y=sin^2x+2sinxcosx+3cos^2x的值域?

分类: 作业答案 • 2022-04-25 10:22:55

问题描述：

答

y=sin^2x+2sinxcosx+3cos^2x
=1-cos^2x+sin2x+3cos^2x
=sin2x+2cos^2x+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
所以值域是[2-√2,2+√2]