20. In like manner dividing one by another, some of the formula of No. 18, we find 1+sin Q_ sin2 (45° + Q)_sin (45°+ Q) sin2 (45°-Q cos2 (45°+Q) 1-sin Q = tang (45°+Q) =cot2 P 1-cos P sin P 1+sin P sin2 (45°+Q) 1+cos P cos P 1-sin Q_ coversin Q_sin (45°-Q) 1-cos Q versin QsinQ 21. Resuming the values of sin (A+B), sin (A-B, cos (A+B), cos(A-B), we deduce from them tang(A+B) tang A+ tang B cot A cot B-1 cot A+cot B tang A-tang B = Therefore cot (A-B)_1+tang A tang B_cot B cot A +1 tang A-tang B 22. Let A-45°, and we shall have 1+tang B_cot B+1 cot B-cot A tang (45°+B)= ; tang (45°-B)= 1-tang B cot B-1 cot B-1 =cot (45°+B)= 1+tang B cot B+1 If we make A=B= C we shall have 2 tang A cot C= cot C-tang C and cot C=2 cot C+tang C. - 24. Let A-B, and we shall have cosec 2 A cosec2 A 1+cot2 A cot A+tang A But cot A2 cot A + therefore cosec A=cot A+tang A =cot A-cot A, by writing for tang A its value, cot A-2 cot A. We have also sec 2 A= 1+tang2 A sec2 A 1-tang2 A 1-tang2 A (1+tang A) 2 tang A 1+tang A_2 tang A 1-tang A 1--tang A1-tang A 1-tang A tang (45°+A,) and 2 tang A =tang 2 A But 1+tang A 1-tang A Therefore sec 2 A=tang (45°+A)-tang 2A; and zec A=tang (458+A)-tang A-cot (45°- A)-tang A Since sec A 1 cos A and cosec A- ; we have sec A 1 sin A tang A cosec A; and substituting all the values of cosec A found tang A above, we shall have sec A- (cot A+tang A)=tang A (cot A+tang A)=1+tang A tang cot A)=tang A cot A-1-tang A tang A A-tang A (cot 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.) CALCULATION OF THE TABLES OF SINES, BY SERIES. 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 duce tang (A+B+C)_tang (A+B)+tang C 1-tang C tang (A+B) 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_abe 1-ab-ac-be Similarly, if a, b, c, d, are the respective tangents of the arcs A, B, C, D, we shall have tang (A+B+C+D)=a+b+c+d-abc_abd-acd_bod I-ab-ac-ad-bc-bd-cd+abcd Whence in general if there be any number of arcs A, B, C, D, &c.; then callings the sum of their tangents, sh the products of them two by two, sit their products three by three, we shall have tang (A+B+C+D+&c.) = s-ss-s + &c. 1-ss-s + &c. Suppose for a moment that the arcs A, B, C, &c. are all equal, then if we call the number of them 1, and tang A the tangent of any one of them, we shall have (Page 111) sn.n-1 tang A, 2 We have therefore in general Tang n A tang A, ... 11. n--1. n--2tang3 A+ n.n-in--2. n--3. n--tang A-&c. 2. 3. 4. 5 n tang A 2.3 2.3.4. sin A " the above fraction becomes cos A or, multiplying both numerator and denominator by Cos" A, we Cos A cosn- sin2 A+ n. n-1.n-2. n-3 cosm-4 sin A,&c. 2.3.4 Let N be the numerator of this last quantity, and D its denominator; then by actually performing the calculation we shall find N+D2 cos2n A+n cos2n-2 A sin2 A+ ...... N sin2n A-(cos2 A+sin2 A)"=1 n. n-1 cos2n-4 A sin+ A + But since on the one hand N2+D2=1, and on the other D-cos n A. 2 ; it is clear that N-sinn A, and that D=tangan A 2 sin'n A 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 A-A, because an infinitely small arc does not sensibly differ from its sine; a 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. a Lastly, A -- These values being substituted in the 'pre 26. Let now the arc a be any part of 90°; since the arc of 1 m 90%=1.570796326794896...&c. we shall have, calling c this The value of any cosine will in like manner be found by the following formula: COS 90°1 m C2 C4 -6+&c. 4 2 m 2.3.4 m2 2.3.4.5.6 m which by substituting the values of c gives 1 19 &c. |