已知cos(π/3+α)=-3/5 ,sin(2π/3-β)=5/13,且0<α<π/2<β<π,求cos(β-α).
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已知cos(π/3+α)=-3/5 ,sin(2π/3-β)=5/13,且0<α<π/2<β<π,求cos(β-α).
已知cos(π/3+α)=-3/5 ,sin(2π/3-β)=5/13,且0<α<π/2<β<π,求cos(β-α).
已知cos(π/3+α)=-3/5 ,sin(2π/3-β)=5/13,且0<α<π/2<β<π,求cos(β-α).
0<α<π/2
0+π/3<α+π/3<π/2+π/3
π/3<α+π/3<5π/6
cos(π/3+α)=-3/5 ,
sin(π/3+α)=4/5 ,
π/2+π/3<β+π/3<π+π/3
5π/6<β+π/3<4π/3
cos(π/3+α)=-3/5 ,
sin(2π/3-β)=5/13
sin(π-π/3-β)=5/13
sin(π/3+β)=5/13
cos(π/3+β)=-12/13
cos(β-α)
=cos[π/3+β-(π/3+α)]
=cos(π/3+β)cos(π/3+α)+sin(π/3+β)sin(π/3+α)
=-12/13*(-3/5)+5/13*4/5
=36/65+20/65
=56/65
∵0<α<π/2
∴π/3<π/3+α<5π/6
∵cos(π/3+α)=-3/5
∴sin(π/3+α)=√[I-cos^2(π/3+α)]=√(1-9/25)=√(16/25)=4/5
∵π/2<β<π
∴-π<-β<-π/2
∴-π/3<2π/3-β<π/6
∵sin(2π/3-β)=5/13
∴cos(2π/3-β)=√[I-...
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∵0<α<π/2
∴π/3<π/3+α<5π/6
∵cos(π/3+α)=-3/5
∴sin(π/3+α)=√[I-cos^2(π/3+α)]=√(1-9/25)=√(16/25)=4/5
∵π/2<β<π
∴-π<-β<-π/2
∴-π/3<2π/3-β<π/6
∵sin(2π/3-β)=5/13
∴cos(2π/3-β)=√[I-sin^(2π/3-β)]=√(1-25/169)=√(144/169)=12/13
cos(β-α)=-cos[(2π/3-β)+(π/3+α)]
=-[cos(2π/3-β)cos(π/3+α)-sin(2π/3-β)sin(π/3+α)]
=-[(12/13)*(-3/5)-(5/13)*(4/5)]
=56/65
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