1. 증명[편집]

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<math> \Delta\theta=\mid\theta_1-\theta_2\mid </math>
<math> \Delta\phi=\mid\phi_1-\phi_2\mid </math>
<math> \angle{R_1}{O}{R_2}=\sigma </math>
<math> \overline {{R_1}{R_2}}=c </math>
<math> \overline {{R_2}{R'_2}}=a </math>
<math> \overline {{R_1}{R'_2}}=b </math>
<math> c^2=R^2+R^2-2R\cos \sigma </math>
<math> a^2=R^2+R^2-2R\cos\Delta\theta </math>
<math> b^2=R^2+R^2-2R\cos\Delta\phi </math>
<math>\triangle {R_1}{R_2}{R'_2}</math>은 <math> \angle {R_1}{R'_2}{R_2}</math>가 직각인 직각삼각형이므로,
<math> c^2=a^2+b^2=2R^2-2R\cos \sigma=4R^2-2R(\cos\Delta\theta+\cos\Delta\phi) </math>
<math>-2R^2\cos\sigma=2R^2-2R^2(\cos\Delta\theta+\cos\Delta\phi) </math>
<math> \cos\sigma=\cos\Delta\theta+\cos\Delta\phi-1 </math>
<math> \sigma=\arccos(\cos\Delta\theta+\cos\Delta\phi-1) </math>
호<math>{R_1}{R_2}=\pi R\frac{\sigma}{360}=\pi R\frac{\arccos(\cos\Delta\theta+\cos\Delta\phi-1)}{360} </math>

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