R sin a sin (ab) cos b+cos (ab) sin b, R cos a=cos (ab) cos bsin (ab) sin b. And from these we find R sin (a-b)=sin a cos b—sin b cos a R cos (a-b)=cos a cos b+sin a sin b. XX. If, in the formulas of the preceding Article, we make b=a, the first and the third will give formulas which enable us to find the sine and cosine of the double arc, knowing the sine and cosine of the simple arc. This forms the problem of doubling an arc. Reciprocally, to divide a given arc a into two equal parts, let us, in the same formulas, put a instead of a; we shall have 2 sin la cosa cos a-sin2 a sin a= R COS a " R Now, since we have at once cos2 4a+sin2 =R2, and cos2 asin2 a R cos a, there results cos2 a=R2+R cos a, and sin2 1a—†R2—†R cos a; whence sin §a=√(†R2 —§R cos a) cos &a=√({R2+†R cos a). Thus, making a=100°, or cos a=0, we have sin 50°=cos 50°— √R2=R; next, making a=50°,which gives cos a=R√, we shall have sin 25°=R(√}—}√}), and cos 25°—R√({+}√}). * XXI. The values of sin a and cos da may also be obtained in terms of sin a; which will be useful on many occasions. These values are: 2 sin &a=√(R2+R sin a)—§√(R2—R sin a), 2 cos a=√(R2+R sin a)+}√(R2-R sin a). Accordingly, by squaring the first, we shall 2 have sin2 4a= † (R2 + R sin a) + 4 (R2 — R sin a) — § √ (R — R2 sin2 a)= R2-R cos a; in like manner, we should have cos2 4a=4R2+ R cos a; values which agree with those already found for sin a and cos a. It must be observed, however, that if cos a were negative, the radical (R-R sin a) would require to be taken with a contrary sign in the values of sin a and cos a; and thus the one value would be changed into the other. *XXII. By means of these formulas, it is easy to determine the sines and cosines of all the tenths of the quadrant.+ First, let sin 20°-x; 2 x will be the chord of 40°, or the side of the regular inscribed decagon: now this side (5. IV.) is equal to the greater segment of the radius divided in extreme and mean ratio; hence making the radius=1, we shall have 1:2::2x: 1-2x. Hence we obtain 4a-1-2 x, or x2+x=1; hence (x+4)2=1+1=; hence x+1=1√5 ; and finally x or sin. 20°=1(−1+√5). 5 This value raised to the second power gives sin2 20° 16 Now, if in the formulas of Art. 21. we make R=1, a=20°, and sin a=(-1+5), we shall find sin 10°=√(3+√5)—1√ (5—√5) cos 10°=√(3+√5)+‡√ (5—√5). Next, if in the same formulas, we make a=60° and sin a= (1+√5), we shall have sin 30°=√ (5+√5)—}√(8—1/5) cos 30°=1√ (5+√5)+\√(3−√5). With these values, and the values of sin 50° and sin 100°, which are already known, the following table may be formed: + Only according to the centesimal scale. It can be shewn, that except for the arc 3° and its multiples, no accurate expressions of the sines can be found, even under an incommensurable form; and, as this arc 3° on the centesimal scale corresponds to 3° on the nonagesimal, it follows that 30° and 60° are the only decimal arcs on the latter whose sines are capable of being accurately expressed, the sine of 90° being put 1. For the multiples of this arc 3o, however, by a process exactly similar to that followed in the text, making use of the nonagesimal division, we obtain the following values: sin 0° cos 100°-0 sin 10°=cos 90°=\√(3+√5)—1√ (5—√5) sin 30°=cos 70°=\√(5+√5)—\√ (3—√5) sin 50° cos 50°=√2 sin 60°-cos 40°=1(1+√5) sin 70°-cos 30°=\√(5+√5)+‡√(3—√5) sin 90°=cos 10°=\√(3+√5)+\√(5—√5) cos 0°-1. These values may farther be simplified, since we have √(8+√5)={√10+1√2, and √(3—√5)=√10—{√2; from which it appears that considering 2, 5 and 10 as known quantities, there remain only four extractions of square-roots, till we arrive at the values of the sines and cosines belonging to all the arcs which are multiples of 10°. * XXIII. From these formulas we may draw two remarkable consequences: square= 10-2/5 First, Since 2 sin 40° is the chord of 80°, or the side of the regular inscribed pentagon, this side = √(10—2/5), its The side of the regular decagon (-1+√5), its square= (6-2/5. Now (10—2,√5)=1+1(6-2/5); hence the sum formed by the square = 2 sin 20° sin 54° cos 36°= sin 57° = cos 33°: 1 (√9—1) + √3 +1 √(5+ √5). √3-1 sin 87° cos 3°= (√5-1)+ √3 + 8/2 sin 90° = cos 0° 1. 8 These values are from Cagnoli's Trigonométrie.-Ed. √(5 + √5). of the radius and the square of the decagon's side is equal to the square of the inscribed regular pentagon's side. Secondly, Among the sines of the uneven decimal divisions of the quadrant, we have this relation: sin 90°+sin 30°+sin 10°-sin 50°+sin 70°, and the even divisions in like manner give sin 60°-sin 20°+4. These formulas, however, are only particular cases: we might shew that a being an arc measuring any number of degrees, we should have sin(100°-x)+sin(20°+x)+sin(20°-x)=sin(60—x)+sin(60°+x). For, the formula sin (a+b)+sin(a—b)=2 sin a cos b, gives sin (20°+x)+sin (20°—x)=2 sin 20° cos x sin (60°+x)+sin (60°—x)=2 sin 60° cos x. Therefore, since we have sin 60°-sin 20°, and cos x= sin (100-x), those two equations, by subtraction, will give sin (60°+x)+sin(60°-x)—sin(20+x)—sin(20°-x)=sin(100—x). A formula from which may be derived the equation of the uneven divisions by making x-10°, and which may serve in general for verifying the tables of sines. XXIV. If in the first and third formulas of Art. 19, we put b=2 a, we shall have sin2acosa+cos 2a sina sin 3a= cos Sa R cos 2a cosa-sin 2asina R Instead of sin 2 a and cos 2 a, substitute in these expressions the values found in Art. 20, simplifying the results by means of the equation sin a+cos a=R2; we shall have These formulas, which serve for the triplication of arcs, may likewise serve for their trisection or division into three equal parts. Thus, if we put sin 3 ac, and sin a=x, for determining this a we shall have the equation c R2-3 R2x-4x3. From which it appears, that the problem of trisecting an angle, when considered analytically, is of the third degree. If in the same formulas of Art. 19, we successively put b=3a, b=4 a, &c. we may obtain the sines and cosines of the arcs 4 a, 5 a, &c. that is to say, in general the sines and cosines of all the multiples of a. And conversely, the formulas, which serve for the multiplication of arcs, will give us equations to re |