题目
f(x)=cos(2x-x/3)+2sin(x-π/4)sin(x+π/4)sin(x+π/4)
化简.
化简.
提问时间：2020-10-10
答案
似乎是f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]
=cos(2x-π/3)+2sin(x-π/4)cos(π/4-x)
=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)
=cos(2x-π/3)+sin[2(x-π/4)]
=cos2xcosπ/3+sin2xsinπ/3+sin(2x-π/2)
=cos2x*1/2+sin2xsinπ/3-cos2x
=sin2xsinπ/3-cos2x*1/2
=sin2xsinπ/3-cos2x*cosπ/3
=-cos(2x+π/3)
=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]
=cos(2x-π/3)+2sin(x-π/4)cos(π/4-x)
=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)
=cos(2x-π/3)+sin[2(x-π/4)]
=cos2xcosπ/3+sin2xsinπ/3+sin(2x-π/2)
=cos2x*1/2+sin2xsinπ/3-cos2x
=sin2xsinπ/3-cos2x*1/2
=sin2xsinπ/3-cos2x*cosπ/3
=-cos(2x+π/3)
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