已知函数y=sin^2 x+2sinxcosx+3cos^2 x,x∈R.⑴函数最小正周期是?⑵函数在什么区间上是增...
已知函数y=sin^2 x+2sinxcosx+3cos^2 x,x∈R.
⑴函数最小正周期是?
⑵函数在什么区间上是增函数?
⑶函数的图像可以由y=√2sin2x,x∈R的图像经过怎样的变换得出?
人气:213 ℃ 时间:2019-08-18 10:48:44
解答
关键是第一步,
y=sin^2 x+2sinxcosx+3cos^2 x=1+2cos^2 x+2sinxcosx=cos2x+sin2x=√2sin(2x+π/4)
1)T=2π/w=π
2)-π/2+2kπ≤2x+π/4≤π/2+2kπ,k∈Z(此非常关键,不写Z就会扣一分)
3)这也是一个易错点,y=√2sin(2x+π/4)==√2sin2(x+π/8),要抽2出来
所以函数的图像可以由y=√2sin2x,x∈R的图像左移π/8得来
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