Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 962

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\[1)\ 3\sin\frac{\pi}{6} + 2\cos\frac{\pi}{6} - tg\frac{\pi}{3} =\]

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

\[+ \sqrt{3} - \sqrt{3} = 1,5\]

\[2)\ 5\sin\frac{\pi}{4} + 3\ tg\frac{\pi}{4} - 5\cos\frac{\pi}{4} -\]

\[- 10\ ctg\frac{\pi}{4} = 5 \bullet \frac{\sqrt{2}}{2} + 3 \bullet 1 -\]

\[- 5 \bullet \frac{\sqrt{2}}{2} - 10 \bullet 1 =\]

\[= 3 - 10 = - 7\]

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

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

\[= \left( \frac{2}{\sqrt{3}} - \sqrt{3} \right) \bullet \frac{2}{\sqrt{3}} =\]

\[= \frac{2}{\sqrt{3}} \bullet \frac{2}{\sqrt{3}} - \sqrt{3} \bullet \frac{2}{\sqrt{3}} = \frac{4}{3} - 2 =\]

\[= \frac{4}{3} - \frac{6}{3} = - \frac{2}{3}\]