Question

三角函数的值域与最值

Answer 1

y=2sinxcos^2x/(1+sinx)
=2sinx﹙1-sin^2x﹚/(1+sinx)
=2sinx﹙1-sinx﹚
=-2sin^2x+2sinx
令sinx=t,1＝＞t＞＝-1则y=-2t^2+2t=-2(t-1/2)^2+1/2
当t=1/2时y取最大值1/2
当t=-1时y取最小值-4
故y=2sinxcos^2x/(1+sinx)的值域为【-4，1/2】
y=sin^2x+2sinx*cosx+3cos^2x
=1+2sinx*cosx+2cos^2x
=sin2x+cos2x+2
=根号2sin(2x+π/4)+2
sin(2x+π/4)的值域为【-1,1】
故y=sin^2x+2sinx*cosx+3cos^2x的值域为【-根号2+2,根号2+2】