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4n 3

en (2n-1) r=2n-2, x= cos #+V-1.sin

2 n

3
T

2n

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And forming quadratic factors out of the pairs of simple factors which are equidistant from the extremities of this series of equations, we have

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139. To resolve the equation x2n+1 – 1 = 0 into its quadratic factors.

As in the last two Articles,

(cos 0 + ✓-1.sin 6)2n+1 = cos (2n + 1) 0 +V - 1. sin (2n+1) 8

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2 m .. = COS

T+V - 1. sin

2n +1

7T;

2n + 1

And if m=p. (2n + 1) +r; where p is o or any integer, and r is o or any integer less than 2n + 1, it may be shewn as before, that

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2n + 2

2n + 2 (n + 2) r=n + 1, x = cos T + V-1.sin

2n + 1

2n +1

T

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Hence, one root is 1, and a factor of the equation is x – 1,and by forming quadratic factors out of equations (2) and (2n + 1),(3) and (2n),-... we have

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To resolve the equation x2 +1 +1 = ( into its

140. factors.

Here 22n+1 = -1,

(cos @ +V - 1. sin 0) 2n + 1 = cos (2n + 1) @ +-1. sin (2n + 1)0,

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2 m +1 = COS

7 + 2n +1

2m +1
+V -
V- 1. sin

T.

2n +1 Proceeding as in the last Article by making m=p.(2n+1)+r, we have

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2n-1 .... {– 2 (cos 7). X + 1}.

141.

To resolve Sin 0 and Cos 0 into factors.

The values of 0 which satisfy the equation Sin 0 = 0, are

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Hence, the series which expresses the value of Sin is divisible by 0, 0-, A+, 8-21, , 2 п,

Wood's Algebra, Art. 269.

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Let therefore

Sin 0 = a.0.(0 – ).(0+7):(0 - 27).(+27)...

where a is some constant quantity whose value is to be determined.

.:. Sin ==0.0.(1-0).(+ + 0). (29 - 0).(27 + 1)......

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.:. Sin A = 0.11

1:(. )( )( - ..... ()

.

Again,—the values of which satisfy the equation Cos 0 = 0, are

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(2).

Cor.

... Cos 8 = (1-P) C

If 0 = 5, the expression for Sin 8 becomes, 1-(-) (-)(-) (-)...

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(2 – 1). (2 + 1) (4 - 1). (4 + 1) (6 – 1). (6 + 1) 22

62

2

42

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82

.: T = 2.

1.3 3.5 5.7 7.9

Wallis's theorem for the determination of , in which the successive factors become more and more nearly equal to 1.

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