tan(π/4-A)=3,则cos2A/(1+sin2A)=用tanA=-1/2再tan2A=-4/3然后算出来两个解,这样做错在哪
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![tan(π/4-A)=3,则cos2A/(1+sin2A)=用tanA=-1/2再tan2A=-4/3然后算出来两个解,这样做错在哪](/uploads/image/z/1737905-41-5.jpg?t=tan%28%CF%80%2F4-A%29%3D3%2C%E5%88%99cos2A%2F%281%2Bsin2A%29%3D%E7%94%A8tanA%3D-1%2F2%E5%86%8Dtan2A%3D-4%2F3%E7%84%B6%E5%90%8E%E7%AE%97%E5%87%BA%E6%9D%A5%E4%B8%A4%E4%B8%AA%E8%A7%A3%2C%E8%BF%99%E6%A0%B7%E5%81%9A%E9%94%99%E5%9C%A8%E5%93%AA)
tan(π/4-A)=3,则cos2A/(1+sin2A)=用tanA=-1/2再tan2A=-4/3然后算出来两个解,这样做错在哪
tan(π/4-A)=3,则cos2A/(1+sin2A)=
用tanA=-1/2
再tan2A=-4/3
然后算出来两个解,这样做错在哪
tan(π/4-A)=3,则cos2A/(1+sin2A)=用tanA=-1/2再tan2A=-4/3然后算出来两个解,这样做错在哪
用tanA=-1/2
再tan2A=-4/3
然后算出来两个解,指(1)cos2A=3/5,sin2A=-4/5.和 (2)cos2A=-3/5,sin2A=4/5.吧 这样做错在没有验算得到的解是否符合题目要求.
∵tanA=-1/2
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