作业帮求值[3/(sin^2 140)-1/(cos^2 140)]*1/(2sin10)
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![求值[3/(sin^2 140)-1/(cos^2 140)]*1/(2sin10)](/uploads/image/z/665948-20-8.jpg?t=%E6%B1%82%E5%80%BC%5B3%2F%28sin%5E2+140%29-1%2F%28cos%5E2+140%29%5D%2A1%2F%282sin10%29)
求值[3/(sin^2 140)-1/(cos^2 140)]*1/(2sin10)
求值[3/(sin^2 140)-1/(cos^2 140)]*1/(2sin10)
求值[3/(sin^2 140)-1/(cos^2 140)]*1/(2sin10)
[3/(sin140)^2-1/(cos140)^2]*1/(2sin10)=[3(cos140)^2-(sin140)^2]/(sin140)^2(cos140)^2*2sin10
对分子化简:
分子=2(cos140)^2-1+(cos140)^2-(sin140)^2+1=2cos280+1=2cos(270+10)+1=2(sin10+sin30)=2[sin(20-10)+sin(10+20)]=4sin20cos10
对分母化简:
分母=(sin280)^2*sin10/2=[sin(270+10)]^2*sin10/2=(cos10)^2sin10/2=sin20cos10/4
所以原式=16sin20cos10/sin20cos10=16
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