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(e^ix)^n=(e^ixn) & Trig identities

by complexhuman
Tags: eixneixn, identities, trig
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complexhuman
#1
Aug22-05, 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!!!
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inha
#2
Aug22-05, 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.
lurflurf
#3
Aug22-05, 10:31 AM
HW Helper
P: 2,263
Quote Quote by complexhuman
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!!!
It might be more clear in the trigonometric form where the equation is
[cos(x)+i sin(x)]^n=cos(n x)+i sin(n x)
so if you wanted to know
cos(3 x)=[cos(x)]^3-3cos(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

dx/dy=?
#4
Aug22-05, 10:37 AM
P: 49
(e^ix)^n=(e^ixn) & Trig identities

thanks for the links inha.
complexhuman
#5
Aug24-05, 02:02 AM
P: 22
Thanks guys :)


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