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I have that the Euclidean distance on the surface of a sphere in terms of the angle they subtend at the centre is

$(\sqrt{2})R\sqrt{1-\cos(\theta_{12})}$

(Where $\theta_{12}$ is the angle that the two points subtend at the centre.)

Why is this; what is the proof?

Cheers, Alex

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    $\begingroup$ The distance is $2R\sin(\theta/2)$ (draw the picture). Now use the identity $\sin(\theta/2)=\pm\sqrt{1-\cos\theta\over 2}$. $\endgroup$ Jul 1, 2013 at 13:11

2 Answers 2

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Consider the diagram:

$\hspace{4cm}$enter image description here

Using the identity $\cos(\theta)=1-2\sin^2(\theta/2)$, the distance is $$ 2r\sin(\theta/2)=r\sqrt{2-2\cos(\theta)} $$

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  • $\begingroup$ Thanks so much this is perfect X $\endgroup$
    – apg
    Jul 2, 2013 at 18:20
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Just draw a picture of the intersection of the sphere with a plane containing the center and the two points.

The euclidean distance is the length of a hypoteneuse of a triangle whose other two sides have lengths $R\cdot \sin(\theta_{12})$ and $R\cdot (1-\cos(\theta_{12}))$, respectively. Then apply Pythagorean theorem.

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