When $\cosh yx/2=\pm 1 $?

Question

When $\cosh \frac{xy}{2}=\pm 1 $? is it correct to say $xy/2=cosh^{-1}(\pm1)$ Then $xy=2 \cosh^{-1}(\pm1)$

I think there is better solution for this problem? any idea?

Answer 1

Note: I am assuming that the quantities in your equations are real numbers.

Since $\cosh u \ge 1$, with the minimum only occurring at $u = 0$, you're complicating the calculation too much. You can conclude immediately that $xy = 0$, so either $x = 0$ or $y = 0$ or both. (EDITED, to correct silly error.)

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By the way, since $\cosh$ is an even function, meaning that $\cosh(-u) = \cosh u$ for all $u \in \Bbb{R}$, it is not one-to-one, hence cannot have an inverse. However, if you artificially restrict it's domain, say to $u \ge 0$ (so $\sinh u \ge 0$), then you can solve for the inverse: $$ \cosh^2 u - \sinh^2 u = 1 \quad\Longrightarrow\quad \sinh^2 u = \cosh^2 u - 1 \quad\Longrightarrow\quad \sinh u = \sqrt{\cosh^2 u - 1} $$ Now, \begin{align} \cosh u + \sinh u &= e^u \\ \cosh u + \sqrt{\cosh^2 u - 1} &= e^u \\ \ln \bigl( \cosh u + \sqrt{\cosh^2 u - 1} \bigr) &= u \\ \end{align} so if $x = \cosh u$ and $u \ge 0$, then $$ u = \ln \bigl( x + \sqrt{x^2 - 1} \bigr), $$ and this defines the restricted inverse. Here's a picture.

Answer 2

The minimum of $\cosh x$ occurs when $x=0$, so in this case you require $\frac{xy}{2}=0$ so you can conclude that either $x=0$ or $y=0$ or both