Реши примеры с тригонометрическими функциями: a) sin 0+cos(\frac{\pi}{2})+sin²(\frac{\pi}{4})

Привет! Давай решим эти примеры вместе. a) $sin(0) + cos(\frac{\pi}{2}) + sin^2(\frac{\pi}{4}) = 0 + 0 + (\frac{\sqrt{2}}{2})^2 = \frac{1}{2}$ б) $3sin(\frac{\pi}{6}) + 2cos(\pi) + ctg^2(\frac{\pi}{6}) = 3 \cdot \frac{1}{2} + 2 \cdot (-1) + (\sqrt{3})^2 = \frac{3}{2} - 2 + 3 = \frac{5}{2}$ в) $6sin(\frac{\pi}{6}) - 2cos(0) + tg^2(\frac{\pi}{3}) = 6 \cdot \frac{1}{2} - 2 \cdot 1 + (\sqrt{3})^2 = 3 - 2 + 3 = 4$ г) $3tg(\frac{\pi}{4}) - sin^2(\frac{\pi}{3}) + cos^2(\frac{\pi}{6}) = 3 \cdot 1 - (\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 = 3 - \frac{3}{4} + \frac{3}{4} = 3$ *Перевод:* a) $sin(0) + cos(\frac{\pi}{2}) + sin^2(\frac{\pi}{4}) = 0 + 0 + (\frac{\sqrt{2}}{2})^2 = \frac{1}{2}$ б) $3sin(\frac{\pi}{6}) + 2cos(\pi) + ctg^2(\frac{\pi}{6}) = 3 \cdot \frac{1}{2} + 2 \cdot (-1) + (\sqrt{3})^2 = \frac{3}{2} - 2 + 3 = \frac{5}{2}$ в) $6sin(\frac{\pi}{6}) - 2cos(0) + tg^2(\frac{\pi}{3}) = 6 \cdot \frac{1}{2} - 2 \cdot 1 + (\sqrt{3})^2 = 3 - 2 + 3 = 4$ г) $3tg(\frac{\pi}{4}) - sin^2(\frac{\pi}{3}) + cos^2(\frac{\pi}{6}) = 3 \cdot 1 - (\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 = 3 - \frac{3}{4} + \frac{3}{4} = 3$ **Ответ:** a) $\frac{1}{2}$ б) $\frac{5}{2}$ в) $4$ г) $3$