Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1016

\[\boxed{\mathbf{1016.}}\]

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

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

\[- tg\frac{\pi}{4} =\]

\[= \frac{\sqrt{3}}{2} \bullet \left( - \frac{\sqrt{3}}{2} \right) - 1 = - \frac{3}{4} - 1 =\]

\[= - \frac{3}{4} - \frac{4}{4} = - \frac{7}{4} = - 1,75\]

\[2)\ \frac{1 + tg^{2}\left( - \frac{\pi}{6} \right)}{1 + ctg^{2}\left( - \frac{\pi}{6} \right)} =\]

\[= \frac{1 + \left( - tg\frac{\pi}{6} \right)^{2}}{1 + \left( - ctg\frac{\pi}{6} \right)^{2}} =\]

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

\[= \frac{1 + \frac{1}{3}}{1 + 3} = \frac{\frac{3}{3} + \frac{1}{3}}{4} =\]

\[= \frac{3 + 1}{3 \bullet 4} = \frac{4}{12} = \frac{1}{3}\]

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

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

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

\[+ \left( - \sin\frac{\pi}{4} \right)^{2} = 2 \bullet \left( - \frac{1}{2} \right) \bullet \frac{\sqrt{3}}{2} -\]

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

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

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

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

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

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

\[- ctg\frac{\pi}{4} = - 1 - 0 - 1 - 1 = - 3\]