Davies Salinas
11/11/2023 · Middle School

Si \( \theta \) está entre \( \pi \) y \( 2 \pi, \cos (\theta)=-\frac{\sqrt{3}}{2} \) y \( \phi \) es un ángulo agudo positivo tal \( q \) \( \cos (\phi)=\frac{7}{25} \), determine el valor exacto de \( \cos (\theta+\phi) \).

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Para resolver el problema, utilizaremos la fórmula del coseno de la suma de ángulos: \[ \cos(\theta + \phi) = \cos(\theta) \cos(\phi) - \sin(\theta) \sin(\phi) \] ### Paso 1: Determinar \(\cos(\theta)\) y \(\sin(\theta)\) Dado que \(\cos(\theta) = -\frac{\sqrt{3}}{2}\) y \(\theta\) está en el intervalo \((\pi, 2\pi)\), podemos determinar \(\sin(\theta)\). En este intervalo, el seno es negativo. Usamos la identidad fundamental: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Sustituyendo \(\cos(\theta)\): \[ \sin^2(\theta) + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1 \] \[ \sin^2(\theta) + \frac{3}{4} = 1 \] \[ \sin^2(\theta) = 1 - \frac{3}{4} = \frac{1}{4} \] Por lo tanto, \(\sin(\theta) = -\frac{1}{2}\) (ya que \(\theta\) está en el tercer o cuarto cuadrante, donde el seno es negativo). ### Paso 2: Determinar \(\cos(\phi)\) y \(\sin(\phi)\) Dado que \(\cos(\phi) = \frac{7}{25}\) y \(\phi\) es un ángulo agudo, podemos encontrar \(\sin(\phi)\) usando la misma identidad: \[ \sin^2(\phi) + \cos^2(\phi) = 1 \] Sustituyendo \(\cos(\phi)\): \[ \sin^2(\phi) + \left(\frac{7}{25}\right)^2 = 1 \] \[ \sin^2(\phi) + \frac{49}{625} = 1 \] \[ \sin^2(\phi) = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625} \] Por lo tanto, \(\sin(\phi) = \frac{24}{25}\) (ya que \(\phi\) es agudo, el seno es positivo). ### Paso 3: Sustituir en la fórmula del coseno de la suma Ahora que tenemos todos los valores necesarios: - \(\cos(\theta) = -\frac{\sqrt{3}}{2}\) - \(\sin(\theta) = -\frac{1}{2}\) - \(\cos(\phi) = \frac{7}{25}\) - \(\sin(\phi) = \frac{24}{25}\) Sustituyendo en la fórmula: \[ \cos(\theta + \phi) = \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{7}{25}\right) - \left(-\frac{1}{2}\right) \left(\frac{24}{25}\right) \] Calculamos cada término: \[ \cos(\theta + \phi) = -\frac{7\sqrt{3}}{50} + \frac{12}{25} \] Convertimos \(\frac{12}{25}\) a un denominador de 50: \[ \frac{12}{25} = \frac{24}{50} \] Por lo tanto: \[ \cos(\theta + \phi) = -\frac{7\sqrt{3}}{50} + \frac{24}{50} = \frac{24 - 7\sqrt{3}}{50} \] ### Resultado Final El valor exacto de \(\cos(\theta + \phi)\) es: \[ \boxed{\frac{24 - 7\sqrt{3}}{50}} \]

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El valor exacto de \(\cos(\theta + \phi)\) es \(\frac{24 - 7\sqrt{3}}{50}\).
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