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1066. Найдите значение выражения \( \sqrt{50} \cos ^{2} \frac{9 \pi}{8}-\sqrt{50} \sin ^{2} \frac{9 \pi}{8} \).

Ask by Murray Potter. in Russia
Mar 12,2025

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Answer

Значение выражения равно 5.

Solution

Calculate or simplify the expression \( \sqrt{50} \cos ^{2} \frac{9 \pi}{8}-\sqrt{50} \sin ^{2} \frac{9 \pi}{8} \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{50}\times \cos^{2}\left(\frac{9\pi }{8}\right)-\sqrt{50}\times \sin^{2}\left(\frac{9\pi }{8}\right)\) - step1: Simplify the root: \(5\sqrt{2}\times \cos^{2}\left(\frac{9\pi }{8}\right)-\sqrt{50}\times \sin^{2}\left(\frac{9\pi }{8}\right)\) - step2: Simplify the root: \(5\sqrt{2}\times \cos^{2}\left(\frac{9\pi }{8}\right)-5\sqrt{2}\times \sin^{2}\left(\frac{9\pi }{8}\right)\) - step3: Transform the expression: \(5\cos^{2}\left(\frac{9\pi }{8}\right)\times \sqrt{2}-5\sin^{2}\left(\frac{9\pi }{8}\right)\times \sqrt{2}\) - step4: Transform the expression: \(5\cos^{2}\left(\frac{9\pi }{8}\right)\times \sqrt{2}-5\sqrt{2}+5\sqrt{2}\times \cos^{2}\left(\frac{9\pi }{8}\right)\) - step5: Reorder the terms: \(5\sqrt{2}\times \cos^{2}\left(\frac{9\pi }{8}\right)-5\sqrt{2}+5\sqrt{2}\times \cos^{2}\left(\frac{9\pi }{8}\right)\) - step6: Add the terms: \(10\sqrt{2}\times \cos^{2}\left(\frac{9\pi }{8}\right)-5\sqrt{2}\) - step7: Transform the expression: \(10\sqrt{2}-10\sqrt{2}\times \sin^{2}\left(\frac{9\pi }{8}\right)-5\sqrt{2}\) - step8: Subtract the numbers: \(5\sqrt{2}-10\sqrt{2}\times \sin^{2}\left(\frac{9\pi }{8}\right)\) - step9: Transform the expression: \(5\sqrt{2}-10\sin^{2}\left(\frac{9\pi }{8}\right)\times \sqrt{2}\) - step10: Transform the expression: \(5\sqrt{2}-10\sqrt{2}+10\sqrt{2}\times \cos^{2}\left(\frac{9\pi }{8}\right)\) - step11: Subtract the numbers: \(-5\sqrt{2}+10\sqrt{2}\times \cos^{2}\left(\frac{9\pi }{8}\right)\) - step12: Transform the expression: \(-5\sqrt{2}+5\sqrt{2}+5\) - step13: Remove the opposites: \(5\) Значение выражения \( \sqrt{50} \cos ^{2} \frac{9 \pi}{8}-\sqrt{50} \sin ^{2} \frac{9 \pi}{8} \) равно 5.

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Для того чтобы вычислить выражение \( \sqrt{50} \cos ^{2} \frac{9 \pi}{8}-\sqrt{50} \sin ^{2} \frac{9 \pi}{8} \), можно воспользоваться формулой приведения и свойствами тригонометрических функций. Значения косинуса и синуса угла \( \frac{9 \pi}{8} \): \[ \cos \frac{9\pi}{8} = -\cos \frac{\pi}{8}, \quad \sin \frac{9\pi}{8} = -\sin \frac{\pi}{8}. \] Теперь посчитаем \( \cos^2 \frac{9 \pi}{8} \) и \( \sin^2 \frac{9 \pi}{8} \): \[ \cos^2 \frac{9\pi}{8} = \left( -\cos \frac{\pi}{8} \right)^2 = \cos^2 \frac{\pi}{8}, \] \[ \sin^2 \frac{9\pi}{8} = \left( -\sin \frac{\pi}{8} \right)^2 = \sin^2 \frac{\pi}{8}. \] Подставляем эти значения в выражение: \[ \sqrt{50} \cos^{2} \frac{9 \pi}{8} - \sqrt{50} \sin^{2} \frac{9 \pi}{8} = \sqrt{50} \cos^{2} \frac{\pi}{8} - \sqrt{50} \sin^{2} \frac{\pi}{8}. \] Это можно переписать как: \[ \sqrt{50} (\cos^{2} \frac{\pi}{8} - \sin^{2} \frac{\pi}{8}). \] Используя формулу для разности квадратов: \[ \cos^{2} x - \sin^{2} x = \cos(2x), \] на данном этапе мы получаем: \[ \sqrt{50} \cos \left( 2 \cdot \frac{\pi}{8} \right) = \sqrt{50} \cos \frac{\pi}{4}. \] Значение \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \), поэтому: \[ \sqrt{50} \cos \frac{\pi}{4} = \sqrt{50} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{100}}{2} = \frac{10}{2} = 5. \] Таким образом, окончательное значение выражения равно: \[ \boxed{5}. \]

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