已知tanα=1/2,则((1/4)sin^2)((1/4)sin^2A+(3/4)cos^2A)/(sin^2A-cos^2A)
已知tanα=1/2,则((1/4)sin^2)((1/4)sin^2A+(3/4)cos^2A)/(sin^2A-cos^2A)
已知tanα=1/2,则((1/4)sin^2)
((1/4)sin^2A+(3/4)cos^2A)/(sin^2A-cos^2A)
已知tanα=1/2,则((1/4)sin^2)((1/4)sin^2A+(3/4)cos^2A)/(sin^2A-cos^2A)
[(1/4)sin^2A+(3/4)cos^2A]/(sin^2A-cos^2A)分子分母同时除以cos^2A
=[(1/4)sin^2A/cos^2A+(3/4)cos^2A/cos^2A]/(sin^2A/cos^2A-cos^2A/cos^2A)
=[(1/4)tan^2A+3/4]/(tan^2A-1)
=[(1/4)*(1/2)^2+3/4]/[(1/2)^2-1]
=[1/4*1/4+3/4]/[1/4-1]
=[1/4*1/4+3/4]/[-3/4]
=[1/4+3]/(-3)
=(13/4)/(-3)
=-13/12
由tana=sina/cosa=1/2得cosa=2sina,代入sina^2加cosa^2=1得sina^2=1/5,故原式=1/4*1/5=1/20
sinα=1/(根号下5), 则 (1/4)*((sinα)^2)=1/20.
tan2A=2tanA/(1-tan²A)=1/(1-1/4)=4/3
原式=[1/4*tan²2A+3/4]/[tan²2A-1]=[1/4*16/9+3/4]/[16/9-1]=43/28,
P.S.其次式都应该用同除cos²x去处理,有时候添加分母1,因为1=sin²x+cos²x。
祝你好运~_~