已知tanx=3,求sin^2x+4sinxcosx-2cos^2x的值

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jacob_junbo 春芽

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tanx =3
sin2x = 2sinxcosx =2(3/10) = 3/5
cos2x = (cosx)^2 - (sinx)^2 = 1/10 - 9/10 = -4/5
(sinx)^2+4sinxcosx - 2(cosx)^2
=[1-cos(2x)]/2 +2sin(2x) - 2( 1+ cos(2x) )/2
=-1/2 - (3/2)cos2x + 2sin(2x)
=-1/2 -(3/2)(-4/5) +2(3/5)
=-1/2 +6/5 +6/5
= (24-5)/10
=19/10

1年前

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