Question

решите
sin $\frac{\pi }{3}$·cos $\pi$-cos $\frac{\pi }{3}$·sin $\frac{\pi }{6}$
2cos 0°+sin90°-4tg 180°
Ag $\frac{\pi }{2}$+3cos $\frac{\pi }{2}$-4sin $\frac{3\pi }{2}$
tg 45° cos 30°·Ag 60°

Answer 1

$1)Sin\frac{\pi }{3}Cos\pi -Cos\frac{\pi }{3}Sin\frac{\pi }{6}=\frac{\sqrt{3}}{2}*(-1)-\frac{1}{2}*\frac{1}{2}=-\frac{\sqrt{3}}{2}-\frac{1}{4}=-\frac{2\sqrt{3}+1 }{4}\\\\Otvet:\boxed{-\frac{2\sqrt{3}+1 }{4}}$

$2)2Cos0^{o}+Sin90^{0}-4tg180^{o}=2*1+1-4*0=3\\\\Otvet:\boxed{3}$