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Napier's Analogies

Let a Spherical Triangle have sides $a$, $b$, and $c$ with $A$, $B$, and $C$ the corresponding opposite angles. Then

$\displaystyle {\sin[{\textstyle{1\over 2}}(A-B)]\over\sin[{\textstyle{1\over 2}}(A+B)]}$ $\textstyle =$ $\displaystyle {\tan[{\textstyle{1\over 2}}(a-b)]\over\tan({\textstyle{1\over 2}}c)}$ (1)
$\displaystyle {\cos[{\textstyle{1\over 2}}(A-B)]\over\cos[{\textstyle{1\over 2}}(A+B)]}$ $\textstyle =$ $\displaystyle {\tan[{\textstyle{1\over 2}}(a+b)]\over\tan({\textstyle{1\over 2}}c)}$ (2)
$\displaystyle {\sin[{\textstyle{1\over 2}}(a-b)]\over\sin[{\textstyle{1\over 2}}(a+b)]}$ $\textstyle =$ $\displaystyle {\tan[{\textstyle{1\over 2}}(A-B)]\over\cot({\textstyle{1\over 2}}C)}$ (3)
$\displaystyle {\cos[{\textstyle{1\over 2}}(a-b)]\over\cos[{\textstyle{1\over 2}}(a+b)]}$ $\textstyle =$ $\displaystyle {\tan[{\textstyle{1\over 2}}(A+B)]\over\cot({\textstyle{1\over 2}}C)}.$ (4)

See also Spherical Trigonometry




© 1996-9 Eric W. Weisstein
1999-05-25