tang 2 A=2 tang A cot 2 A-l-tang? A -COS C we have sin (A-B) cot B-cot A tang A-tang B - tang (A-B) = COS (A-B) 1+tang A tang Bcot B cot A +1 Therefore cot (A—B)=1+teng A tang B__cot B cot A +1 tang A--tang B cot B-cot A 22. Let A=45, and we shall have 1+tang B_cot B+1 tang (45° +B)= ; B-cot B-1 cot B-1 cot B+1 If we make A=B=C we shall have ; or tang C 2 tang 5 C 1-tang A 3C And 2 tang A={cot A–tang A, therefore cot C=cot C-1 tang $C and cot įC=2 cot C+tang C. 1-cos C But (12) tang C=V (1-сos C)? _1 1 +cos CV 1-cos2 C sin c 23. Since sec A= 1 1 sin A’ 1 I sec (A+B)= cos A cos B cos A cos B-sin A sin Bsin A sin B cos A cos B sec A sec B cosec A cosec B 1-tang A tang B-cot A cot B-1 2 sec A sec B sec (A-B= 1 +tang A tang B cosec A cosec B cosec (A+B)= cot B+cot A cosec A cosec B cosec (A-B)= cot B-cot A 24. Let A=B, and we shall have cosec? A 1+cot? A cosec 2 A cot A +tang A 2 Therefore cosec A= A 2 tang A by (22) therefore cosec A=cot A+tang ; } A cot A+tang A =cot -cot A, by writing for tang A its 2 value, cot } A_2 cot A. We have also sec 2 A= sec? A 1+tang? A 1--tang? A 1---tang? A (1+tang A) 2 tang A 1 +tang A_2 tang A 1-tang A 1.-tang* A 1-tang Al-tang' A ; cot 1 A+tang ! A. But cot jA=2 coť A+ cos A tang A But 2 tang A =tang 2 A 1-tang A 1-tang' A Therefore sec 2 A=tang (45° +A)—tang 2 A; and tec A=tang (458 +! A)—tang Acot (45°- A)—tang A Since sec AS 1 1 and cosec AS ; we have sec A sin A tang A cosec A; and substituting all the values of cosec A found above, we shall have sec AS (cot į A+tang 1 A)=tang A (cot A+tang A)=1+tang A tang A=tang A (cot Acot A)=tang A cot, A-1tang A -1 These formulas may be varied in an infinity of ways by adding, subtracting, dividing them, &c. But it is useless to dwell longer upon so easy a matter. See the Introduction to the Analysis of Infinities, by EULER.) tang ! A CALCULATION OF THE TABLES OF SINES, BY SERIES. duce tang (A+B+C)=1_tang C tang (A+B) The same thing has occurred with respect to 'Tables of Sines as had before taken place with the Tables of Logarithms. The first calculators had already completed their labours, when means were found to simplify them. These means are not the less ingenious on this account, as we may judge from the method proposed by John Bernoulli, in the second volume of his works. We shall give the analysis of it. 25. If we turn back to the values of tang (A+B) we shall de tang (A+B) +tang C 1 tang Let then a, b, c be the respective tangents of the arcs A, B, C; and we shall have (by writing for tang (A+B) its value) tang (A+B+C)= a+b+c-abc 1-ab-ac_bc Similarly, if a, b, c, d, are the respective tangents of the arcs A, B, C, D, we shall have a+b+c+d-abc-abd-acd_bod tang (A+B+C+D)= 1-ab-ac-ad-bc-bd-cd + abcd Whence in general if there be any number of arcs A, B, C, D, &c. ; then calling s the sum of their tangents, s'i the products of them two by two, sili their products three by three, we shall have tang (A+B+C+D+&c.) = "+s'-ght 1-8" + sind '-5" + &c. Suppose a moment that the arcs A, B, C, &c. are all equal, then if we call the number of them ri, and tang A the tangent of --g" + &c. for n. n + &c. + &c. , s4= 2 Su n. 0-1. n--2 n. 1-1. n-2. n-3 tang: A tang4 A, 2. 3 2. 3. 4 &c. We have therefore in general Tang nA= n tang An. n-1. n-3 n. n-in--2. n--3. n--4 tang: A+ tang' A-&c. 2. 3 2. 3. 4. 5 1n. n-1 -1.n-2. n-3 tang? A+ tang4 A-&c. 2 2. 3. 4. sin A or, writing for tang A its value the above fraction becomes n Sin A n. n--1. n-2 sin' A 2 3 cos) A n. n-1 sin? A 1 2 cos? A or, multiplying both numerator and denominator by Cos" A, we have finally Tang nA= a cos- A sin A n. n-1. n-2 cos-o sin A + &c. 2. 3 n. n1 cosa Acos- sinA+ n. n-1.0-2. n-3 costsin A,&c. 2 2. 3. 4 Let N be the numerator of this last quantity, and D its denomi. nator; then by actually performing the calculation we shall find N+D’=cos?" A +n cos?n? A sin? A+ n. n-1 cog2n-* A sinA+ 2 ... sinn A=(cos? A +sin? A)"=1 But since on the one hand N+D=1, and on the other N sin n A Da =tangʻ n A= ; it is clear that N=sin n A, and that cos? n A D=cos n A. We have therefore in general sin a A=n cos no-1 A sin A n. 1-1, 1-2 cos3 A sin' A+ 2. 3 n, 0-1. n-2. n-3. n-4 cos" -- A sin A-&c. and 2. 3. 4. 5 n. 0-1 cos n A=cos" A cos?-? A sin? A+ n. n-1. n-2.0-3 2 2. 3. 4 cog* A sin 4 A-&c. Suppose now that the arc A is infinitely small, so that n must be infinite, in order that the arc n A may be of a finite magnitude a, we shall have 1st. Sin ASA, because an infinitely small arc does not sensibly differ from its sine ; a A а n sin a= + as а — 2 а at 26 + + +& sin a 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 'pre ceding formulas give a? as a? &c. a“ + & a? + t&c. sin a 2. 3 2. 3. 4. 5 2. 3. 4. 5. 6. 7 =tang a = cos a 1: + &c. 2 2. 3. 4 2. 3. 4. 5. 6 ao 2 2. 3. 4" 2. 3. 4. 5. 6 a? +&c. 2.3 2. 3. 4. 5 2. 3. 4. 5. 6.7 1 26. Let now the arc a be any part of 90°; since the arc of 900-1.570796326794896...&c. we shall have,' calling c this number cs c? +&c. = 2. 3. m3 2. 3. 4. 5 m 2. 3. 4. 5. 6. 7 m? Positive Terms. • Negative Terms. : -1.570796326794896 -0.645964097506246 m' 1 0.079092626246167 -0.004681754185318 ms m? 1 0.000160441164787 -0.000003598848235 mo m 1 0.000000056921729 0.000000000668303 M"3 0.000000000006066 0.000000000000043 &c. &c. The value of any cosine will in like manner be found by the following formula: c? co 2. 3. 4 m4 2. 3. 4. 5. 6 m which by substituting the values of c gives m 1 14 m m m By m m + mis 0,000000000065659 0,000000006386603 2 + 0.000000000000003 0.000000000000529 &c. &c. 27. It is the same with regard to tang 90° 90° and cot means of these sines we might calculate the sines and tangents of my arcs; we have only to substitute the proper values of m. For example, to calculate the sine of the arc of 30°, we must make m=3 and we shall have Sin 30o= 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. all the rest; the fraction will always be less than.. So thai 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=1; 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 m |