Question

设函数f（x）=2sin（wx+π/3）（w＞0，x∈R），且以π为最小正周期

1.求f（π/2）的值
2.已知f（a/2+π/12）=10/13，a∈（-π/2,0）求sin（a-π/4）的值

Answer 1

函数f（x）=2sin（wx+π/3）（w＞0，x∈R），且以π为最小正周期
则T=2π/w=π
w=2
所以 f(x)=2sin(2x+π/3)
(1) f(π/2)=2sin(π+π/3)=-2sin(π/3)=-√3
(2)f（a/2+π/12）=2sin(a+π/6+π/3)=2cosa=10/13
cosa=5/13
a∈（-π/2,0）
sina=-12/13

sin（a-π/4）=sinacos(π/4)-cosasin(π/4)=-12/13*(√2/2)-(5/13)*(√2/2)=-17√2/26

Answer 2

(1) f(x)=2sin(wx-π/6)•sin(wx+π/2-π/6)
=2sin[π/2+(wx-π/6)]•sin(wx-π/6)
=2cos(wx-π/6)•sin(wx-π/6)
=sin(2wx-π/3)
因周期T=2π/2w=π，则w=1
所以f(x)=sin(2x-π/3)
在三角形ABC中，若A〈B，且f(A)=f(B)=1/2
则由f(x)=sin(2x-π/3)=1/2，
知2A-π/3=π/6，A=π/4
2x-π/3=π-π/6，B=7π/12
所以C=π-A-B=π-π/4-7π/12=π/6
由正弦定理BC/AB=sinA/sinC
=sin(π/4)/sin(π/6)
=(√2/2)/(1/2)
=√2
希望能帮到你，祝学习进步O(∩_∩)O