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(e^ix)^n=(e^ixn) & Trig identities 
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#1
Aug2205, 10:01 AM

P: 22

ok...this was meant to be a fun problem but looks like I don't deserve to have fun!!!
How am I meant to derive trig identities like sin(x)cos^3(x) from some complex **** like [tex]\left( {e}^{{\it ix}} \right) ^{n}={e}^{{\it ixn}}[/tex]!!! I just don't get the idea!!! 


#2
Aug2205, 10:11 AM

P: 576

http://mathworld.wolfram.com/EulerFormula.html
http://mathworld.wolfram.com/deMoivresIdentity.html That should get you started. It's all about manipulating the exponential expressions and identifying the trig identities from there. 


#3
Aug2205, 10:31 AM

HW Helper
P: 2,264

[cos(x)+i sin(x)]^n=cos(n x)+i sin(n x) so if you wanted to know cos(3 x)=[cos(x)]^33cos(x)[sin(x)]^2 you could consider [cos(x)+i sin(x)]^3=cos(3 x)+i sin(3 x) so expand the left side find its real part and you have the identitiy the binomial theorem can be helpful here just be aware that using lots of odd identities at intermediate steps will mess things up 


#4
Aug2205, 10:37 AM

P: 49

(e^ix)^n=(e^ixn) & Trig identities
thanks for the links inha.



#5
Aug2405, 02:02 AM

P: 22

Thanks guys :)



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