y=2sin^2x-sinxcosx+3cos^2x 求值域
问题描述：
y=2sin^2x-sinxcosx+3cos^2x 求值域
答
y=2sin²x-sinxcosx+3cos²x 求值域
y=2(sin²x+cos²x)-(1/2)sin2x+cos²x=2-(1/2)sin2x+(1+cos2x)/2
=5/2-(1/2)(sin2x-cos2x)=5/2-(√2/2)sin(2x-π/4)；
故(5-√2)/2≦y≦(5+√2)/2
答
y=2sin^2x-sinxcosx+3cos^2x
=2sin^2x+2cos^2x-sinxcosx+cos^2x=2+cos^2x-1/2sin2x=2+(cos2x+1)/2-1/2sin2x
=5/2+√/2/4cos(2x+π/4)
值域 [(10-√/2)/4,(10+√/2)/4]
答
解y=2sin²x-sinxcosx+3cos²x =(2sin²x+2cos²x)-1/2(2sinxcosx)+cos²x =2-1/2sin2x+1/2(2cos²x-1)+1/2 =1/2cos2x-1/2sin2x+5/2 =√2/2(cos2xcosπ/4+sinπ/4sin2x)+5/2 =√2/2cos...
答
y=2sin^2x-sinxcosx+3cos^2x
=2-sinxcosx+cos^2x
=2-1/2 sin2x+1/2 (cos2x+1)
=5/2-1/2 【sin2x+sin(2x-π/2)】
=5/2-1/2 【sin(2x-π/4）cosπ/2】
=5/2-√2/4*sin(2x-π/4）
5/2-√2/4