I have two hyperbolic cosine functions that I would like to determine the intersection point of:
$$f_1(x) = (a \cdot cosh(x))+b$$
$$f_2(x) = (a \cdot cosh(x+c))+d$$
The best I can do by setting the functions equal to one and other is this:
$$x=ln\left(\frac{b-d}{2 \cdot a \cdot sinh(\frac{c}{2})} + \sqrt{\left(\frac{b-d}{2 \cdot a \cdot sinh(\frac{c}{2})}\right)^2 +1}\right) -\frac{c}{2}$$
Is there a way to further reduce the above or a preferable method?
Many thanks.
UPDATE
Regarding comment made by @AnneBauval:
The quadratic equation produces a similarly bulky solution:
$$cosh(x+c) - cosh(x) = \frac{b-d}{a} $$ $$ \left( \frac{(e^x \cdot e^c) + (e^{-x} \cdot e^{-c})}{2} \right) - \left(\frac{e^x + e^{-x}}{2}\right) = \frac{b-d}{a} $$ $$((e^c - 1) \cdot e^x) + ((e^{-c} - 1) \cdot e^{-x})= \frac{2 \cdot (b-d)}{a} $$ $$((e^c - 1) \cdot e^{2x}) - \left(\left( \frac{2 \cdot (b-d)}{a} \right) \cdot e^x\right) + (e^{-c} - 1) = 0$$ $$e^x = \frac{\left( \frac{2 \cdot (b-d)}{a} \right) \pm \sqrt{\left( \frac{2 \cdot (b-d)}{a} \right)^2 - (4 \cdot (e^c - 1) \cdot (e^{-c} - 1))}}{2 \cdot (e^{-c} - 1)}$$ $$x = ln\left(\frac{\left( \frac{2 \cdot (b-d)}{a} \right) \pm \sqrt{\left( \frac{2 \cdot (b-d)}{a} \right)^2 + (8 \cdot (cosh(c)-1))}}{2 \cdot (e^{-c} - 1)}\right)$$
I was hoping there was some sort of clever trig relationship that I was missing that would simplify the solution... particularly the component within the square root. For example, something similar to $cosh(\frac{x}{2})= \sqrt{\frac{cosh(x) + 1}{2}}$
Yes, the "clever trig relationship" is
$$\cosh y-\cosh z=2\sinh\frac{y+z}2\;\sinh\frac{y-z}2$$
which follows from setting $y=u+v$ and $z=u-v$ and using the addition formula for $\cosh(u+v)$.
$$\cosh(x+c)-\cosh(x)=\frac{b-d}a$$ $$2\sinh(x+\tfrac c2)\sinh\tfrac c2=\frac{b-d}a$$ $$\sinh(x+\tfrac c2)=\frac{b-d}{2a\sinh\tfrac c2}$$ $$x=-\tfrac c2+\text{arsinh}\left(\frac{b-d}{2a\sinh\tfrac c2}\right)$$