Помоги закончить преобразование, упростить выражение и вычислить значение функции.

Конечно, давай помогу! 1. а) $3cos \alpha - sin \alpha \cdot ctg \alpha = 3cos \alpha - sin \alpha \cdot \frac{cos \alpha}{sin \alpha} = 3cos \alpha - cos \alpha = 2cos \alpha$ б) $sin^2 \alpha - sin^2 \alpha \cdot cos^2 \alpha = sin^2 \alpha (1 - cos^2 \alpha) = sin^2 \alpha \cdot sin^2 \alpha = sin^4 \alpha$ в) $4 - 3sin^2 \alpha - 3cos^2 \alpha = 4 - (3sin^2 \alpha + 3cos^2 \alpha) = 4 - 3(sin^2 \alpha + cos^2 \alpha) = 4 - 3 \cdot 1 = 1$ г) $sin^4 \alpha - cos^4 \alpha - sin^2 \alpha = (sin^2 \alpha - cos^2 \alpha)(sin^2 \alpha + cos^2 \alpha) - sin^2 \alpha = (sin^2 \alpha - cos^2 \alpha) \cdot 1 - sin^2 \alpha = sin^2 \alpha - cos^2 \alpha - sin^2 \alpha = -cos^2 \alpha$ 2. а) $5 - sin^2 \alpha - cos^2 \alpha = 5 - (sin^2 \alpha + cos^2 \alpha) = 5 - 1 = 4$ б) $(1 - sin^2 \alpha)(1 + tg^2 \alpha) = cos^2 \alpha \cdot \frac{1}{cos^2 \alpha} = 1$ в) $1 + cos^2 \alpha - sin^2 \alpha = 1 + cos^2 \alpha - (1 - cos^2 \alpha) = 1 + cos^2 \alpha - 1 + cos^2 \alpha = 2cos^2 \alpha$ г) $tg \alpha (ctg \alpha + tg \alpha) = tg \alpha \cdot ctg \alpha + tg \alpha \cdot tg \alpha = 1 + tg^2 \alpha = \frac{1}{cos^2 \alpha}$ 3. а) $sin(150°) = sin(180° - 30°) = sin(30°) = 0.5$ б) $cos(210°) = cos(180° + 30°) = -cos(30°) = -\frac{\sqrt{3}}{2}$ в) $sin(\frac{5\pi}{6}) = sin(\pi - \frac{\pi}{6}) = sin(\frac{\pi}{6}) = 0.5$ г) $tg(\frac{3\pi}{4}) = tg(\pi - \frac{\pi}{4}) = -tg(\frac{\pi}{4}) = -1$ д) $sin(-120°) = -sin(120°) = -sin(180° - 60°) = -sin(60°) = -\frac{\sqrt{3}}{2}$ е) $cos(-\frac{5\pi}{4}) = cos(\frac{5\pi}{4}) = cos(\pi + \frac{\pi}{4}) = -cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$