Help with hyperbolic trig

Question

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!

Answer 1

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!