Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1048

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

\[= \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

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

\[= \cos\left( 2 \bullet \frac{\pi}{8} \right) = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}\]

\[3)\ \frac{2\ tg\frac{\pi}{8}}{1 - tg^{2}\frac{\pi}{8}} = tg\left( 2 \bullet \frac{\pi}{8} \right) =\]

\[= \text{tg}\frac{\pi}{4} = 1\]

\[4)\ \frac{\sqrt{2}}{2} - \left( \cos\frac{\pi}{8} + \sin\frac{\pi}{8} \right)^{2} =\]

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

\[= \frac{\sqrt{2}}{2} -\]

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

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

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

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