求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/24 02:46:15

求函数y=[sin2x+sin(2x+π/3)-cos1/2﹙π-4x﹚]/[cos2x+cos﹙2x+π/3﹚-sin1/2(π-4x)]的最小正周期

这里需要公式：cos(π-x)=sinx,sin(π/2-x)=cosx
y=[sin2x+sin(2x+π/3)-cos(π/2-2x﹚]/[cos2x+cos﹙2x+π/3﹚-sin(π/2-2x)]
=[sin2x+sin(2x+π/3)-sin2x]/[cos2x+cos﹙2x+π/3﹚-cos2x]
=sin(2x+π/3)/cos﹙2x+π/3﹚
=tan(2x+π/3)
即y=tan(2x-π/3)
所以最小正周期为：π/2.

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