Question

函数f(x)=1+cos2x-2sin²(x-π/6）怎么化简，详细一点

Answer 1

我的看法是这样的 估计别人有更好的想法
f(x)=1+cos2x-[1-cos(2x-π/3)]
=cos2x-cos(2x-π/3)
=cos2x-[cos2xcos(-π/3)-sinxsin(-π/3)]
=1/2cos2x-√3/2sinx
=sin(π/6)cos2x-cos(π/6)sin2x
=sin(π/6-2x)

Answer 2

f(x) = 1+cos2x-2sin²(x-π/6）= 1+cos2x - [1-cos(2x-π/3)]
= cos2x + cos(2x-π/3) = cos2x + (1/2)cos2x + (√3/2)sin2x
= (3/2)cos2x + (√3/2)sin2x = √3sin(2x+π/3)