Question

当tgα=√7时,求【cos4α—cos2α】/【(sinα—cosα)的二次方】

Answer 1

2tan(α) sin2α＝——————=√7/4 1＋tan^2(α) 1－tan^2(α) cos2α＝——————=-3/4 1＋tan^2(α) (cos4α—cos2α)÷(sinα—cosα)^2 =(2cos^2(2α)-1-cos2α)÷(1-sin2α) =(2×0.75^2-1+0.75)÷(1-√7/4) =7(4+√7)/18