Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1064

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\[1)\ 2\cos^{2}\frac{\pi}{8} - 1 =\]

\[= 2 \bullet \frac{1 + \cos\left( 2 \bullet \frac{\pi}{8} \right)}{2} - 1 = 1 +\]

\[+ \cos\frac{\pi}{4} - 1 = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

\[2)\ 1 - 2\sin^{2}\frac{\pi}{12} = 1 -\]

\[- 2 \bullet \frac{1 - \cos\left( 2 \bullet \frac{\pi}{12} \right)}{2} = 1 -\]

\[- \left( 1 - \cos\frac{\pi}{6} \right) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\]

\[3)\ \frac{\sqrt{3}}{2} + 2\sin^{2}{15{^\circ}} = \frac{\sqrt{3}}{2} +\]

\[+ 2 \bullet \frac{1 - \cos(2 \bullet 15{^\circ})}{2} =\]

\[= \frac{\sqrt{3}}{2} + \left( 1 - \cos{30{^\circ}} \right) =\]

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

\[4)\ - \frac{\sqrt{3}}{2} + 2\cos^{2}{15{^\circ}} = - \frac{\sqrt{3}}{2} +\]

\[+ 2 \bullet \frac{1 + \cos(2 \bullet 15{^\circ})}{2} =\]

\[= - \frac{\sqrt{3}}{2} + 1 + \cos{30{^\circ}} =\]

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