Question

已知f(x)=2sinxcosx-2根号3(cosα)^2+根号3,若f(α)=10/13,且α属于【π/2,π】,求s

Answer 1

f(x)=2sinxcosx-2√3(cosα)^2+√3

=sin2x-√3(2(cosα)^2-1)

=sin2x-√3cos2x

=2sin(2x-π/3)

f(α)=10/13

∴2sin(2α-π/3)=10/13

sin(2α-π/3)=5/13,∵α属于[π/2,π],2α-π/3为第二象限角

∴cos(2α-π/3)=-12/13

sin2α=sin(2α-π/3+π/3)

=sin(2α-π/3)*cosπ/3+cos(2α-π/3)*sinπ/3

=(5/13)*(1/2)+(-12/13)*(√3/2)

=(5-12√3)/26.