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where 0 is the angle nearest to (2r + 1) + whose cosine = cos e and whose tangent = tan 0. This result includes those of the preceding two articles.

2r +1 For, if 9. is odd, we may put 2r +1=9, so that cos pa=(-1)",

9 2r +1 and sin por=0. Hence the series for cos po/q and sin polq reduce

9 to the forms given in Art. 620.

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

2r +1 pt=0 and sin

po=(-1)2. Hence the series reduce 9

9 to the forms given in Art. 621.

. Moreover, if and 9 are both even, we cannot make either cos P (2r+1)*/9, or sin p (2r+1)19 vanish, so that cos pola and sin po/q cannot be expressed in a single series.

623. The result of the last article follows immediately from Art. 619. For, if 0 (2r + 1) + is acute, the multipliers of

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

cos {0 – (2r + 1) Tr} p/q and sin {8 – (2r + 1) *} P/9. Hence the expressions given for cos p/q and sin po/q are equal respectively to 2r + 1

Po
2r + 1

2r + 1

P

2r+1 PTT+

PT

pa). a 9 9

9

9 94 624.

Lastly suppose tan 6 > 1. Then (cos + i sin ø) = (i sin 0) (1 – i cot esé. 0

? If sin is positive, by Art. 608,

4r + 1
PTT

4r + 1 (i sin o) é = 2 (sin”) (c

рт

+1. Pa
+ i sin )
9
2

9

2 If sin 0 is negative, by Art. 608,

4r + 3
PTT

4r + 3 (i sin ø) * = % (= sin øy (cos

рп
i ) = - )

+ i sin
9
2

)
9

2 where g (sin() and V (- sin 0)” are taken as real positive.

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pa) and sin

PT +

р

P

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P

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P

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+..

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V ” o) (

(c.

S.

Po,

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(c.

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P (P-9)
Write C for the series 1

1.2
p--)

(cot A13 and S for the series p

9

1.2.3 Here, when cot 0 = 0, C = 1 and S= 0. This gives us the clue for discovering their general value. Thus, if sin 6 is positive,

4r + 1
рп

4r + 1

PTT
(sin? 0) (C.

+ S. sin 5) = case,
9
2

9

2
4r + 1
PTT

4r + 1

рп and Y (sin”) (C. sin

sin
9
2

9
2

9 where @ is the angle nearest to } (4r + 1) + whose sine = sin 0 and whose cot = cot 0. But, if sin 0 is negative, 4r + 3 PTT

4r + 3

PTT (- sin ) (C.cos

S. sin

= cos - 0,

2
9
2

9
2

9
4r + 3 рп

4r + 3

рп P and 7 (-sin )" (C. sin

= sin?, 0,
9
2

9
2

2 where 0 is the angle nearest to 1 (4r + 3) , whose sine = sin 0 and

TT whose cot coto. Thus

(A) If p is even, q of the form 4r + 1, and sin o positive, 0.9 (sin” ) = (-1) cos LA and S. Y (sin() =(-1) "* sin .

2 0

P 9

9 (B) If p is even, 9 of the form 4r + 3, and sin 0 negative,

P C:\/(-sin 6) = (-1) cose and S. Y(-sin @)P =(-1) = sin . P

P 9 (C) If p is odd, q of the form 4r + 1, and sin 0 positive, C.V (sin” 6) = (-1) = sin Ło and S. 7 (sin°C) = (-1)"#" cosłe.

р P V
9

9 (D) If p is odd, q of the form 4r + 3, and sin 0 negative, C./(-sin 6)2 =(-1)=sin Lo and S. 9(-sin Øy? =(-1)"2" cos

)P

9
In (A), (B), (C), (D), O is the angle nearest to $97.

-S.c

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p+2

P
2

P

p+2
2

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h for x.

+

+

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)+ + x}" in powers of x, where x < 1.
We
may

write
1 +nx + a_x2 + açao +...+ a,x" +...={(1 + x2)+ + x}" ...(I),

"

* ) the first two terms being immediately derivable from the binomial theorem. Now write x +

The left-hand becomes 1 +n (+ h) + a2 (x + h) +...+ a, (x + h)" +... The right-hand becomes {(1 + x2 + 2hx + h2)* + x +h}"={(1+x2)} + 1. 2hx(1+x2)-$++h+...}" =-{{1+2*)*+w}"{1+h(1+x)=+...}"={(1+x2)++x}"{1+nh(1+2+)-+*+...}.

Thus, equating the coefficients of h, after substituting from (1), (1+2°)*(n+209x+...+ra,x"=1+...)=n(1+na+ax&*+...+amx*+...)...(II).

Again in (II) write x + 1 for x. Then {(1 + x2)+ + (1 + 2?)–+...}{1+ 2a, (a+h) +...+ ra, (a+h)r-1 +...}

En{1+n (c + h) + a, (x + h) +...+ ay (ac + h)" +...}. Equating the coefficients of h again, we have (1+x2)}{2a3+...+r(r-1)a,"-2+...}+w(1+x2)+(n+2d2+...+ray"–1+...) = n(n+ 20,8 +...+ranac"-1 +...)

. = n° (1 + x^)- *(1 + nx + 0,2° +...+ ax" +...) by (II). Multiply by (1 + a2)Thus 1 (1 + x®) {20, +...+? (r - 1) a,x-? +...} + x (n + 2a, +...+ ra,«"-1 +...)

r
= né (1 + nx + azaca +...+ ay.c" +..

)

..(III)
In (III) equate the coefficients of ". Thus
(r + 2) (r + 1) Ar+2+r (r – 1) az + ra, = n'a, ;

na-p2
72

ap.
(r + 1) (r + 2)

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+

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+

.

2

.. Ar+2

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

n2 n (n° - 19) m(no – 22) {(1 + x2)* + X}" =1+na+

2C3 +
1.2 1.2.3 1.2.3.4
(n-r+2)(n-r+4)... (n +9 — 2)

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24 t...

+

nx" +....

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Х

[1

1 + x +

+

Dat]

. +...]

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Also substituting for an, Az... in (II), we have

na 12

n (n° -29)

na {(1 + x2)} + x}" = (1 + x2)* x

x2 +

1.2 1.2.3 n-r+ 1)(n-r+ 3)...(n +r- - 1)

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

22 – n2 terms is

22, the limit of which is – 2*. Hence, if (r+l) (r + 2) w is wholly real and lies between + 1 and – 1, or if x is wholly imaginary and if ix lies between +1 and -1, the series are convergent.

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 + 2c2)* + x =%; then (1+x2)* – =Z-,

and 2. =%-%-1. If « lies between – 1 and + 1, 2 lies between ✓2 - 1 and 12 + 1; and we have

n?

n (na – 12)
=
2
2.4

2.4.6
Hence any power of z may be expressed in ascending powers

1 of a in a series, which is convergent if lies between 2-1 and 12+ 1.

It would thus appear that, by expanding the powers of %-2-1 by the binomial theorem, we could expand any power of :

=

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t...

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 %-%-1.

630. B. Let a = cot 0; then /(1+2c%) + x=coseco+coto=cot 10.

If then x lies between – 1 and +1, we may take between it and it, so that cot 30 is positive. Thus

na

m (m? – 12) cot" 10 = 1 +n cot 0 +

cot 0 +

cot3 O +.... 1.2

1.2.3

= cos no.

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631. To expand cos no and sin no in powers of sin 0, when n has any

real value. In the formulæ of Art. 626, put x=i sin 0. Then J(1 + x^) + x = cos 0 + i sin 0. Hence the series represent some value of (cos 0 + i sin 6)", i.e. of cos nd + i sin no where $ is some angle whose cosine and sine are equal to those of 0 respectively.

If 0 = 0, 3 = 0; and the series become = 1
Hence, by the principle of continuity, if 0 is acute,

na mở (m? 2°)
1-
sin? 0 +

sino 0 -= cos no;
12

4 n (n-12) m (n? - 1)(x - 3*) n sin sin3 0 +

sin5 A

= sin no, 3

5
na – 12 (no - 12) (no – 3)
sino A +

sino 0.
2

4
n (no -- 22) (na – 42)

sin no

sin 3

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

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1

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ne (no – 22) sin3 +

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n sin A.

cos o

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.

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