2

sin (a+b) + sin (a−b)

sin a cos 6

R

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

2
sin 2 - sin l' cos l'

R

observing that cos l'= VR? - sin? 1' (Art. 72); we have

? thus the value of the sin 2. Making a=2' and b=l', we have, transposing sin (a−b),

2
sin 3 sin 2 cos 1'- sin 1'

R

Thus we find sin 3' in terms of sin 2 before determined, and so on. The cosines are calculated from the sines by the formula

COS = VR sin? the tangents by the formula (Art. 32.)

The sines, cosines, &c., may be calculated by series, a specimen of which is given at Prob. III., 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 Woodhause, 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 1,

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

PROBLEM 1.

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

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

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

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

log tan 6=log 2+1 log a+ log b+log sin c—log (a—b) log c=log (a—b)+10—log cos

Given a =

EXAMPLE,

562, b 320, and c 128° 4', to find c.

=

AD

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.)

COS B = √ } s (}s—b)

ac

2

b

.. 2sinвcos B,or(Art.71)sinв==√s({s—b)(}s—a)(}s—c),

ac

B

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

PROBLEM III.

To find the logarithmic sine of a very small arc.

By Article 74 Dif. Cal. (Davies') the expression for the sine of any arc x is

Now, the length of an arc of one degree is so small that, even when x is so great as this, the third term of the above series can have no significant figure in the first ten places of decimals.

Retaining therefore only the first two terms, we have, when a is small,

that is, since the quantity within the brackets = cos x. (Dif.

Calc, Art. 75.)

cos. z;

sin. x x cos. x;

hence, by introducing the radius so as to render the expres

hence, by introducing the radius,

lóg. x = log. n + log. 3.14159, &c., + 10 log. 180 × 602

(2);

log. sin. x=log. n+4.685575— ar. comp. log. cos x hence this rule. To the logarithm of the arc reduced into seconds, with the decimal annexed, add the constant quantity 4.685575, and from the sum subtract one third of the arithmetical complement of the log. eosine; the remainder will be the logarithmic sine of the given arc.

This rule will determine the log. sine of a very small arc with great accuracy; it was first given, without demon