Activity for Adamâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #280910 |
It's equal to $\frac{a|b|\pi}{(|a|+|b|)\sin(a\pi/(|a|+|b|))}$. You can prove it with series expansions for $\sec$ and the elliptic integral. (more) |
— | over 3 years ago |