已知函数f(x)=sin(π-δx)cosδ x+(cosδ x)^2(δ>0)的最小正周期为π.

(1) f(x)=sin(π-δx)cosδ x+(cosδ x)^2
=sin(δx)cosδ x+(cosδ x)^2
=(1/2)sin2δx+(1+cos2δx)/2
=(√2/2)[(√2/2)sin2δx+(√2/2)cos2δx] +1/2
=(√2/2)sin(2δx+π/4)+1/2
所以 周期T=2π/(2δ)=π,解得 δ=1
f(x)=(√2/2)sin(2x+π/4)+1/2
(2)将函数y=f(x)的图像上各点的横坐标缩短到原来的1/2,纵坐标不变,得到
g(x)=(√2/2)sin(4x+π/4)+1/2
令 -π/2≤4x+π/4≤π/2,得-3π/16≤x≤π/16
所以 g(x)在[0,π/16]上是增函数,最小值为g(0)=(√2/2)sin(π/4)+1/2=1/2+1/2=1