反三角函数题帮我算一下,最好告诉我详细过程1.cos ( arc sin1/3 - arc tan3/2)2.tan (
2个回答

1.cos[arcsin(1/3)-arctan(3/2)]

原式=cos[arcsin(1/3)]sin[arcsin(3/根号13)]-sin[arcsin(1/3)]cos[arcsin(3/根号13)]

=cos[arccos(2(根号2)/3)]*(3/根号13)-(1/3)cos[arccos(2/根号13)]

=[2(根号2)/3]*(3/根号13)-(1/3)*(2/根号13)

=6(根号2)/(3根号13)-[2/(3根号13)]

=(6根号2-2)/(3根号13)

2.tan(2arctan1/2arctan√2/2)

原式=tan(arctan4/3arctan√2/2)

=tan(53度*45度)

=tan[(53派/360)*派/4]

约0.054886

有问题吧?你的题是不是抄错了?