sin(α+π/3)+sinα= -4根号3分之5,-π/2<α<0,则cosα=?
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sin(α+π/3)+sinα= -4根号3分之5,-π/2<α<0,则cosα=?
sin(α+π/3)+sinα= -4根号3分之5,-π/2<α<0,则cosα=?
sin(α+π/3)+sinα= -4根号3分之5,-π/2<α<0,则cosα=?
sin(a+π/3)+sina=-4√3/5
1/2sina+√3/2cosa+sina=-4√3/5
3/2sina+√3/2cosa=-4√3/5
√3/2sina+1/2cosa=-4/5
cos(a-π/3)=-4/5
-π/2-5π/6
cosa=cos(a-π/3+π/3)=cos(a-π/3)cos(π/3)-sin(a-π/3)sin(π/3)
=(-4/5)×(1/2)+(3/5)×(√3/2)
=(3√3-4)/10
sin(α+π/3)+sinα= -4根号3分之5
sin(a+π/3)+sina=-4√3/5
1/2sina+√3/2cosa+sina=-4√3/5
3/2sina+√3/2cosa=-4√3/5
√3/2sina+1/2cosa=-4/5
cos(a-π/3)=-4/5
-π/2-5π/6
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sin(α+π/3)+sinα= -4根号3分之5
sin(a+π/3)+sina=-4√3/5
1/2sina+√3/2cosa+sina=-4√3/5
3/2sina+√3/2cosa=-4√3/5
√3/2sina+1/2cosa=-4/5
cos(a-π/3)=-4/5
-π/2-5π/6
cosa=cos(a-π/3+π/3)=cos(a-π/3)cos(π/3)-sin(a-π/3)sin(π/3)
=(-4/5)×(1/2)+(3/5)×(√3/2)
=(3√3-4)/10
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