Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 617

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\[1)\ arctg\left( \text{ctg}\frac{5\pi}{6} \right) =\]

\[= \text{arctg}\left( \text{ctg}\left( \frac{\pi}{2} + \frac{\pi}{3} \right) \right) =\]

\[= \text{arctg}\left( - tg\frac{\pi}{3} \right) =\]

\[= - arctg\left( \text{tg}\frac{\pi}{3} \right) = - \frac{\pi}{3}\]

\[2)\ arctg\left( \text{ctg}\frac{3\pi}{4} \right) =\]

\[= \text{arctg}\left( \text{ctg}\left( \frac{\pi}{2} + \frac{\pi}{4} \right) \right) =\]

\[= \text{arctg}\left( - tg\frac{\pi}{4} \right) =\]

\[= - arctg\left( \text{tg}\frac{\pi}{4} \right) = - \frac{\pi}{4}\]

\[3)\ arctg\left( 2\sin\frac{5\pi}{6} \right) =\]

\[= \text{arctg}\left( 2\sin\left( \pi - \frac{\pi}{6} \right) \right) =\]

\[= \text{arctg}\left( 2\sin\frac{\pi}{6} \right) =\]

\[= arctg\left( 2 \bullet \frac{1}{2} \right) = arctg\ 1 = \frac{\pi}{4}\]

\[4)\ arctg\left( 2\sin\frac{\pi}{3} \right) =\]

\[= \text{arctg}\left( 2 \bullet \frac{\sqrt{3}}{2} \right) = arctg\ \sqrt{3} =\]

\[= \frac{\pi}{3}\]