where is the angle nearest to (2r+1) π whose cosine and whose tangent = tan 0.

This result includes those of the preceding two articles.

For, if q is odd, we may put 2r+1=q, so that cos

2r+1

= cos

рπ=(-1)", and sin p=0. Hence the series for cos pe/q and sin pe/q reduce

2r+1
q

to the forms given in Art. 620.

COS

And, if q is even and 9 odd, we may put 2r+1=39, so that 2r+1 2r+1 q

Pπ=0 and sin рr=(-1). Hence the series reduce

to the forms given in Art. 621.

Moreover, if q and q are both even, we cannot make either cos p (2r+1)π/q or sin p (2r+1) π/q vanish, so that cos pe/q and sin pe/q cannot be expressed in a single series.

623. The result of the last article follows immediately from Art. 619.

For, if

(2r+ 1) is acute, the multipliers of

π

cos (2r+1) рπ/q and sin (2r + 1)рπ/¶ in the series of Art. 622 are, by Art. 619, respectively

cos {0 − (2r + 1) π} p/q and sin {0 − (2r + 1) π} p/q.

Hence the expressions given for cos p/q and sin p0/q are equal respectively to

where (sin") and (- sin 0) are taken as real positive.

Here, when cot 0 = 0, C=1 and S = 0. clue for discovering their general value.

Thus, if sin is positive,

3

(cot )*+..

This gives us the

where is the angle nearest to 1⁄2 (4r + 3) π, whose sine = sin ✪ and

whose cot = cot 0. Thus

(A) If p is even, q of the form 4r + 1, and sin ◊ positive,

p
2

p+2
2

C. (sino 0) = (-1) cos e and S. (sino 0)=(-1) sin-0.

/

(B) If p is even, q of the form 4r + 3, and sin ◊ negative,

(D) If p is odd, g of the form 4r + 3, and sin 0 negative,

In (A), (B), (C), (D), 0 is the angle nearest to ¶π.

625. In the following article, an important application is made of the Binomial Theorem.. The proof given is equivalent to establishing an equation involving the first and second differential coefficients of the function to be expanded.

626. To expand {(1 +x2)3 +x}o in powers of x, where x < 1. We may write

1 +nx+α ̧ï2 + ɑ ̧ï3 +...+α‚x2 +...= {(1 + x2)3 + x}” ...(I), the first two terms being immediately derivable from the binomial theorem.

Now write x + h for x.

The left-hand becomes

1 + n (x + h) + α2 (x + h)2+...+a, (x + h)" +...

The right-hand becomes

{(1 + x2 + 2hx + h2)1 +x+h}" = {(1+x2)2 + §. 2hx (1+x2) ̄1 +x+h+ ...}" ={(1+x2)2+x}"{1+h(1+x2) ̄3+...}"={(1+x2)3+x}"{1+nh(1+x2) ̄*+.....}.

Thus, equating the coefficients of h, after substituting from (1),

(1+x2)1 (n+2α ̧x+...+ra‚æTM ̄1+...)=n(1+nx+a‚x2+...+α‚æˆ+.....)... (II). Again in (II) write x+h for x. Then

{(1 + x2)2 + hx (1 + x2) ̄ +...} {n + 2α, (x + h) +...+ra, (x + h)r-1 +... } = n {1 + n (x + h) + a2 (x + h)2 +...+ a, (x + h)” +..... }.

Equating the coefficients of h again, we have

(1+x2)3 {2a,+...+r(r−1)a,x2-2+...}+x(1+x2) ̄13 (n+2a2+...+ra,x2-1+.....) = n (n + 2a,x+...+ ra‚„x2−1 +.....)

-1

= n2 (1 + x2) ̄1 (1 + nx + а„x2 +.....+ɑ„x” +...) by (II).

Multiply by (1+x2). Thus

(1 + x2) {2α2+...+r (r− 1) α, x2-2 +...} + x (n + 2a2 + ... + rα,x2-1+.....)

In (III) equate the coefficients of x". Thus

(r+ 2) (r+1) ar +2 + r (r− 1 ) a + ra, = n2a, ;

.....(III)

Putting r in succession 0, 1, 2, 3.... we have

Also substituting for a,, a... in (II), we have

nx2+....

627. In these two expansions, the multiplier of the alternate

628. The above formulæ may be transformed in many ways. If x is wholly real and less than 1, the series are positive (as the student may show): hence the real positive values of the fractional powers must be taken.

629. .
A. Let (1 + x2)3 + x = x; then (1 + x2)* − x = z ̄1,

and 2x-2-2-1.

If x lies between − 1 and + 1, ≈ lies between 2-1 and √2 + 1; and we have

of

Hence any power of z may be expressed in ascending powers

in a series, which is convergent if z lies between √2-1

and √2+1.

It would thus appear that, by expanding the powers of z-z-1 by the binomial theorem, we could expand any power of ≈

in positive and negative integral powers of z. And this is in a sense true. But the resulting series would be divergent, unless we took the terms in groups corresponding to the integral powers of 2 - 2-1.

630. B. Let x = cot 0; then /(1+x2)+x=cosece+cot0=cot10. If then x lies between 1 and +1, we may take between 1 and 2, so that cot 10 is positive. Thus

cot" 10 = 1 + n cot 0 +

n2 1.2

cot2 0 +

n (n2-12)
1.2.3

·cots +...

has

631. To expand cos no and sin no in powers of sin 0, when n real value.

any

some value of

is some angle respectively.

In the formulæ of Art. 626, put x = i sin 0. Then √(1 + x2) + x = cos + i sin 0. Hence the series represent (cos + i sin 0)", i.e. of cos no + i sin no where whose cosine and sine are equal to those of If 0 = 0, x = 0; and the series become = 1 Hence, by the principle of continuity, if 0 is acute,

cos no.

The first and last of these series terminate only if n is an even integer; the middle two terminate only if n is an odd integer. For these special cases, the above formulæ were proved in Art. 375.

632. To show that, if n is positive, the series, obtained by multiplying all the terms in the expansion of (1-1)" by any positive or negative finite quantities, is convergent.