Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 616

\[\boxed{\mathbf{616}\mathbf{.}}\]

\[\text{arctg}a = x,\ если\text{tg}x = a\ \ и\ \]

\[- \frac{\pi}{2} < x < \frac{\pi}{2}.\]

\[Следовательно:\]

\[\text{arctg}\left( \text{tg}x \right) = \text{arctg}a = x.\]

\[1)\ 3\ arctg\left( \text{tg}\frac{\pi}{7} \right) = 3 \bullet \frac{\pi}{7} = \frac{3\pi}{7}\]

\[2)\ 4\ arctg\ (tg\ 0,5) = 4 \bullet 0,5 = 2\]

\[3)\ arctg\left( \text{tg}\frac{7\pi}{8} \right) =\]

\[= \text{arctg}\left( \text{tg}\left( \pi - \frac{\pi}{8} \right) \right) =\]

\[= \text{arctg}\left( \text{tg}\left( - \frac{\pi}{8} \right) \right) = - \frac{\pi}{8}\]

\[4)\ arctg\ (tg\ 13) =\]

\[= \text{arctg\ }\left( \text{tg\ }(4\pi + 13 - 4\pi) \right) =\]

\[= \text{arctg\ }\left( \text{tg\ }(13 - 4\pi) \right) =\]

\[= 13 - 4\pi\]