Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1062

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\[1)\sin{18{^\circ}} = \frac{\sqrt{5} - 1\ }{4}\]

\[4\sin^{2}{18{^\circ}} = \frac{2\sin{18{^\circ}} \cdot \sin{36{^\circ}}}{\cos{18{^\circ}}} =\]

\[= \frac{\cos{18{^\circ}} - \cos{54{^\circ}}}{\cos{18{^\circ}}} =\]

\[= 1 - \frac{\sin{36{^\circ}}}{\cos{18{^\circ}}} =\]

\[= 1 - 2 \cdot \sin{18{^\circ}}\]

\[4\sin^{2}{18{^\circ}} + 2\sin{18{^\circ}} - 1 = 0\]

\[Пусть\ x = \sin{18{^\circ}}:\]

\[4x^{2} + 2x - 1 = 0\]

\[D_{1} = 1 + 4 = 5\]

\[x_{1} = \frac{\sqrt{5} - 1}{4};\ \ \ \]

\[\ x_{2} = \frac{\sqrt{5} + 1}{4} - не\ подходит.\]

\[\sin{18{^\circ}} = \frac{\sqrt{5} - 1\ }{4}.\]

\[Что\ и\ требовалось\ доказать.\]

\[2)\ tg^{2}36{^\circ} \cdot tg^{2}72{^\circ} = 5\]

\[\sin{18{^\circ}} = \frac{\sqrt{5} - 1}{4}\ (см.\ п\ 1);\]

\[\cos{18{^\circ}} = \sqrt{1 - \left( \frac{\sqrt{5} - 1}{4} \right)^{2}} =\]

\[= \sqrt{1 - \frac{5 - 2\sqrt{5} + 1}{16}} =\]

\[= \sqrt{\frac{16 - 5 + 2\sqrt{5} - 1}{16}} =\]

\[= \sqrt{\frac{10 + 2\sqrt{5}\ }{16}} = \frac{\sqrt{10 + 2\sqrt{5}}}{4}\]

\[tg18{^\circ} = \frac{\sin{18{^\circ}}}{\cos{18{^\circ}}} = \frac{\sqrt{5} - 1\ }{\sqrt{10 + 2\sqrt{5}}}\]

\[tg36{^\circ} = \frac{2tg18{^\circ}\ }{1 - tg^{2}18{^\circ}} =\]

\[= \frac{2 \cdot \left( \sqrt{5} - 1 \right)}{\sqrt{10 + 2\sqrt{5}} \cdot \left( 1 - \left( \frac{\sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}} \right)^{2} \right)} =\]

\[= \frac{(\sqrt{3} - 1)(5 + \sqrt{3})}{\sqrt{10 + 2\sqrt{5}} \cdot \left( \sqrt{5} + 1 \right)}\]

\[tg^{2}36{^\circ} = \frac{5 - \sqrt{5}}{3 + \sqrt{5}}\]

\[tg72{^\circ} = \frac{2tg36{^\circ}\ }{1 - tg^{2}36{^\circ}} =\]

\[= \frac{1}{2} \cdot \frac{\left( 5 + \sqrt{5} \right)\left( \sqrt{3} + 1 \right)}{\sqrt{10 + 2\sqrt{5}}}\]

\[tg^{2}72{^\circ} = 5 + 2\sqrt{5}\]

\[tg^{2}36{^\circ} \cdot tg^{2}72{^\circ} =\]

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

\[Что\ и\ требовалось\ доказать.\ \]