Question

[tan(π+α)cos(2π+α)sin(α-3/2π)]/[cos(-α-3π)sin(-3π-α)]

答案负一,
求过程..

Answer 1

解
[tan(π+a)cos(2π+a)sin(a-3/2π)]/[cos(-a-3π)sin(-3π-a)]
=[tanacosasin(-2π+a+π/2)]/[-cos(a+3π)sin(3πa+a)]
=[sinasin(a+π/2)]/[-cos(a+π)sin(a+π)]
=[sinacosa]/[-cosasina]
=-1

Answer 2

tan(π+α)=sinα/cosα
cos(2π+α)=cosα
sin(α-3/2π)=cosα
cos(-α-3π)=-cosα
sin(-3π-α)]=sinα
∴原式=sinα/cosα*cosα *cosα/（-cosα）*sinα
=-1