ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1229

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

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

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

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

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

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

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

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

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

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

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

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