Hyperbolic cosine identity help

[s.perkins]

Homework Statement

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

Homework Equations

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

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??

 

[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 I am of (e^(itheta))^2

 

[tiny-tim]

welcome to pf!

hi s.perkins! welcome to pf!

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

erm …

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

 

[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

 

[s.perkins]

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

 

[lanedance]

Squaring cosh(x) I get 1/4(e^2x + e^-2x +2), which is i believe 1/4(cosh(2x) +2).

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