176. 1) sin 45° sin 15°; 3) cos 50° cos 15°; 5) 4 cos (α + β) cos (α – β); 7) cos (α + β) cos (2α + β); 177.1) 4 cos (π/12-x) cos (π/12 + x); 3) 2 sin (x + a) cos (x – α); 5) sin (3π/10) sin (π/10); 2) cos 35° sin 33°; 4) cos (7π/12) cos (π/12); 6) 12 sin (-9a) sin 4a; 8) 4 cos (α/2) cos a sin (3α/2); 2) 4 cos (π/6 + x) sin (π/3 - x); 4) 4 sin 16a sin 4a; 6) cos 7x cos 5x. 178. 1) sin⁴ x; 2) cos⁴ x; 3) sin⁵ x. 179*. 1) sin 2a sin a = 1/2 (cos a - cos 3α); 2) sin 3a cos α = 1/2 (sin 2x + sin 4g):

Давай разберем эти примеры по тригонометрии. Здесь нужно применить формулы преобразования произведения тригонометрических функций в сумму. 176. 1) \( \sin 45^\circ \sin 15^\circ = \frac{1}{2} [\cos(45^\circ - 15^\circ) - \cos(45^\circ + 15^\circ)] = \frac{1}{2} [\cos 30^\circ - \cos 60^\circ] = \frac{1}{2} [\frac{\sqrt{3}}{2} - \frac{1}{2}] = \frac{\sqrt{3} - 1}{4} \) 2) \( \cos 35^\circ \sin 33^\circ = \frac{1}{2} [\sin(35^\circ + 33^\circ) - \sin(35^\circ - 33^\circ)] = \frac{1}{2} [\sin 68^\circ - \sin 2^\circ] \) 3) \( \cos 50^\circ \cos 15^\circ = \frac{1}{2} [\cos(50^\circ - 15^\circ) + \cos(50^\circ + 15^\circ)] = \frac{1}{2} [\cos 35^\circ + \cos 65^\circ] \) 4) \( \cos \frac{7\pi}{12} \cos \frac{\pi}{12} = \frac{1}{2} [\cos(\frac{7\pi}{12} - \frac{\pi}{12}) + \cos(\frac{7\pi}{12} + \frac{\pi}{12})] = \frac{1}{2} [\cos \frac{\pi}{2} + \cos \frac{2\pi}{3}] = \frac{1}{2} [0 - \frac{1}{2}] = -\frac{1}{4} \) 5) \( 4 \cos(\alpha + \beta) \cos(\alpha - \beta) = 4 \cdot \frac{1}{2} [\cos((\alpha + \beta) - (\alpha - \beta)) + \cos((\alpha + \beta) + (\alpha - \beta))] = 2 [\cos(2\beta) + \cos(2\alpha)] \) 6) \( 12 \sin(-9\alpha) \sin(4\alpha) = 12 \cdot \frac{1}{2} [\cos(-9\alpha - 4\alpha) - \cos(-9\alpha + 4\alpha)] = 6 [\cos(-13\alpha) - \cos(-5\alpha)] = 6 [\cos(13\alpha) - \cos(5\alpha)] \) 7) \( \cos(\alpha + \beta) \cos(2\alpha + \beta) = \frac{1}{2} [\cos((\alpha + \beta) - (2\alpha + \beta)) + \cos((\alpha + \beta) + (2\alpha + \beta))] = \frac{1}{2} [\cos(-\alpha) + \cos(3\alpha + 2\beta)] = \frac{1}{2} [\cos(\alpha) + \cos(3\alpha + 2\beta)] \) 8) \( 4 \cos \frac{\alpha}{2} \cos \alpha \sin \frac{3\alpha}{2} = 2 [\cos(\alpha - \frac{\alpha}{2}) + \cos(\alpha + \frac{\alpha}{2})] \sin \frac{3\alpha}{2} = 2 [\cos \frac{\alpha}{2} + \cos \frac{3\alpha}{2}] \sin \frac{3\alpha}{2} = 2 \cos \frac{\alpha}{2} \sin \frac{3\alpha}{2} + 2 \cos \frac{3\alpha}{2} \sin \frac{3\alpha}{2} = [\sin(\frac{3\alpha}{2} + \frac{\alpha}{2}) - \sin(\frac{\alpha}{2} - \frac{3\alpha}{2})] + \sin(\frac{3\alpha}{2} + \frac{3\alpha}{2}) = [\sin(2\alpha) - \sin(-\alpha)] + \sin(3\alpha) = \sin(2\alpha) + \sin(\alpha) + \sin(3\alpha) \) 177. 1) \( 4 \cos(\frac{\pi}{12} - x) \cos(\frac{\pi}{12} + x) = 4 \cdot \frac{1}{2} [\cos((\frac{\pi}{12} - x) - (\frac{\pi}{12} + x)) + \cos((\frac{\pi}{12} - x) + (\frac{\pi}{12} + x))] = 2 [\cos(-2x) + \cos(\frac{\pi}{6})] = 2 [\cos(2x) + \frac{\sqrt{3}}{2}] = 2 \cos(2x) + \sqrt{3} \) 2) \( 4 \cos(\frac{\pi}{6} + x) \sin(\frac{\pi}{3} - x) = 4 \cdot \frac{1}{2} [\sin(\frac{\pi}{6} + x + \frac{\pi}{3} - x) - \sin(\frac{\pi}{6} + x - (\frac{\pi}{3} - x))] = 2 [\sin(\frac{\pi}{2}) - \sin(2x - \frac{\pi}{6})] = 2 [1 - \sin(2x - \frac{\pi}{6})] \) 3) \( 2 \sin(x + \alpha) \cos(x - \alpha) = 2 \cdot \frac{1}{2} [\sin((x + \alpha) + (x - \alpha)) + \sin((x + \alpha) - (x - \alpha))] = \sin(2x) + \sin(2\alpha) \) 4) \( 4 \sin(16\alpha) \sin(4\alpha) = 4 \cdot \frac{1}{2} [\cos(16\alpha - 4\alpha) - \cos(16\alpha + 4\alpha)] = 2 [\cos(12\alpha) - \cos(20\alpha)] \) 5) \( \sin \frac{3\pi}{10} \sin \frac{\pi}{10} = \frac{1}{2} [\cos(\frac{3\pi}{10} - \frac{\pi}{10}) - \cos(\frac{3\pi}{10} + \frac{\pi}{10})] = \frac{1}{2} [\cos(\frac{\pi}{5}) - \cos(\frac{2\pi}{5})] \) 6) \( \cos(7x) \cos(5x) = \frac{1}{2} [\cos(7x - 5x) + \cos(7x + 5x)] = \frac{1}{2} [\cos(2x) + \cos(12x)] \) 178. 1) \( \sin^4 x = \frac{1}{8} (3 - 4 \cos 2x + \cos 4x) \) 2) \( \cos^4 x = \frac{1}{8} (3 + 4 \cos 2x + \cos 4x) \) 3) \( \sin^5 x = \frac{1}{16} (10 \sin x - 5 \sin 3x + \sin 5x) \) 179*. 1) \( \sin 2\alpha \sin \alpha = \frac{1}{2} (\cos \alpha - \cos 3\alpha) \) 2) \( \sin 3\alpha \cos \alpha = \frac{1}{2} (\sin 2\alpha + \sin 4\alpha) \)

Ответ: Решения выше.

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