107. To exemplify the variety of possible transformations even of simple expressions, we will take two cases which are of frequent occurrence in applications to geometry.

and thus that P is any point equidistant from two fixed points,] may be written

All of these express properties of a sphere. They will be interpreted when we come to geometrical applications.

108. We have seen in § 95 that a quaternion may be divided into its scalar and vector parts as follows:

where

is the angle between the directions of a and ẞ, and

← = UV is the unit-vector perpendicular to the plane of a

α

and ẞ so situated that positive (i. e. left-handed) rotation about it turns a towards ß.

Similarly we have (§ 96)

αβ = δαβ + Γαβ

=TaTB (-cos 0-e sin 0),

and having the same signification as before.

109. Hence, considering the versor parts alone, we have

being the positive angle between the directions of y and ß, and the same vector as before, if a, ß, y be coplanar.

or

cos (+0)+ sin (4+0)

= UL;

a

(cos + sin 4) (cos + sin 0)

= cos cos 0-sin & sin 0 + € (sin cos 0+ cos & sin 0),

from which we have at once the fundamental formulæ for the

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 m0+ e sin m0 = (cos 0+ 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 6. See § 8.

110. To extend this proposition to fractional indices we have

0

n

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

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, y 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.UVaẞUVẞy = cos B, &c.,

TV.UVaẞUVBy = sin B, &c.

Now (§ 90 (1)) we have

SVaẞVBy = S.aV.BVBY

=-SaẞSpy+B'Say.

Remembering that we have

SVaẞVBy = 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 sin pa = sin b sin po sin c sin p. ;

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

Pa

113. Combining the results of the last two sections, we have

Γαβ.βγ = 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.

As another instance, we have

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, B, 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 make EF = AB = Ba1. Then, evidently,

DF

= βα-'γ,

which gives the arcual representation required.

L