Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 766

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\[1)\ y = \cos^{4}x - \sin^{4}x =\]

\[= \left( \cos^{2}x - \sin^{2}x \right)\left( \cos^{2}x - \sin^{2}x \right) =\]

\[= \cos{2x}\]

\[- 1 \leq \cos{2x} \leq 1.\]

\[Ответ:\ \ y_{\min} = - 1\ \ y_{\max} = 1.\]

\[2)\ y = \sin\left( x + \frac{\pi}{4} \right) \bullet \sin\left( x - \frac{\pi}{4} \right) =\]

\[= \left( \sin x \bullet \cos\frac{\pi}{4} + \sin\frac{\pi}{4} \bullet \cos x \right) \cdot\]

\[\cdot \left( \sin x \bullet \cos\frac{\pi}{4} - \sin\frac{\pi}{4} \bullet \cos x \right) =\]

\[= \left( \frac{\sqrt{2}}{2}\sin x + \frac{\sqrt{2}}{2}\cos x \right) \cdot\]

\[\cdot \left( \frac{\sqrt{2}}{2}\sin x - \frac{\sqrt{2}}{2}\cos x \right) =\]

\[= \frac{2}{4}\left( \sin x + \cos x \right)\left( \sin x - \cos x \right) =\]

\[= \frac{1}{2}\left( \sin^{2}x - \cos^{2}x \right) = - \frac{1}{2}\cos{2x}\]

\[- 1 \leq \cos{2x} \leq 1\]

\[- \frac{1}{2} \leq - \frac{1}{2}\cos{2x} \leq \frac{1}{2}.\]

\[Ответ:\ \ y_{\min} = - \frac{1}{2}\text{\ \ }y_{\max} = \frac{1}{2}.\]

\[3)\ y = 1 - 2\left| \sin{3x} \right|\]

\[- 1 \leq \sin{3x} \leq 1\]

\[0 \leq \left| \sin{3x} \right| \leq 1\]

\[- 2 \leq - 2\left| \sin{3x} \right| \leq 0\]

\[- 1 \leq 1 - 2\left| \sin{3x} \right| \leq 1.\]

\[Ответ:\ \ y_{\min} = - 1\ \ y_{\max} = 1.\]

\[4)\ y = \sin^{2}x - 2\cos^{2}x =\]

\[= \left( 1 - \cos^{2}x \right) - 2\cos^{2}x =\]

\[= 1 - 3\cos^{2}x\]

\[- 1 \leq \cos x \leq 1\]

\[0 \leq \cos^{2}x \leq 1\]

\[- 3 \leq - 3\cos^{2}x \leq 0\]

\[- 2 \leq 1 - \cos^{2}x \leq 1.\]

\[Ответ:\ \ y_{\min} = - 2\ \ y_{\max} = 1.\]