This result is known. More generally, a regular $n$-gon with side-length $x$ has area $\frac{n}{4}x^2\cot(\frac{\pi}{n})$.
To see this: by translating and rotating we can assume that the vertices of the regular $n$-gon are at $(r\cos k\theta, r\sin k\theta)$ for $0 \le k < n$, where $\theta = \frac{2\pi}{n}$ and $r$ is the distance from the centre of the $n$-gon to its vertices.
Thus the area is $n$ times the area of the triangle with vertices $(0,0)$, $(r,0)$ and $(r\cos \theta, r\sin \theta)$. Thus
$$A = \frac{n}{2}r^2\sin \frac{2\pi}{n}$$
Now the side length is given by
$$x = \lVert (r\cos \theta, r\sin\theta) - (r,0) \rVert = \sqrt{(r\cos \theta-1)^2+\sin^2\theta} = r\sqrt{2-2\cos \theta}$$
So substituting for $r$ in the formula above gives
$$A = \frac{n}{4(1-\cos \frac{2\pi}{n})}x^2\sin \frac{2\pi}{n}$$
Finally we use $\cos 2\theta \equiv 1-2\sin^2 \theta$ and $\sin 2\theta = 2\sin \theta \cos \theta$ to get
$$A = \frac{n}{8\sin^2\frac{\pi}{n}}x^2 \cdot 2 \sin \frac{\pi}{n} \cos \frac{\pi}{n} = \frac{n}{4}x^2\cot\frac{\pi}{n}$$
Your expression is equivalent in the case when $n=8$.
