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

$\displaystyle \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.