y=sin(x+π/3)cos(π/6-x) 最大值和最小正周期
问题描述:
y=sin(x+π/3)cos(π/6-x) 最大值和最小正周期
答
y=sin(x+π/3)cos(π/6-x)
=1/2[sin(x+π/3+π/6-x)+sin(x+π/3-(π/6-x))]
=1/2(1+sin(2x+π/6))
=1/2sin(2x+π/6)+1/2
∴最大值:1/2+1/2=1
最小正周期:2π/2=π