急!!!请教数学达人
设函数f(x)=cos{2x+(2π/3)}+2(cosx^2)-1,x∈R.(1)求函数f(x)的最小正周期(2)将函数y=f(x)的图象上各点的横坐标变为原来的2倍,...
设函数f(x)=cos{2x+(2π/3)}+2(cosx^2)-1,x∈R. (1)求函数f(x)的最小正周期 (2)将函数y=f(x)的图象上各点的横坐标变为原来的2倍,纵坐标不变,得到函数y=g(x)的图象,求函数y=g(x)的单调递减区间
请给我详细过程~~谢谢 展开
请给我详细过程~~谢谢 展开
1个回答
展开全部
首先,题目提供的f(x)表达式有误,2(cosx^2)应该为2(cosx)^2
解:
1、f(x)=cos{2x+(2π/3)}+2(cosx)^2-1=cos{2x+(2π/3)}+cos2x
=cos2xcos(2π/3)-sin2xsin(2π/3)+cos2x
=-1/2×cos2x-√3/2sin2x+cos2x
=1/2×cos2x-√3/2sin2x=cos(π/3)cos2x-sin(π/3)sin2x
=cos(2x+π/3)=cos[2(x+π/6)]
显然,f(x)最小正周期为π;
2、由f(x)=cos[2(x+π/6)],所以g(x)=cos(x+π/6)
由于余弦函数单调减区间为[2kπ,2kπ+π]
所以有2kπ≤x+π/6≤2kπ+π
∴2kπ-π/6≤x≤2kπ+5π/6
即x∈[2kπ-π/6,2kπ+5π/6],k为整数
解:
1、f(x)=cos{2x+(2π/3)}+2(cosx)^2-1=cos{2x+(2π/3)}+cos2x
=cos2xcos(2π/3)-sin2xsin(2π/3)+cos2x
=-1/2×cos2x-√3/2sin2x+cos2x
=1/2×cos2x-√3/2sin2x=cos(π/3)cos2x-sin(π/3)sin2x
=cos(2x+π/3)=cos[2(x+π/6)]
显然,f(x)最小正周期为π;
2、由f(x)=cos[2(x+π/6)],所以g(x)=cos(x+π/6)
由于余弦函数单调减区间为[2kπ,2kπ+π]
所以有2kπ≤x+π/6≤2kπ+π
∴2kπ-π/6≤x≤2kπ+5π/6
即x∈[2kπ-π/6,2kπ+5π/6],k为整数
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询