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January 已知sinθ1-sinθ2=-(2/3) cosθ1-cos
已知sinθ1-sinθ2=-(2/3) cosθ1-cos
∵θ1，θ2为锐角
∴0<θ1<π/2,0<θ2<π/2
∴-π/2<-θ2<0
∴-π/2<θ1-θ2<π/2
又sinθ1-sinθ2=-(2/3)<0, cosθ1-cosθ2=2/3 >0
即sinθ1
∴θ1<θ2
∴-π/2<θ1-θ2<0
把sinθ1-sinθ2=-(2/3)和 cosθ1-cosθ2=2/3两边平方并相加,得
2-2(sinθ1sinθ2+cosθ1cosθ2)=8/9
∴cos(θ1-θ2)=5/9
∴sin(θ1-θ2)=-√[1-cos^2(θ1-θ2)]=-2√14/9
∴tan(θ1-θ2)＝sin(θ1-θ2)/cos(θ1-θ2)
=(-2√14/9)÷(5/9)
=-2√14/5
最佳答案 - 由提问者2008-06-13 18:13:24选出
[sin(θ1)-sin(θ2)]^2+[cos(θ1)-cos(θ2)]^2
=(-2/3)^2+(2/3)^2=8/9
即:[sin(θ1)]^2+[cos(θ1)]^2+[sin(θ2)]^2+cos(θ2)]^2
-2[sin(θ1)sin(θ2)+cos(θ1)cos(θ2)]
=2-2cos(θ1-θ2)=8/9
得:cos(θ1-θ2)=5/9
因为:θ1、θ2为锐角,又sin(θ1)-sin(θ2)=-2/3<0即θ1<θ2
所以:sin(θ1-θ2)=-√{1-[cos(θ1-θ2)]^2}=-2(√14)/9
得: tan(θ1-θ2)=[sin(θ1-θ2)]/[cos(θ1-θ2)]=-2(√14)/5
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