ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 114

\[x \in \left\lbrack - \frac{\pi}{6};\ \frac{\pi}{3} \right\rbrack.\]

\[1)\ f(x) = \sin x\]

\[f\left( - \frac{\pi}{6} \right) = \sin\left( - \frac{\pi}{6} \right) = - \frac{1}{2};\]

\[f\left( \frac{\pi}{3} \right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2};\]

\[E(y) = \left\lbrack - \frac{1}{2};\ \frac{\sqrt{3}}{2} \right\rbrack.\]

\[2)\ f(x) = \cos x\]

\[f\left( - \frac{\pi}{6} \right) = \cos\left( - \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2};\]

\[f(0) = \cos 0 = 1;\]

\[f\left( \frac{\pi}{3} \right) = \cos\frac{\pi}{3} = \frac{1}{2};\]

\[E(y) = \left\lbrack \frac{1}{2};\ 1 \right\rbrack.\]

\[3)\ f(x) = tg\ x\]

\[f\left( - \frac{\pi}{6} \right) = tg\left( - \frac{\pi}{6} \right) = - \frac{\sqrt{3}}{3};\]

\[f\left( \frac{\pi}{3} \right) = tg\frac{\pi}{3} = \sqrt{3};\]

\[E(y) = \left\lbrack - \frac{\sqrt{3}}{3};\ \sqrt{3} \right\rbrack.\]