Question

函数f(x)=1+cos2x-2sin²(x-π/6）怎么化简，详细一点

Answer 1

f(x)=1+cos2x-[1-cos2(x-π/6).
=1+cos2x-1+cos(2x-π/3)
=cos2x+cos(2x-π/3).
=2cos[(2x+2x-π/3)/2]*cos[2x-(2x-π/3)]/2.
=2cos(2x-π/6)*cosπ/6.
=√3cos(2x-π/6).

Answer 2

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)