设f(x)定义域是R,最小正周期为3π/2的函数,若f(x)=cosx,(-π/2≤x0)sinx,(0≤x

题目

设f(x)定义域是R,最小正周期为3π/2的函数,若f(x)=cosx,(-π/2≤x0)sinx,(0≤x

答案

因为 -15π/4=3π/4-3*3π/2,T=3π/2,
从而 f(-15π/4)=f(3π/4)=sin(3π/4)=√2/2