三角函数y=2sin（1/2x+π/3）-cos（1/2x-π/6）周期怎么求?
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y=2sin（1/2x+π/3）-cos（1/2x-π/6）
=2sin1/2xcosπ/3+2cos1/2xsinπ/3-cos1/2xcosπ/6-sin1/2xsinπ/6
=sin1/2x+cos1/2x*√3-cos1/2x*√3/2-sin1/2x*1/2
=sin1/2x*1/2+cos1/2x*√3/2
=sin1/2xcosπ/3+cos1/2xsinπ/3
=sin(1/2x+π/3)
所以T=2π/(1/2)=4π
=2sin1/2xcosπ/3+2cos1/2xsinπ/3-cos1/2xcosπ/6-sin1/2xsinπ/6
=sin1/2x+cos1/2x*√3-cos1/2x*√3/2-sin1/2x*1/2
=sin1/2x*1/2+cos1/2x*√3/2
=sin1/2xcosπ/3+cos1/2xsinπ/3
=sin(1/2x+π/3)
所以T=2π/(1/2)=4π
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