2 
ON AN EXPRESSION FOR 1 + sin (2p + l) % IN TERMS OF sin U. [630 
To find herein the expression of the factor (1, x)p, write u = ^tt- 0 and consequently 
x = cos 0 ; we have therefore 
1 + cos (2p + 1) 0 = (1 + cos 6) {(1, x) v ) 2 , 
where in the second factor on the right-hand side x is retained to stand for its value 
cos 0. This gives 
2 cos 2 (p +^)0 = 2 cos 2 \9 {(1, xY}*, 
or, what is the same thing, 
a x) p- C0S (P±M 
cos ^0 * 
viz. this is 
which is 
We have 
. . „sin 40 
= cos p0 — sin pa î , 
r r cos^-o 
a • a 1 - COS 9 
= cos p0 — sin pa —;—je— . 
r sin 0 
cos p9 + i sinpd = {x + i V(1 — 
= X + i V(1 — a?) Y, suppose, 
where X, Y are rational and integral functions of x of the orders p and p — 1 
respectively; that is, 
cos pd = X, sin p0 = sin 9. Y, 
and we have therefore 
(1, x) p = X — Y(1 — x), 
which is the required expression for (1, x) p . For instance 
p = 3, X + i y^l — x 2 ) Y = \x + i V(1 — # 2 )} 3 ; 
that is, 
X = —Sx + 4c 8 
Y = — 1 + 4a; 2 , and . — (1 — x) Y= 1 — x — 4a; 2 + 4a? 
so that X — (1 — x) Y= 1 — 4a; — 4a; 2 + 8a?, = (1, x) 3 , 
and hence 
1 - sin 7 u = (1 + x) (1 — 4a; — 4a; 2 + 8a?) 2 , 
which agrees with a result already obtained. 
The foregoing value of (1, x) v may also be written 
(1, 
{sin (p + 1) 0 — sin^?0}, 
which however is not practically so convenient. 
The formula corresponds to a like formula in elliptic functions, viz. writing sinam u = x, 
the numerator of 1 + (—) p sinam (2p + 1) u is 
= (1 + x) {(1, a;) 2 ^ +1 >} 2 , 
which is (1 + x) multiplied by the square of a rational and integral function of x.