题目

已知函数f(x)=2sin(x+π/4)cos(x-π/4)
（1）求f(π/8) （2）若f（x）=1/2,在[0,π]上有2个根,X1、X2 且X1＜X2求f（X2-X1）的值

提问时间：2020-10-16

答案

f(x)=2sin(x+π/4)cos(x-π/4)
=2sin(x+π/4)cos(π/4-x)
=2sin(x+π/4)cos[(π/2)-(x+π/4)]
=2sin²(x+π/4)
=1-cos(2x+π/2)
=sin(2x)+1
（1）f(π/8)=sin(π/4)+1=(√2/2)+1；
（2）令f(x)=1/2,
则sin(2x)+1=1/2,
sin2x= -1/2,
又x∈[0,π],2x∈[0,2π],
∴2x=7π/6,或11π/6,
由题意,x1=7π/6,x2=11π/6,
∴x2-x1=2π/3,
f(x2-x1)=f(11π/6)=sin(11π/3)+1= sin(5π/3)+1=(-√3/2)+1.