# User:Peter Mercator/Draft for spherical triginometry

### The cotangent four-part formulae

The six elements of a triangle may be written in an anti-clockwise order as (aCbAcB\$. The cotangent, or four-part, formulae relate two sides and two angles forming four \emph{consecutive} elements around the triangle, for example (aCbA) or (BaCb). In such a set there are inner and outer elements: for example in the set (BaCb) the inner angle is C, the inner side is a, the outer angle is B, the outer side is b. The cotangent rule may be written as

$\begin{array}{rcl} \cos(\text{inner side}).\cos(\text{inner angle}) &{=}&\;\;\cot(\text{outer side}).\sin(\text{inner side})\\ &&-\cot(\text{outer angle}).\sin(\text{inner angle}) \end{array}$
and the six possible equations are (with the relevant set shown at right):
$\begin{array}{lll} (CT1)\quad& \cos b\,\cos C=\cot a\,\sin b - \cot A \,\sin C ,\qquad&(aCbA)\\[0ex] (CT2)& \cos b\,\cos A=\cot c\,\sin b - \cot C \,\sin A,&(CbAc)\\[0ex] (CT3)& \cos c\,\cos A=\cot b\,\sin c - \cot B \,\sin A,&(bAcB)\\[0ex] (CT4)& \cos c\,\cos B=\cot a\,\sin c - \cot A \,\sin B,&(AcBa)\\[0ex] (CT5)& \cos a\,\cos B=\cot c\,\sin a - \cot C \,\sin B,&(cBaC)\\[0ex] (CT6)& \cos a\,\cos C=\cot b\,\sin a - \cot B \,\sin C,&(BaCb). \end{array}$

### Half-angle and half-side formulae

With 2s=(a+b+c) and 2S=(A+B+C)

$\displaystyle \begin{array}{lll} \sin\frac{A}{2}=\left[\frac{\sin(s{-}b)\sin(s{-}c)}{\sin b\sin c}\right]^{1/2} &\qquad &\sin\frac{a}{2}=\left[\frac{-\cos S\cos (S{-}A)}{\sin B\sin C}\right]^{1/2}\\[2ex] \cos\frac{A}{2}=\left[\frac{\sin s\sin(s{-}a)}{\sin b\sin c}\right]^{1/2} &\qquad &\cos\frac{a}{2}=\left[\frac{\cos (S{-}B)\cos (S{-}C)}{\sin B\sin C}\right]^{1/2}\\[2ex] \tan\frac{A}{2}=\left[\frac{\sin(s{-}b)\sin(s{-}c)}{\sin s\sin(s{-}a)}\right]^{1/2} &\qquad &\tan\frac{a}{2}=\left[\frac{-\cos S\cos (S{-}A)}{\cos (S{-}B)\cos(S{-}C)}\right]^{1/2} \end{array}$

To prove the first formula set $2\sin^2(A/2)=1-\cos A$ and use the cosine rule to express A in terms of the sides. The others follow by using further cosine rules or supplemental cosine rules.

### Delambre (or Gauss) analogies

$\begin{array}{lll} &\\ \frac{\sin{\textstyle\frac{1}{2}}(A{+}B)} {\cos{\textstyle\frac{1}{2}}C} =\frac{\cos{\textstyle\frac{1}{2}}(a{-}b)} {\cos{\textstyle\frac{1}{2}}c} &\qquad\qquad & \frac{\sin{\textstyle\frac{1}{2}}(A{-}B)} {\cos{\textstyle\frac{1}{2}}C} =\frac{\sin{\textstyle\frac{1}{2}}(a{-}b)} {\sin{\textstyle\frac{1}{2}}c} \\[2ex] \frac{\cos{\textstyle\frac{1}{2}}(A{+}B)} {\sin{\textstyle\frac{1}{2}}C} =\frac{\cos{\textstyle\frac{1}{2}}(a{+}b)} {\cos{\textstyle\frac{1}{2}}c} &\qquad & \frac{\cos{\textstyle\frac{1}{2}}(A{-}B)} {\sin{\textstyle\frac{1}{2}}C} =\frac{\sin{\textstyle\frac{1}{2}}(a{+}b)} {\sin{\textstyle\frac{1}{2}}c} \end{array}$

Proved by expanding the numerators and using the half angle formulae.

### Napier's analogies

Follow by dividing the Delambre formulae.

$\begin{array}{lll} &&\\[-2ex]\displaystyle {\tan{\textstyle\frac{1}{2}}(A{+}B)} =\frac{\cos{\textstyle\frac{1}{2}}(a{-}b)} {\cos{\textstyle\frac{1}{2}}(a{+}b)} \cot\frac{1}{2}C &\qquad & {\tan{\textstyle\frac{1}{2}}(a{+}b)} =\frac{\cos{\textstyle\frac{1}{2}}(A{-}B)} {\cos{\textstyle\frac{1}{2}}(A{+}B)} \tan\frac{1}{2}c \\[2ex] {\tan{\textstyle\frac{1}{2}}(A{-}B)} =\frac{\sin{\textstyle\frac{1}{2}}(a{-}b)} {\sin{\textstyle\frac{1}{2}}(a{+}b)} \cot\frac{1}{2}C &\qquad & {\tan{\textstyle\frac{1}{2}}(a{-}b)} =\frac{\sin{\textstyle\frac{1}{2}}(A{-}B)} {\sin{\textstyle\frac{1}{2}}(A{+}B)} \tan\frac{1}{2}c \end{array}$