Страница 160 номер 531, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\cos\frac{23\pi}{4} - \sin\frac{15\pi}{4} - \operatorname{ctg}\left(-\frac{11\pi}{2}\right) =\)
\(= \cos\left(6\pi - \dfrac{\pi}{4}\right) - \sin\left(4\pi - \dfrac{\pi}{4}\right) + \operatorname{ctg}\left(6\pi - \dfrac{\pi}{2}\right) =\)
\(= \cos\dfrac{\pi}{4} + \sin\dfrac{\pi}{4} - \operatorname{ctg}\dfrac{\pi}{2} =\)
\(= \sqrt{2};\)

2) \(\sin\frac{25\pi}{3} - \cos\left(-\frac{17\pi}{2}\right) - \operatorname{tg}\frac{10\pi}{3} =\)
\(= \sin\left(8\pi + \dfrac{\pi}{3}\right) - \cos\left(8\pi + \dfrac{\pi}{2}\right) - \operatorname{tg}\left(3\pi + \dfrac{\pi}{3}\right) =\)
\(= \sin\dfrac{\pi}{3} - \cos\dfrac{\pi}{2} - \operatorname{tg}\dfrac{\pi}{3} =\)
\(= \dfrac{\sqrt{3}}{2} - 0 - \sqrt{3} = -\dfrac{\sqrt{3}}{2};\)

3) \(\sin(-7\pi) - 2 \cos\frac{31\pi}{3} - \operatorname{tg}\frac{7\pi}{4} =\)
\(= -\sin(6\pi + \pi) - 2 \cos\left(10\pi + \dfrac{\pi}{3}\right) - \operatorname{tg}\left(2\pi - \dfrac{\pi}{4}\right) =\)
\(= -\sin \pi - 2 \cos\dfrac{\pi}{3} + \operatorname{tg}\dfrac{\pi}{4} = 0 - 1 + 1 = 0;\)

4) \(\cos(-9\pi) + 2 \sin\left(-\frac{49\pi}{6}\right) - \operatorname{ctg}\left(-\frac{21\pi}{4}\right) =\)
\(= \cos(8\pi + \pi) - 2 \sin\left(8\pi + \dfrac{\pi}{6}\right) + \operatorname{ctg}\left(5\pi + \dfrac{\pi}{4}\right) =\)
\(= \cos \pi - 2 \sin\dfrac{\pi}{6} + \operatorname{ctg}\dfrac{\pi}{4} = -1 - 1 + 1 = -1.\)