(cosA-cos2A)/(sinA+sin2A)=?

网友回答

(cosA-2cos²A+1)/(sinA+2sinAcosA)
=(cosA-cos²A+1-cos²A)/[sinA(1+2cosA)]
=[cosA(1-cosA)+(1+cosA)(1-cosA)]/[sinA(1+2cosA)]
=(1-cosA)(1+2cosA)/[sinA(1+2cosA)]
=(1-cosA)/sinA
=2sin²(A/2)/[2sin(A/2)cos(A/2)]
=sin(A/2)/cos(A/2)
=tan(A/2)
======以下答案可供参考======
供参考答案1:
解利用和差化积公式得
(cosA-cos2A)/(sinA+sin2A)
=[2sin(3A/2)*sin(A/2)]/[2sin(3A/2)*cos(A/2) ]
=tan(A/2)
供参考答案2:
(cosA-cos2A)/(sinA+sin2A)
=-(2cosA+1)(cosA-1)/[sinA(2cosA+1)]
=(1-cosA)/sinA
=2sin²(A/2)/[2sin(A/2)cos(A/2)]
=sin(A/2)/cos(A/2)
=tan(A/2)