I was doing an optics problem and I ended up with the following expression for a reflexion factor:
$$ {r_v ^2 + r^2-2rr_v\cos\delta} \over {1+(rr_v)^2-2rr_v \cos \delta} $$
Now, I could just plug my numbers here, but I couldn't help noticing that the numerator is the opposite side of a triangle with sides $r_v$ and $r$, and an angle $\delta$ between them. Similarly, the denominator is the opposite side of a triangle with sides $1$ and $ { r r_v} $, and an equal angle $\delta$ between them. Taking into account that $1>r>r_v$, I made the following drawing:
This is a really nice geometrical interpretation of the above formula, which makes me suspect that there is a way to simplify it. However I haven't been able to. Completing squares and using $1-\cos\delta=2\sin^2\delta/2$ yields the following:
$$ {(r_v-r)^2+4rr_v\sin^2\delta/2} \over {(rr_v-1)^2+4rr_v\sin^2\delta/2} $$
...which isn't really that simplified. Is there something else I can do?
For example:
$$\begin{cases} r_v^2+r^2-2rr_v\cos\delta=(r-r_v\cos\delta)^2+r_v^2\sin^2\delta=\left|r-r_v\,e^{\pm i\delta}\right|^2\\{}\\ 1+(rr_v)^2-2rr_v\cos\delta=(rr_v-\cos\delta)^2+\sin^2\delta=\left|rr_v-e^{\pm i\delta}\right|^2\end{cases}\;\;\;\;,\,\delta\in\Bbb R$$
So your expression is, using complex numbers, simple the fractions formed with the above, namely
$$\frac{\left|r-r_v\,e^{\pm i\delta}\right|^2}{\left|rr_v-e^{\pm i\delta}\right|^2}=\left|\frac{r-r_v\,e^{\pm i\delta}}{rr_v-e^{\pm i\delta}}\right|^2$$
I don't know if the above simplifies your expression the way you want, but it is at least a shorter one...BTW, as $\;\delta\;$ is a geometrical angle, we can assume it is a positive one (I suppose), so the above can be put a little more succintly as
$$\frac{\left|r-r_v\,e^{i\delta}\right|^2}{\left|rr_v-e^{i\delta}\right|^2}=\left|\frac{r-r_v\,e^{i\delta}}{rr_v-e^{i\delta}}\right|^2$$