设f（x）是定义域为R,最小正周期为3π/2的周期函数,当-π/2_数学解答

设f（x）是定义域为R,最小正周期为3π/2的周期函数,当-π/2

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解答

f(-15π/4)=f(-18π/4 +3π/4)=f[-3•(3π/2)+3π/4]=f(3π/4)=sin3π/4=sin(π-π/4)=sin(π/4)=√2/2