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Encuentra el valor$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}+\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=1$ $

esto siguiendo Mi solución fea: vamos

$$cos{x}+\cos{y}=?$ $ entonces$$\tan{\dfrac{x}{2}}=a,\tan{\dfrac{y}{2}}=b$ $ entonces$$\cos{x}=\dfrac{1-a^2}{1+b^2},\cos{y}=\dfrac{1-b^2}{1+b^2}$ $ y$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}=\dfrac{1}{1+\tan{x}\tan{\dfrac{y}{2}}}=\dfrac{1-a^2}{1-a^2+2ab}$ $ así que$$\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=\dfrac{1-b^2}{1-b^2+2ab}$ $ entonces$$\dfrac{1-a^2}{1-a^2+2ab}+\dfrac{1-b^2}{1-b^2+2ab}=1$ $ entonces$$\Longrightarrow (1-a^2)(1-b^2+2ab)+(1-b^2)(1-a^2+2ab)=(1-a^2+2ab)(1-b^2+2ab)$ $$$(1-b^2)(1-a^2+2ab)=2ab(1-b^2+2ab)$ $ así que$$a^2+b^2=1-3a^2b^2$ $ Mi pregunta: ¿este problema tiene buenos métodos? Gracias, porque Mis métodos son muy feos. Gracias.

Answer 1

Aquí está una ligera simplificación de su método : uno tiene

$$ \dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}=\frac{1}{1-\alpha}, \dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=\frac{1}{1-\beta} \etiqueta{1} $$

con $\alpha=\tan(x)\tan(\frac{y}{2})$$\beta=\tan(y)\tan(\frac{x}{2})$. Su ecuación se convierte entonces en $\frac{1}{1-\alpha}+\frac{1}{1-\beta}=1$, o, equivalentemente, $\alpha\beta=1$, es decir,

$$ \tan(x)\tan(\frac{x}{2})\tan(y)\tan(\frac{y}{2})=1 $$

o (si la ponemos a $a=\tan(\frac{x}{2})$$b=\tan(\frac{y}{2})$),

$$ \frac{2a^2}{1-a^2}\frac{2b^2}{1-b^2}=1 \etiqueta{2} $$

por lo $4a^2b^2=(1-a^2)(1-b^2)$ y por lo tanto $$ 3a^2b^2+a^2+b^2=1 \etiqueta{3} $$

Podemos deducir

$$ (1+a^2)(1+b^2)=1+(a^2)+(b^2)+(a^2b^2)=1+(1-3a^2b^2)+(a^2b^2)= 2(1-a^2b^2) $$

Entonces

$$ \cos(x)+\cos(y)=\frac{1-a^2}{1+a^2}+\frac{1-b^2}{1+b^2}= \frac{2(1-a^2b^2)}{(1+a^2)(1+b^2)}=\frac{2(1-a^2b^2)}{2(1-a^2b^2)}=1 $$

Answer 2

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Ahora,$$\dfrac{\cos x\cos{\dfrac y2}}{\cos\left(x-\dfrac y2\right)}+\dfrac{\cos y\cos{\dfrac x2}}{\cos\left(y-\dfrac{x}{2}\right)}=1$ (multiplicando el numerador y el denominador por$$\implies\dfrac{\cos x\cos{\dfrac y2}}{\cos\left(x-\dfrac y2\right)}=1-\dfrac{\cos y\cos{\dfrac x2}}{\cos\left(y-\dfrac{x}{2}\right)}$)

$$\implies\frac1{1+\tan x\tan\frac y2}=\frac{\sin y\sin \frac y2}{\cos\left(y-\dfrac{x}{2}\right)}=\frac1{\cot y\cot\frac x2+1}$

y de manera similar para$$\implies \tan x\tan\frac y2=\cot y\cot\frac x2\iff \tan x\tan\frac y2\tan y\tan\frac x2=1$

Answer 3

Multiplicando a través de la ecuación dada con 2

$$ \frac{2 Cos(x)Cos(\frac{y}{2})}{Cos(x-\frac{y}{2})} + \frac{2 Cos(y)Cos(\frac{x}{2})}{Cos(y-\frac{x}{2})}=2$$ $ \ implica $

$$\frac{Cos(x+\frac{y}{2})+Cos(x-\frac{y}{2})}{Cos(x-\frac{y}{2})}+\frac{Cos(y+\frac{x}{2})+Cos(y-\frac{x}{2})}{Cos(y-\frac{x}{2})}=2$$ $ \ implica $

$$\frac{Cos(x+\frac{y}{2})}{Cos(x-\frac{y}{2})}+\frac{Cos(y+\frac{x}{2})}{Cos(y-\frac{x}{2})}=0$$ $ \ implica $

$$2Cos(x+\frac{y}{2})Cos(y-\frac{x}{2})+2Cos(y+\frac{x}{2})Cos(x-\frac{y}{2})=0$$ $ \ implica $

$$Cos(\frac{x+3y}{2})+Cos(\frac{3x-y}{2})+Cos(\frac{3x+y}{2})+Cos(\frac{x-3y}{2})=0 $$ $ \ implica $

$$Cos(\frac{x}{2})Cos(\frac{3y}{2})+Cos(\frac{3x}{2})Cos(\frac{y}{2})=0$$ $ \ implica $

$$Cos(\frac{x}{2})Cos(\frac{y}{2}) \left(4Cos^2(\frac{x}{2})+4Cos^2(\frac{y}{2})-6\right)=0$$ $ \ implica $

$$ 4Cos^2(\frac{x}{2})+4Cos^2(\frac{y}{2})-6=0$$ $ \ implica $

$$ 2\left(1+Cos(x)+1+Cos(y)\right)=6$$ $ \ implica $

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