II. When n is a negative integer.

Let n=- m; then m is a positive integer. Then

(cos +V-1 sin 0)" = (cos +√-1 sin 0)-m

0 V

.. (cos +V-1 sin 0)" = cos no +√− 1 sin no,

(by I.)

which proves the theorem when n is a negative integer.

III. When n is a fraction, positive or negative.

(cos +1 sin 0)"=cos po+V-1 sin pe (by I. and II.).

0+

Ρ sin-0

.. (cos +V-1 sin 0) = (cos +V-1 sin

.. (cos +√=1 sin 0) = cos 20+√1 sin20;

q

that is, one of the values of (cos +√1 sin 0)

In like manner,

(cos-V-1 sin 0)" = cos no-√-1 sin no.

Thus, De Moivre's Theorem is completely established. It shows that to raise the binomial cos +√-1 sin to

any power, we have only to multiply the arc by the exponent of the power. This theorem is a fundamental one in Analytic Mathematics.

Р

154. To find All the Values of (cos +V-1 sin ). When n is an integer, the expression (cos +V-1 sin 0)" can have only one value. But if n is a fraction expression becomes

Р

p = the

q'

(cos +√1 sin 0) = √(cos +√1 sin 0)”, which has q different values, from the principle of Algebra (Art. 235). In III. of Art. 153, we found one of the values of (cos +1 sin 0); we shall now find an expression which will give all the q values of (cos +√−1 sin 0)7.

Р

Now both cos and sin remain unchanged when is increased by any multiple of 2; that is, the expression cos +1 sin is unaltered if for we put (0+2), where is an integer (Art. 36).

The second member of (1) has q different values, and no more; these q values are found by putting r=0, 1, 2, ... q-1, successively, by which we obtain the following series of angles.

from which it appears that there are q and only q different

values of cos p(0+2), since the same values afterwards

recur in the same order.

gives all the q values of (cos +√1 sin 0) and no more. And this agrees with the Theory of Equations that there must be q values of x, and no more, which satisfy the equation x = C, where c is either real or of the form a+b√-1.

APPLICATIONS OF DE MOIVRE'S THEOREM.

155. To develop cos no and sin ne in Powers of sin and cos 0.

We shall generally in this chapter write i for V-1 in accordance with the usual notation.

By De Moivre's Theorem (Art. 153) we have

cos no + i sin n✪ = (cos 0 + i sin 0)"

(1)

Let n be a positive integer. Expand the second member of (1) by the binomial theorem, remembering that -1, i, and that +1 (Algebra, Art. 219).

=

Equate

the real and imaginary parts of the two members. Thus,

n

cos"-30 sin3

sin nə = n cos”-10 sin @ — n (n − 1) (n − 2)

+

3

n (n−1) (n−2) (n−3) (n − 4) cos”—50 sin30—etc. (3)

15

n

The last terms in the series for cos no and for sin ne will be different according as n is even or odd.

The last term in the expansion of (cos + i sin 0)" is Ө "sin"; and the last term but one is ni" 1cos 0 sin"-10. Therefore:

When n is even, the last term of cos no is "sin" or

n

2

(-1) sin" 0, and the last term of sin ne is nin—2 cos e sin"-10

n-2

or n(-1) 2 cos e sin"-10.

When n is odd, the last term of cos no is nin-1 cos 0 sin"-10

n-l
2

or n n(−1) cos 0 sin"-10, and the last term of sin no is

EXAMPLES.

Prove the following statements:

1. sin 404 cos3 0 sin

4 cos 0 sin3 0.

2. cos40 cos1 0 - 6 cos20 sin2 0 + sin1 0.

=

156. To develop sin and cos in Series of Powers of 0.

Put no = α in (2) and (3) of Art. 155; and let ʼn be increased without limit while a remains unchanged. Then O must diminish without limit. Therefore the

since

α

n

If n =

= ∞, then 0=0, and the limit of cos 0 and its powers

is 1; also the limit of

(1) and (2) become

(sin )

and its powers is 1. Hence

Sch. In the series for sin a and cos a, just found, a is the

circular measure of the angle considered.