
函数f(x)=1+cos2x-2sin²(x-π/6)怎么化简,详细一点
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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).
=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).
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