¿Cómo puedo probar esta aparentemente simple identidad trigonométrica

Question

$$a = \sin\theta+\sin\phi\\b=\tan\theta+\tan\phi\\c=\sec\theta+\sec\phi$$ Muestran que, $8bc=a[4b^2 + (b^2-c^2)^2]$

Traté de resolver esto por horas y han conseguido nada. Aquí es lo que tengo hasta ahora : $$ \\a= 2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta\phi}{2}) \\ b = \frac{2\sin(\theta+\phi)}{\cos(\theta+\phi)+\cos(\theta\phi)} \\c=\frac{2(\cos\theta+\cos\phi)}{\cos(\theta+\phi)+\cos(\theta\phi)} \\a^2 = \frac{\sin^2(\theta+\phi)[\cos(\theta+\phi)+1]}{\cos(\theta+\phi)+1}\\\cos(\theta-\phi)=\frac{ca}{b}-1\\\sin^2(\frac{\theta+\phi}{2})=\frac{2a^2b}{4(ca+b)}$$

Answer 1

Sugerencia: Divida ambos lados de $8bc=a[4b^2 + (b^2-c^2)^2]$$bc$. Usted va a terminar con esto: $$8 = a[4\frac{b}{c}+bc(\frac{b}{c}-\frac{c}{b})^2]$$ Tome la CARTA y se puede demostrar que es igual a 8.

Proceso: Calcular el $\frac{b}{c}$ de las ecuaciones. $$b = \frac{\sin (\theta + \phi)}{\cos \theta. \cos \phi}$$ $$c = \frac{\cos \theta + \cos \phi}{\cos \theta. \cos \phi}$$ $$\frac{b}{c} = \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})}{2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}$$ $$\frac{b}{c} = \frac{\sin(\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})}$$ Usted va a terminar con: $$\frac{b}{c}-\frac{c}{b}=-\frac{\cos \theta. \cos \phi}{\cos(\frac{\theta - \phi}{2}).sin(\frac{\theta + \phi}{2})}$$ $$bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{\sin (\theta + \phi)(\cos \theta + \cos \phi)}{\cos^2 \theta. \cos^2 \phi}\frac{\cos^2 \theta. \cos^2 \phi}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$ $$ = \frac{2\sin(\frac{\theta + \phi}{2}).\cos(\frac{\theta + \phi}{2})2\cos(\frac{\theta + \phi}{2}).\cos(\frac{\theta - \phi}{2})}{\cos^2(\frac{\theta - \phi}{2}).\sin^2(\frac{\theta + \phi}{2})}$$ $$ = \frac{4\cos^2 (\frac{\theta + \phi}{2})}{\cos(\frac{\theta - \phi}{2})\sin (\frac{\theta + \phi}{2})}$$ $$4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2 = \frac{4}{\cos (\frac{\theta - \phi}{2})}(\sin(\frac{\theta + \phi}{2})+\frac{\cos^2 (\frac{\theta + \phi}{2})}{\sin(\frac{\theta + \phi}{2})})$$ $$= \frac{4}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$a[4\frac{b}{c} + bc(\frac{b}{c}-\frac{c}{b})^2] = 4\frac{\sin \theta + \sin \phi}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$ = 4\frac{2\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}{\cos (\frac{\theta - \phi}{2})\sin(\frac{\theta + \phi}{2})}$$ $$= 8$$

Answer 2

$\color{red}{c^2=\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi\tag{1}}$

$\color{blue}{b^2=\tan^2\theta+\tan^2\phi+2\tan\theta\tan\phi\tag{2}}$

$(1)-(2)$ da,

$\begin{align}\left(c^2-b^2\right) & =\sec^2\theta+\sec^2\phi+2\sec\theta\sec\phi-\tan^2\theta-\tan^2\phi-2\tan\theta\tan\phi\\ &=2(1+\sec\theta\sec\phi-\tan\theta\tan\phi)\\&=2\left(1+\dfrac{1}{\cos\theta\cos\phi}-\dfrac{\sin\theta\sin\phi}{\cos\theta\cos\phi}\right)\\&=\dfrac{2}{\cos\theta\cos\phi}(1+\cos(\theta+\phi))\\&=\dfrac{4\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos\theta\cos\phi}\end{align}$

$\color{darkgreen}{\therefore\left(c^2-b^2\right)^2=\dfrac{16\cos^4\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2\phi}\tag{3}}$

$\color{brown}{4b^2=\dfrac{4\sin^2(\theta+\phi)}{\cos^2\theta\cos^2 \phi}=\dfrac{16\sin^2\left(\dfrac{\theta+\phi}{2}\right)\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}\tag{4}}$

$$\boxed{4b^2+\left(b^2-c^2\right)^2=\dfrac{16\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}}$$

$c=\dfrac{\cos\theta+\cos\phi}{\cos\theta\cos\phi}=\dfrac{2\cos\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)}{\cos\theta\cos\phi}$

$b=\dfrac{\sin(\theta+\phi)}{\cos\theta\cos\phi}=\dfrac{2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta+\phi}{2}\right)}{\cos\theta\cos\phi}$

$a=2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)$

$$\boxed{\dfrac{8bc}{a}=\dfrac{32\sin\left(\dfrac{\theta+\phi}{2}\right)\cos^2\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)}{2\sin\left(\dfrac{\theta+\phi}{2}\right)\cos\left(\dfrac{\theta-\phi}{2}\right)\cos^2\theta\cos^2 \phi}=\dfrac{16\cos^2\left(\dfrac{\theta+\phi}{2}\right)}{\cos^2\theta\cos^2 \phi}}$$

Answer 3

Es mucho más conveniente para simplificar expresiones algebraicas , utilizando solamente dos variables $t$ $T$ juntos por la tangente de cada ángulo mitad que los que luchan con funciones Trigonométricas:

$$ t = \tan(\theta/2); T = \tan(\phi/2); $$

$$ sth = 2 t/(1 + t^2) ; tth = 2 t/(1 - t^2);secth = (1 + t^2)/(1 - t^2); $$

$$ sph = 2 T/(1 + T^2); tph = 2 T/(1 - T^2); secph = (1 + T^2)/(1 - T^2); $$

$$ a = sth + sph; b = tth + tph; c = secth + secph; $$

Cada una de las $ 8\,b\,c $ $ a[4b^2 + (b^2-c^2)^2] $ son llevados a un denominador común, la simplificación de ellos por separado :

$$ \dfrac{32\, (t + T)\, ( t\, T-1)^2 ( t\, T+1)} {(t^2 -1)^2 (T^2 -1)^2} $$

Answer 4

Yo uso la de la tangente de la mitad de ángulo de sustitución:

\begin{align*} t&=\tan\frac\theta2 & \sin\theta&=\frac{2t}{1+t^2} & \tan\theta&=\frac{2t}{1-t^2} & \sec\theta&=\frac{1+t^2}{1-t^2} \\ u&=\tan\frac\phi2 & \sin\phi&=\frac{2u}{1+u^2} & \tan\phi&=\frac{2u}{1-u^2} & \sec\phi&=\frac{1+u^2}{1-u^2} \end{align*}

Entonces usted tiene

\begin{align*} a &= \frac{2 t^{2} u + 2 t u^{2} + 2 t + 2 u}{t^{2} u^{2} + t^{2} + u^{2} + 1} \\ b &= \frac{-2 t^{2} u - 2 t u^{2} + 2 t + 2 u}{t^{2} u^{2} - t^{2} - u^{2} + 1} \\ c &= \frac{-2 t^{2} u^{2} + 2}{t^{2} u^{2} - t^{2} - u^{2} + 1} \end{align*}

\begin{multline*} 8bc=a[4b^2 + (b^2-c^2)^2] =\\ 32\frac{t^{4} u^{3} + t^{3} u^{4} - t^{3} u^{2} - t^{2} u^{3} - t^{2} u - t u^{2} + t + u}{t^{4} u^{4} - 2 t^{4} u^{2} - 2 t^{2} u^{4} + t^{4} + 4 t^{2} u^{2} + u^{4} - 2 t^{2} - 2 u^{2} + 1} \end{multline*}

La gran ventaja aquí es que después de ese primer paso, elegir el uso de la formulación, que casi no tiene que pensar en nada más. A partir de ahí se trata de un sencillo cálculo en funciones racionales. Dejé que mi sistema de álgebra computacional, pero si usted tiene cuidado de usted sin duda puede hacer a mano también. El hecho de que mi CAS ampliado de todos los polinomios de forma automática hace que las fórmulas de arriba quizás el aspecto más complicado de lo que realmente son, no lo he probado.

Answer 5

\begin{align*}4\cos^2(\theta)\cos^2(\phi)b^2 &= 4\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))^2 \\ &= 4(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))^2 \\ &= 4\sin^2(\theta + \phi) \\ &= 16\sin^2((\theta + \phi)/2)\cos^2((\theta + \phi)/2) \\ &= 16(1 - \cos^2((\theta + \phi)/2))\cos^2((\theta + \phi)/2) \\ &= 16\cos^2((\theta + \phi)/2) - 16\cos^4((\theta + \phi)/2)\end{align*} \begin{align*}c^2 - b^2 &= \sec^2\theta + \sec^2\phi + 2\sec(\theta)\sec(\phi) - (\tan^2\theta + \tan^2\phi + 2\tan(\theta)\tan(\phi)) \\ &= 2 + 2\sec(\theta)\sec(\phi) - 2\tan(\theta)\tan(\phi)\end{align*} \begin{align*}\cos(\theta)\cos(\phi)(c^2 - b^2) &= 2(\cos(\theta)\cos(\phi) + 1 - \sin(\theta)\sin(\phi)) \\ &= 2(1 + \cos(\theta + \phi)) \\ &= 4\cos^2((\theta + \phi)/2)\end{align*} $cos^2(\theta)\cos^2(\phi)(b^2 - c^2)^2 = 16\cos^4((\theta + \phi)/2$ $cos^2(\theta)\cos^2(\phi)(4b^2 + (b^2 - c^2)^2) = 16\cos^2((\theta + \phi)/2$ \begin{align*}8\cos^2(\theta)\cos^2(\phi)bc/a &= 8\cos^2(\theta)\cos^2(\phi)(\tan(\theta) + \tan(\phi))(\sec(\theta) + \sec(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= 8(\sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= 8\sin(\theta + \phi)(\cos(\theta) + \cos(\phi))/(\sin(\theta) + \sin(\phi)) \\ &= \frac{8(2\sin((\theta + \phi)/2)\cos((\theta + \phi)/2))(2\cos((\theta + \phi)/2)\cos((\theta - \phi)/2))}{2\sin((\theta + \phi)/2)\cos((\theta - \phi)/2)} \\ &= 16\cos^2((\theta+\phi)/2)\end{align*}

El resultado deseado de la siguiente manera.