1. ## Hyperbolic functions

Hi,

I'm going through a past paper for an exam and found a problem which I don't know how to do, I am not sure if it is still examined any more but I am still interested in how you work it out:

Using definitions $\cosh(x) = \frac{1}{2}(e^x + e^{-x})$ and $\sinh(x) = \frac{1}{2}(e^x - e^{-x})$

$\sinh(x - y) = \sinh(x)\cosh(y) - \cosh(x)\sinh(y)$

I think that it is related to double angle formula but I might be mistaken.

2. Have you tried subbing in the definitions on each side?

3. Well I stuck the terms in but they seem to cancel out each other so I assume I'm doing something wrong. Heres what I've done so far.

$\frac{1}{2}(e^x - e^{-x}) \cosh(y) - \frac{1}{2}(e^x + e^{-x})\sinh(y)$

I then did the same for the ones with the y and got.

$\frac{1}{4}((e^x - e^{-x})(e^y + e^{-y}) - (e^x + e^{-x})(e^y - e^{-y}))$

When i multiplied out i got:

$\frac{1}{4}((e^{xy} + e^{-xy} - e^{xy} - e^{xy}) - (e^{xy} - e^{-xy} + e^{-xy} - e^{xy}))$

and surely these terms all just cancel out, so where am I going wrong

4. Originally Posted by Beard
and surely these terms all just cancel out, so where am I going wrong
all of them?

Edit:

Oh, also, your exponents aren't right

Originally Posted by Beard
$\frac{1}{4}((e^x - e^{-x})(e^y + e^{-y}) - (e^x + e^{-x})(e^y - e^{-y}))$
Fine.

Originally Posted by Beard
When i multiplied out i got:

$\frac{1}{4}((e^{xy} + e^{-xy} - e^{xy} - e^{xy}) - (e^{xy} - e^{-xy} + e^{-xy} - e^{xy}))$
$e^x * e^y = e^{xy}$ ?!

5. Remember that $e^x \cdot e^y=e^{x+y}$, not $e^{xy}$.

6. Sorry I haven't done maths in a looooong time

7. I now get an alternative which I can sort of see is close to the answer but I'm not sure how to do the next step.

$\frac{1}{2}(e^{x - y} - e^{y - x})$

8. $
\frac{1}{2}(e^{x - y} - e^{y - x})
= \frac{1}{2}(e^{x-y} - e^{-(x-y)})
= \sinh (x-y)
$