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
∵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
∴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
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