y=sin^2x+2sinxcosx+3cos^2xy=sin^2x+2sinxcosx+3cos^2x(1)求函数的最小周期(2)求该函数的单调递增区间
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![y=sin^2x+2sinxcosx+3cos^2xy=sin^2x+2sinxcosx+3cos^2x(1)求函数的最小周期(2)求该函数的单调递增区间](/uploads/image/z/7081612-52-2.jpg?t=y%3Dsin%5E2x%2B2sinxcosx%2B3cos%5E2xy%3Dsin%5E2x%2B2sinxcosx%2B3cos%5E2x%EF%BC%881%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0%E7%9A%84%E6%9C%80%E5%B0%8F%E5%91%A8%E6%9C%9F%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%A5%E5%87%BD%E6%95%B0%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
y=sin^2x+2sinxcosx+3cos^2xy=sin^2x+2sinxcosx+3cos^2x(1)求函数的最小周期(2)求该函数的单调递增区间
y=sin^2x+2sinxcosx+3cos^2x
y=sin^2x+2sinxcosx+3cos^2x
(1)求函数的最小周期
(2)求该函数的单调递增区间
y=sin^2x+2sinxcosx+3cos^2xy=sin^2x+2sinxcosx+3cos^2x(1)求函数的最小周期(2)求该函数的单调递增区间
1、
y=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1
=1+sin2x+cos2x+1
=√2sin(2x+π/4)+2
所以T=2π/2=π
2、
sin的系数为正
所以y的单调性和sin(2x+π/4)相同
sinx的增区间是(2kπ-π/2,2kπ+π/2)
所以2kπ-π/2<2x+π/4<2kπ+π/2
2kπ-3π/4<2x<2kπ+π/4
kπ-3π/8
y=sin^2x+2sinxcosx+3cos^2x
=1+2cos^2x+sin2x
=2+cos2x+sin2x=2+√2sin(π/4+2x)
最小周期T=π
单调递增区间2kπ-π/2<π/4+2x<2kπ+π/2
2kπ-3π/4<2x<2kπ-π/4
kπ-3π/8
1.化简原函数
原式=sin^2x+2sinxcosx+cos^2x+2cos^2x
=2cos^2x+1+sin2x
=2+cos2x+sin2x
=2+√2sin(2x+45)
2.最小周期:
T=2π/2=π
3.单调递增区间:
2kπ-π/2<π/4+2x<2kπ+π/2
2kπ-3π/4<2x<2kπ-π/4
kπ-3π/8
y=(1-cos2x)/2+sin2x+3*(1+cos2x)/2
=sin2x+cos2x+2
常数2不影响周期与单调性