题目
化简cos(2π/3-x)+cos(2π/3+x)+cosx
提问时间：2021-02-18
答案
cos(2π/3-x)+cos(2π/3+x)+cosx
= cos(2π/3)cosx+sin(2π/3)sinx+cos(2π/3)cosx-sin(2π/3)sinx+cosx
= -cos(π/3)cosx+sin(π/3)sinx-cos(π/3)cosx-sin(π/3)sinx+cosx
= -cos(π/3)cosx-cos(π/3)cosx+cosx
= -(1/2)cosx-(1/2)cosx+cosx
= 0
= cos(2π/3)cosx+sin(2π/3)sinx+cos(2π/3)cosx-sin(2π/3)sinx+cosx
= -cos(π/3)cosx+sin(π/3)sinx-cos(π/3)cosx-sin(π/3)sinx+cosx
= -cos(π/3)cosx-cos(π/3)cosx+cosx
= -(1/2)cosx-(1/2)cosx+cosx
= 0
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