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is very material to be considered. The reason is, that the sines of angles, which are nearly 90°, or the cosines of angles, which are nearly-0, vary very little for a considerable variation in the corresponding angles, as may be seen from looking into the tables of sines and cosines. The consequence of this is, that when the sine or cosine of such an angle is given, (that is, a sine or cosine nearly equal to the radius,) the angle itself cannot be very accurately found. If, for instance, the natural sine .9998500 is given, it will be immediately perceived from the tables, that the arch corresponding is between 89o, and 89°, 1'; but it cannot be found true to seconds, because the sines of 89° and of 89°, 1', differ only by 50 (in the two last places), whereas the arches themselves differ by 60 seconds. Two arches, therefore, that differ by 1", or even by more than 1", have the same sine in the tables, if they fall in the last degree of the quadrant.

The fourth solution, which finds the angle from its tangent, is not liable to this objection; nevertheless, when an arch approaches very near to 90, the variations of the tangents become excessive, and are too irregular to allow the proportional parts to be found with exactness, so that when the angle sought is extremely obtuse, and its half of consequence very near to 90, the third solution is the best.

It may always be known, whether the angle sought is greater or less than a right angle by the square of the side opposite to it being greater or less than the squares of the other two sides.

SECTION III.

CONSTRUCTION OF TRIGONOMETRICAL TABLES.

In all the calculations performed by the preceding rules, tables of sines and tangents are necessarily employed, the construction of which remains to be explained.

These tables usually contain the sines, &c. to every minute of the quadrant from 1' to 90°, and the first thing required to be done is to compute the sine of 1', or of the least arch in the tables.

1. If ADB be a circle, of which the centre is C, DB any arch of that circle, and the arch DBE double of DB; and if the chords DE, DB be drawn, and also the perpendiculars to them from C, viz. CF, CG, it has been demonstrated, (8. 1. Sup.) that CG is a mean propor

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tional between AH, half the radius, and AF, the line made up of the radius and the perpendicular CF. Now CF is the cosine of the arch BD, and CG the cosine of the half of BD; whence the cosine of the half of any arch BD, of a circle of which the radius =1, is a mean proportional between and 1+cos BD. Or for the greater generality, supposing A any arch, cos A is a mean proportional between and 1+cos A, and therefore (cos A)2 = 1(1 + cos A) or cos A ✓ (1+cos A).

=

=

2. From this theorem, (which is the same that is demonstrated (8. 1. Sup., only that it is here expressed trigonometrically), it is evident, that if the cosine of any arch be given, the cosine of half that arch may be found. Let BD, therefore, be equal to 60°, so that the chord BD=radius, then the cosine or perpendicular CF was shewn (9. 1. Sup.) to be=1, and therefore cos BD, or cos 30o=

In the same manner, cos 15°=

✓ (1+cos 30°), and cos 70, 30'√ } (1+ cos 15o), &c. In this way the cosine of 3o, 45', of 1°, 52', 30", and so on, will be computed, till after twelve bisections of the arch of 60°, the cosine of 52′′. 44′′. 03"". 45v. is found. But from the cosine of an arch its sine may be found, for if from the square of the radius, that is, from the square of the cosine be taken away, the remainder is the square of the sine, and its square root is the sine itself. Thus, the sine of 52′. 44"". 03""". 45v. is found.

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3. But it is manifest, that the sines of very small arches are to one another nearly as the arches themselves. For it has been shewn, that the number of the sides of an equilateral polygon' inscribed in a circle may be so great, that the perimeter of the polygon and the cir

cumference of the circle may differ by a line less than any given line, or which is the same, may be nearly to one another in the ratio of equality. Therefore their like parts will also be nearly in the ratio of equality, so that the side of the polygon will be to the arch which it subtends nearly in the ratio of equality; and therefore, half the side of the polygon to half the arch subtended by it, that is to say, the sine of any very small arch will be to the arch itself, nearly in the ratio of equality. Therefore, if two arches are both very small, the first will be to the second as the sine of the first to the sine of the second. Hence, from the sine of 52. 44"". 03""'. 45. being found, the sine of 1' becomes known; for, as 52". 44". 03""'. 45. to 1, so is the sine of the former arch to the sine of the latter. Thus the sine of 1' is found 0.0002908882.

4. The sine 1' being thus found,. the sines of 2', of 3', or of any number of minutes, found by the following proposition.

THEOREM.

Let AB, AC, AD be three such arches, that BC the difference of the first and second is equal to CD the difference of the second and third; the radius is to the cosine of the common difference BC as the sine of AC, the middle arch, to half the sum of the sines of AB and AD, the extreme arches.

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M

Draw CE to the centre; let BF, CG, and DH perpendicular to AE, be the sines of the arches AB, AC, AD. Join BD, and let it. meet CE in I; draw IK perpendicular to AE, also BL and IM per- / pendicular to DH. Then, because the arch BD is bisected in C, EC is at right angles to BD, and bisects it in ; also BI is the sine, and El the cosine of BC or CD. And, since BD is bisected in I, and IM is parallel to BL, (2. 6.), LD is also bisected in M. Now BF is equal to HL, therefore, BF+DH-DH+HL= DL + 2LH = 2LM + 2LH = 2MH or 2K1; and therefore IK is half the sum of BF and DH. But because the triangles CGE, IKE are equiangular, CE: EI:: CG: IK, and it has been shewn that EI=cos BC, and IK={ (BFDH); therefore R cos BC sin AC (sin AB+sin AD). QE. D.

B

AF GK H F

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COR. Hence, if the point B coincide with A,

R: cos. BC sin. BC: sin. BD, that is the radius is to the cosine of any arch, as the sine of the arch is to half the sine of twice the arch; or if any arch=A, sin. 2A=sin. AXcos. A, or sin. 2A=2

sin. AXcos. A..

Therefore also, sin 2'2' sin 1'x cos 1'; so that from the sine and cosine of one minute the sine of 2' is found.

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Again, 1', 2', 3' being three such arches that the difference between the first and second is the same as between the second and third, R: cos 1':: sin 2': (sin 1' + sin 3'), or sin 1' + sin 3′ = 2 cos 1'X sin 2', and taking sin 1' from both, sin 3' 2 cos 1'X sin 2′-sin 1. In like manner, sin 4'-2' cos 1'X sin 3'--sin 2',

sin 5' 2 cos 1'X sin 4'-sin 3′,

sin 6'2 cos 1' X sin 5'-sin 4', &c.

Thus a table containing the sines for every minute of the quadrant may be computed; and as the multiplier, cos 1' remains always the same, the calculation is easy.

For computing the sines of arches that differ by more than 1', the method is the same. Let A, A+B, A+2B be three such arches, then, by this theorem, R: cos B:: sin (A+B) : ¦ (sin A + sin (A+2B)); and therefore making the radius 1,

sin A+sin (A+2B)=2 cos BXsin (A+B),

or sin (A+2B)=2 cos BXsin (A+B)—sin A.

=

sin A

cos A

By means of these theorems, a table of the sines, and consequently also of the cosines, of arches of any number of degrees and minutes, from 0 to 90, may be constructed. Then, because tan A: the table of tangents is computed by dividing the sine of any arch by the cosine of the same arch. When the tangents have been found in this manner as far as 45°, the tangents for the other half of the quadrant may be found more easily by another rule. For the tangent of an arch above 45° being the co-tangent of an arch as much under 45o; and the radius being a mean proportional between the tangent and co-tangent of any arch, (1. Cor. def. 9.), it follows, if the difference between any arch and 45o be called D, that tan (45° –D) : 1 : : 1: tan (45°+D), so that tan (45°+D)

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1

tan (45°-D)

Lastly, the secants are calculated from (Cor. 2. def. 9.) where it is shewn that the radius is a mean proportional between the cosine and 1 the secant of any arch, so that if A be any arch, sec A

cos A

The versed sines are found by subtracting the cosines from the radius.

5. The preceding Theorem is one of four, which, when arithmetically expressed, are frequently used in the application of trigonometry to the solution of problems.

1mo, If in the last Theorem, the arch AC=A, the arch BC = B, and the radius EC=1, then AD=A+B, and AB = A—B; and by what has just been demonstrated,

1: cos B: sin A : ¦ sin (A+B)+1 sin (A−B),

and therefore

sin Axcos B¦ sin (A+B) + 1⁄2 sin (A—B).

2do, Because BF, IK, DH are parallel, the straight lines BD and FH are cut proportionally, and therefore, FH, the difference of the

straight lines FE and HE, is bisected in K; and therefore, as was shewn in the last Theorem, KE is half the sum of FE and HE, that is, of the cosines of the arches AB and AD. But because of the similar triangles EGC, EKI, EC : EI :: GE : EK; now, GE is the cosine of AC, therefore,

R: cos BC cos AC: cos AD + cos AB, or 1: cos B: cos A : ¦ cos (A+B) + ÷ cos (A−B); and therefore,

cos AXcos B= cos (A+B) + cos (A—B) ;

3tio, Again, the triangles IDM, CEG are equiangular, for the angles KIM, EID are equal, being each of them right angles, and therefore, taking away the angle EIM, the angle DIM is equal to the angle EIK, that is to the angle ECG; and the angles DMI, CGE are also equal, being both right angles, and therefore, the triangles IDM, CGE have the sides about their equal angles proportionals, and consequently, EC: CG :: DI: IM; now, IM is half the difference of the Cosines FE and EH, therefore,'

R sin AC sin BC: or 1 : sin A :: sin B::

cos AB- cos AD,
cos (A-B) — cos (A+B);
and also,

sin AXsin B= cos (A-B)—1 cos (A+B).

4to, Lastly, in the same triangles ECG, DIM, EC: EG: ID: DM; now, DM is half the difference of the sines DH and BF, there

fore,

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cos AXsin B= sin (A+B)- sin (A-B).

6. If therefore A and B be any two arches whatsoever, the radius being supposed 1;

I. sin AXcos B≈1 sin (A+B)+1 sin (A−B).

II. cos AXcos B=

cos (A-B)+ cos (A+B).

III. sin AXsin B=
IV. cos Axsin B=

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From these four Theorems are also deduced other four.

For adding the first and fourth together, sin AXcos B+cos AXsin B-sin (A+B).

Also, by taking the fourth from the first, sin AXcos B-cos AXsin B=sin (A-B.). Again, adding the second and third, cos AXcos B+sin AXsin B=cos (A-B.); And, lastly, subtracting the third from the second, cos AXcos B-sin AXsin B=cos (A+B).

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