Homework Help: Hyperbolic cosine identity help!

1. Jan 24, 2012

s.perkins

1. The problem statement, all variables and given/known data

Show that cosh^2(x) = (cosh(2x) - 1)/2

2. Relevant equations

cosh(x) = (e^x + e^-x)/2

3. The attempt at a solution

I have attempted this multiple times and get the same results every time.
Squaring cosh(x) I get 1/4(e^2x + e^-2x +2), which is i believe 1/4(cosh(2x) +2).

Maybe i just can't see it but how it that equivalent to the identity above??

Last edited: Jan 24, 2012
2. Jan 24, 2012

lanedance

wait, so what are you trying to show?

the first is just normal double angle cos and can be shown by looking at Re and Im of (e^(itheta))^2

Last edited: Jan 24, 2012
3. Jan 24, 2012

tiny-tim

welcome to pf!

hi s.perkins! welcome to pf!
erm

cosh(x) = (e^x plus e^-x)/2

4. Jan 24, 2012

lanedance

looks like the correct form has been used in the square, but also don't forget the factor of 2 as well...
cosh(2x)=(e^(x)+e^(-2x))/2

5. Jan 24, 2012

s.perkins

Sorry fixed the typo. I did include the 1/2, when its squared you get a 1/4 in front.

6. Jan 24, 2012

lanedance

might as well use equals as its a bit clearer what you're trying

here should be

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

which is a valid identity, as shown by adding th two below together
cosh(x)^2- sinh(x)^2=1
cosh(x)^2+sinh(x)^2=cosh(2x)
which gives
2cosh(x)^2=cosh(2x)+1