题目

函数y=cos(x+π/3),x∈(0,π/3]的值域是

提问时间：2020-11-21

答案

y=cosx+cos(x+π/3)
=cosx+cosxcosπ/3-sinxsinπ/3
=3/2cosx-√3/2sinx
=√3(√3/2cosx-1/2sinx)
=√3cos(x+π/6)
∵x∈[0,π]
∴x+π/6∈[π/6,7π/6]
∴-1≤cos(x+π/6)≤√3/2
∴-√3≤√3cos(x+π/6)≤3/2
即函数值域为[-√3,3/2]
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