Анонимно

сравните arcsin (-√2/2) и arccos (-√2/2)

Ответ

Анонимно
[tex]- \frac{\pi }{2}\ \textless \ arcsinx\ \textless \ \frac{\pi }{2}\; \; \; \Rightarrow \; \; \; arcsin(-\frac{\sqrt2}{2})=-\frac{\pi}{4} \\\\0\ \textless \ arccosx\ \textless \ \pi \; \; \; \Rightarrow \; \; \; arccos(-\frac{\sqrt2}{2})=\frac{3\pi}{4}[/tex]

[tex]\frac{3\pi}{4}>-\frac{\pi}{4}[/tex]