Question

已知函数f(x)=sin(ωx+φ)(ω>0,|φ|<Π/2),它的一个对称中心到最近的对称轴之间的距离为Π/4,

且函数f(x)的图像过点(-Π/6,0).
(1)求函数f(x)的解析式并写出单调递增区间;(2)设α∈[Π/6,5Π/3],f(α/2)=3/5,求sinα的值。

Answer 1

解：(1)T/4=Π/4,T=Π=2Π/ω ω=2 f(x)=sin(2x+Φ) 把(-Π/6,0)代入
0=sin(2*(-Π/6)+Φ)=sin(Φ-Π/3)
Φ-Π/3=KΠ Φ=KΠ+Π/3 取K=0 Φ=Π/3
So f(x)=sin(2x+Π/3)
2kΠ-Π/2<=2X+Π/3<=2KΠ+Π/2
解得KΠ-5Π/12<=X<=KΠ+Π/12 (K整数)
(2) f(α/2)=sin(2*α/2+Π/3)=sin(α+Π/3)=3/5 >0
Π/6+Π/3<=Α+Π/3<=5Π/3+Π/3 Π/2<=Α+Π/3<=2Π
so Π/2<=Α+Π/3<=Π
Cos(α+Π/3)=-4/5
Sinα=sin[(α+π/3)-π/3]=3/5*1/2-(-4/5)* (√3/2)=[3+4√3]/10