Substituting 1-sin2a for cos2a, then 1-- cos2a for sina, in (6), and reducing, we have

Dividing (11) by (12), then (12) by (11), we have

Dividing (5) first by (10), then by (9), and transposing, we have

Taking the reciprocal of (16), then of (15), we have

Let the formulas of this article be expressed in words.

S. N. 8.

96. Consequences of (a), (b), (c), (d).

Taking the sum and difference of (a) and (c), (d) and (b), we have

(a - b) = 2 sin a cos b.
(a - b) = 2 cos a sin b.
2 cos a cos b.

(1) sin (a+b) + sin
(2) sin (a+b) — sin
(3) cos (a + b) + cos (a — b) :

=

(4) -
cos (ab) cos (a + b) = 2 sin a sin b.

Substituting the values of a + b, a−b, a, and b, in

(1), (2), (3), and (4), we have

By formula (5) of the preceding article we have

(9) sin (sd) = 2 sin (10) sin (sd) = 2 sin

(8 + d) cos (s + d).
(sd) cos (sd).

Dividing each of these formulas by each of the following, we have

Formula (11) gives the proportion,

sins sind sin s-sin d :: tan (s+d): tan(s-d).

:

Hence, The sum of the sines of two angles is to their difference as the tangent of one-half the sum of the angles is to the tangent of one-half their difference.

Let us apply this principle in solving triangles when two sides and their included angle are given. Article 75.

d)

sin A+ sin B: sin A-sin B :: tan(A+B): tan}(A—B).

... a+ba-b :: tan (A+B) : tan †(A—B).

97. Theorem.

The square of any side of a triangle is equal to the sum of the squares of the other sides, minus twice their product into the co-sine of their included angle.

(2)+(3)=(4) m2 + p2 = b2 + n2 + p2 - 2 bn.

But m2 + p2=a2 and n2+ p2= c2, ... (4) becomes

(5) a2= b2+ c2 - 2 bn.

But nc cos A, which substituted in (5) gives

(6) a2 = b2+c2-2 be cos A.

But m2+p2a2 and n2+ p2 c2, ... (4) becomes

(5) a2= b2+ c2 + 2 bn.

But nc cos BAD: =c cos BAC:

-

C cos A.

.. (6) a2=b2c2-2 be cos A.

98. Problem.

To find the angles of a triangle when the sides are given. From either formula (6) of the last article we have

Hence, The co-sine of any angle of a triangle is equal to the sum of the squares of the adjacent sides, minus the square of the opposite side, divided by twice the rectangle of the adjacent sides.

Formula (1) gives the natural co-sine of A; hence, A can be found. But it is best to place the formula under such a form as to adapt it to logarithmic computation.

Adding 1 to both members of (1) we have

(b+c)2 — a2

(a + b + c) (b + c — a).

2 bc

(2) 1+ cos A= 2 be

But 1 cos A=2 cos2 A. Article 95, (10).

(a+b+c) (b+c-a)

Let a+b+c=p, then

2 bc

p(p—2a).

2 be

Substituting these values in (2), and dividing by 2,