Here's an example in planar euclidean geometry. Consider an equilateral triangle of side $a$ and a general point in the plane distant $b$, $c$, and $d$ from the respective vertices. Then
$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2$.
This is an an awful slog to get by planar trigonometry. Even harder to do by trig in three dimensions is the corresponding result for the regular tetrahedron. However, it's easy to get the $(n - 1)$-dimensional result for a regular $(n - 1)$-dimensional simplex of side $d_0$, with vertex distances $d_1$ ,..., $d_n$ :
$n(d_0^4 + ... + d_n^4) = (d_0^2 + ... + d_n^2)^2$.
You can do this by embedding the euclidean $(n - 1)$-dimensional space as the hyperplane of points $(x_1 ,..., x_n)$ in euclidean $n$-space such that $x_1 + ... +x_n = d_0/\sqrt2$. The vertices of the simplex can then be represented as the points $(d_0/\sqrt2)(1, 0 ,..., 0)$, ... , $(d_0/\sqrt2)(0 ,..., 0, 1)$ in the hyperplane, and the result drops out in a few lines.