Страница 165 номер 553, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить (553—554). 1) \(2 \sin 6\alpha \cos^2 \left(\frac{\pi}{4} + 3\alpha\right) - \sin 6\alpha\) при \(\alpha = \frac{5\pi}{24}\); 2) \(\cos 3\alpha + 2 \cos (\pi - 3\alpha) \sin^2 \left(\frac{\pi}{4} - 1{,}5\alpha\right)\) при \(\alpha = \frac{5\pi}{36}\).
1) \(2 \sin 6\alpha \cos^{2} \left(\frac{\pi}{4} + 3\alpha\right) - \sin 6\alpha\) при \(\alpha = \frac{5\pi}{24}\)
\(2 \sin 6\alpha \cos^{2} \left(\dfrac{\pi}{4} + 3\alpha\right) - \sin 6\alpha = \sin 6\alpha \left(2 \cos^{2} \left(\dfrac{\pi}{4} + 3\alpha\right) - 1\right) =\)
\(= \sin 6\alpha \left(\cos^{2} \left(\dfrac{\pi}{4} + 3\alpha\right) - \sin^{2} \left(\dfrac{\pi}{4} + 3\alpha\right)\right) =\)
\(= \sin 6\alpha \cos \left(\dfrac{\pi}{2} + 6\alpha\right) =\)
\(= -\sin 6\alpha \cdot \sin 6\alpha,\)
\(-\sin^{2} \left(6 \cdot \dfrac{5\pi}{24}\right) = -\sin^{2} \left(\pi + \dfrac{\pi}{4}\right) = -\dfrac{1}{2};\)
2) \(\cos 3\alpha + 2 \cos(\pi - 3\alpha) \sin^{2} \left(\frac{\pi}{4} - 1{,}5\alpha\right)\) при \(\alpha = \frac{5\pi}{36}\)
\(\cos 3\alpha + 2 \cos(\pi - 3\alpha) \sin^{2} \left(\dfrac{\pi}{4} - 1{,}5\alpha\right) =\)
\(= \cos 3\alpha \left(1 - 2 \sin^{2} \left(\dfrac{\pi}{4} - 1{,}5\alpha\right)\right) =\)
\(= \cos 3\alpha \left(\cos^{2} \left(\dfrac{\pi}{4} - 1{,}5\alpha\right) - \sin^{2} \left(\dfrac{\pi}{4} - 1{,}5\alpha\right)\right) =\)
\(= \cos 3\alpha \cos \left(\dfrac{\pi}{2} - 3\alpha\right) =\)
\(= \cos 3\alpha \sin 3\alpha = \dfrac{1}{2} \sin 6\alpha,\)
\(\dfrac{1}{2} \sin \left(6 \cdot \dfrac{5\pi}{36}\right) = \dfrac{1}{2} \sin \left(\pi - \dfrac{\pi}{6}\right) = \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4}.\)
