=1/2sin2x-sin^2x
=12/sin2x-(1-2cos2x)/2
=1/2sin2x+1/2cos2x-1/2
=√2/2sin(2x+π/4)-1/2
(1) T=2π/2=π
(2)X属于(-π/8,3π/8)
01/2sin2x-sin^2x请问是怎么化简的我只化简到1/2SIN2X+1-2SIN*X降幂公式sin^2x=[1-cos2x]/2能写写前面的吗。。我菜看不懂F(X)=SIN(π-X)SIN(π/2-X)+COS*X =sinx*cosx+cos^2x=1/2sin2x+(1+cos2x)/2降幂公式cos^2x=(1+cos2x)/2=1/2sin2x+1/2cos2x+1/2=√2/2[sin2x*cosπ/4+cos2x*sinv/4]+1/2=√2/2sin(2x+π/4)+1/2