# Proving a hyperbolic identity

1. Nov 25, 2008

### thomas49th

\equiv1. The problem statement, all variables and given/known data
Hi, I've been given a hyperbolic identity to prove:

$$2sinhAsinhB \equiv Cosh(A+B) - Cosh(A-B)$$

2. Relevant equations

$$Cos(A\pm B) \equiv CosACosB \mp SinASinB$$

3. The attempt at a solution

I have Cosh(A+B) and - Cosh(A-B) so i can kind of see that there will be two lots of SinhASinhB and the CoshACoshB will cancel, but how do I prove it? I mean how do I know that $$Cosh(A\pm B) \equiv CoshACoshB \mp SinhASinhB$$

Thanks :)

2. Nov 25, 2008

### tiny-tim

Danger!

(have a ± )

Warning, warning, thomas49th!

Cosh(A±B) = coshAcoshB ± sinhAsinhB (the opposite sign to cos).

(this is because, from Euler's formula … cosx = coshix, isinx = sinhix, so i2sinAsinB = sinhAsinhB )

3. Nov 25, 2008

### thomas49th

Errr sorry I don't follow :S. I've only just started hyperbolics today and havn't used an imaginary numbers with them yet?

Thanks :)

4. Nov 25, 2008

### gabbagabbahey

I assume that you have been taught the definitions of these functions:

$$\cosh(x)=\frac{e^x+e^{-x}}{2}$$ and $$\sinh(x)=\frac{e^x-e^{-x}}{2}$$

That is all you need to prove this identity....What are $\sinh A$ and $\sinh B$? What does that make the LHS of your identity? What are $\cosh(A+B)$ and $\cosh(A-B)$? What does that make the RHS of your identity? Can you show that the two expressions are equivalent? If so, then you prove the identity.

5. Nov 25, 2008

### tiny-tim

oops!

Hi thomas49th!
Oops!

In that case, all you need to know is that "hyperbolic" trig functions cosh sinh tanh sech coth and cosech work almost the same as ordinary trig functions (for example, sinh(2x) = 2sinhx coshx), but occasionally you get a + instead of a minus (or vice versa) … I think only when you have two sinh's.

But, to be on the safe side, use gabbagabbahey's method!

6. Nov 25, 2008

### thomas49th

using the identities i got

$$\frac{1}{2} e^{2x} - e^{-2x}$$
which is equilivlent to cosh2x. but where next?

Thanks ;)

7. Nov 25, 2008

### gabbagabbahey

I think you'd better show me your work

8. Nov 25, 2008

### thomas49th

$$2(\frac{e^{2x} - e^{-2x}}{2})(\frac{e^{2x} + e^{-2x}}{2})$$
one 2 cancels so you get a half overall. the difference of two squares acts nicely leaving us with:
$$\frac{1}{2} e^{2x} - e^{-2x}$$
and i was looking back over the notes in class and I saw that we identified cosh2x as that

Right?
Thanks :)

9. Nov 25, 2008

### tiny-tim

oh i see … you're proving 2 sinhx coshx = sinh 2x (not cosh 2x! … cosh is the positive one )

but what about the original problem, with A and B?

10. Nov 25, 2008

### thomas49th

11. Nov 25, 2008

### gabbagabbahey

Well, if $$\sinh(x)=\frac{e^x-e^{-x}}{2}$$....then $$\sinh(A)=\frac{e^A-e^{-A}}{2}$$....so $\sinh(B)=$___? And so $2\sinh(A)\sinh(B)=$___?

12. Nov 25, 2008

### thomas49th

$$\sinh(B)=\frac{e^B-e^{-B}}{2}$$

$2\sinh(A)\sinh(B)= \frac{1}{2} [ e^{AB} - 2e^{-AB} + e^{AB}]$

$$e^{AB} - e^{-AB}$$

I'm almost there?
Thanks :)

13. Nov 25, 2008

### gabbagabbahey

Hmmm... $$e^Ae^B=e^{A+B} \neq e^{AB}$$

14. Nov 25, 2008

### thomas49th

$$e^{A+B} - e^{A-B}$$

Notice how there is an A+B and A-B from JUST like in cos(A+B)

now how do I show that cos(A+B) = $$e^{A+B}$$

15. Nov 25, 2008

### gabbagabbahey

aren't you missing a couple of terms in that expression?

16. Nov 25, 2008

### thomas49th

arrrg it was just starting to look nice:

stick
$$-e^{-A+B} -e^{-A-B}$$

I've goto go to bed... knackard sorry. it's almost midnight ere in merry old england

i'd read any other message people post on here in the morning. thanks for everything!
:)

17. Nov 25, 2008

### gabbagabbahey

Looks good, except your missing a factor of 1/2, and one of your signs is incorrect..... you should have:

$$2\sinh (A) \sinh (B)=\frac{e^{A+B}-e^{A-B}-e^{-A+B}+e^{-A-B}}{2}$$

You also know the definition of cosh: $$\cosh(x)=\frac{e^x+e^{-x}}{2}$$....so $\cosh (A+B)=$___? And $\cosh (A-B)=$__? And so $\cosh (A+B)-\cosh (A-B)=$___?