We know that \(\cos { 2x } =\cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } \)
This is just a proof I have made. I don't know if it already exists but I just wanted to share it with you.
(eix)2=e2ix(eix)2=ei×2x(cosx+isinx)2=cos2x+isin2xcos2x+i2sin2x+2icosxsinx=cos2x+isin2xcos2x+(−1)sin2x+i×2cosxsinx=cos2x+isin2xcos2x−sin2x+isin2x=cos2x+isin2xSo,cos2x=cos2x−sin2x
Easy Math Editor
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This also exists !!
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But I thought it myself.
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Ok ! Did you try all of my problems - new ones ?
Nice observational skill Archit :)
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Thanks!