Question

[sin(pi/2+x)-sinx]^2 +m

Answer 1

[sin(pi/2+x)-sinx]^2 +m

=[cosx-sinx]^2+m

=cos^2 x-2sinxcosx+sin^2 x +m

=-sin2x+1+m

所以最小正周期T=2π/2=π

当-sin2x=-1时有最大值

1+1+m=3

m=1