Ответ:
Решение:
№2. а)
- \( \sin 240^{\circ} \)
\( 240^{\circ} = 180^{\circ} + 60^{\circ} \). \( \sin(180^{\circ} + 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2} \).
- \( \cos \frac{7\pi}{4} \)
\( \frac{7\pi}{4} = 2\pi - \frac{\pi}{4} \). \( \cos(2\pi - \frac{\pi}{4}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).
- \( \operatorname{tg} \frac{11\pi}{6} \)
\( \frac{11\pi}{6} = 2\pi - \frac{\pi}{6} \). \( \operatorname{tg}(2\pi - \frac{\pi}{6}) = -\operatorname{tg} \frac{\pi}{6} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \).
- \( 2 \sin \frac{3\pi}{4} - \cos \frac{9\pi}{4} \)
\( \sin \frac{3\pi}{4} = \sin(\pi - \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).
\( \cos \frac{9\pi}{4} = \cos(2\pi + \frac{\pi}{4}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).
\( 2 \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = \sqrt{2} - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \).
№2. b)
- \( \cos 240^{\circ} \)
\( 240^{\circ} = 180^{\circ} + 60^{\circ} \). \( \cos(180^{\circ} + 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2} \).
- \( \sin \frac{7\pi}{3} \)
\( \frac{7\pi}{3} = 2\pi + \frac{\pi}{3} \). \( \sin(2\pi + \frac{\pi}{3}) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \).
- \( \operatorname{tg} \frac{11\pi}{6} \)
\( \frac{11\pi}{6} = 2\pi - \frac{\pi}{6} \). \( \operatorname{tg}(2\pi - \frac{\pi}{6}) = -\operatorname{tg} \frac{\pi}{6} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \).
- \( 2 \sin \frac{3\pi}{4} - \cos \frac{9\pi}{4} \)
\( \sin \frac{3\pi}{4} = \sin(\pi - \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).
\( \cos \frac{9\pi}{4} = \cos(2\pi + \frac{\pi}{4}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).
\( 2 \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = \sqrt{2} - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \).
Ответ: а) 1) \( -\frac{\sqrt{3}}{2} \); 2) \( \frac{\sqrt{2}}{2} \); 3) \( -\frac{\sqrt{3}}{3} \); 4) \( \frac{\sqrt{2}}{2} \). b) 1) \( -\frac{1}{2} \); 2) \( \frac{\sqrt{3}}{2} \); 3) \( -\frac{\sqrt{3}}{3} \); 4) \( \frac{\sqrt{2}}{2} \).
