Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 698

\[\boxed{\mathbf{698}\mathbf{.}}\]

\[y = \sin x - 5\cos x\]

\[\sin\left( \arccos\frac{1}{\sqrt{26}} \right) = \frac{5}{\sqrt{26}}:\]

\[\sin\left( \arccos\frac{1}{\sqrt{26}} \right) =\]

\[= \sqrt{1 - \cos^{2}\left( \arccos\frac{1}{\sqrt{26}} \right)^{2}} =\]

\[= \sqrt{1 - \left( \frac{1}{\sqrt{26}} \right)^{2}} =\]

\[= \sqrt{\frac{26}{26} - \frac{1}{26}} = \sqrt{\frac{25}{26}} = \frac{5}{\sqrt{26}}.\]

\[Упростим:\]

\[y = \sqrt{26} \bullet \left( \frac{1}{\sqrt{26}}\sin x - \frac{5}{\sqrt{26}}\cos x \right) =\]

\[= \sqrt{26} \bullet \sin\left( x - \arccos\frac{1}{\sqrt{26}} \right) =\]

\[= \sqrt{26} \bullet \sin\varphi;\]

\[где\ \varphi = x - \arccos\frac{1}{\sqrt{26}}.\]

\[Область\ значений:\]

\[- 1 \leq \sin\varphi \leq 1\]

\[- \sqrt{26} \leq 5\sin\varphi \leq \sqrt{26}.\]

\[Ответ:\ \ E(y) = \left\lbrack - \sqrt{26};\ \sqrt{26} \right\rbrack.\]