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

[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)

 

[scurty]

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

 

[cocopops12]

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

 

[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.

 

[Steely Dan]

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.

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.

 

[cocopops12]

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

 

[mathman]

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

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.

 

[cocopops12]

it's clear now! thanks sir!

 

[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)