Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 434

\[\boxed{\mathbf{434.}}\]

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

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

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

\[= 3 - 10 = - 7\]

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

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

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

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

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

\[4)\sin\frac{\pi}{3} \bullet \cos\frac{\pi}{6} - tg\frac{\pi}{4} =\]

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

\[= - 0,25\]