From this equation we have the following practical rule: Subtract the less from the greater of the given logs, the remainder will be the log tan of an angle: from this angle take 45 degrees, and to the log tan of the remainder add the log cotan of half the given angle; the sum will be the log tan of half the difference of the other two angles of the plane triangle.

39. The remaining case is that in which the three sides of the triangle are known, and for which indeed we have already obtained expressions for the angles in arts. 6 and 8. But, as neither of these is best suited for logarithmic computation, (however well fitted they are for instruments of investigation), another may be deduced thus: In the equation for cos A, (given equation 11), viz, cos a = if we substitute, instead of cos A, its value 12 sin2 A, change the signs of all the terms, transpose the 1, and divide by 2, a2 — b2 — c2 + 2bc a2—(b-c)1 we shall have sin2 A =

4bc

b2+ c2 — a2
1

4bc

2bc

Here, the numerator of the second member being the product of the two factors (a+b-c) and (a-b+c), the equa4(a+b-c). (a−b+c) tion will become sin2 =

41c

But, since

(a+b−c)=(a+b+c)−c, and 4(a−b+c)=4(a+b+c)−b; if we put sa 4b+c, and extract the square root, there will result,

These expressions, besides their convenience for logarithmic computation, have the further advantage of being perfectly free from ambiguity, because the half of any angle of a plane triangle will always be less than a right angle.

40. The student will find it advantageous to collect into one place all those formule which relate to the computation of sines, tangents, &c*; and, in another place, those which are of use in the solutions of plane triangles: the former of

* What is here given being only a brief sketch of an inexhaustible subject; the reader who wishes to pursue it further is referred to the copious Introduction to our Mathematical Tables, and the comprehensive treatises on Trigonometry, by Emerson and many other modern writers on the same subject, where he will find his curiosity richly gratified.

these

these are equations v, VIII, IX, X, XI, x, xi, xf, XIII, XIV, XV, XVI, XVII, XVIII, XIX, xx, xxII, xxй, XXIII, XXIV, XXVII; the latter are equa. II, III, IV, VII, XXXII, XXXIII,

XXXIV.

To exemplify the use, of some of these formulæ, the following exercises are subjoined.

EXERCISES.

Er. 1. Find the sines and tangents of 15°, 30°, 45°, 60°, and 75° and show how from thence to find the sines and tangents of several of their submultiples.

First, with regard to the arc of 45°, the sine and cosine are manifestly equal; or they form the perpendicular and base of a right-angled triangle whose hypothenuse is equal to the assumed radius. Thus, if radius be R, the sine and cosine of 45°, will each be RR=R2, If R be equal to 1, as is the case with the tables in use, then

sin 45° cos 45° = 4√√/2 = ·7071068.

sin

COS

tan 45° = == = Cotangent 45o.

COS

Secondly, for the sines of 60° and of 30°: since each angle in an equilateral triangle contains 60°, if a perpendicular be demitted from any one angle of such a triangle on the opposite side, considered as a base, that perpendicular will be the sine of 60°, and the half base the sine of 30°, the side of the triangle being the assumed radius. Thus, if it be R, we shall have R for the sine of 30°, and R-R2R√3, for the

sine of 60°. When R = 1, these become

Thirdly, for the sines of 15° and 75°, the former arc is the half of 30°, and the latter is the compliment of that half arc. Hence, substituting 1 for R and √3, for cos A, in the expression sina = ± ± √2R2 ± 2R COS A

it becomes sin 15° 플 2√32588190.

(equa. XII)

Hence, sin 75° = cos 15°/1-(2-√3) = {√2 + √3 =

-

=

Now, from the sine of 30°, those of 6o, 2o, and 1o, may easily be found. For, if 5A 30°, we shall have, from equation x, sin 5A = 5 sin a 20 sin3 A + 16 sin3 A: or, if sin A, this will become 16.rs 20x35x='5. This equation solved by any of the approximating rules for such equations, will give r = 1045285, which is the sine of 6°. Next, to find the sine of 2°, we have again, from equation x, sin 3A = 3 sin A 4 sin3 A that is, if x be put for sin 2°, 3x 4x31045285. This cubic solved, gives

a0348995 sin 2o.

Then, if s = sin 1o, we shall, from the second of the equations marked x, have 2s/1-s2 = 0348995; whence s is found0174524 sin 1°.

Had the expression for the sines of bisected arcs been applied successively from sin 15°, to sin 7°30°, sin 3°45′, sin 1°524′, sin 564, &c, a different series of values might have been obtained: or, if we had proceeded from the quinquisection of 45°, to the trisection of 9°, the bisection of 3°, and so on, a different series still would have been found. But what has been done above, is sufficient to illustrate this method. The next example will exhibit a very simple and compendious way of ascending from the sines of smaller to those of larger arcs.

Ex. 2. Given the sine of 1°, to find the sine of 2°, and then the sines of 3°, 4o, 5o, 6o°, 7°, 8o, 9o, and 10o, each by a single proportion.

Here, taking first the expression for the sine of a double arc, equá. x, we have sin 2°2 sin 1°√1 — sin2 1°='0348995. Then it follows from the rule in equa. xx, that

:

sin 1o sin 2°- sin 1° :: sin 2o + sin 1° : sin 3° = ·0523360 sin 2o: sin 3o-sin 1° :: sin 3° + sin 1° : sin 4° = '0697565 sin 3° sin 4-sin 1°:: sin 4o + sin 1°: sin 5° : =0871557 sin 4° : sin 5° — sin 1° :: sin 5o + sin 1° : sin 6° = ·1045285 sin 5°: sin 6o- sin 1o :: sin 6o + sin 1° : sin 7° 1218693 sin 6° sin 7° -sin 1° :: sin 7° + sin 1°: sin 8° =1391731 sin 7° sin 8°-sin 1° :: sin 8° + sin 1°: sin 9° sin 8° sin 9° - sin 1° :: sin 9° + sin 1° sin10° 1736482 To check and verify operations like these, the proportions should be changed at certain stages. Thus,

:

:

=1564375

sin 1° sin 3° - şin 2° :: sin 3° + sin 2° : sin 5°,

sin 1o : sin 4o — sin 3° : : sin 4° + sin 3o : sin 7o, sin 4° sin 7°— sin 3° :: sin 7° + sin 3o : sin 10°. The coincidence of the results of these operations with the analogous results in the preceding, will manifestly establish the correctness of both.

Cor.

Cor. By the same method, knowing the sines of 5o, 10o, and 15, the sines of 20°, 25°, 35°, 55°, 65°, &c, may be found, each by a single proportion. And the sines of 1o, 9°, and 10°, will lead to those of 19°, 29°, 39o, &c. So that the sines may be computed to any arc: and the tangents and other trigonometrical lines, by means of the expressions in art. 4, &c.

Ex. 3. Find the sum of all the natural sines to every minute in the quadrant, radius = 1.

In this problem the actual addition of all the terms would be a most tiresome labour: but the solution by means of equation XXVII, is rendered very easy. Applying that theorem to the present case, we have sin (A+B) = sin 45°, sin (n+1)B= sin 45°0′30′′, and sin B sin 30′′. Therefore

sin 45° x sin 45° 0' 30" = 3438-2467465 the same sum required.

sin 30"

From another method, the investigation of which is omitted here, it appears that the same sum is equal to (cot 30"+1). Ex. 4. Explain the method of finding the logarithmic sines, cosines, tangents, secants, &c, the natural sines, cosines, &c, being known.

The natural sines and cosines being computed to the radius unity, are all proper fractions, or quantities less than unity, so that their logarithms would be negative. To avoid this, the tables of logarithmic sines, cosines, &c, are computed to a radius of 10000000000, or 101; in which case the logarithm of the radius is 10 times the log of 10, that is,

it is 10.

Hence, if s represent any sine to radius 1, then 101×s= sine of the same arc or angle to rad 10. And this, in logs is, log 100s 10 log 10+ log s = 10 + log s.

The log cosines are found by the same process, since the cosines are the sines of the complements.

The logarithmic expressions for the tangents, &c, are deduced thus:

Tan rad

sin

COS

=

Theref. log tan log rad + log sin-log cos 10+ log sin - log cos.

rad2

tan
rad2

COS

rad2 Cosec= --

sin

Therf. log cot2 lograd-log tan=20-log tan.

Therf. log sec=2 log rad―log cos=20—log cos.

Therf. l.cosec=2 log rad-log sin=20-log sin.
chord2 (2 sinarc)2
diam
2 rad

Versed sine =

=

2 x sin2 arc

rad

Therefore, log vers sin = log 2+2 log sin are 10.

Er. 5. Given the sum of the natural tangents of the angles A and B of a plane triangle 3.1601988, the sum of the tangents of the angles B and C = 3·8765577, and the continued product, tan A. tan B. tan c = 5.3047057; to find the angles A, B, and c.

It has been demonstrated in art. 36, that when radius is unity, the product of the natural tangents of the three angles of a plane triangle is equal to their continued product. Hence the process is this:

From tan Atan B + tan c5·3047057

Take tan Atan B

Remains tan c

Take tan B + tan c

=21445069=tan 65°.

From tan Atan B + tan c = 5·3047057

Remains tan ▲

=3.8765577

= 14281480=tan 55°.

Consequently, the three angles are 55°, 60°, and 65°.

Er. 6. There is a plane triangle, whose sides are three consecutive terms in the natural series of integer numbers, and whose largest angle is just double the smallest. Required the sides and angles of that triangle?

If A, B, C, be three angles of a plane triangle, a, b, c, the sides respectively opposite to A, B, C; and s = a + b + c. Then from equa. 111 and XXXIV, we have

2

sin Ass-a). (is—b).
√ s ( 1 s − a) . ( s − b ).(is—c),

and sin c = √

(s-a). (s-b)

ab

Let the three sides of the required triangle be represented by x, x + 1, and r+2; the angle a being supposed opposite to the side r, and c opposite to the side +2: then the preceding expressions will become

(+1). (x+2)
(x + 1). (x+3)
4x(x+1)

3x + 3

Assuming these two expressions equal to each other, as they ought to be, by the question; there results, after a little reduction, x3- 2x2 — 11 x − 2 — 0, a cubic equation, with one positive integer root = 4. Hence 4, 5, and 6, are the x sides of the triangle.

6

• 4

şin B = 7; sin c = 7; sincv7.

The angles are, A = 41°409603 = 41° 24′ 84" 34",

B = 55° 771191 55 46 16 18,

82.819206 = 82 49 9 8.

Any