Plane trigonometry, by S.L. Loney.

142 TRIGONOMETRY.
Hence the solutions of the equation are respectively
2n'7r + QOP and 2n7r + Q'OP.
The construction clearly fails if c be > f1a o + b2, for then
the point N would fall outside the circle.
131. As a numerical example let us solve the equation
5 cos 0 - 2 sin 0 =2.

given that tan 21~ 48'=.
Dividing both sides of the equation by
/52 + 22 i.e. V29,
we have
5 2 2
--- cos 6 - -9 sin 2
V29.s29 V. 29'
Hence cos 0 cos 21~ 48' - sin 0 sin 21~ 48'
= sin 21~ 48'= sin (90 - 68~ 12')
= cos 680 12'... cos (0 + 21~ 48')= cos 68~ 12'.
Hence 0 + 21~ 48' = 2n7r = 68~ 12'... 8 =2nr - 21~ 48' - 68~ 12'
= 2n7r - or 2n7r + 46~ 24',
where n is any integer.

~212
1 (. 48' 8
5
(Art. 83)

EXAMPLES. XXI.

Solve the equations
1. sin 0+ sin 79=sin 40. 2. cos 0 cos 7= cos 48.
3. cos + cos 30 =2 cos 2. 4. sin 4sin4 sin 20= cos 38.
5. cos 8 - sin 30 = cos 20. 6. sin 78=sin + sin 30.
7. cos 8+cos28+cos38=0. 8. sin 0+sin38+sin 5=0.
9. sin 20 - cos 28 - sin 0 +cos 6 =0.
10. sin (30 + a) + sin (30 - a) + sin (a - 0) - sin (a + 0) =cos a.