
已知sinα=4分之3,α∈(2分之π,π)求cos(6分之π+α),cos(6分之π-α)
2个回答
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sina=3/4
a∈(π/2,π)
cosa=-√(1-9/16)=-√7/4
cos(π/6+a)
=cosacos(π/6)-sinasin(π/6)
=(-√7/4)*(√3/2)-(3/4)*(1/2)
=-√21/8-3/8
cos(π/6-a)
=cos(π/6)cosa+sin(π/6)sina
=(√3/2)*(-√7/4)+(1/2)*(3/4)
=-√21/8+3/8
a∈(π/2,π)
cosa=-√(1-9/16)=-√7/4
cos(π/6+a)
=cosacos(π/6)-sinasin(π/6)
=(-√7/4)*(√3/2)-(3/4)*(1/2)
=-√21/8-3/8
cos(π/6-a)
=cos(π/6)cosa+sin(π/6)sina
=(√3/2)*(-√7/4)+(1/2)*(3/4)
=-√21/8+3/8
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