Question

已知函数f(x)=2sin[(π/2)x-π/3],当x∈[1/2,5/2]时，求函数y=f(x-1)+f(x)的值域.

Answer 1

y=2sin[π(x-1)/2-π/3]-2sin(πx/2-π/3)
=2sin(πx/2-5π/6)-2sin(πx/2-π/3)
=2(sinπx/2cos5π/6-cosπx/2sin5π/6-sinπx/2cosπ/3+cosπx/2sinπ/3)
=2(-sinπx/2*√3/2-cosπx/2*1/2-sinπx/2*1/2+cosπx/2*√3/2)
=(1-√3)(sinπx/2-cosπx/2)
=(1-√3)*√2sin(πx/2-π/4)

1/2<=x<=5/2
0<=πx/2-π/4<=π
所以0<=sin(πx/2-π/4)<=1
0*(1-√3)*√2<=(1-√3)*√2sin(πx/2-π/4)<=1*(1-√3)*√2
所以值域是[0，√2-√6]