ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 553

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\[= \sin{6a} \bullet \cos\left( \frac{\pi}{2} + 6a \right) =\]

\[= \sin{6a} \bullet \left( - \sin{6a} \right) = - \sin^{2}{6a}\]

\[при\ a = \frac{5\pi}{24}\ :\]

\[- \sin^{2}\left( 6 \bullet \frac{5\pi}{24} \right) = - \sin^{2}\frac{5\pi}{4} =\]

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

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

\[= \cos{3a} \bullet \cos\left( \frac{\pi}{2} - 3a \right) =\]

\[= \cos{3a} \bullet \sin{3a} =\]

\[= \frac{1}{2} \bullet 2\sin{3a} \bullet \cos{3a} = \frac{1}{2}\sin{6a}\]

\[при\ a = \frac{5\pi}{36}\ :\]

\[\frac{1}{2} \bullet \sin\left( 6 \bullet \frac{5\pi}{36} \right) = \frac{1}{2} \bullet \sin\frac{5\pi}{6} =\]

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

\[= \frac{1}{2} \bullet \frac{1}{2} = \frac{1}{4}.\]