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Again, "(cos +/-1 sin 0)"= cos no + √-1 sin no,

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170. To express cos" ✪ in terms of cosines of multiples of O, when n is a positive integer.

Writing x for cos 0+ √-1 sin 0, we have

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substituting, and dividing both sides by 2,

2n-1 cos"0= cos ne+ncos (n-2)0+

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n (n-1)... (n−r+1)

n(n-1)

cos (n-4) 0

1.2

cos (n-2r) 0+

169. To express sin" ✪ in terms of sines or cosines of multiples of 0, according as n is odd or even.

Let x = cos 0+ √−1 sin 0, then 2-1 sin 0:

n

and since (√−1)” = (—1)2, therefore

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1

=x- ;

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1

Therefore, since "- = 2-1 sin no, &c., substi

xn

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1. Obtain one value of /(cos 4a+/-1 sin 4a).

2. Find the values of 3/1.

3. Find the values of (−1)3.

4. Expand, in terms of the cosines of multiples of e, cos and sin6 0.

5. Expand, in terms of the sines of multiples of e, sin5 and sin7 0.

6. Expand, in terms of the powers of sine and cos 0, sin 40 and cos 70.

7. Expand (sin 0)+2 in terms of cosines of multiples of 0.

8. Given sin (+) = 51, find approximately the value of e,

neglecting powers of above the second.

9. Find the values of (1)3.

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n

show, by Demoivre's Theorem, that

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(a-b-1)+(a− b −1)" = 2 (a2+b2)*
(a−b √✅/−1)" + (a-

m

2n

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Arranging the terms of this development according to the powers of x, suppose

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Then, since x is any quantity,

a=1+Ay+By2+Cy3+

(2).

(3),

and a2+ = 1+A(x+y)+B(x+y)2+C(x+y)3+... (4).

But axa = ax+y. Therefore the series (4) is equal to the product of (2) and (3); and hence, by the principle of indeterminate coefficients, the coefficient of y in (4) is equal to the coefficient of y in the product of (2) and (3). Therefore

A+2Bx+3Cx2+4Dx3+...=A+A2x+ABx2+ACic3+...

Applying a second time the principle of indeterminate coefficients,

2B A2, 3C A. B, 4D = 4. C, &c.;

=

=

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Substituting these values in (2), we have

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171. Logarithmic Series.-By the exponential theorem,

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Since log, a = A, therefore

log, a = (a−1)— (a−1)2 (a−1)3

+ 2

3

+

+

(3).

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