symbolic solution to trig equation for a variable

Is it possible to solve the following singular transcendental equation in $a$ for the variable $a$? Any symbolic solution will do.

$$\sqrt{s^2 - v^2} = 2a \, \sinh \left( \frac{h}{2a} \right)\,\,\,$$

If so, I'd like to see the resulting equation.

Even given additional computation time, Wolfram Alpha can't.

More about the equation on Wikipedia: http://en.wikipedia.org/wiki/Catenary#Determining_parameters

• These tags and titles are confusing. Do you want a numerical solution or a symbolic solution? Commented Sep 9, 2014 at 19:26
• @Arkamis wikipedia says "This is a transcendental equation in a and must be solved numerically", but if a symbolic solution exists, that's even better. Commented Sep 9, 2014 at 19:29
• A numerical solution is a numerical approximation to a solution to an arbitrary degree of precision. A symbolic solution is an exact solution coded in mathematical symbols. If an equation must be solved numerically, you cannot expect a solution in terms of your parameters; accordingly, you must prescribe some values for $s, v, h$. Being transcendental does not mean symbolically unsolvable; however, it means it is generally unlikely for symbolic solutions in terms of elementary functions to exist. Commented Sep 9, 2014 at 19:33
• @Arkamis Because I don't have a given s,v, and h, I am wanting a symbolic solution. I don't mind the use of non-elementary functions at all in your solution. Question edited to reflect this. Commented Sep 9, 2014 at 20:35
• I highly doubt one exists. Commented Sep 9, 2014 at 20:51

The inverse of $a\sinh(A/a)=B$ does not possess a closed form expression. Neither does that of the much simpler $a\sinh a=C$, for that matter. However, assuming a very small value for a, we can then easily express the approximate solution in terms of Lambert's W function, since one of the two exponential terms of $\sinh(A/a)$ will be approximately $0$, yielding $\exp(A/a)\approx B/a$, which in its turn gives $a=-A/W(-A/B)$, where $A=h/2$, and $B=\tfrac12\sqrt{s^2-v^2}$. But if a were to be large instead of small, we'd have $\sinh(A/a)\approx A/a+\tfrac16(A/a)^3$, which yields a quadratic equation in a, whose roots are $a=\pm A\Big/\!\sqrt{6~(B/A-1)}$. I'm afraid that, analytically, this is as good as it gets.