[961 
961] 
A TRIGONOMETRICAL IDENTITY. 
539 
or, say a, b, 2c are equal to these values; and then, substituting in the first equation, 
we have 
X (1 + y-Z 2 ) (x 2 — yz) + x (y 2 + z 2 ) (— 1 + x 2 yz) + (1 — ж 4 ) (y-z + yz 2 ) = 0, 
which is 
(x-y)(x—z){x + y + z- (yz + zx + xy) (x + у 4- z)} = 0, 
viz. our relation between x, y, z is 
x+ у + z — (yz + zx + xy) xyz = 0, 
or, what is the same thing, 
1 1 1 / V л 
1 1 (yz + zx + xy) = 0. 
yz zx xy w 
Then 
a + h = x (— 1 + x 2 ) (1 + yz), 
a — b = x( l+x 2 )(l—yz), 
2c = ( 1 —x 4 )(y + z), 
a 2 — b 2 = x 2 (1 — ж 4 ) (y 2 z 2 — 1), 2he = x (— 1 + x 2 yz) (1 — ж 4 ) (у + z). 
The equation to be verified is 
x (ifz 2 - 1) = (1 - x 2 yz) (y + z), 
that is, 
x + y + z — (yz + zx + xy) xyz — 0, 
as it should be. 
A somewhat different form of the proof is as follows:—We have identically 
cos (/3 + 7), cos (/3 - 7), 1 
cos (7 + a), cos (7 — a), 1 
cos (a + /3), cos (a — /3), 1 
= 4 sin %(/3- 7) sin \ (7 - a) sin £ (a - /3) {sin (/3 + 7) + sin (7 + a) + sin (a + /3)}, 
and therefore the relation between the angles is 
sin (/3 + 7) + sin (7 + a) + sin (a + /3) = 0. 
From the second and third equations, 
a : b : c = sin {|(/3 + 7) — a} : — sin (/3 + 7) + a] : 2 sin a cos a cos £ (/3 — 7), 
or say 
a = sin £ (¡3 + 7) cos a — cos | (/3 + 7) sin a, 
b = — sin | (/3 + 7) cos a — cos £ (/3 + 7) sin a, 
c = 2 sin a cos a cos ^ (/3 — 7), 
therefore 
a 2 — b 2 = — 4 sin a cos a sin (/3 + 7) cos (/3 + 7) = - 2 sin a cos a sin (/8 + 7), 
be — 2 sin a cos a {- cos (/3 — 7) sin £ (/3 + 7) cos a — cos |(/3 — 7) cos £ 08 + 7) sin a}, 
= 2. sin a cos a {— (sin /3 + sin 7) cos a — (cos /3 + cos 7) sin a], 
= sin a cos a {— sin (7 + a) — sin (a + /3)} = sin a cos a sin (/3 + 7), 
68—2