Plane trigonometry, by S.L. Loney.

ANGLES OF ANY SIZE AND SIGN.

67

direction through the angle 0, and let the position of the
revolving line be then OP'.
The angle AOP' is then 90 - 0.
Take OP' equal to OP and draw P'M' and PM perpendicular to OA, produced if necessary. Also draw P'N'
perpendicular to OB, produced if necessary.
In each figure the angles AOP and BOP' are numerically equal, by construction.
Hence in each figure
z MOP = Z N'OP' = z OP'M',
since ONi' and JM'P' are parallel.
Hence the triangles MOP and M'P'O are equal in all
respects, and therefore OM= M'P' numerically,;and OM' = MP numerically.
Also in each figure OM and M'P' are of the same,sign, and so also are MP and OM',
i.e. OM = + M'P', and 1O1 = + MP.
Hence
sin (90~-) = sin AOP' = op- = -P = cos 0
OM' M31P
cos (90 - ) = cos AOP'= O' OP = sin,
tan (90~ - 0) = tan A OP' = MI O P= cot 0,
OM' MP
cot (90~ - ) = cot AOP'=,p = OM tan 0,
OP' OP
see (90~ - 0) = sec A OP' = O MP = cosec 0,
OP' OP
and cosec (90~ - 0) = cosec A OP' = lP' = OM = see 0.
5-2