Solve for a hyperbolic Laplace Transform by expressing as exponents and shiftig on s-axis (5.3-21) I cannot get past a certain point on this problem as shall be shown. I need guidance in order to complete the problem.
The exercise as stated in the text:
Represent the hyperbolic function in terms of exponential functions and apply the $s$-shifting theorem to show that:
$$\mathcal{L}\{\cosh(at)\cos(at)\}=\frac{s^3}{s^4+4a^4}$$
My Solution:
For clarity we state what the formula to the $s$-shifting theorem is:
$$\mathcal{L}\{e^{at}f(t)\}=F(s-a)$$
We then fragment $\cosh(at)$ into its exponential constituents:
$$\cosh(at)=\frac{e^{at}+e^{-at}}{2}=\frac{e^{at}}{2}+\frac{e^{-at}}{2}$$
We can now express our original function as follows:
$$\cosh(at)\cos(at)=\frac{e^{at}}{2}\cos(at)+\frac{e^{-at}}{2}\cos(at)$$
For clarity we state what the Laplace Transform for $\cos(at)$ is:
$$\mathcal{L}\{\cos(at)\}=\frac{s}{s^2+a^2}$$
Our next step is to find the Laplace Transform for our new expression using the s-shifting theorem:
$$F(s)=\frac{1}{2}\frac{s-a}{(s-a)^2+a^2}+\frac{1}{2}\frac{s+a}{(s+a)^2+a^2}$$
$$=\frac{s-a}{2s^2-4as+4a^2}+\frac{s+a}{2s^2+4as+4a^2}$$
This is where I get stumped. I can't figure out how to join these two fractions and its just basic algebra. How embarrassing!

$$\frac{1}{2}\left(\frac{s-a}{(s-a)^2+a^2}\times\frac{(s+a)^2+a^2}{(s+a)^2+a^2}+\frac{s+a}{(s+a)^2+a^2}\times\frac{(s-a)^2+a^2}{(s-a)^2+a^2}\right)$$
$$\frac{1}{2}\frac{(s-a)((s+a)^2+a^2)+(s+a)((s-a)^2+a^2)}{((s+a)^2+a^2)((s-a)^2+a^2)}$$
$$\frac{1}{2}\frac{(s-a)(s^2+2as+2a^2)+(s+a)(s^2-2as+2a^2)}{(s^2+2as+2a^2)(s^2-2as+2a^2)}$$
$$\frac{1}{2}\frac{2s^3}{(s^2+2a^2)^2-4a^2s^2}$$
$$\frac{s^3}{s^4+4a^2s^2+4a^4-4a^2s^2}$$
$$\frac{s^3}{s^4+4a^4}$$
