Iv
MATU • Trigonometria
MATU_TRI_344
Guía de Ejercicios
Enunciado
Paso 1:
Si $a^3 = \csc \theta - \sin \theta$ y $b^3 = \sec \theta - \cos \theta$, demuestre que $a^2b^2(a^2 + b^2) = 1$.
Si $a^3 = \csc \theta - \sin \theta$ y $b^3 = \sec \theta - \cos \theta$, demuestre que $a^2b^2(a^2 + b^2) = 1$.
Solución Paso a Paso
1. Simplificación de las condiciones:
$$ a^3 = \frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta} \Rightarrow a^2 = \left(\frac{\cos^2 \theta}{\sin \theta}\right)^{2/3} $$
$$ b^3 = \frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \Rightarrow b^2 = \left(\frac{\sin^2 \theta}{\cos \theta}\right)^{2/3} $$
2. Cálculo de $a^2b^2$:
$$ a^2b^2 = \left(\frac{\cos^2 \theta}{\sin \theta}\right)^{2/3} \cdot \left(\frac{\sin^2 \theta}{\cos \theta}\right)^{2/3} = \left(\frac{\cos^2 \theta \sin^2 \theta}{\sin \theta \cos \theta}\right)^{2/3} = (\sin \theta \cos \theta)^{2/3} $$
3. Cálculo de $a^2 + b^2$:
$$ a^2 + b^2 = \frac{\cos^{4/3} \theta}{\sin^{2/3} \theta} + \frac{\sin^{4/3} \theta}{\cos^{2/3} \theta} = \frac{\cos^{6/3} \theta + \sin^{6/3} \theta}{\sin^{2/3} \theta \cos^{2/3} \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{(\sin \theta \cos \theta)^{2/3}} = \frac{1}{(\sin \theta \cos \theta)^{2/3}} $$
4. Sustitución en la expresión final:
$$ a^2b^2(a^2 + b^2) = (\sin \theta \cos \theta)^{2/3} \cdot \frac{1}{(\sin \theta \cos \theta)^{2/3}} = 1 $$
$$ \boxed{1 = 1} $$
$$ a^3 = \frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta} \Rightarrow a^2 = \left(\frac{\cos^2 \theta}{\sin \theta}\right)^{2/3} $$
$$ b^3 = \frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \Rightarrow b^2 = \left(\frac{\sin^2 \theta}{\cos \theta}\right)^{2/3} $$
2. Cálculo de $a^2b^2$:
$$ a^2b^2 = \left(\frac{\cos^2 \theta}{\sin \theta}\right)^{2/3} \cdot \left(\frac{\sin^2 \theta}{\cos \theta}\right)^{2/3} = \left(\frac{\cos^2 \theta \sin^2 \theta}{\sin \theta \cos \theta}\right)^{2/3} = (\sin \theta \cos \theta)^{2/3} $$
3. Cálculo de $a^2 + b^2$:
$$ a^2 + b^2 = \frac{\cos^{4/3} \theta}{\sin^{2/3} \theta} + \frac{\sin^{4/3} \theta}{\cos^{2/3} \theta} = \frac{\cos^{6/3} \theta + \sin^{6/3} \theta}{\sin^{2/3} \theta \cos^{2/3} \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{(\sin \theta \cos \theta)^{2/3}} = \frac{1}{(\sin \theta \cos \theta)^{2/3}} $$
4. Sustitución en la expresión final:
$$ a^2b^2(a^2 + b^2) = (\sin \theta \cos \theta)^{2/3} \cdot \frac{1}{(\sin \theta \cos \theta)^{2/3}} = 1 $$
$$ \boxed{1 = 1} $$