求函数y=sin²x+2sinxcosx+3cos²x的值域

问题描述:

求函数y=sin²x+2sinxcosx+3cos²x的值域

y=sin²x+cos8x+2sinxcosx+2cos²x
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
所以值域[-√2+2,√2+2]

y=sin²x+cos8x+2sinxcosx+2cos²x
=1+sin2x+(1+cos2x)
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
-1