已知cos(π/3+α)=-3/5,sin(2π/3-β)=5/13,且0
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已知cos(π/3+α)=-3/5,sin(2π/3-β)=5/13,且0
已知cos(π/3+α)=-3/5,sin(2π/3-β)=5/13,且0
已知cos(π/3+α)=-3/5,sin(2π/3-β)=5/13,且0
∵0＜α＜π/2
∴π/3＜α+π/3 ＜5π/6
又∵cos(π/3 +α)=-3/5 ＜ 0
∴π/2＜α+π/3 ＜5π/6
sin(π/3 +α)=4/5
∵π/2＜β＜π
∴-π/3＜2π/3 -β＜π/6
又∵sin(2π/3 -β)=5/13 ＞ 0
∴0＜2π/3 -β＜π/6
cos(2π/3 -β)=12/13
cos(β-α)=-cos(π-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
0<α<π/2<β<π
cos(π/3+α)=-3/5 π/3<π/3+α<5π/6 sin(π/3+α)>0 sin(π/3+α)=4/5
sin(2π/3-β)=5/13 -π/3<2π/3-β<π/6 cos(2π/3-β)>0 cos(2π/3-β)=12/13
cos(β-α)=cos(α-β)
=-cos(π+(α-β...
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0<α<π/2<β<π
cos(π/3+α)=-3/5 π/3<π/3+α<5π/6 sin(π/3+α)>0 sin(π/3+α)=4/5
sin(2π/3-β)=5/13 -π/3<2π/3-β<π/6 cos(2π/3-β)>0 cos(2π/3-β)=12/13
cos(β-α)=cos(α-β)
=-cos(π+(α-β))
=-cos(π/3+α)+(2π/3-β))
=-cos(π/3+α)cos(2π/3-β)+sin(π/3+α)sin(2π/3-β)
=-(-3/5)(12/13)+4/5*5/13
=36/65+20/65
=56/65
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