ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 53

\[y = \frac{1}{\sin^{2}x};\ \ \ x = a.\]

\[1)\ a = - \frac{\pi}{2}:\]

\[\frac{1}{\sin^{2}\left( - \frac{\pi}{2} \right)} = \frac{1}{\sin^{2}\frac{\pi}{2}} = \frac{1}{1^{2}} = \frac{1}{1} = 1;\]

\[2)\ a = \frac{3\pi}{4}:\]

\[\frac{1}{\sin^{2}\frac{3\pi}{4}} = 1\ :\left( \frac{\sqrt{2}}{2} \right)^{2} = \left( \frac{2}{\sqrt{2}} \right)^{2} =\]

\[= \frac{4}{2} = 2;\]

\[3)\ a = - \frac{10\pi}{3}:\]

\[\frac{1}{\sin^{2}\left( - \frac{10\pi}{3} \right)} = \frac{1}{\sin^{2}\frac{2\pi}{3}} =\]

\[= 1\ :\left( \frac{\sqrt{3}}{2} \right)^{2} = \left( \frac{2}{\sqrt{3}} \right)^{2} = \frac{4}{3};\]

\[4)\ a = - \frac{19\pi}{4}:\]

\[\frac{1}{\sin^{2}\left( - \frac{19\pi}{4} \right)} = \frac{1}{\sin^{2}{\frac{5\pi}{4}\ }} =\]

\[= 1\ :\left( - \frac{\sqrt{2}}{2} \right)^{2} = \left( \frac{2}{\sqrt{2}} \right)^{2} = \frac{4}{2} = 2.\]