ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1018

\[\boxed{\mathbf{1018.}}\]

\[1)\ \frac{2 - \sin^{2}\left( - \frac{\pi}{6} \right) + \cos^{2}\left( - \frac{\pi}{3} \right)}{2\cos\left( - \frac{\pi}{3} \right) + \sin\left( - \frac{\pi}{6} \right)} =\]

\[= \frac{2 - \left( - \sin\frac{\pi}{6} \right)^{2} + \cos^{2}\frac{\pi}{3}}{2\cos\frac{\pi}{3} - \sin\frac{\pi}{6}} =\]

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

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

\[= 2 \bullet 2 = 4\]

\[2)\ \sqrt{3}\sin\left( - \frac{\pi}{3} \right) - 2\ ctg\left( - \frac{\pi}{4} \right) +\]

\[+ 4\cos\left( - \frac{3\pi}{2} \right) =\]

\[= - \sqrt{3}\sin\frac{\pi}{3} + 2\ ctg\frac{\pi}{4} +\]

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

\[+ 4 \bullet 0 = - \frac{3}{2} + 2 =\]

\[= - 1,5 + 2 = 0,5\]