设a=arcsin(-1/3)b=arcos(-2√2/3)c=arctg（-√2/4）比较abc的大小,

Question

设a=arcsin(-1/3) b=arcos(-2√2/3) c=arctg（-√2/4） 比较a b c的大小,

Answer 1

由 a = arcsin(-1/3)知,－π/2 ≤a ≤0.
由 b = arccos(-2√2/3)知,π/2 ≤b≤π
由 c =arctg（-√2/4）知,－π/2 ≤a ≤0
且 由 a = arcsin(-1/3)可得 sina = - 1/3
tana = -1/(2√2)=-√2/4
所以a = c < b