# How to prove that Cosh(x) = Cos(ix)

1. Apr 17, 2012

### cocopops12

i know that Cosh(x) = (E^x + E^-x ) / 2

how do i prove that Cosh(x) = Cos(ix) , where i is the imaginary unit
similarly Cosh(ix) = Cos(x)

and Sin(ix) = i Sinh(x)
similarly Sinh(ix) = i Sin(x)

2. Apr 17, 2012

### scurty

Do you know any formulas for cosine and sine that relate to exponentials?

3. Apr 17, 2012

### cocopops12

yeah of course Euler's identity
e^ix = cos(x) + isin(x)
but i still can't do it :(

4. Apr 17, 2012

### scurty

I should have been more clear, sorry. I meant a formula that only involves cosine and exponentials and a formula that only involves sines and exponentials. The cosine formula won't have sines in it and vice versa. Once you have those formulas, this becomes trivial to prove as you'll see.

5. Apr 17, 2012

### Steely Dan

Best of all, you can derive these from Euler's identity: write down $e^{ix}$ and $e^{-ix}$ using it, and then add the two equations or subtract them to derive the relevant equations.

6. Apr 17, 2012

### cocopops12

i have those formulas too, but i don't understand how they got them

7. Apr 17, 2012

### mathman

e^ix = cos(x) + isin(x)
e^-ix = cos(x) - isin(x)

Add to get 2cos(x), subtract to get 2isin(x).

I presume you can do the rest.

8. Apr 17, 2012

### cocopops12

it's clear now! thanks sir!

9. May 25, 2013

### Fer Duarte

Easy! Just need a Laplace transformation sheet; then look at the transformation of cos(a*t), it says that L{cos(a*t)}=s/(s^2+a^2). That means that the inverse Laplace transformation of s/(s^2+a^2) is cos(at), right.
L^(-1){s/(s^2+a^2)}=cos(a*t)

Then ask yourself... Which is the inverse Laplace transformation of s/(s^2-1)?
that means your a^2 would be -1 so a=sqrt(-1)=i
then L^(-1){s/(s^2-1)}=L^(-1){s/(s^2+i)}=cos(i*t)

Now let do that by partial fractions:
L^(-1){s/(s^2-1)}=L^(-1){A/(s-1)+B/(s+1)}
s/(s^2-1)=A/(s-1)+B/(s+1); then
s/[(s-1)*(s+1)]=A/(s-1)+B/(s+1); then if you multiplie that equation for (s-1)(s+1) you got
s=A*(s+1)+B*(s-1); then
s=A*s+A+B*s-B=(A+B)*s+(A-B); now you got a equation system;
s=(A+B)*s and 0=A-B
A+B=1 and A-B=0; if you resolve
A=1/2=B; so
L^(-1){A/(s-1)+B/(s+1)}=L^(-1){(1/2)/(s-1)+(1/2)/(s+1)}=L^(-1){(1/2)/(s-1)}+L^(-1){(1/2)/(s+1)}=
(1/2)*L^(-1){1/(s-1)}+(1/2)*L^(-1){1/(s+1)}=(1/2)*(L^(-1){1/(s-1)}+L^(-1){1/(s+1)})

If you look at the inverse Laplace transformation formula for 1/(s-a) you will see e^(a*t) so
L^(-1){1/(s-1)}=e^t
and
L^(-1){1/(s+1)}=e^-t
so
(1/2)*(L^(-1){1/(s-1)}+L^(-1){1/(s+1)})=(1/2)*(e^t+e^-t)=(e^t+e^-t)/2=cosh(t)
so cos(i*t)=cosh(t)

It's the same process to proof sen(i*t)=senh(t)