题目

求证,sin²α+cosαcos(π/3+α)-sin²(π/6+α)的值与α无关

提问时间：2020-11-04

答案

sin²α+cosαcos（π/3+α）-sin²（π/6-α）
=sin²α+cosαcos（π/3+α）-sin²[π/2-(π/3+α)]
=sin²α+cosαcos（π/3+α）-cos²(π/3+α)
=sin²α+cos(π/3+α)*[cosα-cos(π/3+α)]
=sin²α+cos(π/3+α)*[-2sin(-π/6)sin（π/6+α）]
=sin²α+cos(π/3+α)*sin（π/6+α）
=sin²α+cos(π/3+α)*cos（π/3-α）
=sin²α+[cos(π/3)cosα-sin(π/3)sinα]*[cos(π/3)cosα+sin(π/3)sinα]
=sin²α+cos²(π/3)cos²α-sin²(π/3)sin²α
=sin²α+1/4*cos²α- 3/4*sin²α
=1/4*sin²α+1/4*cos²α
=1/4
sin²α+cosαcos(π/3+α)-sin²(π/6-α)的值与α无关