I have been working on a problem in my physics homework and I am stuck on a particular line of many trying to show that the following hyperbolic trig expressions are equivalent. Can someone please help? I want to show: $$ -\frac{1}{2}(x^2+y^2)\coth(\gamma)+\frac{xy}{\sinh(\gamma)}=-\frac{1}{4}\left[(x+y)^2\tanh\left(\frac{\gamma}{2}\right)+(x-y)^2\coth\left(\frac{\gamma}{2}\right)\right] $$ where I am told this identity has been used: $$ \tanh\left(\frac{\gamma}{2}\right)=\frac{\cosh(\gamma)-1}{\sinh(\gamma)}=\frac{\sinh(\gamma)}{1+\cosh(\gamma)} $$ I am just going in circles and getting lost in the steps. Please help! Thank you!
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$\begingroup$ A common strategy with trig identities is to convert all teig functions to sines and cosines. Have you tried converting everything in sight into hyperbolic sines and hyperbolic cosines, then multiply out the binomials, then start matching terms? $\endgroup$– Eric TowersMar 26, 2021 at 23:00
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1$\begingroup$ you ought to expand out the $(x+y)^2$ and the $(x-y)^2$ into $x^2 + y^2$ and the appropriate from $\pm 2xy$ Then you collect the coefficient of $(x^2 + y^2),$ show that is zero, and the total coefficient of $xy,$ show that is zero $\endgroup$– Will JagyMar 26, 2021 at 23:08
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1$\begingroup$ Thank you for your help! I have gotten it now. There is a factor of 4 missing in one of the definitions. Everything is working now. I will edit the problem for others if they are interested. $\endgroup$– AnneMar 26, 2021 at 23:29
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$\begingroup$ @Anne: "I have gotten it now." ... Please provide your solution as an answer so that we can up-vote your success (and to remove the question from the Unanswered queue). :) $\endgroup$– BlueMar 26, 2021 at 23:38
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After writing the problem on the screen, things seemed to become clear and I found a missing factor of 4 in a defined constant. Now everything seems okay. I start with the expression in brackets on the right-hand side and rewrite using the given indentities. $$ -\left[(x+y)^2\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+(x-y)^2\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$ $$ =-\left[(x^2+2xy+y^2)\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+(x^2-2xy+y^2)\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$ $$ =-\left[(x^2+y^2)\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+\frac{2xy\sinh(\gamma)}{1+\cosh(\gamma)}+(x^2+y^2)\frac{\sinh(\gamma)}{\cosh(\gamma)-1}-2xy\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$ $$ =-\left[(x^2+y^2)\left(\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right)+2xy\frac{\sinh(\gamma)}{\cosh+1}-2xy\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$ Take last two terms on the right. $$ 2xy\frac{\sinh(\gamma)}{\cosh(\gamma)+1}-2xy\frac{\sinh(\gamma)}{\cosh(\gamma)-1} $$ $$ =\frac{2xy\sinh(\gamma)(\cosh(\gamma)-1)-2xy\sinh(\gamma)(\cosh(\gamma)+1)}{(\cosh(\gamma)+1)(\cosh(\gamma)-1)} $$ $$ =4xy\frac{\sinh(\gamma)}{\cosh^2(\gamma)-1}=-\frac{4xy}{\sinh(\gamma)} $$ Now look at first two terms on the left of the expression. $$ -(x^2+y^2)\left[\frac{\sinh(\gamma)}{1+\cosh(\gamma)}+\frac{\sinh(\gamma)}{\cosh(\gamma)-1}\right] $$ $$ =-(x^2+y^2)\left[\frac{\sinh(\gamma)\cosh(\gamma)-\sinh(\gamma)+\sinh(\gamma)+\sinh(\gamma)\cosh(\gamma)}{\sinh^2(\gamma)}\right] $$ $$ =-2(x^2+y^2)\coth(\gamma) $$ Then plugging it all, we have $$ -\frac{1}{2}(x^2+y^2)\coth(\gamma)+\frac{xy}{\sinh(\gamma)}=-\frac{1}{4}\left[2\coth(\gamma)-\frac{4xy}{\sinh(\gamma)}\right] $$
and we arrive at the desired result. Thank everyone for their useful comments!