Show that the $n$-th Fibonacci number is given by $\frac{\cosh na}{\cosh a}$ or $\frac{\sinh na}{\cosh a}$, where $\sinh a=1/2$ This question is taken from book: Advanced Calculus: An Introduction to Classical Analysis, by Louis Brand. The book is concerned with introductory real analysis.
I request to help find the solution.

Show that the general term of Fibonacci sequence $1,1,2,3,5,\cdots$, is given by :
$f_n = \frac{\cosh\, n\alpha}{\cosh\, \alpha}$ ($n$ odd), $f_n = \frac{\sinh\, n\alpha}{\cosh\, \alpha}$ ($n$ even),
where $\sinh \alpha= \frac 12$, and that $\lim \frac{f_{n+1}}{f_n} = e^\alpha.$

As given here:
The hyperbolic functions $\sinh$, $\cosh $ are given by:
$$\sinh\,z = \frac{e^z - e^{-z}}{2}, \cosh\,z = \frac{e^z + e^{-z}}{2}$$
where $\,z= x+iy\,\,$ is a complex variable.
These functions satisfy the identities:
$$\cosh^2\,z-\sinh^2\,z = 1,\,\, \cosh\,iy = \cosh\,y,\,\, \sinh\,iy = i\,\sinh\,y$$
If $\sinh\,z = \frac{e^z - e^{-z}}{2} = \frac 12$, then $\cosh^2\,z-\sinh^2\,z = 1\,\,$ gives $\cosh^2\,z=\frac 54$
This yields nothing, so consider opposite approach of taking help by another means of generating Fibonacci sequence terms.
Say, the polynomial $f(x) = x+1$ yields terms for successive values for $x \in \mathbb{N}$ as $2,3,\dots$.
So, took help from paper here that concerns with fibonacci polynomials, with text portion from page 1 copied below:

The Fibonacci polynomials $\{F_n (x)\}$ are defined by
(1.1) $F_1(x) = 1, F_2(x) = x$, and $F_{n+1}(x) = xF_n(x) + F_{n-1}x$.
Notice that, when $x = 1, F_n(1) = F_n$ , the $n$ Fibonacci number. It is easy
to verify that the relation 
(1.2) $F_{-n}(x) = (-1)^{n+1} F_n(x)$
extends the definition of Fibonacci polynomials to all integral subscripts.

But, this also doesn't help.
 A: Hint for induction. By the addition formula for $\cosh$ (see wiki),
$$\cosh((n+1)\alpha)=\cosh(n\alpha)\cosh(\alpha)+\sinh(n\alpha)\sinh(\alpha)$$
and 
$$\cosh((n-1)\alpha)=\cosh(n\alpha)\cosh(\alpha)-\sinh(n\alpha)\sinh(\alpha).$$
Hence 
$$\cosh((n+1)\alpha)-\cosh((n-1)\alpha)=2\sinh(n\alpha)\sinh(\alpha)$$
and, after dividing by $\cos(\alpha)$, if $n$ is even we get
$$f_{n+1}-f_{n-1}=2f_{n}\sinh(\alpha)=f_n\implies f_{n+1}=f_n+f_{n-1}.$$
In a similar way, by using the addition formula for $\sinh$, we verify that the same recurrence holds when $n$ is odd.
As regards the limit you may use the unified formula
$$f_n=\frac{e^{n\alpha} - (-1)^ne^{-n\alpha}}{e^{\alpha}+e^{-\alpha}}$$
Since $\alpha>0$, it follows that, as $n\to \infty$,
$$\frac{f_{n+1}}{f_n}=\frac{e^{(n+1)\alpha} - (-1)^{n+1}e^{-(n+1)\alpha}}{e^{n\alpha} - (-1)^ne^{-n\alpha}}\to e^{\alpha}.$$
P.S. Note that $e^{\alpha}=\frac{1+\sqrt{5}}{2}=\varphi$ is the Golden Ratio and therefore the above unified formula can be written as
$$f_n=\frac{\varphi^n-(-1)^n\varphi^{-n}}{\varphi+\varphi^{-1}}=\frac{\varphi^n-(-\varphi)^{-n}}{\sqrt{5}}$$
which is the usual closed-form expression for the Fibonacci numbers.
