ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1181

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\[1)\ 6\ arctg\ \sqrt{3} -\]

\[- 4\arcsin\left( - \frac{1}{\sqrt{2}} \right) = 6 \bullet \frac{\pi}{3} +\]

\[+ 4\arcsin\frac{1}{\sqrt{2}} = 2\pi + 4 \bullet \frac{\pi}{4} = 3\pi\]

\[2)\ 2\ arctg\ 1 + 3\arcsin\left( - \frac{1}{2} \right) =\]

\[= 2 \bullet \frac{\pi}{4} - 3\arcsin\frac{1}{2} = \frac{\pi}{2} -\]

\[- 3 \bullet \frac{\pi}{6} = \frac{\pi}{2} - \frac{\pi}{2} = 0\]

\[3)\ 5\ arctg\left( - \sqrt{3} \right) -\]

\[- 3\arccos\left( - \frac{\sqrt{2}}{2} \right) =\]

\[= - 5\ arctg\ \sqrt{3} -\]

\[- 3\left( \pi - \arccos\frac{\sqrt{2}}{2} \right) =\]

\[= - 5 \bullet \frac{\pi}{3} - 3\pi + 3\frac{\pi}{4} =\]

\[\text{=}\frac{- 20\pi - 36\pi + 9\pi}{12} = - \frac{47\pi}{12}\]