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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 $-1$?

user144464
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  • $s^2=a\implies s\color{Red}{\ne}\pm a$ ?? – anon May 05 '14 at 18:18
  • I put that point because $\sinh$ only takes ONE value. – user144464 May 05 '14 at 18:19
  • And $\ne$ signifies that? – anon May 05 '14 at 18:21
  • @seaturtles well I've answered your question so I hope that signifies it... – user144464 May 05 '14 at 18:23
  • Between editing the original post to make sense versus leaving it as it stands while only clarifying it in the comments you chose the latter option? People will understand what's supposed to be written so it's not a big deal, but now we've wasted five comments discussing a very simple matter because you're indifferent to being clear and straightforward combined with my vocally being curious as to why. I find that puzzling, so I'm going to leave this alone now. – anon May 05 '14 at 18:27
  • (I think the conventional interpretations of $x=\pm a$ and $x\ne\pm a$ are "$x=a$ or $x=-a$" and "$x\ne a$ and $x\ne -a$" respectively. Apparently I lied about leaving it alone.) – anon May 05 '14 at 18:34

1 Answers1

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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 $x$.

anon
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