Sign of $\cosh(\frac{t}{2}\cosh(\theta))\cosh(\frac{t}{2})-\cosh(\frac{t}{2}\cosh(\theta))-\cosh(\frac{t}{2})$ on $t\ge 0$ For fixed $\theta>0$  we consider the function:
$f(t)=\cosh(\frac{t}{2}\cosh(\theta))\cosh(\frac{t}{2})-\Big(\cosh(\frac{t}{2}\cosh(\theta))+\cosh(\frac{t}{2})\Big)$ for all $t\ge 0$
I want to study the sign of  $f$ in $[0,+\infty[$
I calculated the derivative of  $f$ and i find :
$f'(t)=\frac{1}{2}\cosh(\theta)sh(\frac{t}{2}\cosh(\theta))\Big(\cosh(\frac{t}{2})-1\Big)+\frac{1}{2}\sinh(\frac{t}{2})\Big(\cosh(\frac{t}{2}\cosh(\theta))-1\Big)\ge 0$ for all  $t\ge 0$
Then $f(t)\ge f(0)=-2$
But i woud like to find  $t$ for which $f(t)\ge 0$ and $t$ for which $f(t)\le 0$
Please help me. Thanks
 A: 
Key-tool: $\cosh(2x)=1+2\sinh^2(x)$

This is a nice example where computing the derivative only obscures things...
Let $u=\cosh\theta$, then, for every $t$, $$f(4t)=\cosh(2ut)\cosh(2t)-\cosh(2ut)-\cosh(2t)=4\sinh^2(ut)\sinh^2(t)-1$$ hence $f(4t)$ has the sign of $g(t)-1$ where $$g(t)=2\sinh(ut)\sinh(t)$$ The function $g$ increases from $g(0)=0$ to $g(+\infty)=+\infty$ hence, for every $u$, there exists a unique $t^*$ such that $$g(t^*)=1$$ Thus, $f(0)=f(\frac14t^*)=0$, $f(t)<0$ for every $t$ in $(0,\frac14t^*)$, and $f(t)>0$ for every $t$ in $(\frac14t^*,+\infty)$, where $t^*>0$ is the unique solution of the identity 

$$2\sinh(\cosh(\theta)t^*)\sinh(t^*)=1$$

A: So your $f(t) = \cosh(t/2)\cosh(At/2) - (\cosh(t/2) + \cosh(At/2)$ where $A=\cosh(\theta)>1$, and you want to find $t$ solving $$\cosh(At/2) = \frac{\cosh(t/2)}{\cosh(t/2)-1}.$$ I cannot believe there is a closed form solution, except possibly for a few special values of $\theta$.  But the monotony of $f$ should make numerical calculations easy and accurate, if that is any consolation. Or perhaps your $\theta$ is very small, or very large, and asymptotic expansions in powers of $\theta$ would be useful to you.
