Page:Encyclopædia Britannica, Ninth Edition, v. 23.djvu/586

These formulæ (1) may be regarded as the fundamental equations connecting the sides and angles of a spherical triangle; all the other relations which we shall give below may be deduced analytically from them; we shall, however, in most cases give independent proofs. By using the polar triangle transformation we have the formulæ

$$\left.{\begin{aligned}\cos A&=-\cos B\cos C+\sin B\sin C\cos a\\\cos B&=-\cos C\cos A+\sin C\sin A\cos b\\\cos C&=-\cos A\cos B+\sin A\sin C\cos b\\\end{aligned}}\right\rbrace .$$

In the figure we have $AM=AL\sin C=r\sin b\sin C$, where r denotes the radius of the sphere. By drawing a perpendicular from A on OB, we may in a similar manner show that $AM=r\sin c\sin B$, therefore

$$\sin B\sin c=\sin C\sin b.$$

By interchanging the sides we have the equation ${\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}$ we shall find below a symmetrical form for k.

If we eliminate $\cos b$ between the first two formulae of (1) we have

$$\cos a\sin ^{2}c=\sin b\sin c\cos A+\sin c\cos c\sin a\cos B;$$

therefore

$${\begin{aligned}\cot a\sin c&=\cos A+\cos c\cos B\\&=\sin B\cot A+\cos c\cos B.\end{aligned}}$$

We thus have the six equations

$$\left.{\begin{aligned}\cot a\sin b&=\cot A\sin C+\cos b\cos C\\\cot b\sin a&=\cot B\sin C+\cos a\cos C\\\cot b\sin c&=\cot B\sin A+\cos c\cos A\\\cot c\sin b&=\cot C\sin A+\cos b\cos A\\\cot c\sin a&=\cot C\sin B+\cos a\cos B\\\cot a\sin c&=\cot A\sin B+\cos c\cos B\\\end{aligned}}\right\rbrace .$$

When $C={\frac {\pi }{2}}$formula (1) gives

$$\cos c=\cos a\cos b,$$

and (3) gives

$$\left.{\begin{aligned}\sin b=\sin B\sin c\\\sin a=\sin A\sin c\\\end{aligned}}\right\rbrace ;$$

from (4) we get

$$\left.{\begin{aligned}\tan a&=\tan A\sin b&=\tan c\cos B\\\tan b&=\tan B\sin a&=\tan c\cos A\\\end{aligned}}\right\rbrace .$$

The formulae

$$\cos c=\cot A\cot B$$

and

$$\left.{\begin{aligned}\cos A&=\cos a\sin B\\\cos B&=\cos b\sin A\end{aligned}}\right\rbrace .$$

follow at once from (α), (β), (γ). These are the formulas which are used for the solution of right-angled triangles. Napier gave mnemonical rules for remembering them.

The following proposition follows easily from the theorem in equation (3) : If AD, BE, CFare three arcs drawn through A, B, V to meet the opposite sides in D, E, F respectively, and if these arcs pass through a point, the segments of the sides satisfy the relation sin BD sin CE sin AF= sin CD sin AEsin BF; and conversely if this relation is satisfied the arcs pass through a point. From this theorem it follows that the three perpendiculars from the angles on the opposite sides, the three bisectors of the angles, and the three arcs from the angles to the middle points of the opposite sides, each pass through a point.

Formulae for sine and cosine of half-angles; If D be the point of intersection of the three bisectors drawn perpendicular to BC, it may be shown that BE=$(a + c-b) and CE=(a + b-c), and that the angles BDE, ADC are supplementary. We have sine sin ADB sin b sin ADC , . A sinBD s[n A sinCD s ^A 2 2 _siuBDsinCD sinCDEsinBDE n ._ . . 2 BDE=smBE a + b-c sin b sin c ( . a+c-b . a+6-c i . (5). A 1 sm 9 " 9 fore -in ) 2 f sin o sin c

Apply this formula to the associated triangle o

• " - B, C are the angles and ir - a, ir-b, c are the s

a ; n b+c a n . a + b + c 

which IT -A, des ; we obtain i (6). tlifi formula <>n s A < - z 2 / sin & sin c By division we have ( . a + c-b . a + b-c^ I ...a>, tan^J 2 2 | ^ J . b+c-a . a+b+c b . a+b-cl ( 2 Sm 2 J and by multiplication sinA- 2 i in a + 6 + C flin 6 + C ~%iT, c + a - sinftsiacl 811 2 8m 2 2 sm 2 j-

os b cos c} i-

"sin&sinc 1 Cos2a cos s 6-cos z c + 2cosa(

Hence the quantity k in (3) is sina sin 6 sine* 1 -coa a-cos 6-cosc + 2 cosacosb cose}* (8).

Of half-sides.Apply the polar triangle transformation to the formulaj (5), (6), (7), (8) and we obtain ( A + O-B A+B-Cl " cos cos sin B sin C B + C-A A + B + C - cos cos - sin B sin C J B+C-A A+B+C - cos 2 cos A+C-B A+B-C cos cos 2 ..(10); .(11).

If j. - 1 sinAsiiiBsinC we have kk = -2cos^4cos J BcosC ! ^, (12).

Delambre's formulæLet E be the middle point of AB ; draw ED at right angles to AB to meet AC in D ; then DE bisects ,? the angle ADB. Let CF bisect the angle /]r DCB and draw FG perpendicular to BC, then =90 -5. 3Q

From the triangle CFG we have cos = cos (7<? sin JF 1 ^, and from the triangle FEE cos JEFB = cos EB sin .F&E. Now Fig. 7. the angles CFG, EFB are each supplementary to the angle DFB, therefore a-b C . A + B c cos - cos 2= sm cos .(13).

Also sin CG = sin CF sin CFG and sin EB = sin BF sin EFB ; .a-b C A- B . c therefore sm jr cos - = sm - sm- (14). 22 22

Apply the formulae (13), (14) to the associated triangle of which a, IT - b, rr - c, A, IT - B, tr - (7 are the sides and angles, we then have . a+b . C A-B . c sin sin = cos - sin -. Si 22 a + b . C cos -- sin = A + B c - cos .(15), .(16).

The four formulae (13), (14), (15), (16) were first given by Delambre in the Connaissance des Temps for 1808. Formulae equivalent to these were given by Mollweide in Zach's Monatliche Correspondenz for November 1808. They were also given by Gauss (Theoria motus, 1809), and are usually called after him.

Napier's analogies. From the same figure we have tan FG = tan FCG sin CG = tan FBG sin BG ; therefore .0 . a-b , A-B . cot -sin -g- =tan ^ sm a-b tan A-B cot 2 (17).

Apply this formula to the associated triangle (IT - a, b, tr-c, ir-A, B, v-C], and we have a + b S -2- C tan a-b A+B cos -r- cos .(18).

If we apply these formulae (17), (18) to the polar triangle, we have . A-B , sin - a- b _ 2 c tan-g - sm , 1f ., (19 ; A-B . COS jr a + b 2 tan~2~- = .(20). cos- 2 - The formulae (17), (18), (19), (20) are called Napier s "Analogies" ; they were given in the Mirif. logar. canonis descriptio.