a n sin a=2 + + а ein a 2 cos a а + + &c. a 1 +&C lo 2dly. cos A=1; because the cosine of an infinitely small arc is equal to radius. 3dly. n-1=n=n-2-n-3, &c. because n is infinite: Lastly, A= These values being substituted in the 'preceding formulas give a as a? + + &c. 2. 3 2.3.4.5 2. 3. 4. 5. 6.7. . 2. 3. 4. 5. 6. 7. 8. 9. a? a4 a6 as cola=1 &c a? t&c. 2. 3 2. 3. 4. 5 2. 3. 4. 5. 6. 7 ==tang a= at a ao + cos a 2 2. 3. 4" 2. 3. 4. 5. 6 a? 1 26. Let now the arc a be any part of 90° ; since the arc of 90° -1.570796326794896...&c. we shall have,' calling c this number 90° cs c? sin +&c. = 2. 3. m3 ' 2. 3. 4. 5 ms 2. 3. 4. 5. 6. 7 m Positive Terms. • Negative Terms. 1: 1.570796326794896 1 -0.645964097506 46 m; m 1 1 0.079092626246167 -0.004681754185318 m 1 -0.000160441164787 -0.000003598843235 mao m 1 1 7,0.000000056921729 0.000000000668303 a + +&c. mis 1 mi10.000000000006066 0.000000000000043 &c. &c. The value of any cosine will in like manner be found by the following formula: cos 90°;=:1– c? co 2. 3. 4 m4 2. 3. 4. 5. 6 m which by substituting the values of c gives c4 +&c. 2 6 m + m m m m of 0.000000000000003 0.000000000000529 &c. &c. 90° 90° 27. It is the same with regard to tang and cot Ву means of these sines we might calculate the sines and tangents my arcs; we have only to substitute the proper For values of m. example, to calculate the sine of the arc of 30°, we must make m=3 and we shall have Sin 3095 0.523598775598299 +0.000327953194428 -0.023926736944019 Therefore sin 30°=0,5, as we already know (4). And since it is sufficient to calculate the sines up to 30° to obtai. 1 all the rest ; the fraction will always be less than , the sines equal to sin 90° will be very convergent. For example, do we require the sine of 9°? Make m=10, and we find immediately sin 9°=0.156434465040231. If it were required to find the sine of an arc of a certain number of degrees with minutes, seconds, &c. it might evidently be done by the same method. 28. The sines which we obtain by this calculation belong to a circle whose radius is=l; consequently to obtain those which belong to a circle whose radius is a, we must multiply the first by a. In the common tables, the radius is supposed to be= 10,000000000, and to facilitate the calculations, there are inserted the logarithms of the sines, cosines, tangents, and cotangents of all so thai m m COS A cos A’ arcs from l' to 90° ; excepting those arcs which contain seconds, because the sines, cosines, &c. of these arcs are easily found, as is explained in the tables themselves. 29. Particular tables of secants and cosecants have not been made. They are seldom used, and besides it is easy to calculate 1 them by means of the formulas sec A-- and cosec A= R? ,which for the radius R of the tables, becomes sec A= sin A R2 and cosec A== ; whence we deduce log sec A=2 log R—log cos sin A A–20.0000000—log cos A; and log cosec A=20.0000000—log sin A 30. After having resolved this problem generally, given any arc to find ils sine, cosine, tangent, &c. it remains for us to give the solution of the inverse problem, viz. given the sine, cosine, tangen', or cotangent of an arc to find the length of that arc. If the cosine or cotangent is given, we can immediately obtain the sine and tangent. Therefore the problem may be reduced to the finding the length of an arc whose sine or tangent is given. 1. If we return to the value of sin A, we easily deduce by the inverse method of series that sin' A , 3 sin A 3. 5. sin? A, 3. 5. 7 sino A A=sin A+ + &c. 2. 3 2. 4. 5 2. 4. 6. 7 2. 4. 6. 8. 9 II. If we represent the tangent of the arc A by t, we shall have (25) a3 a5 at &c. 2 2. 3. 4 or multiplying by the denominator and arranging the terms according to the powers of a, at a3 at t t=8+ + &c. Let a-At+Bt+Ct' + &c. and 2 2. 3 2. 3. 4 by substituting this value of a, and determining the unknown coefficient, A, B, C, &c. we shall find that A=1, B=-3, C=}, &c. tangs A tang' A +&c. 5 7. 31. These two series give the solution of the proposed problem. Let us now apply them to the determination of the ratio of the diameter to the circumference. we shall have the length of the arc of + + a + +&c. 3 If we make sin A== + Let B = 1 and we shall have tang A= Hence the sum of This quantity, multiplied by 6, would give the semi-circumference, and consequently the ratio required. But as this series, though very convergent, is still tedious to calculate, it is better to employ the second series, which supposing the arc A to be 458. gives 1 1 1 1 arc A=1 + + &c. 3 5 7 9 11 And as this progression is still too slow for our purpose, a more expeditious method has been invented for discovering the length of the arc of 45'. 32. This method consists in decomposing the arc of 45° into two other arcs which we shall call A and B, and in determining separately their lengths. On this supposition tang (A+B)=1= tang A+tang B Therefore tsog A= 1-tang B 1-tang A tang B A 1+tang B 1 3 2 the two ares A and B, or the quarter of the semi-circumferwce * will be 1 1 1 1 1 + &c. 1 1 1 1 1 + -&c. 3 3. 33 5. 35 7. 37 The sum of these two series will be found =0.7853981633974483...&c. whence r=3.1415916535897932 ... &c. the ratio of the diameter to the circumference, 33. Before we conclude this subject we shall remark that the formula already found for the values of tang n A, sine n A, &c. serve to find the sines, cosines, tangents, and cotangents of multiple arcs. For making sin A-s, cos A=c, tang A=t, we shall have sin A= cos A=0 sin 2 A=2 sc cos 2 A=csin 3 A=3 sc_83 cos 3 A=-3 e se sin 4 A=4 sc_4c 33 cos 4 A=cf_6c 92 +s4 sin 5 A=5 sc4 _10 s c+s4 cos 5 A=(-10 : goo +5 C 84 &e. &c. 9 9. 29 } 9. 39 and 2 t tang 2 A = 1-tt 3t-t3 tang 3 A = 1-3 t? tang 4 A 41-4 t 1-6t+t tang 5 A = 5t-10t+ts 1-t? cot? A= 2 t 1-3 ta cot 3 AM 3 t-ts 1-6t+ cot 4 AS -4 t-4t3 1-10 t? +5 to cot 5 A 51-10 t +ts 3t for 2 parts for 3 parts for 4 parts 3 34. By means of the same formulas we may obtain equations which serve to divide an arc or an angle into a given number of equal parts. For then sin (n A) is known, and sin A required. Let sin (n A)=b, sin A=x, cos A=z, and we shall have to solve this equation n. n-. n-2. n. n-1. n-2. n-3. n–422-5 X!_&c. benz"-'x Z"-" x + 2. 3 2. 3. 4. 5 Whence making n successively 2, 3, 4, 5, &c. the following equations will serve to divide an arc into the corresponding numbe of equal parts. b=2 2 X=2 XVI–XX for 5 parts 35. By way of applying these principles, we shall give a method of solving by approximation any equation of the third degree falling under the irreducible case. From what we have seen, if A is an arc whose sine is b, we shall 1 3 have sin 1, or x by solving this equation x Xt-b=0 3 And when the radius of the circle instead of unity is r, we shali have the equation 3 r? 4 4 We may now observe that the arcs 1809-A, and—1809 +A, have the same sine as the arc A, so that to divide them into three 3 equal parts, we have the same equation xxx + 4bra , x3- -rx + 4 to solve. Hence it follows that the three roots of this equation are 180° -A sin. A, sin (180 and-sin 180°+A); or sin A, sin 3 3 (60°— A), and—sin (60°+ĻA). 1 -A. 3 This being premised, let the proposed equation be x'-px +q=0, (if it were x-px+q=0, it could easily be reduced to the foregoing form by making x3-y); comparing this equation with the 3 equation x3 +1 brzo, we have r=p, & br? 4 =9; whence r=2 and b- =39. Therefore if we describe р a circle whose radius=2v p, the circular arc whose sine is 89 being called A, we shall have sin. A, sin (60° -A -sin р brzo ( 3 r? xt 3 1 ---A), --ஸ் 3 |