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I am doing a problem which requires me to find the arclength of the hypotenuse of an isosceles right triangle. (The book calls it a 2 Dimensional Sphere but I hope that is a typo)

I start at the north pole $\theta=0^\circ$ and make the two legs at the end points ($\theta,\phi$) as $(\theta_0,0)$ and $(\theta_0,\frac{\pi}{2})$. The hypotenuse is now an arc of radius $R$ and angle $\theta'$ so my arclength is $R\theta'$. Working back from the final answer, the result should be $$\theta' = \cos^{-1}\cos^2\theta_0$$ which I am unable to get.

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It sounds like you should use the spherical law of cosines. (en.wikipedia.org/wiki/Spherical_law_of_cosines) –  Jim Belk Jun 15 '11 at 16:12
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So this is one of the topics that no one teaches you but one is expected to know! A look at the wikipedia article on spehrical trigonometry clears it up. For the record: pythogorean theorem reduces to $\cos c = \cos a \cos b$ for a right spherical triangle and the rest follows. Thanks @Jim –  kuch nahi Jun 15 '11 at 16:18
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Re: $2$-dimensional sphere: A $2$-dimensional sphere is a sphere in $\mathbb{R}^3$ and is itself a two-dimensional gadget (it has a $2$-dimensional surface). To make this precise, you can use any of the classical constructions (e.g. Mercator projection or the stereographic projection) to get $2$-dimensional charts of it. –  t.b. Jun 15 '11 at 16:19
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if you are now satisfied with what you have found, you should answer yourself and accept the answer –  Henry Jun 15 '11 at 17:55

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up vote 3 down vote accepted

Following Henry's comment, here's the answer I got:

The spherical law of cosines says that for a triangle on a unit sphere:

enter image description here

$$\cos c = \cos a \cos b + \sin a \sin b \cos \gamma$$

So for a sphere of radius $R$ $$\cos \left(\frac{c}{R}\right)=\cos \left(\frac{a}{R}\right)\cos \left(\frac{b}{R}\right) +\sin\left(\frac{a}{R}\right) \sin\left(\frac{b}{R}\right) \cos \gamma$$ In the problem $\frac{a}{R} = \frac{b}{R} = \theta_0$ and $\gamma = \phi_b-\phi_a = \frac{\pi}{2}$ So we get $$ \cos \theta' = \cos\theta_0 \cos \theta_0$$

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Now you can accept your own answer –  Henry Jun 15 '11 at 20:58
    
@Henry The server would not allow me to, until the next two days :) I think ethically I should let it be, or flag it for CW. –  kuch nahi Jun 15 '11 at 21:00

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