Задание 14. Найдите значение выражения: 1) 26√2 cos(π/4) cos(4π/3), 2) 28√2 cos(3π/4) cos(π/3), 3) 18√2 tg(π/4) sin(π/4), 5) √2 sin(7π/8) cos(7π/8), 6) 7√2 sin(15π/8) cos(15π/8), 7) 10√3 sin(7π/6) cos(7π/6), 9) 5 sin(13π/12) cos(13π/12), 10) 7 sin(17π/12) cos(17π/12)

1) $26\sqrt{2} \cos \frac{\pi}{4} \cos \frac{4\pi}{3} = 26\sqrt{2} \cdot \frac{\sqrt{2}}{2} \cdot \left(-\frac{1}{2}\right) = 26 \cdot \frac{2}{2} \cdot \left(-\frac{1}{2}\right) = 26 \cdot \left(-\frac{1}{2}\right) = -13$ 2) $28\sqrt{2} \cos \frac{3\pi}{4} \cos \frac{\pi}{3} = 28\sqrt{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) \cdot \frac{1}{2} = 28 \cdot \left(-\frac{2}{2}\right) \cdot \frac{1}{2} = -28 \cdot \frac{1}{2} = -14$ 3) $18\sqrt{2} \text{tg} \frac{\pi}{4} \sin \frac{\pi}{4} = 18\sqrt{2} \cdot 1 \cdot \frac{\sqrt{2}}{2} = 18 \cdot \frac{2}{2} = 18$ 5) Используем формулу синуса двойного угла $\sin 2\alpha = 2 \sin \alpha \cos \alpha$: $\sqrt{2} \sin \frac{7\pi}{8} \cos \frac{7\pi}{8} = \frac{\sqrt{2}}{2} \cdot \left(2 \sin \frac{7\pi}{8} \cos \frac{7\pi}{8}\right) = \frac{\sqrt{2}}{2} \sin \frac{7\pi}{4} = \frac{\sqrt{2}}{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{2}{4} = -0,5$ 6) $7\sqrt{2} \sin \frac{15\pi}{8} \cos \frac{15\pi}{8} = \frac{7\sqrt{2}}{2} \sin \frac{15\pi}{4} = \frac{7\sqrt{2}}{2} \sin \left(4\pi - \frac{\pi}{4}\right) = \frac{7\sqrt{2}}{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{7 \cdot 2}{4} = -3,5$ 7) $10\sqrt{3} \sin \frac{7\pi}{6} \cos \frac{7\pi}{6} = 5\sqrt{3} \sin \frac{7\pi}{3} = 5\sqrt{3} \sin \left(2\pi + \frac{\pi}{3}\right) = 5\sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{5 \cdot 3}{2} = 7,5$ 9) $5 \sin \frac{13\pi}{12} \cos \frac{13\pi}{12} = 2,5 \sin \frac{13\pi}{6} = 2,5 \sin \left(2\pi + \frac{\pi}{6}\right) = 2,5 \cdot \frac{1}{2} = 1,25$ 10) $7 \sin \frac{17\pi}{12} \cos \frac{17\pi}{12} = 3,5 \sin \frac{17\pi}{6} = 3,5 \sin \left(3\pi - \frac{\pi}{6}\right) = 3,5 \cdot \frac{1}{2} = 1,75$ **Ответ:** 1) -13; 2) -14; 3) 18; 5) -0,5; 6) -3,5; 7) 7,5; 9) 1,25; 10) 1,75.