ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 774

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\[1)\ y = 12\sin x - 5\cos x\]

\[\sin\left( \arccos\frac{12}{13} \right) = \frac{5}{13}\]

\[\sin\left( \arccos\frac{12}{13} \right) =\]

\[= \sqrt{1 - \cos^{2}\left( \arccos\frac{12}{13} \right)} =\]

\[= \sqrt{1 - \left( \frac{12}{13} \right)^{2}} = \sqrt{\frac{169}{169} - \frac{144}{169}} =\]

\[= \sqrt{\frac{25}{169}} = \frac{5}{13}.\]

\[y = 13\left( \frac{12}{13}\sin x - \frac{5}{13}\cos x \right) =\]

\[= 13 \bullet \sin\left( \arccos\frac{12}{13} - x \right) =\]

\[= 13\sin\varphi\ \ \varphi = \arccos\frac{12}{13} - x.\]

\[Область\ значений:\]

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

\[- 13 \leq 13\sin\varphi \leq 13.\]

\[Ответ:\ \ E(y) = \lbrack - 13\ 13\rbrack.\]

\[2)\ y = \cos^{2}x - \sin x =\]

\[= 1 - \sin^{2}x - \sin x =\]

\[= - \sin^{2}x - \sin x - \frac{1}{4} + \frac{5}{4} =\]

\[= - \left( \sin^{2}x + 2 \bullet \frac{1}{2}\sin x + \frac{1}{4} \right) + \frac{5}{4} =\]

\[= - \left( \sin x + \frac{1}{2} \right)^{2} + \frac{5}{4}.\]

\[Область\ значений:\]

\[- 1 \leq \sin x \leq 1\]

\[- \frac{1}{2} \leq \sin x + \frac{1}{2} \leq \frac{3}{2}\]

\[0 \leq \left( \sin x + \frac{1}{2} \right)^{2} \leq \frac{9}{4}\]

\[- \frac{9}{4} \leq - \left( \sin x + \frac{1}{2} \right)^{2} \leq 0\]

\[- 1 \leq - \left( \sin x + \frac{1}{2} \right)^{2} \leq \frac{5}{4}.\]

\[Ответ:\ \ E(y) = \left\lbrack - 1\ \frac{5}{4} \right\rbrack.\]