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?