求下列函数的最小正周期,递增区间及最大值（1）y=2sin2xcos2x（2）y=2cos的平方x/2+1（3）√3cos4x+sin4x

周期为pi/2, 递增区间为[-pi/12,pi/24][5pi/48,5pi/12] 最大值为2 y=sin2xcos2x=2sin2xcos2x/2=sin4x/2 最小正周期：π/2 递增区间：[k

y = 2 sin2x cos2x
y = sin4x
T = 2π/4 = π/2
ymin = -1 at 4x = 2kπ - π/2
ymax = 1 at 4x = 2kπ + π/2
递增区间[kπ/2 - π/8,kπ/2 + π/8],k∈Z
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y = 2 cos²(x/2) + 1
y = 1 + cosx + 1
y = cosx + 2
T = 2π
ymin = 2 - 1 = 1 at x = 2kπ - π
ymax = 2 + 1 = 3 at x = 2kπ
递增区间[2kπ - π,2kπ],k∈Z
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y = sin4x + √3 cos4x
y = 2 sin(4x + π/3)
T = 2π/4 = π/2
ymin = -2 at 4x + π/3 = 2kπ - π/2,x = kπ/2 - 5π/24
ymax = 2 at 4x + π/3 = 2kπ + π/2,x = kπ/2 + π/24
递增区间[kπ/2 - 5π/24,kπ/2 + π/24],k∈Z