Plane trigonometry, by S.L. Loney.

324 TRIGONOMETRY.
By equating real and imaginary parts, we have
n? (n- _ 1)
cos n0 = cos 0 -- ) cos91-2 sin +.....(1),
1.2
and
sin nO = n coSn-l 0 sin 0 _n (~ 1) ( 2) cos 1-3 0 sin 0
1.o9. --
n (n - 1) (n- 2) (n- 3) (n - 4) co sin... (2)
- -- -.C....(2).
'1.. 3.4. 5
The terms in each of these series are alternately
positive and negative. Also each series continues till one
of the factors in the numerator is zero and then ceases.
275. From equations (1) and (2) of the last article
we have, by division,
sin nO
tan n =cos Bn
cos-l 0 sin 0 - (n ( 1) (n 2) os 0 sin3 0 +......
1.2.3
cos,0 n(n-l) (n.2)(n-1 ) (no-2)(nn-3)
COn - s-2 sin n ) (n 2) (n 3) osn-4 sin......
1.2 1.2.3.4
Divide the numerator and denominator of the righthand member of this equation by cosn 0, and we have
tan nO =
n tO t - ( n1 ) t n -1)( -2)(n-3) (-4)tan5
1.2.3 a
1. 2. L 5..........
( - tan + (n-) (n- 2) (a -3 ) tan 0......
1.2 14
276. The values for cos n0 and sin n10 in Art. 274 may also be
obtained, by Induction, without the use of imaginary quantities.
For assume (1) and (2) to be true for any value of n. Then, since
cos (n + 1) 0 = cos n0 cos 0 - sin nO sin 0,