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
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![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](/uploads/image/z/11293540-52-0.jpg?t=%E6%B1%82%E4%B8%8B%E5%88%97%E5%87%BD%E6%95%B0%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%2C%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%E5%8F%8A%E6%9C%80%E5%A4%A7%E5%80%BC%EF%BC%881%EF%BC%89y%3D2sin2xcos2x%EF%BC%882%EF%BC%89y%3D2cos%E7%9A%84%E5%B9%B3%E6%96%B9x%2F2%2B1%EF%BC%883%EF%BC%89%E2%88%9A3cos4x%2Bsin4x)
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
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
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
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
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