ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 684

\[z^{2} = \left( \cos{15{^\circ}} + i\sin{15{^\circ}} \right)^{2};\]

\[z^{2} = \cos{30{^\circ}} + i\sin{30{^\circ}} =\]

\[= \frac{\sqrt{3}}{2} + \frac{1}{2}i = (x + iy)^{2} =\]

\[= x^{2} + 2xyi - y^{2}\]

\[2xy = \frac{1}{2}\]

\[y = \frac{1}{4x}.\]

\[x^{2} - y^{2} = \frac{\sqrt{3}}{2}\text{\ \ \ }\]

\[x^{2} - \frac{1}{16x^{2}} = \frac{\sqrt{3}}{2}\]

\[x^{4} - \frac{\sqrt{3}}{2}x^{2} - \frac{1}{16} = 0\]

\[16x^{4} - 8\sqrt{3}x^{2} - 1 = 0\]

\[D = 192 + 64 = 256\]

\[x^{2} = \frac{8\sqrt{3} \pm \sqrt{256}}{2 \bullet 16} =\]

\[= \frac{8\sqrt{3} \pm 16}{32} = \frac{\sqrt{3} \pm 2}{4};\]

\[x = \sqrt{\frac{\sqrt{3} + 2}{4}} =\]

\[= \sqrt{\frac{1}{2} + 2 \bullet \frac{1}{\sqrt{2}} \bullet \frac{\sqrt{3}}{\sqrt{2}} + \frac{3}{2}}\ :2 =\]

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

\[= \frac{\sqrt{2} + \sqrt{6}}{2}\ :2 = \frac{\sqrt{6} + \sqrt{2}}{4};\]

\[y = \frac{1}{4}\ :\frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{\sqrt{6} + \sqrt{2}} =\]

\[= \frac{\sqrt{6} - \sqrt{2}}{6 - 2} = \frac{\sqrt{6} - \sqrt{2}}{4}.\]

\[Ответ:\ \cos{15{^\circ}} = \frac{\sqrt{6} + \sqrt{2}}{4};\ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\sin{15{^\circ}} = \frac{\sqrt{6} - \sqrt{2}}{4}.\]