Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 599

\[\boxed{\mathbf{599}\mathbf{.}}\]

\[\arcsin a = x,\ если\sin x = a\ \ и\ \]

\[- \frac{\pi}{2} \leq x \leq \frac{\pi}{2};\]

\[a - положительное\ число:\]

\[\sin\left( \arcsin a \right) = \sin x = a;\]

\[\sin\left( \arcsin( - a) \right) =\]

\[= \sin\left( - \arcsin(a) \right) = \sin( - x) =\]

\[= - \sin x = - a.\]

\[1)\sin\left( \arcsin\frac{1}{7} \right) = \frac{1}{7}\]

\[2)\sin\left( \arcsin\left( - \frac{1}{5} \right) \right) = - \frac{1}{5}\]

\[3)\sin\left( \pi + \arcsin\frac{3}{4} \right) =\]

\[= - \sin\left( \arcsin\frac{3}{4} \right) = - \frac{3}{4}\]

\[4)\cos\left( \frac{3\pi}{2} - \arcsin\frac{1}{3} \right) =\]

\[= - \sin\left( \arcsin\frac{1}{3} \right) = - \frac{1}{3}\]

\[5)\cos\left( \arcsin\frac{4}{5} \right) =\]

\[= \sqrt{1 - \sin^{2}\left( \arcsin\frac{4}{5}\ \right)} =\]

\[= \sqrt{1 - \left( \frac{4}{5} \right)^{2}} = \sqrt{\frac{25}{25} - \frac{16}{25}} =\]

\[= \sqrt{\frac{9}{25}} = \frac{3}{5}\]

\[6)\ tg\left( \arcsin\frac{1}{\sqrt{10}} \right) =\]

\[= \frac{\sin\left( \arcsin\frac{1}{\sqrt{10}} \right)}{\cos\left( \arcsin\frac{1}{\sqrt{10}} \right)} =\]

\[= \frac{\frac{1}{\sqrt{10}}}{\sqrt{1 - \sin^{2}\left( \arcsin\frac{1}{\sqrt{10}} \right)}} =\]

\[= \frac{1}{\sqrt{10}}\ :\ \sqrt{1 - \frac{1}{10}} = \frac{1}{\sqrt{10}}\ :\ \sqrt{\frac{9}{10}} =\]

\[= \frac{1}{\sqrt{10}} \bullet \frac{\sqrt{10}}{3} = \frac{1}{3}\]