\sen b = \frac{\sqrt{2}}{2} + \sin\pecado b\pecado c.\la etiqueta{4} $$ Si añadimos nca. $(3)$ $(4)$ , obtenemos $$ \sin(a+b)\cos c + \sin(a+c)\cos b + \sin(b+c)\cos a = \sqrt{2} + 2\sin\pecado b\pecado c, $$ lo que se reduce a (usando también se $(1)$) $$ \begin{multline} \sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \sin\left(\frac{\beta}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\beta}{2}\right) +\\ \sin\left(\frac{\gamma}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\gamma}{2}\right) = \sqrt{2} + 2\cos a\cos b\cos c. \tag{5} \end{multline} $$ Por último, nos resta eq $(5)$ de eq $(2)$: $$ \begin{multline} \left(\cos\left(\frac{\alpha}{2}\right)-\sin\left(\frac{\alpha}{2}\right)\right) \cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) + \ldots = - \frac{\sqrt{2}}{2}, \end{multline} $$ que es $$ \begin{multline} \frac{\sqrt{2}}{2}\left(\cos\left(\frac{\alpha}{2}\right)-\sin\left(\frac{\alpha}{2}\right)\right)\left(\cos\left(\frac{\alpha}{2}\right)+\sin\left(\frac{\alpha}{2}\right)\right) + \ldots + \frac{\sqrt{2}}{2} =0, \end{multline} $$ así que, finalmente, $$ \cos\alpha +\cos\beta+\cos\gamma + 1 = 0. $$