Anyone knows the reference for this formula or can derive them in Spherical coordinates? I have used the formula which is available on a website to calculate the destination longitude and latitude based-on current long/lat, distance to destination and bearing. For example,long/lat of a point on earth is given and the goal is calculating long/lat of a point which is located 1km away from start point. Here is the formula:
φ2 = asin( sin φ1 ⋅ cos δ + cos φ1 ⋅ sin δ ⋅ cos θ )
λ2 = λ1 + atan2( sin θ ⋅ sin δ ⋅ cos φ1, cos δ − sin φ1 ⋅ sin φ2 )
I implemented the formula with my desired programming language and the formula works with no problem, but I need the reference for this formula. In other words, I want to understand how this formula is derived analytically. I already sent the writer an email to ask about the reference but no reply yet. I'm wondering anyone knows the reference for this formula or any similar formula? Thanks
 A: Knowns: start point $(\varphi_1$, $\lambda_1)$, heading $\theta$, radius $r$, distance $d$.
By the law of spherical cosines, we know that:
\begin{equation}\tag{1}
\cos(\delta) = \sin(\varphi_1)\sin(\varphi_2) + \cos(\varphi_1)\cos(\varphi_2)\cos(\Delta\lambda)
\end{equation}
defines the angular distance, which means the arclength formula on a sphere gives 
$\delta = d/r $, which is easily computed.
Next, comes a tougher part. (see here).
Consider the spherical triangle generated between the two points and the north pole. Let $\psi_1$ and $\psi_2$ be the angles between the north pole (with which the heading is defined) and the points. The law of spherical cosines then gives:
$$
\cos(\psi_2) = \cos(\psi_1)\cos(\delta)+\sin(\psi_1)\sin(\delta)\cos(\theta)
$$
$$
\therefore\; \cos(\theta) = \frac{\cos(\psi_2) - \cos(\psi_1)\cos(\delta)}{\sin(\psi_1)\sin(\delta)}
$$
But then using $\psi_i = \pi/2 - \varphi_i$, we get:
$$\tag{2}
\cos(\theta) = \frac{\sin(\varphi_2) - \sin(\varphi_1)\cos(\delta)}{\cos(\varphi_1)\sin(\delta)}
$$
This lets us solve for $\varphi_2$ as:
$$\tag{$\ast$}
\varphi_2 = \arcsin\left(
\cos(\varphi_1)\sin(\delta)\cos(\theta) + \sin(\varphi_1)\cos(\delta)
\right)
$$
Note it is also possible to express $\theta$ without the angular distance via:
$$ \tag{3}\tan(\theta) = \frac{\sin(\Delta\lambda)\cos(\varphi_2)}{\cos(\varphi_1)\sin(\varphi_2)-\sin(\varphi_1)\cos(\varphi_2)\cos(\Delta\lambda)} $$
Now for $\lambda_2$. The first equation we wrote tells us that:
$$\tag{4}
\cos(\Delta\lambda) = \frac{\cos(\delta) - \sin(\varphi_1)\sin(\varphi_2)}{
\cos(\varphi_1)\cos(\varphi_2)}
$$
Furthermore, using the expression for $\tan(\theta)$ (eq 3) we can get $\sin(\Delta\lambda)$ as follows:
\begin{align}
\sin(\theta)\left[
\cos(\varphi_1)\sin(\varphi_2) - \sin(\varphi_1)\cos(\varphi_2)\cos(\Delta\lambda)\right]
 &=
\sin(\Delta\lambda)\cos(\varphi_2)\cos(\theta)
\end{align}
Next, substitute in eqs 2 and 4, to get:
$$
\sin(\Delta\lambda) = \frac{\sin(\theta)\sin(\delta)}{\cos(\varphi_2)}
$$
Thus, we get a formula for the tangent via:
$$
\tan(\Delta\lambda) = \frac{\sin(\Delta\lambda)}{\cos(\Delta\lambda)}
=
\frac{\cos(\varphi_1)\sin(\theta)\sin(\delta)}{\cos(\delta) - \sin(\varphi_1)\sin(\varphi_2)}
$$
Solving for $\lambda_2$ gives us:
$$\tag{$\dagger$}
\lambda_2 = \lambda_1 + \arctan\left(\frac{\cos(\varphi_1)\sin(\theta)\sin(\delta)}{\cos(\delta) - \sin(\varphi_1)\sin(\varphi_2)}\right)
$$
The formulas we wished to derive are exactly those seen in eqs $\ast$ and $\dagger$.
