Question

f(x) = 2sin^2(π/4+x)-根号3cos2x-1 详细要求在下面

X∈R 若函数h(x)= f(x+a)的图像关于点(-π/3,0)对称，且a属于（0，π），则a等于？

Answer 1

f(x) = 2sin²(π/4+x)-√3cos2x-1

= 1 - cos{π/2+2x) - √3cos2x - 1
= - cos{π/2+2x) - √3cos2x
= sin2x - √3cos2x
= 2(sin2xcosπ/3-cos2xsinπ/3)
= 2sin(2x-π/3）

h(x) = f(x+a) = 2sin(2x+a-π/3）
图像关于点（-π/3，0）对称
2sin{2*(-π/3)+a-π/3}=0
2sin(-π+a}=0
sina=0
a=0或π

追答

f(x) = 2sin²(π/4+x)-√3cos2x-1

= 1 - cos{π/2+2x) - √3cos2x - 1
= - cos{π/2+2x) - √3cos2x
= sin2x - √3cos2x
= 2(sin2xcosπ/3-cos2xsinπ/3)
= 2sin(2x-π/3）

h(x) = f(x+a) = 2sin(2x+2a-π/3）
图像关于点（-π/3，0）对称
2sin{2*(-π/3)+2a-π/3}=0
2sin(-π+2a}=0
sin2a=0
a=π/2