已知sin(2a+b)=3sinb,设tana=x,tanb=y,记y=f(x)若a为三角形最小的内角求f(x)的值域
人气:437 ℃ 时间:2020-04-07 18:44:32
解答
sin(2A+B)=3sinB
sin[(A+B)+A]=3sin[(A+B)-A]
sin(A+B)cosA+cos(A+B)sinA=3[sin(A+B)cosA-cos(A+B)sinA]
sin(A+B)+cos(A+B)tanA=3[sin(A+B)-cos(A+B)tanA]
sin(A+B)+cos(A+B)=3[sin(A+B)-cos(A+B)]
4cos(A+B)=2sin(A+B)
tan(A+B)=2
tan(A+B)=(tanA+tanB)/(1-tanAtanB)=2
tanB+2tanAtanB=2-tanA
tanB=(2-tanA)/(1+2tanA)
y=(2-x)/(1+2x)
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