Question

已知函数f(x)=Asin(ωx+α)的图象如图所示，f(π/2)=-2/3，则f(0)=( )

图

Answer 1

由图像知：函数的半个周期是11π/12-7π/12=π/3.
所以2π/ω=2π/3. ω=3.

图像过点（7π/12，0），代入得：Asin(7π/4+α)=0,
sin(7π/4+α)=0, sin(2π-π/4+α)=0,
sin(-π/4+α)=0, -π/4+α=0, α=π/4.

又f(π/2)=-2/3，所以Asin(3π/2+α)= -2/3，
将α=π/4代入得：Asin(3π/2+π/4)= -2/3
A•(-√2/2)= -2/3, A=2√2/3.

则f(x)= 2√2/3 sin(3x+π/4),
F(0)= 2√2/3• sinπ/4=2/3.