已知sin(π⼀2+α)=-4⼀5,α∈(π,3⼀2π),则cos(π⼀3-α)=

2025-01-02 17:59:51
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回答1:

α∈(π,3/2π), 所以可得:3π/2<π/2+a<2π
所以可得: cos(π/2+a)=3/5
cos(π/3-a)
=cos[2π/3-(π/2+a)
=cos2π/3cos(π/2+a)+sin2π/3sin(π/2+a)
=-1/2x3/5+√3/2x(-4/5)
=-(3+4√3)/10