题目内容：

求y=sin(2x+4分之π)+cos(2x-4分之π)的最小正周期

优质解答

y=sin(2x+4分之π)+cos(2x-4分之π)
=sin(2x+π/4)+cos[-π/2+(π/4+2x)]
=sin(2x+π/4)+sin(2x+π/4)
=2sin(2x+π/4)
∴ 求y=sin(2x+4分之π)+cos(2x-4分之π)的最小正周期是T=2π/2=π