Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 438

\[\boxed{\mathbf{438.}}\]

\[1)\sin\frac{\pi}{4} \bullet \cos\frac{\pi}{4} - \sin\frac{\pi}{3} \bullet \cos\frac{\pi}{6} =\]

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

\[= - \frac{1}{4} = - 0,25\]

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

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

\[= 3 - 0,25 = 2,75\]

\[= \left( 1 - \frac{1}{\sqrt{3}} \right)\left( 1 + \frac{1}{\sqrt{3}} \right) = 1 - \frac{1}{3} =\]

\[= \frac{2}{3}\]

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

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

\[= \frac{9}{12} + \frac{4}{12} = \frac{13}{12} = 1\frac{1}{12}\]