- 高一数学题?已知函数f(x)=2cosxsin(x+pai/3)
- 已知f(x)=2cosxsin(x+pai/3)-genhao3sin^x+sinxcosx
1 求函数的最小正周期
2 求函数f(x)的最大值及最大值时x的值
3 求函数f(x)的单调区间
^表示平方
- 1.
f(x)=2cxsin(x+π/3)-√3(sinx)^2+sinxcosx
=2cosx[sinxcos(π/3) +cosxsin(π/3)]-√3(sinx)^2+sinxcosx
=2cosx[(1/2)sinx+(√3/2)cosx]- √3(sinx)^2+sinxcosx
=sinxcosx+√3(cosx)^2-√3(sinx)^2+sinxcosx
=2sinxcosx+√3[(cosx)^2-√3(sinx)^2]
=sin2x+√3cos2x
=2sin(2x+π/3)
即:f(x)=2sin(2x+π/3)
所以f(x)的最小正周期T=2π/2=π
2.
-1≤sin(2x+π/3)≤1,所以
-2≤f(x)≤2,fmax=2,fmin=-2
3.
单增区间:
2kπ-π/2≤2x+π/3≤2kπ+π/2
即:kπ-5π/12≤x≤kπ+π/12
单减区间:
2kπ+π/2≤2x+π/3≤2kπ+3π/2
即:kπ+π/12≤x≤kπ+7π/12