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ON THE N ON-EUCLIDIAN GEOMETRY.
[528
I return to the expression for cosh a; in explanation of its meaning, let the
distances OB, 00 be q, r respectively and let the angle BOO be a; to find q we
have only to take 0 at 0, that is, in the formula for cosh a to write r = 0, we thus
1 1
find cosh q = : and similarly cosh r = , whence also
* cos q J cos r
cos q = sech q, sin q = i tanh q,
cos r = sech r, sin r = i tanh r,
also, as seen above, a = a; the formula thus is
. _ 1 + tanh q tanh f cos a
cosh a = t-== r—=
sech q sech r
= cosh q cosh r + sinh q sinh r cos a,
or, what is the same thing, it is
cosh a, — cosh q cosh r
cos a = . . _ . t _
smh q sinh r
viz. as will presently appear, this is the formula for cos BOG in the triangle BOG.
From the above formulae
and
. _ 1 — sm q sm r cos a.
cosh a =
cos qcos r
. -r cos v sm A —r cos A 4- sm b sm c
sm A = —±7 , cos A = t
cos b cos c cos b cos c
and the like formulae for b, c, B, G, it may be shown that in the triangle ABG we have
. _ cos A + cos B cos G
cosh a = ; —=— ; —= .
sin B sin C
In fact, substituting the foregoing values, this equation becomes
(1 — sin 2 a) (cos A + sin b sin c) + (cos B + sin c sin a) (cos G + sin a sin b) _ 1 — sin q sin r cos a
sin B sin G cos b cos c cos q cos r
that is,
cos A + cos B cos G — sin 2 a cos A + sin a sin b cos B + sin a sin c cos C + sin b sin c
= sin B sin (7(1 — sin q sin r cos a),
or, what is the same thing,
sin 2 a (cos B cos G — sin B sin G) + sin a sin b cos B + sin a sin c cos G + sin b sin c
= — sin B sin G sin q sin r cos a,
that is,
(sin a cos B -l- sin c) (sin a cos G + sin b) = sin B sin C (sin 2 a — sin q sin r cos a),
a relation which I proceed to verify.