FORMULAS EXPRESSING RELATIONS BETWEEN THE CIRCULAR FUNCTIONS OF DIFFERENT ARCS.

65. Let MB and BA represent two arcs, having the common radius 1; denote the first by a, and the second by b: then, MA=a+b. From M draw MP perpendicular to CA, and MV perpendicular to CB; from N draw NP' perpendicular to CA, and NL parallel to AC.

Then, by definition, we shall have,

B IN

PP'A

PM = sin (a + b), NM = sin a, and CN= cos a.

From the figure, we have,

PM = ML + LP.

(1).

Since the triangle MLN is similar to CP'N (B. IV, P. 21), the angle LMN is equal to the angle P'CN; hence, from the right-angled triangle MLN, we have,

MLMN cos b = sin a cos b.

From the right-angled triangle CP'N (Art. 37), we have,

NP' CN sin b;

or, since NP' LP, =

LP cos a sin b.

Substituting the values of PM, ML, and LP, in Equation (1), we have,

sin (a + b) = sin a cos b + cos a sin b; (A.).

that is, the sine of the sum of two arcs, is equal to the sine of the first into the cosine of the second, plus the cosine of the first into the sine of the second.

Since the above formula is true for any values of a and - b, for b; whence,

b, we may substitute

sin (a - b) = sin a cos (b) + cos a sin (b) ;

that is, the sine of the difference of two arcs, is equal to the sine of the first into the cosine of the second, minus the cosine of the first into the sine of the second.

If, in Formula (B), we substitute (90° — a), have,

for a, we

sin (90°—a—b) = sin (90°-a) cos b-cos (90°-a) sin b;

but (Art. 63),

sin (90°— a—b) = sin [90°— (a + b)] = cos (a + b),

and,

sin (90° - a) = cos α,

cos (90° - a)

= sin a;

hence, by substitution in Equation (2), we have,

cos (a + b) = cos a cos b sin a sin b;

that is, the cosine of the sum of two arcs, is equal to the rectangle of their cosines, minus the rectangle of their sincs.

cos (a - b) = cos a cos b + sin a sin b

that is, the cosine of the difference of two arcs, is equal to the rectangle of their cosines, plus the rectangle of their sines.

It we divide Formula (A) by Formula (@), member Ly nember, we have,

Dividing both terms of the second member by cos a cos b, recollecting that the sine divided by the cosine is equal to the tangent, we find,

that is, the tangent of the sum of two arcs, is equal to the sum of their tangents, divided by 1 minus the rectangle of their tangents

If, in Formula (), we substitute - b, for b, recollecting that

tan (-b)

=

tan b, we have,

that is, the tangent of the difference of two arcs, is equal to the difference of their tangents, divided by 1 plus the rectangle of their tangents.

In like manner, dividing Formula () by Formula (A), member by member, and reducing, we have,

and thence, by the substitution of b, for b

66. If, in Formulas (A), (C), (B), and (G), we

Dividing Equation (A'), first by Equation (4), and then by Equation (3), member by member, we have,

Taking the reciprocals of both members of the last two

67. If Formulas (A) and (B) be first added, member to member, and then subtracted, and the same operations be performed upon () and (D), we shall obtain,