b,

thus, when ab, the negative arc ba has a negative sign. If we construct fig. 4* on this supposition by taking AM = MN = a, and carrying this last arc below the point M, in order to represent the operation to be performed according to art. 11, the arc AN will be found below AC instead of being above it; the sine QN then will change its direction, as well as the arc. As to the cosine, it will remain in the same direction; and we find also by the formulas, that cos (b — a) = cos (a — b).

27. There are many other conclusions to be drawn from the proposition demonstrated in art. 11, some of which will be necessary in the subsequent part of this treatise; I will therefore put them down in this place.

1. By adding together the two equations

sin a cos b+ sin b cos a

sin (a + b)

=

sin (a - b)

=

R

2. By subtracting the second equation from the first, we ob

it being recollected, that the cosine is equal to radius, when the arc is nothing.

4. By subtracting the first equation from the second, the result becomes

5. If we make a + b = a', a − b = b', we find, by adding these two equations, 2a = a ond from the first, 2 b = a' —

+ b' and by subtracting the sec

b' ;

it follows from this that

Putting these values of a and b in the expression for sin a cos b, sin b cos a, cos a cos b, sin a sin b, obtained above, we find

Dividing the second of these formulas by the first, we have

We infer in like manner from the last two formulas above

6. By dividing the expression for the sin (a + b) by that for the cos (ab), we have

then, dividing the numerator and denominator of the second mem

hence, dividing R2 by tang (a + b) and by its equal in the

tang (a' + b') sin a' + sin b'

from which

we infer, that the sum of the sines of two arcs is to their differ ence, as the tangent of half the sum of these arcs is to the tangent of half their difference, is obtained immediately by a very elegant geometrical construction.

AM and AN (fig. 11), being two arcs represented by a' and Fig. 11. b', we have MP = sin a', NQ = sin b'; drawing NC parallel to the diamater AB, and producing MP to M', we deduce

MR MP+NQ sin a' + sin b' (14).
M'R=M'P

This being done, if from the point C, as a centre, and with a ra-
dius CD equal to that of the circle ACB, we describe an arc
EDG, and draw, through the point D of this arc, a tangent
meeting the straight lines CM and CM', it is evident, that DF
and DH will be the tangents of the arcs DE and DG, which
measure the angles MCN, NCM'; and as these angles have
their vertex in the circumference of the circle ACB, they will
have for their measure, respectively, (Geom. 126),

} NM = } (AM — AN) = } (a' — - b'),

NM' =

we have then ·

= } (AM' + AN) = } (a' + b′);

DF = tang (a'

tang (a' — b'), DH = tang (a' + b'). But on account of the parallels MM' and FH, we have this proportion,

that is,

:

MR MR:: DF: DH,

sin a' sin b': sin a' + sin b':: tang (ab): tang (a + b'), which is the same as the equation above given.

It would be easy to modify the construction, so as to deduce from it the different equations analogous to that just demonstrated.

29. As we have often occasion to make use of the formulas, which we have already obtained, I have put them together in the following table with others, which may be deduced by a process easy to be imagined. The number against each formula marks the article, in which it may be found, or from which it may be obtained.

cos a2 = R (R+ cos 2 a) (27)

sin asin b2cos b2cos asin (a + b) sin (ab) (11,10)

sin b2 = cos (a + b) cos (a — b) (11, 10)

cos a2