Question

已知tanx=3,求sin^2x+4sinxcosx-2cos^2x的值

Answer 1

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