Страница 166 Проверь себя, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

Вторая четверть.

\(\sin \alpha = \sqrt{1 - \dfrac{16}{25}} = \dfrac{3}{5},\)
\(\operatorname{tg} \alpha = -\dfrac{3}{5} \cdot \dfrac{5}{4} = -\dfrac{3}{4},\)
\(\cos 2\alpha = \cos^{2} \alpha - \sin^{2} \alpha = \dfrac{16}{25} - \dfrac{9}{25} = \dfrac{7}{25}.\)

1) \(\cos 135^{\circ} = \cos(90^{\circ} + 45^{\circ}) = -\sin 45^{\circ} = -\frac{\sqrt{2}}{2};\)

2) \(\sin \frac{8\pi}{3} = \sin \left(3\pi - \frac{\pi}{3}\right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2};\)

3) \(\operatorname{tg} \frac{7\pi}{3} = \operatorname{tg} \left(2\pi + \frac{\pi}{3}\right) = \operatorname{tg} \frac{\pi}{3} = \sqrt{3};\)

4) \(\cos^{2} \frac{\pi}{8} - \sin^{2} \frac{\pi}{8} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}.\)

1) \(3 \cos 2\alpha + \sin^{2} \alpha - \cos^{2} \alpha = 2 \cos 2\alpha\)
\(3 \cos 2\alpha + \sin^{2} \alpha - \cos^{2} \alpha = 3 \cos^{2} \alpha - 3 \sin^{2} \alpha + \sin^{2} \alpha - \cos^{2} \alpha =\)
\(= 2(\cos^{2} \alpha - \sin^{2} \alpha) = 2 \cos 2\alpha;\)

2) \(\frac{\sin 5\alpha - \sin 3\alpha}{2 \cos 4\alpha} = \sin \alpha\)
\(\dfrac{\sin 5\alpha - \sin 3\alpha}{2 \cos 4\alpha} = \dfrac{2 \sin \frac{5\alpha - 3\alpha}{2} \cos \frac{5\alpha + 3\alpha}{2}}{2 \cos 4\alpha} = \dfrac{2 \sin \alpha \cos 4\alpha}{2 \cos 4\alpha} = \sin \alpha.\)

1) \(\sin(\alpha - \beta) - \sin \left(\frac{\pi}{2} - \alpha\right) \cdot \sin(-\beta) =\)
\(= \sin \alpha \cos \beta - \cos \alpha \sin \beta + \cos \alpha \cdot \sin \beta = \sin \alpha \cos \beta;\)

2) \(\cos^{2}(\pi - \alpha) - \cos^{2} \left(\frac{\pi}{2} - \alpha\right) = \cos^{2} \alpha - \sin^{2} \alpha = \cos 2\alpha;\)

3) \(2 \sin \alpha \sin \beta + \cos(\alpha + \beta) =\)
\(= 2 \sin \alpha \sin \beta + \cos \alpha \cos \beta - \sin \alpha \sin \beta =\)
\(= \sin \alpha \sin \beta + \cos \alpha \cos \beta =\)
\(= \cos(\alpha - \beta).\)