(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)化简,跪求
问题描述:
(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)化简,跪求
化简啊啊啊啊~~~~~~~~~~~急
答
(3-4cos 2A+cos 4A)/(3+4cos 2A+cos 4A)
=(2cos 2A^2-4cos 2A+2)/(2cos 2A^2+4cos 2A+2)
=(cos 2A-1)^2/(cos 2A+1)^2
=tanA^4