showing $$\sinh(x/2) = \epsilon \sqrt{\frac{1}{2}(\cosh(x) -1)}$$
and I was told to determine the value of $\epsilon$.
From identities I reached $ \sinh^2(x) = \dfrac{1}{2}(\cosh(x) -1)$ however when taking the square root, I understand that $$\sinh(x/2) \not= \pm \sqrt{\frac{1}{2}(\cosh(x) -1)}$$ (as $\sinh(x)$ can only take one value) but why does it have $\epsilon = +1$ not $-1$?
Since the left-hand side is continuous, the sign must be constant on any interval in which the radicand is nonzero. Therefore it suffices to check the sign by plugging in values for $x$ in a finite number of intervals. You should find the sign of $\epsilon$ depends on the sign of $x$. Which should make sense, because by inspection from the definition of $\sinh$ it's clear that $\sinh$ shares sign with $x$.