请教高中数学题目,请写出详细过程,谢谢! (71⼀3)

因为2cos4π/3=（-1）
所以原式=cos²α + cos²(π/3+α)-cosα cos(π/3+α) 然后凑平方差
cos²α + cos²(π/3+α)-2cosα cos(π/3+α) +cosα cos(π/3+α)=【cosα-cos(π/3+α)】²+cosα cos(π/3+α) 然后就要自己计算了cosα-cos(π/3+α）=1/2cosα+根号下3倍的sinα，再平方一下。后面减去cosα cos(π/3+α)就出来3/4了
算cos(π/3+α)时候利用公式cos(A+B)=cosAcosB-sinAsinB，计算认真点就可以了，可能还有更简单的方法，不过我就只能想到这里了，主要是凑平方差

原式=cos²α + cos²(π/3+α)+ 2cos2π/3 cosα cos(π/3+α)
=cos²α + cos²(π/3+α)-cosα cos(π/3+α)
=cos²α + cos(π/3+α)*( cos(π/3+α)-cosα) 和差化积
=cos²α -2 cos(π/3+α)*sin(π/6+α)sinπ/6
=cos²α - cos(π/3+α)*sin(π/6+α) 积化和差
=cos²α -1/2(sin(π/2+2α)-sin(π/6))
=cos²α -1/2cos2α+1/4 倍角公式
=cos²α -1/2(1-2sin²α)+1/4
=cos²α+sin²α -1/2+1/4
=3/4
希望对你有帮助：）

cos²α + cos²(π/3+α)+ 2cos4π/3 cosα cos(π/3+α)
= cos²α + cos²(π/3+α)- cosα cos(π/3+α)
= cos²α+ cos(π/3+α)[ cos(π/3+α) – cosα]
= cos²α+ cos(π/3+α)[1/2 cosα-√3/2sinα- cosα]
= cos²α+(1/2 cosα-√3/2sinα)(- 1/2 cosα-√3/2sinα)
= cos²α+3/4sin²α-1/4 cos²α
=3/4 cos²α+3/4sin²α
=3/4(cos²α+sin²α)
=3/4