
已知函数f(x)=2sinxcosx-2cos^2 (x)+1+根号3,求 (1)f(π/4) (2)函数f(x)的最小正周期及最大值
1个回答
展开全部
f(x)=2sinxcosx-2cos^2 (x)+1+根号3
=sin2x-cos2x+根号3
=根号2*sin(2x- π/4)+根号3
(1)f(π/4)=根号2*sin(π/2- π/4)+根号3
=根号2*sin(π/4)+根号3
=1+根号3
(2)函数f(x)的最小正周期T=2π/2=π;
当2x-π/4=π/2+2kπ即x=3π/8 +kπ,k属于Z时,函数f最大值(x)max=根号2+根号3
=sin2x-cos2x+根号3
=根号2*sin(2x- π/4)+根号3
(1)f(π/4)=根号2*sin(π/2- π/4)+根号3
=根号2*sin(π/4)+根号3
=1+根号3
(2)函数f(x)的最小正周期T=2π/2=π;
当2x-π/4=π/2+2kπ即x=3π/8 +kπ,k属于Z时,函数f最大值(x)max=根号2+根号3
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