which proves the theorem for fractional indices.

COR. By the theorem just proved,

n

m

n

-0;

(cos + √ sin p)" = cos m±√-1 sin mp, m being positive or negative, whole or fractional.

Let 2p+ 0, where p is any integer;
ф = 2рп

1.

Let the index of (cos +/-1 sin 0) be integral, as m;

Then p, being an integer, may be represented by qn +r, where q may be 0 or any integer, and r may be 0 or any integer less than n;

From these two cases therefore it appears that the theorem might have been thus enunciated;

If the index be an integer,

(cos +- 1 sin 0)" = cos m◊ ± √ − 1 sin me;

If the index be fractional,

m

(cose-1 sin()" = COS

m

n

(2rπ+0)±√−1 sin — (2 rπ +0) ; sinTM rπ+0);

where r is 0, or any integer less than n.

n

NOTE. It is to be observed that by giving r all values from 0 to n 1, we obtain from the second member of the second equation, the n different values of

m

(cos 0 ± √ - 1 sin ()".

1

107. If 2 cose be represented by x+, then 2 cos me will

X

:

(cos + √1 sin 0) = cos m0+√- 1 sin m 0,

m

(cos-1 sin 0) = cos m0-√-1 sin m0,

COR. By making use of the equations of the corollary to

where r is 0 or any integer less than n.

108. To express any positive integral power of the cosine of an angle in terms of the cosines of the multiples of the angle.

1. If ʼn be even; the last term of (2), or the (n + 1)th

2. If ʼn be odd; the sum of the two middle terms of (1), viz.

the {(n - 1) + 13th and the {(n − 1) + 2}th,

}

(which have the same coefficients), is

{}

.`. 2ṛ-1. (cos 0)” = cos n✪ + n . cos (n − 2)0

2 cos ;

n odd.

2

n

1

n

2

1 n + 1
1 n

;

n even.

109.

To express any positive integral power of the sine

of an angle in terms of the sines and cosines of the multiples of the angle.

the upper or lower signs being taken as n is even or odd,

2

xn−4

FN.

-2

n

then (√/− 1)” = (√/ − 1)2 • * = {(√//−−1)2} * = (− 1)* ;

and the middle, or

(v

n

(213 + 1 )

according as this middle, or

n

or even term, i. e. as is even or odd. Wherefore it is

2

n

of the same sign as (− 1); and therefore when ʼn is even, the series (2) becomes

n

(− 1)". 2′′ . (sin 0)” = 2 (cos ne) – n. 2 cos (n − 2)0