IV) Преобразование числовых тригонометрических выражений Задание 14. Найдите значение выражения 1) \(26\sqrt{2}cos\frac{\pi}{4}cos\frac{4\pi}{3}\) 2) \(28\sqrt{2}cos\frac{3\pi}{4}cos\frac{\pi}{3}\) 3) \(18\sqrt{2}tg\frac{\pi}{4}sin\frac{\pi}{4}\) 4) \(12\sqrt{2}tg\frac{5\pi}{4}sin\frac{3\pi}{4}\) 5) \(\sqrt{2}sin\frac{7\pi}{8}cos\frac{7\pi}{8}\) 6) \(7\sqrt{2}sin\frac{15\pi}{8}cos\frac{15\pi}{8}\) 7) \(10\sqrt{3}sin\frac{7\pi}{6}cos\frac{7\pi}{6}\) 8) \(6\sqrt{3}sin\frac{5\pi}{6}cos\frac{5\pi}{6}\) 9) \(5sin\frac{13\pi}{12}cos\frac{13\pi}{12}\) 10) \(7sin\frac{17\pi}{12}cos\frac{17\pi}{12}\)

1) \(26\sqrt{2}cos\frac{\pi}{4}cos\frac{4\pi}{3} = 26\sqrt{2} \cdot \frac{\sqrt{2}}{2} \cdot (-\frac{1}{2}) = 26 \cdot \frac{2}{2} \cdot (-\frac{1}{2}) = -13\)

2) \(28\sqrt{2}cos\frac{3\pi}{4}cos\frac{\pi}{3} = 28\sqrt{2} \cdot (-\frac{\sqrt{2}}{2}) \cdot \frac{1}{2} = 28 \cdot (-\frac{2}{2}) \cdot \frac{1}{2} = -14\)

3) \(18\sqrt{2}tg\frac{\pi}{4}sin\frac{\pi}{4} = 18\sqrt{2} \cdot 1 \cdot \frac{\sqrt{2}}{2} = 18 \cdot \frac{2}{2} = 18\)

4) \(12\sqrt{2}tg\frac{5\pi}{4}sin\frac{3\pi}{4} = 12\sqrt{2} \cdot 1 \cdot \frac{\sqrt{2}}{2} = 12 \cdot \frac{2}{2} = 12\)

5) \(\sqrt{2}sin\frac{7\pi}{8}cos\frac{7\pi}{8} = \frac{\sqrt{2}}{2} \cdot 2sin\frac{7\pi}{8}cos\frac{7\pi}{8} = \frac{\sqrt{2}}{2} sin(2 \cdot \frac{7\pi}{8}) = \frac{\sqrt{2}}{2} sin\frac{7\pi}{4} = \frac{\sqrt{2}}{2} \cdot (-\frac{\sqrt{2}}{2}) = -\frac{1}{2}\)

6) \(7\sqrt{2}sin\frac{15\pi}{8}cos\frac{15\pi}{8} = \frac{7\sqrt{2}}{2} \cdot 2sin\frac{15\pi}{8}cos\frac{15\pi}{8} = \frac{7\sqrt{2}}{2} sin(2 \cdot \frac{15\pi}{8}) = \frac{7\sqrt{2}}{2} sin\frac{15\pi}{4} = \frac{7\sqrt{2}}{2} sin(4\pi - \frac{\pi}{4}) = \frac{7\sqrt{2}}{2} \cdot (-\frac{\sqrt{2}}{2}) = -\frac{7}{2}\)

7) \(10\sqrt{3}sin\frac{7\pi}{6}cos\frac{7\pi}{6} = 5\sqrt{3} \cdot 2sin\frac{7\pi}{6}cos\frac{7\pi}{6} = 5\sqrt{3} sin(2 \cdot \frac{7\pi}{6}) = 5\sqrt{3} sin\frac{7\pi}{3} = 5\sqrt{3} sin(2\pi + \frac{\pi}{3}) = 5\sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{15}{2}\)

8) \(6\sqrt{3}sin\frac{5\pi}{6}cos\frac{5\pi}{6} = 3\sqrt{3} \cdot 2sin\frac{5\pi}{6}cos\frac{5\pi}{6} = 3\sqrt{3} sin(2 \cdot \frac{5\pi}{6}) = 3\sqrt{3} sin\frac{5\pi}{3} = 3\sqrt{3} \cdot (-\frac{\sqrt{3}}{2}) = -\frac{9}{2}\)

9) \(5sin\frac{13\pi}{12}cos\frac{13\pi}{12} = \frac{5}{2} \cdot 2sin\frac{13\pi}{12}cos\frac{13\pi}{12} = \frac{5}{2} sin(2 \cdot \frac{13\pi}{12}) = \frac{5}{2} sin\frac{13\pi}{6} = \frac{5}{2} sin(2\pi + \frac{\pi}{6}) = \frac{5}{2} \cdot \frac{1}{2} = \frac{5}{4}\)

10) \(7sin\frac{17\pi}{12}cos\frac{17\pi}{12} = \frac{7}{2} \cdot 2sin\frac{17\pi}{12}cos\frac{17\pi}{12} = \frac{7}{2} sin(2 \cdot \frac{17\pi}{12}) = \frac{7}{2} sin\frac{17\pi}{6} = \frac{7}{2} sin(2\pi + \frac{5\pi}{6}) = \frac{7}{2} \cdot \frac{1}{2} = \frac{7}{4}\)

Ответ: 1) -13; 2) -14; 3) 18; 4) 12; 5) -1/2; 6) -7/2; 7) 15/2; 8) -9/2; 9) 5/4; 10) 7/4