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

y=sin(2x+π/6)+cos(2x+π/3)的最小正周期和最大值是?
写下过程，谢谢哈~~

sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]

y=sin(2x+π/6)+cos(2x+π/3)
=sin(2x+π/6)+sin[π/2-(2x+π/3)]
=sin(2x+π/6)+sin(-2x+π/6)
=2sin(π/6)cos(2x)
=cos(2x)
y=sin(2x+π/6)+cos(2x+π/3)的最大值=1
最小正周期T=2∏/2=∏.

y=sin(2x+π/6）+cos(2x+π/3） =sin2xcosπ/6+cos2xsinπ/6+cos2xcosπ/3-sin2xsinπ/3 = cos2xsinπ/6+cos2xcosπ/3 =1/2* cos2x+1/2* cos2x =cos2x, 所以函数的最小正周期是π，最大值是1.