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

[Since (x-cos+√.sin). (cos.sin

2n

π

1.

2n

139. To resolve the equation x2+1 − 1 = 0 into its

quadratic factors.

As in the last two Articles,

(cos + √1. sin 0)2n+1 = cos (2n + 1) 0 + √ - 1. sin (2n+1) 0

And if m = p . (2n + 1) + r; where p is 0 or any integer, and ris O or any integer less than 2n + 1, it may be shewn as before, that

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

140. To resolve the equation x2+1+1=0 into its

factors.

Here 2+1 — — 1,
x2n

=

(cos +√1. sin 0)2n + 1 = = cos (2n + 1) 0 + √ −1 . sin (2 n + 1)0,

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

141. To resolve Sine and Cose into factors.

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

Hence, the series which expresses the value of Sin is divisible by

Sin 0 = α.0. (0 – π) · (0 + π) · (0 − 2 π) · (0 + 2π)...

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

... Sin 0= ± α.0. (π − 0). (π + 0). (2 π − 0). (2 π + 0)......

π (2 − 1) . (2 + 1) (4 − 1) . (4 + 1) (6 − 1) . (6 + 1)

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

to 1.