已知tanα=-2,则(3tan2α-sinαcosα+cos2α)/(2sinαcosα+5cos2α)的值为＿＿＿＿

已知tanα=-2,则(3tan2α-sinαcosα+cos2α)/(2sinαcosα+5cos2α)的值为＿＿＿＿
已知tanα=-2,则(3tan2α-sinαcosα+cos2α)/(2sinαcosα+5cos2α)的值为＿＿＿＿

已知tanα=-2,则(3tan2α-sinαcosα+cos2α)/(2sinαcosα+5cos2α)的值为＿＿＿＿
(3tan2α-sinαcosα+cos2α)/(2sinαcosα+5cos2α)
=(3tan2a-1/2sin2a+cos2a)/(sin2a+5cos2a)
tan2a=2tana/1-tan^2a=-4/-3=4/3
tan2a=sin2a/cos2a
sin2a=4/3cos2a
(3tan2a-1/2sin2a+cos2a)/(sin2a+5cos2a)
=(3tan2a+1/3cos2a)/(19/3cos2a)
cos2a=-0.6
(3tan2a+1/3cos2a)/(19/3cos2a)
=(-6-1/5)/(-19/5)
=31/5*5/19
=31/19
∴(3tan2α-sinαcosα+cos2α)/(2sinαcosα+5cos2α)的值为31/19

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