設f(x)定義域是R,最小正週期為3π/2的函數,若f(x)=cosx,(-π/2≤x0)sinx,(0≤x

設f(x)定義域是R,最小正週期為3π/2的函數,若f(x)=cosx,(-π/2≤x0)sinx,(0≤x

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