Question

y=sin(2x+π/6)+cos(2x-π/3)的最小正周期和最大值是多少

详细过程

Answer 1

cos(2x－π/3)=sin[(π/2)＋(2x－π/3)]=sin(2x＋π/6)，则：
y=sin(2x＋π/6)＋sin(2x＋π/6)
=2sin(2x＋π/6)
最小正周期是2π/2=π
最大值是2

Answer 2

y=sin(2x+π/6)+cos(2x-π/3)
=sin2xcosπ/6+cos2xsinπ/6+cos2xcosπ/3+sin2xsinπ/3
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
最小正周期T=2π/2=π
最大值为2