Page:Encyclopædia Britannica, Ninth Edition, v. 23.djvu/585

The values of $\sin {\frac {1}{2}}A$, $\cos {\frac {1}{2}}A$, $\tan {\frac {1}{2}}A$ are given in terms of $\cos A$ by the formulae

$$\sin {\frac {1}{2}}A=(-1)^{p}({\frac {1-\cos A}{2}})^{\frac {1}{2}},\cos {\frac {1}{2}}A=(-1)^{q}({\frac {1+\cos A}{2}})^{\frac {1}{2}},\tan {\frac {1}{2}}A=(-1)^{r}({\frac {1-\cos A}{1+\cos A}})^{\frac {1}{2}},$$

where '$p$ is the integral part of ${\frac {A}{2\pi }}$, $q$ the integral part of ${\frac {A}{2\pi }}+{\frac {1}{2}}$, and $r$ the integral part of ${\frac {A}{\pi }}$.

$\sin {\frac {1}{2}}A$, $\cos {\frac {1}{2}}A$ are given in terms of $\sin A$ by the formulae

$$2\sin {\frac {1}{2}}A=(-1)^{p'}(1+\sin A)^{\frac {1}{2}}+(-1)^{q'}(1-\sin A)^{\frac {1}{2}}$$

$$2\cos {\frac {1}{2}}A=(-1)^{p'}(1+\sin A)^{\frac {1}{2}}-(-1)^{q'}(1-\sin A)^{\frac {1}{2}}$$

where $p'$ is the integral part of ${\frac {A}{2\pi }}+{\frac {1}{4}}$ and $q'$ the integral part of ${\frac {A}{2\pi }}-{\frac {1}{4}}.$

Properties of triangles. In any plane triangle ABC we will denote the lengths of the ties of sides BC, CA, AB by a, b, c respectively, and the angles BAC, ABC, ACB by A, B, C respectively. The fact that the projections of b and c on a straight line perpendicular to the side a are equal to one another is expressed by the equation $b\sin C=c\sin B$; this equation and the one obtained by projecting c and a on a straight line perpendicular to a may be written ${\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}$. The equation $a=b\cos C+c\cos B$ expresses the fact that the side a is equal to the sum of the projections of the sides b and c on itself ; thus we obtain the equations

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

If we multiply the first of these equations by $-a$, the second by $b$, and the third by $c$, and add the resulting equations, we obtain the formula $b^{2}+c^{2}-a^{2}=2bc\cos A$ or $\cos A={\frac {b^{2}+c^{2}-a^{2}}{2bc}}$, which gives the cosine of an angle in terms of the sides. From this expression for $\cos A$ the formulæ $\sin {\frac {1}{2}}A=\left\{{\frac {(s-b)(s-c)}{bc}}\right\}^{\frac {1}{2}}$, $\cos {\frac {1}{2}}A=\left\{{\frac {s(s-a)}{bc}}\right\}^{\frac {1}{2}}$, $\tan {\frac {1}{2}}A=\left\{{\frac {(s-b)(s-c)}{s(s-a)}}\right\}^{\frac {1}{2}}$, $\sin A={\frac {2}{bc}}\{s(s-a)(s-b)(s-c)\}^{\frac {1}{2}}$ where $s$ denotes ${\frac {1}{2}}(a+b+c)$, can be deduced by means of the dimidiary formula. From any general relation between the sides and angles of a triangle other relations may be deduced by various methods of transformation, of which we give two examples.

(α) In any general relation between the sines and cosines of the angles A, B, C of a triangle we may substitute $pA+qB+rC$, $rA+pB+qC$, $qA+rB+pC$ for A, B, C respectively, where p, q, r are any quantities such that $p+q+r+1$ is a positive or negative multiple of 6, provided that we change the signs of all the sines. Suppose $p+q+r+1=6n$, then the sum of the three angles $2n\pi -(pA+qB+rC)$, $2n\pi -(rA+pB+qC)$, $2n\pi -(qA+rB+pC)$ is $\pi$; and, since the given relation follows from the condition $A+B+C=\pi$, we may substitute for A, B, C respectively any angles of which the sum is $\pi$; thus the transformation is admissible.

(β) It may easily be shown that the sides and angles of the triangle formed by joining the feet of the perpendiculars from the angular points A, B, C on the opposite sides of the triangle ABC are respectively $a\cos A$, $b\cos B$, $c\cos C$, $\pi -2A$, $\pi -2B$, $\pi -2C$; we may therefore substitute these expressions for a, b, c, A, B, C respectively in any general formula. By drawing the perpendiculars of this second triangle and joining their feet as before, we obtain a triangle of which the sides are $a\cos A\cos 2A$, $b\cos B\cos 2B$, $c\cos C\cos 2C$ and the angles are $4A-\pi$, $4B-\pi$, $4C-\pi$; we may therefore substitute these expressions for the sides and angles of the original triangle; for example, we obtain thus the formula

$$\cos 4A={\frac {a^{2}\cos ^{2}A\cos ^{2}2A-b^{2}\cos ^{2}B\cos ^{2}2B-c^{2}\cos ^{2}C\cos ^{2}2C}{2bc\cos B\cos C\cos 2B\cos 2C}}$$

Solution of triangles.This transformation obviously admits of further extension.

(1) The three sides of a triangle ABC being given, the angles can be determined by the formula and two corresponding formulae for the other angles.

(2) The two sides a, b and the included angle C being given, the angles A, B can be determined from the formulae L tan (A -B)= log (a - b) - log (a+ b) +L cot C, and the side c is then obtained from the formula logc= loga + isin(7-isin^.

(3) The two sides a, b and the angle A being given, the value of sin B may be found by means of the formula L sin B=L sin A + log b - log a ; this gives two supplementary values of the angle B, if b sin A<a. If bsinA>a there is no solution, and if isin^=a there is one solution. In the case b sin A < a, both values of B give solutions provided b>a, but the acute value only of B is admissible if b<a. The other side c can then be determined as in case (2).

(4) If two angles A, B and a side a are given, the angle C is de termined from the formula C=ir- A -B and the side b from the formula log b = log a + L sin B - L sin A.

The area of a triangle is half the product of a side into the per- Areas pendicular from the opposite angle on that side ; thus we obtain of tri- the expressions bcsmA, {s(s-a)(s-b)(s-c)}l for the area of a angles triangle. A large collection of formulae for the area of a triangle and are given in the Annals of Mathematics for 1885 by M. Baker. quadri-

Let a, b, c, d denote the lengths of the sides AB, BC, CD, DA laterals. respectively of any plane quadrilateral and A + C=2a; we may obtain an expression for the area S of the quadrilateral in terms of the sides and the angle a. We have 2S=adsinA + bcsin(2a-A) and $(a? + d?-b 2 -c-) = ad cos A -be cos (2a- A) ; hence 4S 2 + 1( 2 + d 2 - & 2 - c 2 ) 2 = a 2 cF + & 2 c 2 - 2abcd cos 2a. If 2s=a + b + c + d, the value of S may be written in the form S= {s(s - a)(s - b)(s - c)(s - d) - abed cos 2 a} *.

Let R denote the radius of the circumscribed circle, r of the in- Radii of scribed, and r lt r 2 , r 3 of the escribed circles of a triangle ABC ; the circum- values of these radii are given by the following formulae. Radii of circumscribed, inscribed, and escribed circles of a triangle.

_ _ a be _ a ~ = n r=- = (s-a) tan A = s Of sin B sinj C, = - = s tan A = iB sinj A cosj B cos J C. s a

Spherical Trigonometry.

We shall throughout assume such elementary propositions in spherical geometry as are required for the purpose of the investiga tion of formulae given below.

A spherical triangle is the portion of the surface of a sphere Defini- bounded by three arcs of great circles of the sphere. If BC, CA, tion of AB denote these arcs, the circular measure of the angles subtended spherical by these arcs respectively at the centre of the sphere are the sides triangle. a, b, c of the spherical triangle ABC ; and, if the portions of planes passing through these area and the centre of the sphere be drawn, the angles between the portions of planes intersecting at A, B, C respectively are the angles A, B, C of the spherical triangle. It is not necessary to consider triangles in which a side is greater than IT, since we may replace such a side by the remaining arc of the great circle to which it belongs. Since two great circles intersect each Asso- other in two points, there are eight triangles of which the sides are ciated arcs of the same three great circles. If we consider one of these triangles, triangles ABC as the fundamental one, then one of the others is equal in all respects to ABC, and the remaining six have each one side equal to, or common with, a side of the triangle ABC, the opposite angle equal to the corresponding angle of ABC, and the other sides and angles supplementary to the corresponding sides and angles of ABC. These triangles may be called the associated Transfor- triangles of the fundamental one ABC. It follows that from any mation. general formula containing the sides and angles of a spherical triangle we may obtain other formulae by replacing two sides and the two angles opposite to them by their supplements, the remain ing side and the remaining angle being unaltered, for such formulae are obtained by applying the given formulae to the associated triangles.

If A ,S, (7 are those poles of the arcs BC, CA, ^^respectively which lie upon the same sides of them as the opposite angles A, B, C, then the triangle A B C is called the polar triangle of the triangle ABC. The sides of the polar triangle are v - A, ir-B, ir-C, and the angles IT -a, w-b, TT-C. Hence from any general formula connecting the sides and angles of a spherical triangle we may obtain another formula by changing each side into the supplement of the opposite angle and each angle into the supplement of the opposite side.

Let O be the centre of the sphere on which is the spherical triangle ABC. Draw AL per pendicular to OC and AM perpendicular to o* the plane OBC. Then the projection of OA on OB is the sum of the projections of OL, LM, MA on the same straight Jane. Since AM has no projection on any straight line in the plane OBC, this gives angles. OA cos c = OL cos a + LM sin a. Now OL = OA cos b, LM= AL cos C= OA sin b cos C ; therefore cos c cos a cos b + sin a sin b cos C. We may obtain similar formulae by interchanging the letters a, b, c, thus cos a = cos b cos c + sin 6 sin c cos A ^j cos& = cosc cosa + sinc sina cos 2? j- (1). cos c = cos a cos b + sin a sin b cos C ) Fig. 5.