Sine Rule on a Sphere

Rearrange \(\cos(c)=\cos(a)\cos(b)+\sin(a)\sin(b)\cos(C),\) to get:

\[\cos(C)=\frac{\cos(c)-\cos(a)\cos(b)}{\sin(a)\sin(b)}.\]

Now we use \(\sin^2(C)=1-\cos^2(C)\) substituting the above in for \(\cos^2(C)\):

\[\sin^2(C)=1-\left(\frac{\cos(c)-\cos(a)\cos(b)}{\sin(a)\sin(b)}\right)^2,\]

\[\sin^2(C)=\frac{\sin^2(a)\sin^2(b)-(\cos(c)-\cos(a)\cos(b))^2}{\sin^2(a)\sin^2(b)}.\]

Now again use \(\sin^2(C)=1-\cos^2(C)\) but for \(a\) and \(b\) to get:

\[\sin^2(C)=\frac{(1-\cos^2(a))(1-\cos^2(b))-(\cos(c)-\cos(a)\cos(b))^2}{\sin^2(a)\sin^2(b)}.\]

Times everything out and cancel to get:

\[\sin^2(C)=\frac{1-\cos^2(a)-\cos^2(b)-\cos^2(c)+2\cos(a)\cos(b)\cos(c)}{\sin^2(a)\sin^2(b)}.\]

Finally divide everything by \(\sin^2(c)\) to get the required form:

\[\frac{\sin^2(C)}{\sin^2(c)}=\frac{1-\cos^2(a)-\cos^2(b)-\cos^2(c)+2\cos(a)\cos(b)\cos(c)}{\sin^2(a)\sin^2(b)\sin^2(c)}.\]

Which is what we wanted, since the right hand side is symmetrical in \(a\), \(b\) and \(c\), i.e.

\[\frac{\sin^2(A)}{\sin^2(a)} = \frac{\sin^2(B)}{\sin^2(b)} = \frac{\sin^2(C)}{\sin^2(c)}.\]

But since our angle must be between \(0\) and \(\pi\), all the terms are positive, so we can take the square root to get the sine rule on a sphere:

\[\frac{\sin(A)}{\sin(a)}=\frac{\sin(B)}{\sin(b)}=\frac{\sin(C)}{\sin(c)}.\]