Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 475

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

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

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

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

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

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

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

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

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

\[= - 1 - 0 - 1 - 1 = - 3\]

\[5)\ \frac{3 - \sin^{2}\left( - \frac{\pi}{3} \right) - \cos^{2}\left( - \frac{\pi}{3} \right)}{2\cos\left( - \frac{\pi}{4} \right)} =\]

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

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

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

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

\[= - 2\sin\frac{\pi}{6} + 3 - 7,5\ tg\ \pi + \frac{1}{8} \bullet 0 =\]

\[= - 2 \bullet \frac{1}{2} + 3 - 7,5 \bullet 0 =\]

\[= - 1 + 3 = 2\]