that ratio to a greater sphere LMN: Therefore, by inversion, the sphere LMN has to the sphere ABC, the triplicate ratio of that which the diameter EF has to the diameter BC. But as the sphere LMN to ABC so is the sphere DEF to some sphere, which must be less (14. 5.) than the sphere ABC, because the sphere LMN is greater than the sphere DEF: There fore the sphere DEF has to a sphere less than ABC the triplicate ratio of that which EF has to BC; which was shown to be impossible: Therefore the sphere ABC has not to any sphere greater than DEF the triplicate ratio of that which BC has to ÈF: And it was demonstrated, that neither has it that ratio to any sphere less than DEF. Therefore the sphere ABC has to the sphere DEF, the triplicate ratio of that which BC has to EF. Q. E.D. The uses of Plane Geometry are too numerous to be inserted in this place; we may, however, mention a few. Every branch of mathematics which regards lines, surfaces, and solids, are entirely dependent on its principles. Mensuration, and the whole of that curious and entertaining branch of science called Trigonometry, is nothing but the application of Geometry; the Conic Sections cannot be established without a knowledge of its ele ments. It is indispensible in navigation, geography, astronomy, projection, projection of the sphere, perspective, dialing, &c. There is no mechanical profession that does not derive considerable advantage from it; and even one workman in the line of his profession, is as much superior to another as he understands more of geometry. By its means the architect lays down his plans, and erects his edifice, and the engineer directs canals to be cut, and bridges to be erected over large rivers. In short it is impossible to view the landscape before us with advantage, or to give a just account of things which we see, while we are ignorant of Geometry. TRIGONOMETRY is that part of mathematics which teaches us to investigate the relations that obtain between the sides and angles of triangles. It is usually divided into two branches. PLANE TRIGONOMETRY, which determines the relation between the parts of plane triangles, or triangles formed by right lines only; and SPHERICAL TRIGONOMETRY, which concerns triangles formed by arcs of circles. Both are of very great importance; for on one or either of them depend the theories of surveying, astronomy, navigation, dialling, and many other branches of mixed mathematics. We shall commence with plane Trigonometry, and as a preliminary step, explain the Theory of Sines. THEORY AND ARITHMETIC OF SINES. 1. The sine of an arc or of an angle, is a line which, proceeding from one of the extremities of the arc or angle, falls perpendicularly on the radius or diameter, passing through the other extremity. Thus the perpendicular BD, drawn from the extremity B, of the arc BA, upon the radius CA, which M passes through the other extremity, is called the sine of the arc AB, or of the angle ACB measured by that arc. 2. If the arc ÈB= 90o-arc BA, or in other words, if the 'arc EB is the complement of the arc BA, its sine GB is the sine of the complement, or the cosine of the arc AB. It is evident that CD= BG, and that BD=GC. 3. The perpendicular AT, drawn from the extremity A, of the radius CA, till it meets with the radius CB prolonged, is called the tangent of the arc AB, and CT is the secant of that arc. Similarly the tangent EM of the arc EB is the tangent of the complement or the cotangent of the arc AB; and CM is the cosecant. The lines AD, EG are called the versed sine and coversed sine: they are seldom used. By way of abbreviation we shall write sin, cos, tang, col, sec, cosec, versin, coversin, instead of sine, cosine, tangent, cotangent, secant, cosecant, versed sine, and co-versed sine. 4. From the definition of the sine, it follows that sine of any arc is the half of the chord of twice that arc. For if we prolong BD to F, BD will be the half of BF, the chord of the double arc BA. Since the sine of an arc is half the chord subtended by twice that arc, it evidently follows that the sine of an arc of 30° is half the radius. 5. The definition of the sine also shews that the sine of any acute angle BCA is the same as that of the obtuse angle a CB=180° -angle BCA, which is called its supplement. 6. The cosine of an obtuse angle is always negative. The same is also the case with its tangent cotangent, secant, &c. 7. We easily perceive that the sines increase from 0°, where the cosine is equal to radius till 90°. The sine then equals radius, and the cosine vanishes. The sines then decrease from 90° to 180°, where they become nothing, while at that point the cosine equals the radius taken negatively. 8. We may also observe that the tangents and secants increase from 0o, where the tangent is zero, and the secant equal to radius, till 90o, at which point the tangent and secant become equal, parallel, and consequently infinite. The cotangent is then zero, and the consecant equal to the radius. From 90° to 180° they continually decrease; but are negative. When they arrive at 180o, the tangent vanishes, and the secant is equal to the radius taken negatively. At this same point the cotangent and cosecant become equal, and negatively infinite. 9. The tangent of 45° is equal to radius, as also the cotangent. For then the right-angled triangles CEM, CTA became equal and isosceles. 10. These things premised, let the arc BA=A; (preceding figure) the radius CB-CA-1 (which supposition we shall make in future, in order to render the calculations more simple), because of the right angled and similar triangles CBD, CGB, CTA, CEM, we shall have the following proportions and equations. I. CD2+BD2=CB2; or sin2 A+ cos2 A=1=sin' B+cos2 B (B being any other arc). Therefore Sin A+ sin B: cos B+cos A:: cos B-cos A: sin A-sin B. III. CE2-CM2—EM2. tang 2A: IV. CD: BD:: CA: AT......or cos A: sin A:: Therefore sin A=cos Axtang A............or cos A= tang Asin A cos A V. CB: CD :: CT: AT....or 1: sin A:: sec A: tang A. 1 Cos A VI: CG: GB:: CE: EM....sin A: cos:: 1: cot A. sin A Therefore cot A= cos A 1 = and cot Ax tang 11. Let it now be required to determine the sine and cosine of the sum of two given arcs AB and BE. Calls the sine BD of the arc AB; c its cosine; the sine EG of the arc EB; c its cosine; and lastly, the cosine required-CF= cos (AB+BE), and y the sine EF-sine (AB+BE). This being premised, because of the similar triangles CBD, CFH, EGH, we shall have B H F DA CD (c) : CF (x) : : CB (1): CH=:: BD (s): FH== Again, C CD (c) : EG (s') : : BD (s) : GH (c'—): : CB (1): EH(y--**) C C Therefore s's-cc-x and s'-cy-sx. The first equation gives x=cc-ss', and substituting this value in the second, we have (observing that 1-s-c2) y=sc'+s'c. In general therefore, A and B being any two arcs, we have sin (A+B)=sin A cos B+sin B cos A cos (A+B)=cos A cos B-sin A sin B 12. Let (A+B)=C, and by substitution we shall have sin C sin A cos (C-A)+cos A sin (C-A) cos C=cos A cos (C-A)-sin A sin (C-A) And treating sin (C-A) and cos (C-A) as unknown quantities, we shall find Sin (C-A)-sin C cos A-sin A cos C, and cos (C-A)=cos C cos A+ sin A sin B. Therefore in general, Sin (A-B)=sin A cos B-sin B cos A Cos (A-B)=cos A cos B+sin A sin B. Making A=B, we have sin 2 A=2 sin A cos A; and cos 2 A =cos A-sin A=2 cos2 A-1 (by writing for sin A, its value 1-cos2 A). It is therefore easy to obtain the sine and cosine of twice any arc of which we already know the sine and cosine. With the same facility we may find the sine and cosine of the half of any arc. For if we make 2 A=C, we shall have sin C= 2 sin C cos C, and cos C+1=2 cos2 C. Therefore cos C= But as these formulas suppose that the gines and cosines are already known, before we proceed farther we must learn to calculate them. 13. And first, it is clear that if we calculate the sines of all the arcs comprized in the quadrant or quarter of a circle, from 1′′ v. to the arc of 90°, we shall be in possession of all the sines from 1" up to the arc of 180°. (See 5). Now from 180° to 360° the since are the same, as from 0 to 180, excepting the sign which is negative. Therefore the computation of the sines is reduced to those sines contained within the quadrant. Again it is clear that the cosines may easily be determined from the formula, Cos A(i-sin A). We shall therefore apply ourselves only to the calculation of the sines. It will hereafter be shewn that if the radius be 1, the arc of 90o will be represented by 1.570796326794896, &c. ; therefore the arc of 1" is 0.000004848, &c. parts of the radius; and as so small an arc does not sensibly differ from its sine, we may assume 0.000004848, &c. for the sine of the arc of 1". Double, triple, &c. this decimal fraction, and we shall have the sine of 2′′, 3", &c. We might have calculated the sines 2", and then that of 3", &c. by the formulas sin 2 A-2 sin A and sin (A+B)=sin A cos B+sin B cos A; but it was fou. the difference between such small arcs and their sines was so inconsiderable, that any of these arcs might be taken instead of their respective sines. ୯ Ascending in this manner from seconds to minutes, and continuing the calculation from minutes to degrees, by means of the two preceding formula, we arrive at the sine of 30°. As this sine is known to be equal to half the radius, it serves to verify all the anterior calculations, and we shall then have all the sines from one second up to that of 30°. But not to swell the size of the Tables beyond all bounds, it is usual to insert in them only the sines of minutes and degrees. 14. Suppose now that A=30°, we shall have Sin (30+B)=sin 30° cos B+ cos 30° sin B. But Sin 30°, and cos 30=√1-3. Therefore Sin (30o +B)=sin B √ 3+ cos B. And Sin (30°-B)= cos B-sin B√3. Therefore Sin (30°+B)—sin (30°—B)+sin B. √ 3. Hence it follows that if we know all the sines from 0° to 30o, we easily obtain all those between 30, and 60o. 15. This being premised, let A=60°, and we shall have sin (60°+B)= cos B 3+ sin B, and sin (60°-B)= cos B √3 -sin B. Therefore Sin (60°+B)=sin (60° —B)+sin B. For example, sin 662-sin 54+ sin 6o-If therefore we know the sines of all the arcs between 30o and 60o, we shall immediately obtain all those between 60° and 90o, which will complete this sort of calculation. 16 Resume the two formulas sin (A+B)=sin A cos B+sin B cos A, and sin (A—B)—sin A cos B-sin B cos A, and add them together, and we shall have sin (A+B) + sin (A-B). Subtract the second from the first, and we shall find Sin A cos B |