John Gonzalez
03/20/2024 · Middle School

10. Con los valores de los ángulos notables solucione las siguientes operaciones con todo el proceso: \( \frac{\operatorname{sen}{ }^{2} 45^{\circ}+\cos 60^{\circ}}{\csc 30^{\circ}} \) \( \left(\operatorname{tg} 60^{\circ}+\sec 30^{\circ}-\operatorname{sen} 60^{\circ}\right)^{\sec 60} \) \( \frac{4 \cdot \operatorname{sen} 30^{\circ}+\sqrt{3} \cdot \operatorname{tg} 60^{\circ}}{10 \cdot \cos 37^{\circ}+\sqrt{2} \cdot \sec 45^{\circ}} \)

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Para resolver las operaciones con los valores de los ángulos notables, primero recordemos los valores de las funciones trigonométricas para los ángulos que vamos a utilizar: - \( \sin 30^{\circ} = \frac{1}{2} \) - \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \) - \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) - \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) - \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \) - \( \cos 60^{\circ} = \frac{1}{2} \) - \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \) - \( \tan 60^{\circ} = \sqrt{3} \) - \( \sec 30^{\circ} = \frac{2}{\sqrt{3}} \) - \( \sec 60^{\circ} = 2 \) - \( \csc 30^{\circ} = 2 \) Ahora, resolvamos cada operación paso a paso. ### 1. \( \frac{\sin^2 45^{\circ} + \cos 60^{\circ}}{\csc 30^{\circ}} \) Primero, calculamos los valores de las funciones trigonométricas: - \( \sin^2 45^{\circ} = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \) - \( \cos 60^{\circ} = \frac{1}{2} \) - \( \csc 30^{\circ} = 2 \) Ahora sustituimos en la expresión: \[ \frac{\sin^2 45^{\circ} + \cos 60^{\circ}}{\csc 30^{\circ}} = \frac{\frac{1}{2} + \frac{1}{2}}{2} = \frac{1}{2} \] ### 2. \( \left(\tan 60^{\circ} + \sec 30^{\circ} - \sin 60^{\circ}\right)^{\sec 60} \) Calculamos los valores: - \( \tan 60^{\circ} = \sqrt{3} \) - \( \sec 30^{\circ} = \frac{2}{\sqrt{3}} \) - \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) - \( \sec 60^{\circ} = 2 \) Sustituyendo en la expresión: \[ \tan 60^{\circ} + \sec 30^{\circ} - \sin 60^{\circ} = \sqrt{3} + \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} \] Para simplificar, primero convertimos todos los términos a un denominador común. El denominador común es \( 2\sqrt{3} \): \[ \sqrt{3} = \frac{2\sqrt{3}}{2}, \quad \frac{2}{\sqrt{3}} = \frac{4}{2\sqrt{3}}, \quad \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \] Entonces: \[ \sqrt{3} + \frac{2}{\sqrt{3}} - \sin 60^{\circ} = \frac{2\sqrt{3}}{2} + \frac{4}{2\sqrt{3}} - \frac{\sqrt{3}}{2} = \frac{2\sqrt{3} + 4 - \sqrt{3}}{2} = \frac{\sqrt{3} + 4}{2} \] Ahora elevamos esto a la potencia de \( \sec 60^{\circ} = 2 \): \[ \left(\frac{\sqrt{3} + 4}{2}\right)^{2} = \frac{(\sqrt{3} + 4)^{2}}{4} = \frac{3 + 8\sqrt{3} + 16}{4} = \frac{19 + 8\sqrt{3}}{4} \] ### 3. \( \frac{4 \cdot \sin 30^{\circ} + \sqrt{3}

Quick Answer

1. \( \frac{\sin^2 45^{\circ} + \cos 60^{\circ}}{\csc 30^{\circ}} = \frac{1}{2} \) 2. \( \left(\tan 60^{\circ} + \sec 30^{\circ} - \sin 60^{\circ}\right)^{\sec 60} = \frac{19 + 8\sqrt{3}}{4} \) 3. \( \frac{4 \cdot \sin 30^{\circ} + \sqrt{3} \cdot \tan 60^{\circ}}{10 \cdot \cos 37^{\circ} + \sqrt{2} \cdot \sec 45^{\circ}} = \frac{2 + \sqrt{3}}{5} \)
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