f(x)=2cosx*sin(x+派/6)+根号3sinx*cosx-sin^2x.求f(x)的单调递增区间;

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/15 02:13:21

f(x)=2cosx*sin(x+派/6)+根号3sinx*cosx-sin^2x.求f(x)的单调递增区间;
f(x)=2cosx*sin(x+派/6)+根号3sinx*cosx-sin^2x.
求f(x)的单调递增区间;

f(x)=2cosx*sin(x+派/6)+根号3sinx*cosx-sin^2x.求f(x)的单调递增区间;
为了你能看明白,我尽量不跳步
f(x)=2cosxsin(x+π/6)+√3sinxcosx-sin²x
= 2cosx[sinxcos(π/6)+cosxsin(π/6)]+√3sinxcosx-sin²x
= 2cosx(√3sinx/2+1/2cosx)+√3sinxcosx-sin²x
=√3sinxcosx+cos²x+√3sinxcosx-sin²x
=2√3sinxcosx+cos²x-sin²x
=√3sin(2x)+cos(2x)
=2[√3/2sin(2x)+1/2sin(2x)]
=2sin(2x+π/6)
由-π/2+2kπ

化简后就是cos2x,然后还不会?

这个简单:
f(x)=2cosx(sinxcos(pi/3)+cosxsin(pi/3))-根号33sin^2x+sinx*cosx
=2sinxcosx+根号3cos2x=2sin(x+pi/3)
所以:最小正周期为2pi;
最小值是-2,此时x=-5pi/6+2kpi(其中k=整数)

f(x)=2cosx*sin(x+派/6)+根号3sinx*cosx-sin^2x
=2cosx*[sinxcos(π/6)+cosxsin(π/6)]+√3sinx*cosx-(1-cos^2x)
=sinxcosx+√3cos^2x+√3sinx*cosx+cos^2x-1
=(√3+1)/2*(2sinxcosx+2cos^2x)-1

全部展开

f(x)=2cosx*sin(x+派/6)+根号3sinx*cosx-sin^2x
=2cosx*[sinxcos(π/6)+cosxsin(π/6)]+√3sinx*cosx-(1-cos^2x)
=sinxcosx+√3cos^2x+√3sinx*cosx+cos^2x-1
=(√3+1)/2*(2sinxcosx+2cos^2x)-1
=(√3+1)/2*[sin(2x)+cos(2x)+1]-1
=(√6+√2)/2*sin(2x+π/4)+(√3-1)/2
令a=2x+π/4,sina的单调递增区间为[2kπ-π/2,2kπ+π/2]
2kπ-π/2≤2x+π/4≤2kπ+π/2
kπ-3π/8≤x≤kπ+π/8
因此,f(x)的单调递增区间为[kπ-3π/8,kπ+π/8]

收起