Алгебра 2 года назад lena11112

sin(4arccos(-1/2)-2arctg(sqrt3/3)

Ответ

ATLAS

$<var>sin(4arccos(-\frac{1}{2})-2arstg\frac{\sqrt{3}}{3})=sin(4*\frac{2\pi}{3}-2*\frac{\pi}{6})=</var>$

$<var>=sin(\frac{8\pi}{3}-\frac{\pi}{3})=sin\frac{7\pi}{3}=sin(2\pi+\frac{\pi}{3})=sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}</var>$