Perform the same operations on the other two formulas cos A+B) cos A cos B-sin A sin B, and cos (A-B)=cos A cos B +sin A sin B, and we shall deduce Cos A cos B cos (A+B) + 1⁄2 cos (A-B) cos (A+B) — cos (A+B) These four last formulas are useful when we wish to transform products of sines into simple sines. The four following forms enable us to substitute for the sums or differences of sines, the product of other sines, in order that the logarithmic calculus may apply to them. P+Q 17. Let A+B=P, A—B=Q, we shall have A+, and B=P÷Q Therefore from the preceding article 2 Sin P+sin Sın P—sin O=2'sin PQ P+Q P-Q sin COS 2 2 2 Cos P+cos Q=2 cos P+G Cos Q-cos P⇒2 sin 18. Suppose that in the two first of the foregoing formula P= 90°, and in the two last that Q=0; and we shall have 1+sin Q-2 sin (45° + Q) cos (45°-Q)-2 sin2 (45° + 4Q) 1-sin Q-2 sin (45°— Q) cos (45° + Q)=2 sin2 (45°-Q) =2 cos2 (450+Q) coversed sine of Q 1+cos P-2 cos2 P P-versed sine of P. 19. Divide the formulas of No. 17, one by another, and we 20. In like manner dividing one by another, some of the formula of No. 18, we find 1+sin Q sin2 (45o + Q) __sin2 (45°+} Q). =tang2 (45° + } Q) 1-sin Q 1+ cos P 1-cos P sin2 (45°-Q ̄cos2 (45° + & Q) cos P cot P sinP versin Qsin & Q coversin Q_sin2 (45°—§ Q) 2 21. Resuming the values of sin (A+B), sin (A-B, cos (A+B), cos (A-B), we deduce from them sin (A-B)__sin A cos B-sin B cos A_cot B-cot A = cos (A+B) cos A cos B-sin A sin B 1-tang A tang B_cot A-tang B 1+tang A tang B cot A+tang B sin (A+B) cos (A+B) = tang (A+B)=; Therefore, cot (A+B)=; 1 = cot A+ cot B tang(A+B) tang A+tang B 2 But cotA 2 cot A+ A by (22) therefore cosec A=cot A+tang A cot A-cot A, by writing for tang A its value, cot A-2 cot A. tang cot 2 But 2 tang A 1+tang A tang (45°+A,) and. =tang 2 A 1-tang A 1-tang A Therefore sec 2 A=tang (45° +A)—tang 2 A; and rec Atang (458 +A)—tang A-cot (45°-A)-tang A and cosec A- ; we have sec A 1 sin A Since sec A cos A tang A cosec A; and substituting all the values of cosec A found above, we shall have sec Atang A (cot A+tang A)=tang A (cot A+tang A)=1+tang A tang A-tang A (cot Acot A)=tang A cot A-1-tang A tang 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 deduce 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) a+b+c-abc tang (A+B+C)=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)=1-ab-ac-ad-be-bd-cd+abcd a+b+c+d-abc-abd-acd-bod Whence in general if there be any number of arcs A, B, C, D, &c.; then calling s the sum of their tangents, s the products of them two by two, si their products three by three, we shall have s- +s- -s" + &c. tang (A+B+C+D+&c.) = 1 iii vii +s"-s" + &c. Suppose for a moment that the arcs A, B, C, &c. are all equal, then if we call the number of them, and tang A the tangent of n. n-1 any one of them, we shall have (Page 111) s" tang2 A, ... 2 or, multiplying both numerator and denominator by Cos" A, we have finally Tang n A= Let N be the numerator of this last quantity, and D its denomi nator; then by actually performing the calculation we shall find N2 sin2 A=(cos2 A+sin2 A)"=1 But since on the one hand N2+D2=1, and on the other Datang n A D-cos n A. sin' n A ; it is clear that N-sin n A, and that We have therefore in general sin a A-n cos na-1 A sin A n. n-1. n-2. n-3. n-4 cos"-5 A sin5 A-&c. and Suppose now that the arc A is infinitely small, so that ʼn must be infinite, in order that the arc n A may be of a finite magnitude a, we shall have 1st. Sin AA, because an infinitely small arc does not sensibly differ from its sine; a A |