Question

помогите, срочно, очень срочно ​

Answer 1

$arccos\frac{\sqrt{2}}{2}+arctg(\frac{\sqrt{3}}{3}-arcctg(-1)=\frac{\pi}{4}+\frac{\pi}{6}-(-\frac{\pi}{4})= \frac{\pi}{4}+\frac{\pi}{6}+\frac{\pi}{4}=\\\\=\frac{2\pi}{4}+\frac{\pi}{6}=\frac{\pi}{2}+\frac{\pi}{6}=\frac{3\pi}{6}+\frac{\pi}{6}=\frac{4\pi}{6}=\frac{2\pi}{3}\\\\sin(arccos\frac{\sqrt{3}}{2})=sin\frac{\pi}{6}=\frac{1}{2}$

Answer 2

Ответ:
a) П/4+П/6+П/4= П/2+П/6= 4П/6=2П/3
б) sin30°= 1/2