已知tana=3,a属于(π/4,π/2),求sin2a,cos2a,tan2a的值
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已知tana=3,a属于(π/4,π/2),求sin2a,cos2a,tan2a的值
已知tana=3,a属于(π/4,π/2),求sin2a,cos2a,tan2a的值
已知tana=3,a属于(π/4,π/2),求sin2a,cos2a,tan2a的值
解析:
因为tana=sina/cosa=3,所以:sina=3cosa
又sin²a+cos²a=1,则:9cos²a+cos²a=1
易得cos²a=1/10
所以:sin2a=2sinacosa=6cos²a=3/5
cos2a=2cos²a-1=1/5 -1=-4/5
tan2a=sin2a/cos2a=-3/4
或tan2a=2tana/(1-tan²a)=6/(1-9)=-3/4
sin2a=2tana/(1+tan^2a)
=6/(1+9)
=3/5
cos2a=(1-tan^2a)/(1+tan^2)
=(1-9)/(1+9)
=-4/5
tan2a=sin2a/cos2a=-3/4