已知tana=-1/2 ,则5cos^2 a + 3sina cosa - 2sin^2 a

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已知tana=-1/2,则5cos^2a+3sinacosa-2sin^2a已知tana=-1/2,则5cos^2a+3sinacosa-2sin^2a已知tana=-1/2,则5cos^2a+3si

已知tana=-1/2 ,则5cos^2 a + 3sina cosa - 2sin^2 a
已知tana=-1/2 ,则5cos^2 a + 3sina cosa - 2sin^2 a

已知tana=-1/2 ,则5cos^2 a + 3sina cosa - 2sin^2 a
5cos^2 a + 3sina cosa - 2sin^2 a
=(5cos^2 a + 3sina cosa - 2sin^2 a)/(cos^2 a + sin^2 a)
=(5 + 3tana - 2tan^2 a)/(1 + tan^2 a)
=(5 - 3/2 - 1/2) / (1+ 1/4)
=3/(5/4)
=12/5

5(cosa)^2+3sina*cosa-2(sina)^2=[5(cosa)^2+3sina*cosa-2(sina)^2]/[(sina)^2+(cosa)^2]
=[5+3tana-2(tana)^2]/[(tana)^2+1]
=[5+3*(-1/2)-2*(-1/2)^2]/[(-1/2)^2+1]
=[5+(-3/2)-1/2]/(1/4+1)
=(5-2)/(5/4)
=3*(4/5)
=12/5

原式化为=(cosa)^2(5+3tana-2(tana)^2)
再把(cosa)^2化为倒数，并把1＝(sina)^2+(cosa)^2引出即可
＝12/5

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