ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1125

\[\boxed{\mathbf{1125}\mathbf{.}}\]

\[1)\ 2\sin{6a} \bullet \cos^{2}\left( \frac{\pi}{4} + 3a \right) -\]

\[- \sin{6a} = \sin{6a} \bullet\]

\[\bullet \left( 2\cos^{2}\left( \frac{\pi}{4} + 3a \right) - 1 \right) =\]

\[= \sin{6a} \bullet (2\cos^{2}\left( \frac{\pi}{4} + 3a \right) -\]

\[- \left( \cos^{2}\left( \frac{\pi}{4} + 3a \right) + \sin^{2}\left( \frac{\pi}{4} + 3a \right) \right) =\]

\[= \sin{6a} \bullet (\cos^{2}\left( \frac{\pi}{4} + 3a \right) -\]

\[- \sin^{2}\left( \frac{\pi}{4} + 3a \right) = \sin{6a} \bullet\]

\[\bullet \cos\left( 2 \bullet \left( \frac{\pi}{4} + 3a \right) \right) =\]

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

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

\[= - \sin^{2}{6a}\]

\[при\ a = \frac{5\pi}{24}\ :\]

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

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

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

\[2)\cos{3a} + 2\cos{(\pi - 3a}) \bullet\]

\[\bullet \sin^{2}\left( \frac{\pi}{4} - 1,5a \right) =\]

\[= \cos{3a} - 2\cos{3a} \bullet\]

\[\bullet \sin^{2}\left( \frac{\pi}{4} - 1,5a \right) = \cos{3a} \bullet\]

\[\bullet (1 - 2\sin^{2}\left( \frac{\pi}{4} - 1,5a \right) =\]