Plane trigonometry, by S.L. Loney.

MULTIPLE ANGLES. 107
Ex. To find the values of sin 15~ and cos 15~.
Let the angle 2A be 30~, so that A is 15~.
Let the radius CP be 2a, so that we have
CN =2a cos 30=~a^/3,
and NP = 2a sin 30~= a.
Hence ON OC+ CN= a (2 +/3),
and NQ = CQ - CN=a(2 - 3).. OP2=ON. OQ=a (2+,/3) x4a (Euc. vI. 8),
so that OP = a/2 (/3 + 1),
and PQ2= QN. QO=a (2 - 3) x 4a,
so that PQ = aV2 (/3 - 1).
Hence sin 15~=PQ,/2 (,/3 - 1) _ J3 -1
Hence sin 15 =) =3=
Hec4 2Q '
and cos 15= OP = 2 (^3 + 1) 3 +1
oQ 4 2d2'
107. To find the trigonometrical functions of 3A in
terms of those of A.
By Art. 88, putting B equal to 2A, we have
sin 3A = sin (A + 2A) = sin A cos 2A + cos A sin 2A
= sin A (1 - 2 sin A) + cos A. 2 sin A cos A
= sin A (1 - 2 sin2 A)+ 2 sin A (1 - sin2 A).
Hence sin 3A = 3 sin A - 4sin3 A.........(1).
So
cos 3A = cos (A + 2A) = cos A cos 2A - sin A sin 2A
= cos A (2 cos2 A - 1) - sin A. 2 sin A cos A
= cos A (2 cos2 A - 1) - 2 cos A (1 - cos2 A).
Hence cos 3A = 4 cos3A - 3 cos A.........(2).