Question

Алгебра,даю 35 баллов

Answer 1

Объяснение:

$\cos \alpha + \ctg \alpha \cdot \sin \alpha = \\ = \cos \alpha + \frac{\cos \alpha}{ \cancel{\sin \alpha} } \cdot \cancel{\sin \alpha} = \\ = \cos \alpha + \cos \alpha =2 \cos \alpha$

$5 \sin^{2} \alpha - 1 = 5(1 - \cos^{2} \alpha) - 1 = \\ 5 - 5\cos^{2} \alpha - 1 = 4 - 5\cos^{2} \alpha = ... \\ \ \small{При} \: \cos^{2} \alpha = 0.2 \\ ... = 4 - 5 \cdot0.2 = 4 - 1 = 3$

$4 \sin^{2}120 \degree - 2 \cos 600 \degree + \sqrt{27} \tg660 \degree = \\ = 4 \sin^{2} \tfrac{2\pi}{3} - 2 \cos \tfrac{10\pi}{3} + \sqrt{9 \cdot3 \: } \tg \tfrac{11\pi}{3} = \\ \small{ = 4 \sin^{2}(\pi - \tfrac{\pi}{3}) - 2 \cos (3\pi + \tfrac{\pi}{3} ) + 3\sqrt{3 \: } \tg(4\pi - \tfrac{\pi}{3}) = }\\ \small{ = 4 \sin^{2}(\tfrac{\pi}{3}) + 2 \cos (\tfrac{\pi}{3} ) - 3\sqrt{3 \: } \tg( \tfrac{\pi}{3}) = } \\ = 4( \tfrac{ \sqrt{3} }{2} )^{2} + 2 \cdot \tfrac{1}{2} - 3\sqrt{3 \: } \cdot \sqrt{3} = \\ = 3 + 1 - 3\cdot3 = - 5$

1