Question

已知函数f(x)=(sin2x+cos2x)²-2sin²2x

求f(x)的最小正周期
若函数y=g(x)的图像由y=f(x)的图像向右移π/8个单位长度，当x属于【0，π/4】时，求y=g(x)的最大值和最小值

Answer 1

f(x)=(sin2x+cos2x)²-2sin²2x
=sin²2x+cos²2x+2sin2xcos2x-2sin²2x
=cos²2x-sin²2x+sin4x
=cos4x+sin4x
=√2(sin4x+π/4)
所以，最小正周期=2π/4=π/2

函数y=g(x)的图像由y=f(x)的图像向右移π/8个单位长度
y=f(x-π/8)
=√2sin(4x-π/2+π/4)
=√2sin(4x-π/4)
x∈[0，π/4)
4x-π/4∈[-π/4，3π/4]
sin(4x-π/4)的最小值=-√2/2；最大值=1
所以，
y=g(x)的最小值=√2*(-√2/2)=-1
最大值=√2

Answer 2

f(x)=(sin2x+cos2x)²-2sin²2x
=1+2sin2xcos2x-2sin²2x
=1+sin4x+cos4x-1
=sin4x+cos4x
=√2sin(4x+π/4)
所以最小正周期为2π/4=π/2

y=f(x)的图像向右移π/8个单位长度
g(x)=√2sin[4(x+π/16-π/8)]
=√2sin(4x-π/4)
x∈[0,π/4]
所以
4x-π/4∈[-π/4,3π/4]
所以
g(x)∈[-1,√2]
最大值√2 最小值为 -1

Answer 3

f(x) = 1+sin4x - (1-cos4x) = sin4x+cos4x = sqrt(2)sin(4x+pi/4)
最小正周期=pi/2

右移后
g(x) = sqrt(2)sin(4x+pi/8)
g(0) = sqrt(2) sin(pi/8) = sqrt(1-1/sqrt(2))
g(pi/4) = sqrt(2) * sin(pi+pi/8) = -sqrt(2)sin(pi/8) = -sqrt(1-1/sqrt(2))最小值
最大值g(3pi/32) = sqrt(2)