1)1/2 sin 717/12 cos 717/12 2)√3'(sin² 25П/12-cos² 25П/12) 3) √8 '(sin² 9П/8 -cos² 9П/8) 4) √50' sin² 11П/8 -√50 cos² 11П/8 5) √48' -√192'cos² 23П/12 6)√3'-√12' sin² 7П/12

Решение:

1) \(\frac{1}{2} \sin{\frac{7\pi}{12}} \cos{\frac{7\pi}{12}}\)

Используем формулу синуса двойного угла: \(\sin{2x} = 2 \sin{x} \cos{x}\)

\(\frac{1}{2} \sin{\frac{7\pi}{12}} \cos{\frac{7\pi}{12}} = \frac{1}{4} \cdot 2\sin{\frac{7\pi}{12}} \cos{\frac{7\pi}{12}} = \frac{1}{4} \sin{\frac{14\pi}{12}} = \frac{1}{4} \sin{\frac{7\pi}{6}} \)

\(\sin{\frac{7\pi}{6}} = \sin{(\pi + \frac{\pi}{6})} = -\sin{\frac{\pi}{6}} = -\frac{1}{2}\)

Тогда: \(\frac{1}{4} \sin{\frac{7\pi}{6}} = \frac{1}{4} \cdot (-\frac{1}{2}) = -\frac{1}{8}\)

2) \(\sqrt{3}(\sin^2{\frac{25\pi}{12}} - \cos^2{\frac{25\pi}{12}})\)

Используем формулу косинуса двойного угла: \(\cos{2x} = \cos^2{x} - \sin^2{x}\)

Тогда: \(\sqrt{3}(\sin^2{\frac{25\pi}{12}} - \cos^2{\frac{25\pi}{12}}) = -\sqrt{3} \cos{\frac{50\pi}{12}} = -\sqrt{3} \cos{\frac{25\pi}{6}}\)

\(\cos{\frac{25\pi}{6}} = \cos{(4\pi + \frac{\pi}{6})} = \cos{\frac{\pi}{6}} = \frac{\sqrt{3}}{2}\)

Тогда: \(-\sqrt{3} \cos{\frac{25\pi}{6}} = -\sqrt{3} \cdot \frac{\sqrt{3}}{2} = -\frac{3}{2}\)

3) \(\sqrt{8}(\sin^2{\frac{9\pi}{8}} - \cos^2{\frac{9\pi}{8}})\)

Используем формулу косинуса двойного угла: \(\cos{2x} = \cos^2{x} - \sin^2{x}\)

Тогда: \(\sqrt{8}(\sin^2{\frac{9\pi}{8}} - \cos^2{\frac{9\pi}{8}}) = -\sqrt{8} \cos{\frac{9\pi}{4}} = -\sqrt{8} \cos{\frac{\pi}{4}}\)

\(\cos{\frac{9\pi}{4}} = \cos{(2\pi + \frac{\pi}{4})} = \cos{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}\)

Тогда: \(-\sqrt{8} \cos{\frac{9\pi}{4}} = -\sqrt{8} \cdot \frac{\sqrt{2}}{2} = -\frac{\sqrt{16}}{2} = -\frac{4}{2} = -2\)

4) \(\sqrt{50} \sin^2{\frac{11\pi}{8}} - \sqrt{50} \cos^2{\frac{11\pi}{8}} = -\sqrt{50} (\cos^2{\frac{11\pi}{8}} - \sin^2{\frac{11\pi}{8}}) = -\sqrt{50} \cos{\frac{11\pi}{4}}\)

\(\cos{\frac{11\pi}{4}} = \cos{(2\pi + \frac{3\pi}{4})} = \cos{\frac{3\pi}{4}} = -\frac{\sqrt{2}}{2}\)

Тогда: \(-\sqrt{50} \cos{\frac{11\pi}{4}} = -\sqrt{50} \cdot (-\frac{\sqrt{2}}{2}) = \frac{\sqrt{100}}{2} = \frac{10}{2} = 5\)

5) \(\sqrt{48} - \sqrt{192} \cos^2{\frac{23\pi}{12}}\)

\(\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}\)

\(\sqrt{192} = \sqrt{64 \cdot 3} = 8\sqrt{3}\)

Тогда: \(4\sqrt{3} - 8\sqrt{3} \cos^2{\frac{23\pi}{12}} = 4\sqrt{3}(1 - 2\cos^2{\frac{23\pi}{12}} + \cos^2{\frac{23\pi}{12}}) = 4\sqrt{3}(1 - \cos^2{\frac{23\pi}{12}})-4\sqrt{3} \cos^2{\frac{23\pi}{12}} \)

\(\cos{\frac{23\pi}{12}} = \cos{(2\pi - \frac{\pi}{12})} = \cos{(-\frac{\pi}{12})} = \cos{\frac{\pi}{12}}\)

\(\cos{\frac{\pi}{12}} = \cos{15^\circ} = \cos{(45^\circ - 30^\circ)} = \cos{45^\circ} \cos{30^\circ} + \sin{45^\circ} \sin{30^\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

Тогда: \(4\sqrt{3} - 8\sqrt{3} \cos^2{\frac{23\pi}{12}} = 4\sqrt{3} - 8\sqrt{3} (\frac{\sqrt{6} + \sqrt{2}}{4})^2 = 4\sqrt{3} - 8\sqrt{3} \cdot \frac{6 + 2\sqrt{12} + 2}{16} = 4\sqrt{3} - 8\sqrt{3} \cdot \frac{8 + 4\sqrt{3}}{16} = 4\sqrt{3} - \frac{\sqrt{3}(8 + 4\sqrt{3})}{2} = 4\sqrt{3} - \frac{8\sqrt{3} + 12}{2} = 4\sqrt{3} - 4\sqrt{3} - 6 = -6\)

6) \(\sqrt{3} - \sqrt{12} \sin^2{\frac{7\pi}{12}} = \sqrt{3} - \sqrt{4 \cdot 3} \sin^2{\frac{7\pi}{12}} = \sqrt{3} - 2\sqrt{3} \sin^2{\frac{7\pi}{12}}\)

\(\sin{\frac{7\pi}{12}} = \sin{({\frac{\pi}{3} + \frac{\pi}{4}})} = \sin{\frac{\pi}{3}} \cos{\frac{\pi}{4}} + \cos{\frac{\pi}{3}} \sin{\frac{\pi}{4}} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

Тогда: \(\sqrt{3} - 2\sqrt{3} \sin^2{\frac{7\pi}{12}} = \sqrt{3} - 2\sqrt{3} (\frac{\sqrt{6} + \sqrt{2}}{4})^2 = \sqrt{3} - 2\sqrt{3} \cdot \frac{6 + 2\sqrt{12} + 2}{16} = \sqrt{3} - 2\sqrt{3} \cdot \frac{8 + 4\sqrt{3}}{16} = \sqrt{3} - \frac{\sqrt{3}(8 + 4\sqrt{3})}{8} = \sqrt{3} - \frac{8\sqrt{3} + 12}{8} = \sqrt{3} - \sqrt{3} - \frac{3}{2} = -\frac{3}{2}\)