Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 437

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\[1)\ 2\sin a + \sqrt{2}\cos a\ \]

\[a = \frac{\pi}{4}:\]

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

\[= 2 \bullet \frac{\sqrt{2}}{2} + \sqrt{2} \bullet \frac{\sqrt{2}}{2} = \sqrt{2} + \frac{2}{2} =\]

\[= \sqrt{2} + 1.\]

\[2)\ 0,5\cos a - \sqrt{3}\sin a\]

\[a = 60{^\circ}:\]

\[0,5\cos{60{^\circ}} - \sqrt{3}\sin{60{^\circ}} =\]

\[= 0,5 \bullet \frac{1}{2} - \sqrt{3} \bullet \frac{\sqrt{3}}{2} = \frac{1}{4} - \frac{3}{2} =\]

\[= \frac{1}{4} - \frac{6}{4} = - \frac{5}{4} = - 1,25.\]

\[3)\sin{3a} - \cos{2a}\ \]

\[a = \frac{\pi}{6}:\]

\[\sin\frac{3\pi}{6} - \cos\frac{2\pi}{6} =\]

\[= \sin\frac{\pi}{2} - \cos\frac{\pi}{3} = 1 - \frac{1}{2} =\]

\[= 1 - 0,5 = 0,5.\]

\[4)\cos\frac{a}{2} + \sin\frac{a}{3}\text{\ \ \ }\]

\[a = \frac{\pi}{2}:\]

\[\cos\frac{\pi}{2 \bullet 2} + \sin\frac{\pi}{3 \bullet 2} =\]

\[= \cos\frac{\pi}{4} + \sin\frac{\pi}{6} = \frac{\sqrt{2}}{2} + \frac{1}{2} =\]

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