求下列函数的最小正周期 递增区间和最大值 (1) y=sin2xcos2x (2)y= 2cos

1个回答

  • 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