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y=2sinxcosx-2sin^2x+1
=2sinxcosx+1-2sin^2x
=sin2x+cos2x
=√2(√2/2sin2x+√2/2cos2x)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
-1<=sin(2x+π/4)<=1
-√2<=√2sin(2x+π/4)<=√2
所以y=2sinxcosx-2sin^2x+1的值域:[-√2,√2]
y=2sin^2 X + 2sinxcosx-1 = 1 - cos 2x + sin 2x - 1 = sin 2x - cos 2x = sin (2x - 45°)最小正周期:2π/2 = π
=2sinxcosx+1-2sin^2x
=sin2x+cos2x
=√2(√2/2sin2x+√2/2cos2x)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
-1<=sin(2x+π/4)<=1
-√2<=√2sin(2x+π/4)<=√2
所以y=2sinxcosx-2sin^2x+1的值域:[-√2,√2]
y=2sin^2 X + 2sinxcosx-1 = 1 - cos 2x + sin 2x - 1 = sin 2x - cos 2x = sin (2x - 45°)最小正周期:2π/2 = π
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