解法(1)
y=(1-cos2x)/2+√3/2sin 2x-1
=-cos2x/2+√3/2sin 2x-1/2
=sin2xcos30度-cos2xsin30度-1/2
=sin(2x-30度)-1/2
由于-1=则-3/2=即,y的值域为{y|-3/2<=y<=1/2}
解法(2)
y=(1-cos2x)/2+√3/2sin 2x-1
=-cos2x/2+√3/2sin 2x-1/2
=-(1/2cos2x-√3/2sin2x)-1/2
=-(cos60度cos2x-sin60度sin2x)-1/2
=-cos(2x+60度)-1/2
由于-1<=-cos(2x+60度)<=1,所以-3/2<=-cos(2x+60度)-1/2<=1/2
即,y的值域为{y|-3/2<=y<=1/2}
注：sin^2x=(1-cos2x)/2 解法(1)
y=(1-cos2x)/2+√3/2sin 2x-1
=-cos2x/2+√3/2sin 2x-1/2
=sin2xcos30度-cos2xsin30度-1/2
=sin(2x-30度)-1/2
由于-1=则-3/2=即,y的值域为{y|-3/2<=y<=1/2}
解法(2)
y=(1-cos2x)/2+√3/2sin 2x-1
=-cos2x/2+√3/2sin 2x-1/2
=-(1/2cos2x-√3/2sin2x)-1/2
=-(cos60度cos2x-sin60度sin2x)-1/2
=-cos(2x+60度)-1/2
由于-1<=-cos(2x+60度)<=1,所以-3/2<=-cos(2x+60度)-1/2<=1/2
即,y的值域为{y|-3/2<=y<=1/2}
注：sin^2x=(1-cos2x)/2