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I know the formula for calculating the heading between to given points (latitude, longitude) is $$ \tag{1} \theta = \arctan2(\sin(\Delta\lambda)*\cos(\varphi_2), \cos(\varphi_1)*\sin(\varphi_2) − \sin(\varphi_1)*\cos(\varphi_2)*\cos(\Delta\lambda)) $$ where $\theta$ is the heading from the starting point $P_1(\varphi_1, \lambda_1)$ (latitude, longitude) to the target point $P_2(\varphi_2, \lambda_2)$ and the difference in longitude is $\Delta\lambda = \lambda_2 - \lambda_1$ unless the great circle between the points crosses longitude $\pi$ or $-\pi$, in which case you have to correct. Equation 1 comes from the formula $$\tag{2} \tan(\theta) = \frac{\sin(\Delta\lambda)\cos(\varphi_2)}{\cos(\varphi_1)\sin(\varphi_2)-\sin(\varphi_1)\cos(\varphi_2)\cos(\Delta\lambda)} $$ but I don't know where this one comes from. Can anyone provide a derivation as I want to understand how this formula came to be.

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  • $\begingroup$ Your first formula appears on page 23 of the old (1886) book "Spherical Trigonometry: For the Use of Colleges and Schools" by I. Todhunter, available for free download from Project Gutenberg. gutenberg.org/ebooks/19770 $\endgroup$ – awkward Jan 14 '18 at 14:31
  • $\begingroup$ I didn't find the formula on page 23... $\endgroup$ – thomasfermi May 8 '18 at 18:06
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Have a look at this sketch of a triangle on a sphere. B denotes point 1, i.e., $(\phi_1, \lambda_1)$ as well as the angle in the depicted triangle at that point. C denotes point 2, i.e., $(\phi_2, \lambda_2)$ as well as the depicted angle in the triangle at that point. A denotes the north pole as well as the angle in the triangle at that point. We want to find the angle B, which you denoted by $\theta$ in your question.

We can read off these sides and angles from the sketch: \begin{align} c&=(\pi/2 - \phi_1) \\ b&=(\pi/2-\phi_2) \\ A &= (\lambda_2 - \lambda_1) =: \Delta \lambda \end{align}

To start off, let's use one of the cosine rules of spherical trigonometry \begin{align} \cos a&= \cos b \cos c + \sin b \sin c \cos A \\ &=\sin \phi_2 \sin \phi_1 + \cos \phi_2 \cos \phi_1 \cos (\Delta \lambda) \end{align} Note that I used $\sin(\pi/2-x)=\cos(x)$ and $\cos(\pi/2-x)=\sin(x)$ for the last step.

Now that we have $\cos a$, let's solve the following cosine rule of spherical trigonometry for $\cos B$: \begin{align} \cos b&= \cos c \cos a + \sin c \sin a \cos B \\ \cos B &= \frac{\cos b -\cos c \cos a }{\sin c \sin a} \end{align} The formula you are looking for is given in terms of $\tan B$. So we want to divide $\sin B$ by $\cos B$. For $\sin B$ we will use the sine rule of spherical trigonometry, which states that \begin{align} \frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c} \end{align} Hence \begin{align} \sin B = \sin b \frac{\sin A }{\sin a} \end{align} Now we have a formula for $\cos B$ and a formula for $\sin B$. We can use that to find \begin{align} \frac{\sin B}{\cos B} &= \sin b \frac{\sin A }{\sin a} \frac{1}{\cos B} \\ \tan (B) &= \cos \phi_2 \frac{\sin(\Delta \lambda)}{\sin a} \frac{\sin c \sin a} {\cos b -\cos c \cos a }\\ &= \cos \phi_2 \sin(\Delta \lambda) \frac{\sin c} {\cos b -\cos c \cos a }\\ &= \cos \phi_2 \sin(\Delta \lambda) \frac{\cos \phi_1} {\sin \phi_2 -\sin \phi_1 \cos a } \\ &= \cos \phi_2 \sin(\Delta \lambda) \frac{\cos \phi_1} {\sin \phi_2 -\sin \phi_1( \sin \phi_2 \sin \phi_1 + \cos \phi_2 \cos \phi_1 \cos (\Delta \lambda) ) } \\ &= \frac{ \sin(\Delta \lambda) \cos \phi_2 \cos \phi_1} {\sin \phi_2(1-\sin^2\phi_1) -\sin \phi_1 \cos \phi_2 \cos \phi_1 \cos (\Delta \lambda) ) } \\ &= \frac{\sin(\Delta \lambda) \cos \phi_2 } {\sin \phi_2 \cos \phi_1 -\sin \phi_1 \cos \phi_2 \cos (\Delta \lambda) ) } \end{align}

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