Im looking for a proof of the derivative of the trig functions... FROM WHICH I can derive cosh x = [ e^x + e^-x ] / 2
Can anyone help with this?
The hyperbolic functions existed independent from the normal trig functions, and prior to complex analysis and their subsequently proven relationship.
Im looking for a way to prove:
cos ix = [ e^i(ix) + e^-i(ix) ] / 2 = [ e^x + e^-x ] / 2 = cosh x
... without merely defining cosh x = [ e^x + e^-x ] / 2.
My point is that how do we know cosh x = [ e^x + e^-x ] / 2 ?
cos ix = [ e^x + e^-x ] / 2... but how does that happen to be cosh also?
2007-12-06
23:50:04
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5 answers
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asked by
Anonymous
in
Mathematics