I was trying to derive by myself the distance on a great circle.
To confirm my calculations are correct I was comparing against the formula (4) in this link, which is something like
$$ \cos(\delta_1)\cos(\delta_2) \cos(\lambda_1 - \lambda_2) + \sin(\delta_1) \sin(\delta_2) $$
According to wikipedia instead the formula looks more:
$$ hav(\Theta) = hav(\varphi_2 - \varphi_1) + \cos(\varphi_1) \cos(\varphi_2) hav(\lambda_2 - \lambda_1) $$
Where
$$ hav(\theta) = \sin^2\left( \frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{2} $$
They don't seem equivalent to me, I've tried to use some trig to prove they're an identity, but unless I'm making mistakes it doesn't seem the case.
My derivation leads exactly to the first formula I've shown. The question is... which one is correct, are they the same?
Note that $$ \operatorname{hav}\Theta = \operatorname{hav}(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)\operatorname{hav}(\lambda_1-\lambda_2) $$ iff $$ 1-\cos\Theta = 1-\cos(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)[1-\cos(\lambda_1-\lambda_2)] $$ iff $$ \cos\Theta = \cos(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)[\cos(\lambda_1-\lambda_2)-1], $$ and recall $\cos(\varphi_1-\varphi_2)=\cos\varphi_1\cos\varphi_2-\sin\varphi_1\sin\varphi_2$.