设a=arcsin(-1/3) b=arcos(-2√2/3) c=arctg(-√2/4) 比较a b c的大小,
问题描述:
设a=arcsin(-1/3) b=arcos(-2√2/3) c=arctg(-√2/4) 比较a b c的大小,
答
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 = - 1/3tana = -1/(2√2)=-√2/4所以a = c