The elements of non-Euclidean plane geometry and trigonometry, by H. S. Carslaw.

84, 85, 86] A FUNCTIONAL EQUATION 145
~ 85. We shall now show that
0(x+y) + (x- y) = 2(x)0(y).
With the figure of ~ 84, let the perpendicular at c to Cc
meet Dd and Bb in p and q. From cd cut off cr=cb, and
join pr.
Then we have cp = cq and ipr =qb.
We shall presently suppose Ss to become infinitesimal. In
this case the angles at p and q differ infinitesimally from right
angles, and L dpr becomes infinitesimal.
It follows that dr is infinitesimal as compared with pd; *
and that if Ss is an infinitesimal of the first order, dr is at
least of the second order.
But dp - qb = dp - pr < dr.
And dp - qb = (Dp - Dd) - (Bb - Bq).
Therefore we have
t p( Cc Dd Bb Bq CcG
\Cc Ss Ss Ss Cc Ss
But Lt (p) = (y) = Lt (Cc)
\Cc/ <w/ \Cc/
And Lt () = p(x), Lt ( = ( + Y),
and Lt (B) =(x-y).
Thus we have
~(P + y) 4+ 0(- - y)= s0(x) (y).
~ 86. We proceed to the equation
~ (x +y) + 9 (x - y) = 2 0(x) (y).
We are given that 0(x) is a continuous function, which is
equal to unity when x=0, and when x >0, (x) < 1.
Let xc be a value of x in the interval to which the equation
applies.
Xl
Then we can find k, so that ( (x) = cos.
*Cf. Coolidge, loc. cit. p. 49.
N,-EQ. K