ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 655

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

\[= 2 \bullet \frac{\pi}{3} - 3\arcsin\frac{1}{2} = \frac{2\pi}{3} - 3 \bullet \frac{\pi}{6} =\]

\[= \frac{4\pi}{6} - \frac{3\pi}{6} = \frac{\pi}{6};\]

\[2)\arcsin\frac{1}{\sqrt{2}} - 4\arcsin 1 =\]

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

\[= \frac{\pi}{4} - \frac{8\pi}{4} = - \frac{7\pi}{4};\]

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

\[= \pi - \arccos\frac{1}{2} - \frac{\pi}{3} =\]

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

\[4)\arccos( - 1) - \arcsin( - 1) =\]

\[= \pi - \arccos 1 + \arcsin 1 =\]

\[= \pi - 0 + \frac{\pi}{2} = \frac{3\pi}{2};\]

\[5)\ 2\ arctg\ 1 + 3\ arctg\left( - \frac{1}{\sqrt{3}} \right) =\]

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

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

\[6)\ 4\ arctg( - 1) + 3\ arctg\ \sqrt{3} =\]

\[= - 4\ arctg\ 1 + 3 \bullet \frac{\pi}{3} =\]

\[= - 4 \bullet \frac{\pi}{4} + \pi = 0.\]