cosine and sine of the sum of two arcs, by equating separately the scalar and vector parts of these quaternions.

And we see, as an immediate consequence of the expressions above, that

cos mo+e sin me = (cos + e sin 0)m

if m be a positive whole number. For the left-hand side is a versor which turns through the angle me at once, while the right-hand side is a versor which effects the same object by m successive turnings each through an angle 0. See § 8.

110. To extend this proposition to fractional indices we have 0

only to write for 0, when we obtain the results as in ordinary

trigonometry.

n

From De Moivre's Theorem, thus proved, we may of course deduce the rest of Analytical Trigonometry. And as we have already deduced, as interpretations of self-evident quaternion transformations (§§ 97, 104), the fundamental formulæ for the solution of plane triangles, we will now pass to the consideration of spherical trigonometry, a subject specially adapted for treatment by quaternions; but to which we cannot afford more than a very few sections. The reader is referred to Hamilton's works for the treatment of this subject by quaternion exponentials.

111. Let a, ẞ, γ be unit-vectors drawn from the centre to the corners A, B, C of a triangle on the unit-sphere. Then it is evident that, with the usual notation, we have (§ 96),

Also UVaß, UVẞy, UVya are evidently the vectors of the corners

of the polar triangle.

Hence

S.UVaẞUVBY = cos B, &c.,

TV.UVaẞUVẞy = sin B, &c.

Remembering that we have

SVaẞVẞy = TVaẞTVẞyS.UVaßUVßy,

we see that the formula just written is equivalent to sin a sin c cos B-cos a cos c+cos b,

or cos b = cos a cos c+ sin a sin c cos B.

or sin a sin c sin B = sin a sinp, sin b sin po sin c sin pe;

where p is the arc drawn from 4 perpendicular to BC, &c.

113. Combining the results of the last two sections, we have TaB.TBy = sin a sin c cos B-ß sin a sin c sin B

These are therefore versors which turn the system negatively or positively about OB through the angle B.

The interpretation of each of these forms gives a different theorem in spherical trigonometry.

114. A curious proposition, due to Hamilton, gives us a quaternion expression for the spherical excess in any triangle. The following proof, which is very nearly the same as one of his, though by no means the simplest that can be given, is chosen here because it incidentally gives a good deal of other information. We leave the quaternion proof as an exercise.

Let the unit-vectors drawn from the centre of the sphere to A, B, C, respectively, be a, ß, y. It is required to express, as an arc and as an angle on the sphere, the quaternion

The figure represents an orthographic projection made on a plane perpendicular to y. Hence C is the centre of the circle DEe. Let the great circle through A, B meet DEe in E, e, and let DE be a quadrant. Thus DE represents y (§ 72). Also Then, evidently,

make EF = AB = Ba-1. βατι.

DF=
= βα'γ,

which gives the arcual representation required.

L

Let DF cut Ee in G. Make Ca= EG, and join D, a, and a, F. Obviously, as D is the pole of Ee, Da is a quadrant; and since EG Ca, Ga = EC, a quadrant also.

=

Hence a is the pole of DG, and therefore the quaternion may be represented by the angle DaF.

Make Co Ca, and draw the arcs Paß, Pba from P, the pole Cb = of AB. Comparing the triangles Eba and eaß, we see that Ea eß. But, since P is the pole of AB, Fßa is a right angle: and therefore as Fa is a quadrant, so is FB. Thus AB is the complement of Ea or ße, and therefore

=

aß = 2 AB.

Join A and produce it to c so that Ac = b; join c, P, cutting AB in o. Also join c, B, and B, a.

Since P is the pole of AB, the angles at o are right angles ; and therefore, by the equal triangles ba A, coA, we have

and therefore the triangles coB and Baß are equal, and c, B, a lie on the same great circle.

Produce cA and CB to meet in H (on the opposite side of the sphere). H and c are diametrically opposite, and therefore cP, produced, passes through H.

Now Pa = Pb = PH, for they differ from quadrants by the equal arcs aß, ba, oc. Hence these arcs divide the triangle Hab into three isosceles triangles.

But

Also

LPHb+2 PHa = LaHb = Lbca.

LPab-Leab-L PaH,

LPba L Pab-Lcba-LPbH.

Adding, 2 Pab 2π-Lcab-Leba- Lbca

= (spherical excess of abc).

But, as ▲ Faß and ▲ Dae are right angles, we have

angle of Ba1y = L FaD Lẞae = L Pab

[Numerous singular geometrical theorems, easily proved ab initio by quaternions, follow from this: e. g. The arc AB, which bisects two sides of a spherical triangle abc, intersects the base at the distance of a quadrant from its middle point. All spherical triangles, with a common side, and having their other sides bisected by the same great circle (i. e. having their vertices in a small circle parallel to this great circle) have equal areas, &c., &c.]

115. Let Oaa', Ob = B, Oc=y, and we have

ά,

But FG is the complement of DF. Hence the angle of the quaternion

な

is half the spherical excess of the triangle whose angular points are at the extremities of the unit-vectors a′, ß',

[In seeking a purely quaternion proof of the preceding propositions, the student may commence by showing that for any three unit-vectors we have

The angle of the first of these quaternions can be easily assigned; and the equation shows how to find that of Ba-1y. But a still simpler method of proof is easily derived from the composition of rotations.]