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Suppose that you have have two points $x = (\phi_1, \lambda_1)$ and $y = (\phi_2, \lambda_2)$, where $\phi$ and $\lambda$ represent latitude and longitude. Let $\Delta\phi = \phi_2 - \phi_1$ and $\Delta\lambda = \lambda_2 - \lambda_1$, then the distance between the two points in meters, $d$, is given by the following system

\begin{align} &a = \sin²(\Deltaφ/2) + \cos φ_1 ⋅ \cos φ_2 ⋅ \sin²(\Delta\lambda/2)\\ &c = 2 ⋅ atan2( \sqrt{a}, \sqrt{(1−a)} )\\ &d = R ⋅ c \end{align}

or alternatively by

$$ d = cos^{-1}( \sin \phi_1 ⋅ \sin \phi_2 + \cos \phi_1 ⋅ \cos \phi_2 ⋅ \cos \Delta\lambda ) ⋅ R\tag{1}$$

(Source for these equations can be found here)

Question: Suppose that $\lambda_1 = \lambda_2$, and suppose that I know $\phi_1$. How do I determine $\phi_2$ such that the distance between $(\phi_1, \lambda_1)$ and $(\phi_2, \lambda_2)$ is $d$?

What I've tried: If $\lambda_1 = \lambda_2$ then equation $(1)$ above can be reduced to $$ \cos\big( \frac{d}{R}\big) = \sin\phi_1\sin\phi_2\\ \Leftrightarrow \phi_2 = sin^{-1}\bigg(\dfrac{\cos(d/R)}{\sin(\phi_1)}\bigg) $$ This doesn't work and I would like to know why!

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I believe you have mistakenly reduced the expression $\cos{\phi_1}\cos{\phi_2}\cos{\Delta \lambda}$ to $0$ instead of $1$, as $\cos{0}=1$. Then we have $$\cos{\frac{d}{R}} = \cos{(\phi_2-\phi_1)}$$ $$\phi_2 = \frac{d}{R}+\phi_1+2k\pi, k \in \mathbb{Z}$$

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