化简(1+sin2A+cos2A)(sinA-cosA)/根号(2+2cos2A

问题描述:

化简(1+sin2A+cos2A)(sinA-cosA)/根号(2+2cos2A
本来原题2A是A,A是A/2的,嫌输入麻烦,擅自改了...不要被迷惑....

(1+sin2A+cos2A)(sinA-cosA)/根号(2+2cos2A)
=2cosa(sina+cosa)(sina-cosa)/|2cosa|
=-cos2a或cos2a