Задание 4. Найдите значение выражения: 1) 4 sin(π/3) − 2 cos(π/6) + tg(π/4); 2) 3 cos(π/3) − 2 ctg(π/6) + sin(π/2); 3) (2 sin(π/4) + tg(π/3)) : cos(π/6); 4) ctg(π/4) · cos(π/6) − sin(π/3); 5) 5 sin(π/6) + 4 cos(π/3) − 3 ctg(π/4).

Question

Вопрос:

Accepted Answer

Подставим табличные значения тригонометрических функций.

\(4\sin\frac{\pi}{3}-2\cos\frac{\pi}{6}+ g\frac{\pi}{4}=4\cdot\frac{\sqrt3}{2}-2\cdot\frac{\sqrt3}{2}+1=2\sqrt3-\sqrt3+1=\sqrt3+1\).
\(3\cos\frac{\pi}{3}-2\ctg\frac{\pi}{6}+\sin\frac{\pi}{2}=3\cdot\frac12-2\sqrt3+1=\frac52-2\sqrt3\).
\(\left(2\sin\frac{\pi}{4}+ g\frac{\pi}{3}\right):\cos\frac{\pi}{6}=(\sqrt2+\sqrt3):\frac{\sqrt3}{2}=\frac{2(\sqrt2+\sqrt3)}{\sqrt3}=2+\frac{2\sqrt6}{3}\).
\(\ctg\frac{\pi}{4}\cdot\cos\frac{\pi}{6}-\sin\frac{\pi}{3}=1\cdot\frac{\sqrt3}{2}-\frac{\sqrt3}{2}=0\).
\(5\sin\frac{\pi}{6}+4\cos\frac{\pi}{3}-3\ctg\frac{\pi}{4}=5\cdot\frac12+4\cdot\frac12-3\cdot1=\frac32\).

Ответ: 1) \(\sqrt3+1\); 2) \(\frac52-2\sqrt3\); 3) \(2+\frac{2\sqrt6}{3}\); 4) 0; 5) \(\frac32\).