Show that cosh(2z)=cosh^2(z)+sinh^2(z)
Do you know Osborn's rule? ( http://en.wikipedia.org/wiki/Osborn's_Rule#Similarities_to_circular_trigonometric_functions )
Yeah. I tried using the standard form for these expressions, when considering the RHS. I am then left with a quarter e^2z. Could you check this please?
Hmm, I was just considering z as a random variable label, could just as easily be a, theta, or x.
So comparing to cos(2z) == cos2z - sin2z, there is a product of 2 sines, which you flip the sign of when comparing to hyperbolics, so cosh(2z) == cosh2z + sinh2z
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You should get some e-2z also.
Show us what you got for the RHS.
Re: Welcome to PF!
Oh yeah you are correct, my mistake. I can't even read my own working :)
How exactly do you go about solving thi problem?
I leave it to you.
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