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.
Best Answer
Following Henry's comment, here's the answer I got:
The spherical law of cosines says that for a triangle on a unit sphere:
$$\cos c = \cos a \cos b + \sin a \sin b \cos \gamma$$
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 $$ \cos \theta' = \cos\theta_0 \cos \theta_0$$