FP2 Hyperbolic Functions - Proof

Prove that cosh 2x = 2 cosh² x - 1

My working:

cosh x = 1/2(e^x + e^-x)
cosh² x = 1/4(e^2x + 2e⁰ + e^-2x)
2cosh² x = 1/2(e^2x + e^-2x + 2)
2cosh² x = 1/2(e^2x + e^-2x) + 1
2cosh² x - 1 = 1/2(e^2x + e^-2x)
= cosh 2x

Is this enough? Are there any other ways to prove it?

Reply 1

generalebriety

You could note that coshx = cos(ix), and thus the identity cos2a = 2cos²a - 1 holds.

Note though that sinhx = -isin(ix), and so cosh²x = cos²(ix) = 1 - sin²(ix) = 1 + (-i)²sin²(ix) = 1 + sinh²x. So it follows that cosh²x - sinh²x = 1, and also that cosh2x = cos2(ix) = cos²(ix) - sin²(ix) = cosh²x + sinh²x.

This can be verified using your method, of course:

We know that coshx = 1/2(e^x + e^-x), so cosh2x = 1/2(e^2x + e^-2x).
cosh²x = 1/4(e^2x + e^-2x + 2)
sinh²x = 1/4(e^2x + e^-2x - 2)
Adding,
cosh²x + sinh²x = 1/2(e^2x + e^-2x) = cosh2x.

Bit irrelevant, but meh, useful. :biggrin:

Reply 2

Juwel

Looks like you did alright mate. I might have started with cosh(2x) = [e^(2x) + e^(-2x)]/2 and kept working.

Reply 3

CalculusMan

Juwel

Looks like you did alright mate. I might have started with cosh(2x) = [e^(2x) + e^(-2x)]/2 and kept working.

I tried this way but couldn't get actually get anywhere. Any ideas on how to start it off?

Cheers guys for your help.

Reply 4

generalebriety

CalculusMan

I tried this way but couldn't get actually get anywhere. Any ideas on how to start it off?

Cheers guys for your help.

cosh(2x) = [e^(2x) + e^(-2x)]/2
= [(e^x)^2 + (e^-x)^2]/2
= [(e^x)^2 + (e^-x)^2 + 2 - 2]/2
= [(e^x + e^-x)^2 - 2]/2
= [(e^x + e^-x)^2]/2 - 1
= cosh^2 x - 1.