Question

求值[3/(sin^2 140)-1/(cos^2 140)]*1/(2sin10)

Answer 1

[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