Ii
MATU • Trigonometria
MATU_TREC_043
Compendio de Trigonometría
Enunciado
Paso 1:
Si: $\frac{1 - 2 \cos^2 \theta}{sen \, \theta + \cos \theta} = \frac{1}{2}$, determine: $E = sen \, \theta \cos \theta$
Si: $\frac{1 - 2 \cos^2 \theta}{sen \, \theta + \cos \theta} = \frac{1}{2}$, determine: $E = sen \, \theta \cos \theta$
Solución Paso a Paso
1. Simplificación de la ecuación:
Sabemos que $\cos 2\theta = 2 \cos^2 \theta - 1$, por lo tanto $1 - 2 \cos^2 \theta = -\cos 2\theta$.
Además, $\cos 2\theta = (\cos \theta - sen \, \theta)(\cos \theta + sen \, \theta)$.
$$\frac{-(\cos \theta - sen \, \theta)(\cos \theta + sen \, \theta)}{sen \, \theta + \cos \theta} = \frac{1}{2}$$
$$-( \cos \theta - sen \, \theta ) = 1/2 \implies sen \, \theta - \cos \theta = 1/2$$
2. Elevando al cuadrado:
$$(sen \, \theta - \cos \theta)^2 = (1/2)^2$$
$$sen^2 \theta - 2 sen \, \theta \cos \theta + \cos^2 \theta = 1/4$$
$$1 - 2 sen \, \theta \cos \theta = 1/4$$
3. Despeje de E:
$$2 sen \, \theta \cos \theta = 1 - 1/4 = 3/4$$
$$E = sen \, \theta \cos \theta = 3/8$$
Sabemos que $\cos 2\theta = 2 \cos^2 \theta - 1$, por lo tanto $1 - 2 \cos^2 \theta = -\cos 2\theta$.
Además, $\cos 2\theta = (\cos \theta - sen \, \theta)(\cos \theta + sen \, \theta)$.
$$\frac{-(\cos \theta - sen \, \theta)(\cos \theta + sen \, \theta)}{sen \, \theta + \cos \theta} = \frac{1}{2}$$
$$-( \cos \theta - sen \, \theta ) = 1/2 \implies sen \, \theta - \cos \theta = 1/2$$
2. Elevando al cuadrado:
$$(sen \, \theta - \cos \theta)^2 = (1/2)^2$$
$$sen^2 \theta - 2 sen \, \theta \cos \theta + \cos^2 \theta = 1/4$$
$$1 - 2 sen \, \theta \cos \theta = 1/4$$
3. Despeje de E:
$$2 sen \, \theta \cos \theta = 1 - 1/4 = 3/4$$
$$E = sen \, \theta \cos \theta = 3/8$$