Ответ: Решение ниже
475. Вычислить:
1) \(\cos(-\frac{\pi}{6}) \cdot \sin(-\frac{\pi}{3}) + \operatorname{tg}(-\frac{\pi}{4});\)
\(\cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2};\)
\(\sin(-\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2};\)
\(\operatorname{tg}(-\frac{\pi}{4}) = -\operatorname{tg}(\frac{\pi}{4}) = -1;\)
Тогда:
\(\frac{\sqrt{3}}{2} \cdot (-\frac{\sqrt{3}}{2}) - 1 = -\frac{3}{4} - 1 = -\frac{7}{4} = -1\frac{3}{4}.\)
2) \(\frac{1 + \operatorname{tg}^2(-\frac{\pi}{6})}{1 + \operatorname{ctg}^2(-\frac{\pi}{6})};\)
\(\operatorname{tg}(-\frac{\pi}{6}) = -\operatorname{tg}(\frac{\pi}{6}) = -\frac{\sqrt{3}}{3};\)
\(\operatorname{ctg}(-\frac{\pi}{6}) = -\operatorname{ctg}(\frac{\pi}{6}) = -\sqrt{3};\)
Тогда:
\(\frac{1 + (-\frac{\sqrt{3}}{3})^2}{1 + (-\sqrt{3})^2} = \frac{1 + \frac{3}{9}}{1 + 3} = \frac{1 + \frac{1}{3}}{4} = \frac{\frac{4}{3}}{4} = \frac{1}{3}.\)
3) \(2 \sin(-\frac{\pi}{6}) \cos(-\frac{\pi}{6}) + \operatorname{tg}(-\frac{\pi}{3}) + \sin^2(-\frac{\pi}{4});\)
\(\sin(-\frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2};\)
\(\cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2};\)
\(\operatorname{tg}(-\frac{\pi}{3}) = -\operatorname{tg}(\frac{\pi}{3}) = -\sqrt{3};\)
\(\sin^2(-\frac{\pi}{4}) = (-\sin(\frac{\pi}{4}))^2 = (-\frac{\sqrt{2}}{2})^2 = \frac{2}{4} = \frac{1}{2};\)
Тогда:
\(2 \cdot (-\frac{1}{2}) \cdot \frac{\sqrt{3}}{2} - \sqrt{3} + \frac{1}{2} = -\frac{\sqrt{3}}{2} - \sqrt{3} + \frac{1}{2} = -\frac{3\sqrt{3}}{2} + \frac{1}{2} = \frac{1 - 3\sqrt{3}}{2}.\)
4) \(\cos(-\pi) + \operatorname{ctg}(-\frac{\pi}{2}) - \sin(-\frac{3\pi}{2}) + \operatorname{ctg}(-\frac{\pi}{4});\)
\(\cos(-\pi) = \cos(\pi) = -1;\)
\(\operatorname{ctg}(-\frac{\pi}{2}) = -\operatorname{ctg}(\frac{\pi}{2}) = 0;\)
\(\sin(-\frac{3\pi}{2}) = -\sin(\frac{3\pi}{2}) = -(-1) = 1;\)
\(\operatorname{ctg}(-\frac{\pi}{4}) = -\operatorname{ctg}(\frac{\pi}{4}) = -1;\)
Тогда:
\(-1 + 0 - 1 - 1 = -3.\)
5) \(\frac{3 - \sin^2(\frac{\pi}{3}) - \cos^2(\frac{\pi}{3})}{2 \cos(-\frac{\pi}{4})};\)
\(\sin^2(\frac{\pi}{3}) = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4};\)
\(\cos^2(\frac{\pi}{3}) = (\frac{1}{2})^2 = \frac{1}{4};\)
\(\cos(-\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2};\)
Тогда:
\(\frac{3 - \frac{3}{4} - \frac{1}{4}}{2 \cdot \frac{\sqrt{2}}{2}} = \frac{3 - 1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}.\)
6) \(2 \sin(-\frac{\pi}{6}) + 3 + 7.5 \operatorname{tg}(-\pi) + \frac{1}{8} \cos(\frac{3}{2}\pi);\)
\(\sin(-\frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2};\)
\(\operatorname{tg}(-\pi) = -\operatorname{tg}(\pi) = 0;\)
\(\cos(\frac{3}{2}\pi) = 0;\)
Тогда:
\(2 \cdot (-\frac{1}{2}) + 3 + 7.5 \cdot 0 + \frac{1}{8} \cdot 0 = -1 + 3 = 2.\)
Ответ: Решение выше
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