cosh 3A=...?

Question

I was asked to prove that this is equal to 4(cosh A)^3 -3(cosh A). This is easy using the usual trig rules and applying Osborn's Rule, but can it be proved by rewriting cosh 3A in its exponential form and working from there? (Like proving sinh 2A = 2 sinh A cosh A by turning sinh 2A into e^2A etc. and using difference of two squares.)

Answer 1

LHS:
cosh3A={e^3A+e^-3A}/2

RHS:
4(cosh A)^3 -3(cosh A)

4(coshA)^3=4(e^A+e^-A)^3
=4[e^3A+e^-3A+3e^A+3e^-A]/8
=1/2[e^3A+e^-3A
+3e^A+3e^-A]
=1/2[e^3A+e^-3A]
+1/2[3e^A+3e^-A]
=cosh3A+3coshA

but RHS is
4(coshA)^3-3coshA
=(cosh3A+3coshA)-3coshA
=cosh3A=LHS

therefore,

cosh3A=4(cosh A)^3 -3(cosh A)
QED

if you can't follow my working,
please contact me

Answer 2

The proof can be done in exponential form, and will closely parallel similar proofs for trigonometric identities, except for differences in sign.