still be true, for substituting 90°+-m for a, we have sin (90° +m+b) in place of the first member, which is equal to cos (m+b)*; for the second member by the same substitution, we have

sin (90°+m) cos b+sin b cos (90+m)

R

but sín (90°+m) cos m and cos (90°+m)= hence equation (1) becomes

cos (m+b) cos m cos b - sin m sin b

which, since m and b are less than 90°, we know to be true, by Art. 69; hence (1), from which it is derived, is true also.

Assuming (1) to be true with a > 90°, which we have just proved, make b=90°+m, and in a similar manner the truth of the formula may be established on the supposition of both a and b > 90°.

Afterwards make a = 180°+m and observe that sin (180° +m+b)=sin m+b and cos (180°+m)=— cos m, and you will show that the formula extends to the third quadrant, and so on.

121. As we were not sufficiently advanced in the theory of trigonometrical lines to explain the construction of the tables of sines, tangents, &c., at Art. 38, having once been obliged to use them without explanation, we thought best to defer that to this place, so as not to interrupt, more than was necessary, the train of reasoning relative to the solution of triangles.

3.1415926

The diameter of a circle being multiplied by we have the length of its circumference; this divided by 360 gives us the length of one degree, and this by 60 the length of one minute of the circumference. So small an arc as 1' may be considered as equal to its sine, without sensible error. Having thus found the sine of l' we may find the sines of other arcs by formula (1), of Art. 74.

* By referring to either of the diagrams in which a sine is drawn, it will be evident that sin (90o+a), a being any arc less than a quadrant, is equal in length to sin (90°—a) = cos a. Also that cos (90o+a) · -- cos (90°—a) = — sin a,

sin (a+b)+ sin (a—b) =

2

=-
R

sin a cos ¿

making a and b each equal to 1', this becomes

Thus we find şin 3' in terms of sin 2' before determined, and so on.

The cosines are calculated from the sines by the formulą

The sines, cosines, &c., may be calculated by series, a specimen of which is given at Prob. 11., Art. 123. These series are most conveniently derived by the aid of the Differential Calculus, and as those of our readers, who will wish to investigate them, are likely to become acquainted with the mode, in the study of that branch of Analysis, we have taken the liberty of quoting one of them at the article above mentioned.

PART VI.

MISCELLANEOUS TRIGONOMETRICAL INQUIRIES,

122. We now come to the final part of our subject, in which we propose to bring together several miscellaneous particulars which properly come under consideration in a treatise on Trigonometry. Some of these might have been introduced much earlier, but we have preferred to leave their consideration for a supplementary chapter, agreeing with Woodhouse, that it is better for the student first "to attend solely to the general solutions, and to postpone to a time of leisure and of acquired knowledge the consideration of the methods that are either more expeditious or are adapted to particular exigencies."

CHAPTER I.

ON THE SOLUTIONS OF CERTAIN CASES OF PLANE TRIANGLES, AND ON DETERMINING THE TRIGONOMETRICAL LINES OF SMALL ARCS.

PROBLEM I.

123. Given two sides and the included angle of a pla

triangle, to determine the third side, without finding the re

maining angles.

The general expression for the side c, in terms of the two sides a, b, and the included angle c, is (Art. 68) on the supposition of R = 1,

c2 = a2 + b2 — 2 ab cos c

= (a - b)2 + 2 ab (1

c

cos c)

(ab)2+2 ab .2 sin 2

Assume the second term within the brackets equal to tan,20,

Hence c is determined by these two formulas, viz.,

log tan 0=log 2+1 log a+ log b+log sin c—log (a—b) log c=log (a—b)+10—log cos 0

EXAMPLE,

Given a 562, b = 320, and c = 128° 4', to find c.

log, 2

0.301030

log, 562

1.374868

log, 320

1.252575

log. sin 64° 2′

9.953783

ar. comp. log, 242

7.616185 log. 242+10. 12,383815

To determine the area of a plane triangle when any three

arts except the three angles are given.

Let two sides, a, c, and the included angle в, be given. fig. Art. 64.)

The area of the triangle is expressed by BC. AD; but AD AB sin. B; hence the expression for the area, in terms of the given quantities, is

2. Let two angles, A, B, and the interjacent side c, be given. Then, since

Also, by adding (5) and (6) of Art. 72, and proceeding as for the above, may be found. (See Art. 129.)

Consequently, by substituting this value of sin в in the first expression, we have

areas (sa) (†s—b) (§s—c); which formula furnishes the well known rule, given in all books on mensuration, for the area of a triangle when the three sides are given.

These expressions for the area of a plane triangle are all adapted to logarithmic computation.