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I have the following transcendental equation which may very well lack an analytic solution. I would, at the least, like an expression for the relationship between $\theta$ and $\phi$ in some closed form:

$\cos\left((n+\frac{1}{2})\theta\right)\cosh\left((n-\frac{1}{2})\phi\right) = \cos\left((n-\frac{1}{2})\theta\right)\cosh\left((n+\frac{1}{2})\phi\right)$

Both $\cos((x-1/2)\theta)$ and $\cosh((x-1/2)\phi)$ are eigenvectors defined on the interval $x\in (1,2,...,n)$.

It might help that it seems that the derivatives with respect to x match at $x=n$, but I currently lack proof.

Any special functions that could be used to re-express this or a known method of solution would be useful.

Michael Jarret
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  • What happens when you transform $\cos$ and $\cosh$ into their $e^x$ forms? – abiessu Jul 23 '13 at 16:29
  • Not much useful, honestly. Things get ugly pretty fast, but I do know that both $\theta$ and $\phi$ are real. – Michael Jarret Jul 23 '13 at 16:30
  • And I presume that things are similarly ugly if you have first cross-divided the $\cos$ and $\cosh$ to separate sides? – abiessu Jul 23 '13 at 16:31
  • I'm not quite sure, but I haven't found a good way to even formally expand the series and simplify it. The approaches I've tried were to do that and then use half-angle and multiple-angle formulae to simplify the problem in terms of series solutions. – Michael Jarret Jul 23 '13 at 16:33
  • If anyone is interested in the above result, I can post them, but I don't know how helpful they'll be. For whatever the reason I have not tried expanding the two series and then doing term-wise multiplication under the double sum. I will give that a shot too but am skeptical it will lead anywhere. – Michael Jarret Jul 23 '13 at 16:45
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    I can get this to the point that ${\cos(a+b) \over \cos(a-b)} = {\cosh(c+d) \over \cosh(c-d)}$, and I can see that if $f(n) = \cos((n + {1 \over 2})\theta)$ then ${\cos((n + {1 \over 2})\theta) \over \cos((n - {1 \over 2})\theta)} = {f(n) \over f(-n)}$ but I haven't been able to get much farther yet. – abiessu Jul 23 '13 at 17:59

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