Question

y=sin^2x+2sinxcosx+3cos^2x的值域?

过程! 谢谢了!

Answer 1

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]

Answer 2

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

Answer 3

y=sin^2x+2sinxcosx+3cos^2x
＝（1－cos2x)/2+sin2x+3(1+cos2x)/2
=cos2x+sin2x+2
=√2（√2/2 cos2x+√2/2sin2x）+2
=√2sin(2x+π/4)+2
∴函数的值域是〔－√2＋2，√2＋2〕

Answer 4

原式=1-cos^2 x+sin2x+3cos^2 x
= 1-2cos^2 x+sin2x
= sin2x-cos2x
= sqrt(2)(sin(2x-pi/4))
所以值域为(-sqrt(2), sqrt(2))
sqrt表示根号下

Answer 5

由cos2x=1-2sin^2x=2cos^2x-1
得sin^2x=1/2-1/2cos2x, cos^2x=1/2+1/2cos2x
即 原式=1/2-1/2cos2x+sin2x+3/2+3/2cos2x =2+sin2x+cos2x =2+√2sin(2x+派/4)
函数的值域为[2-√2，2+√2]