Plane trigonometry, by S.L. Loney.

426 TRIGONOMETRY.
We shall first shew that xn+-o - 2 cos na is divisible by
1
x + —2 cos a.
x
1 1
Let xn + — 2 cos na be denoted by 0 (n), and x +- - 2 cos a by X, so
X
that we have to shew that q5 (n) is divisible by X, for all positive integral
values of n.
Assume that this is true for 0 (n - 1) and 0 (n - 2).
We have then, by ordinary multiplication,
( + l) x(n-1)= =+l i Xn-l + xl -2cos(n-1) a
= (a )+( 2Jr )-2 cos(1n-1)ax (>x+)
= Cn+ I - 2 cos na
+ -2 —, 2 cos(n-2) a -2 cs(n )a x 2osa
since 2 cos na + 2 cos (n- 2) a =4 cos a cos (n - 1) a.
Hence (x +-) x (n -1) = (n)+0(n-2)-2Xcos(n-1)a,. ) - 0 (n=-+ 1) - l (n -2)+2 cos (n- 1) a......(1).
1
Now (1)=x + — 2 cos a=X,
X
0(2) =x2 - 2 cos 2a= x+ 2 cosa) (x+-+2cosa)
() =2 x-~-cs X ~x X
=X x+- +2cosa),
so that 0 (1) and 0 (2) are divisible by X.
Hence, putting n=3 in (1), we see that b (3) is divisible by X.
Similarly putting, in (1), n=4, 5, 6...... in succession we see that,
by induction, 0 (n) is divisible by X for all values of n..i x1+ -
x + - — 2 cos na is divisible by x+- - 2 cos a.
xn x