高中三角函数练习题

显示全部楼层 · 2010-6-2 13:41:36

y=sin2x+3sinxcosx+4cos2x
求这个函数最值，要过程

千问 · 2010-6-2 13:41:36

y=sin2x+3sinxcosx+4cos2x =sin2x+(3/2)sin2x+4cos2x =(5/2)sin2x+4cos2x最大值为√[(5/2)^2+4^2]=√89/2最小值为-√89/2

千问 · 2010-6-2 13:41:36

y=sin2x+3sinxcosx+4cos2x =sin2x+(3/2)sin2x+4cos2x =(5/2)sin2x+4cos2x（公式asinx+bcosx=√(a^2+b^2)sin(x+dd),
tan(dd)=b/a） =√89/2sin(2x+dd) (tan(dd)=b/

千问 · 2010-6-2 13:41:36

y=5/2sin2x+4cos2xy=开根号(25/4+16)sin(x/2+?)所以最大值为 +开根号(25/4+16)最小值为 -开根号(25/4+16)

千问 · 2010-6-2 13:41:36

y=sin2x+3sinxcosx+4cos2x =sin2x+(3/2)sin2x+4cos2x =(5/2)sin2x+4cos2x最大值为√89/2最小值为-√89/2