把cos^2 x+cos^2 (2派/3-x)化简成单角

问题描述:

把cos^2 x+cos^2 (2派/3-x)化简成单角

cos²x+cos²(2π/3-x)
=cos²x+[(1/2)cosx+(√3/2)sinx]²
=cos²x+(1/4)cos²x+(√3/2)sinxcosx+(3/4)sin²x
=(5/4)cos²x+(3/4)sin²x+(√3/2)sinxcosx
=(1/2)cos²x+(√3/2)sinxcosx+3/4
=(1/4)(1+cos2x)+(√3/4)sin2x+3/4
=(1/2)sin(2x+π/6)+1

楼上错了:“=cos²x+[(1/2)cosx+(√3/2)sinx]²”这一句的1/2应该是负的才对,而且过程显得复杂.我的做法:
cos²x+cos²(2π/3-x)=1/2(cos2x+1)+1/2[cos(4π/3-2x)+1]=1/2[cos2x-1/2cos2x-(√3/2)sin2x+2]
=1/2[1/2cos2x-(√3/2)sin2x+2]
=(1/2)sin(-2x+π/6)+1