.. (− 1)~ 2′′−1. (sin ()" = cos n ( − n . cos (n − 2) 4

2.

If n be odd, then (√-1)=√ — 1. (√ − 1)" ~1

+ or

according as there is an odd or an even number of binomials in (2), i. e. as 1 (n + 1) is odd or even; the sign

n+1

therefore of this term is the same as that of (-1), or (-1), —and therefore, when n is an odd number, (2) becomes,

= 2 √ - 1 . sin n 0 - n. 2√ - 1. sin (n − 2) 0 + ...
1.

110. To expand cos ne in terms of cos alone, n being

Expanding these logarithms, Appendix 1. 14. iii, and changing the signs of both sides of the equation, we have

The general term being the coefficient of that term of

which involves a";-and this being that term in the expansion

N

(b − a)"-" which involves a', must be the (r+ 1)th in that expansion; also this term is + or

Therefore the general term is

as r is even or odd;

the equation becomes, when both sides are multiplied by n,

In determining the series from the general term by giving different values to r, r must be taken of every positive integral

n

value not greater than ; for since none but positive integral

2

powers of (ba) appear, n - 2r, the index of b in the general term, must be a positive integer, or r must not be greater

than

NOTE. In the proof of this Article the expansion of l。 (1 + x) is used, in which it might be shewn that none but positive and integral powers of a can enter. This method of proof therefore can only be applied to those cases in which n is a positive integer. In fact, it can be demonstrated that cos no cannot be expressed in a series of this form when n is either negative or fractional.

111. To express cos ne, n being a positive integer, in a series ascending by integral powers of cos 0.

In the last Article cos no has been expressed in a series of descending integral powers of cos 0: by means of the general term there found, cos no may be expressed in a series of ascending integral powers of cos 0.

1. Let n be even.

Since the limitation of the value of r is that it shall not

The number of terms is n + 1; and since r is diminished successively by 1, the terms are alternately positive and negative: the first term being of the same sign as

... 2 cos no

(n − ↓ n − 1 ) . ( n − 1 n − 2) ... 3 . 2 . 1

= (− 1)3 {n.

(− 1)" ;

+n.

{n−({n−1)−2}...{n−2({n−1)+1}. (2 cos()2

1.2.3...(n-1)

{n−({n−2)−1} . {n−(†n−2) −2} ... {n−2 (n−2)+1} ̧ (2 cos@)' — ..

1.2.3...(n-2)

=

− ( − 1)3 {m 1.2... / n

n. n. ( n − 1)... 4.3

1.2... ( n − 1)

n . ( . —
n + 1) . 1 n . ( n − 1)... 6.5

+

1.2.3 (n-2)

...

(2 cos 0)' — ...}

(2 cos 0)2

1)

(2 cos 0)' - ...},

2.

-

1.2

n2. (n2 − 22) . (n2 — 43)

1.2.3.4.5.6

(cos 0)2 + ...}.

n; it

n.

Let n be odd; r must not be greater than may therefore be (n − 1), the integer next less than The number of terms is 1⁄2 (n + 1), and the terms are alternately positive and negative.

For r writing 1⁄2 (n − 1), ↓ (n − 3), † (n − 5)...3, 2, 1, 0, successively, we have as before,

2 cos n o =

(− 1)

(n. {n − } (n − 1) — 1}. {n − } (n−1)−2}...4.3.2

+

And reducing the terms in the same manner as in the last case, when n was even, we have