三角函数y=2sin(1/2x+π/3)-cos(1/2x-π/6)周期怎么求?

问题描述:

三角函数y=2sin(1/2x+π/3)-cos(1/2x-π/6)周期怎么求?

y=2sin(1/2x+π/3)-cos(1/2x-π/6)
=2cos(π/2-1/2x-π/3)-cos(1/2x-π/6)
=2cos(π/6-1/2x)-cos(1/2x-π/6)
=2cos(1/2x-π/6)-cos(1/2x-π/6)
=cos(1/2x-π/6)
可知其w=1/2, 所以它的周期就是2π/w=4π
希望能帮到你,请采纳,谢谢

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