3. By multiplication, sin. (A + B) sin. (A — B) = sin.2 A cos.2 B-sin.2 B cos.2 A cos. (A+B) cos. (AB)= cos. A cos.2 B-sin. A sin.2 B. Or eliminating cos.2 A, cos.2 B, from each of these equations by means of the conditions sin.2 A+ cos.2 A = 1 sin.2B+cos.2 B=1; the second members of them become, respectively, sin.2 Asin. A sin.2 B- sin.2 B + sin.2 B sin.2 A, or sin.2 A-sin.2 B; and, I-sin. B-sin.2 A + sin. A sin.2B-sin. A sin.2B, or cos.2 B - sin.2 A; so that sin.(A+B)sin.(A-B)=sin.2A—sin.2B=(sin.A+sin.B)(sin.A―sin.B) cos. (A+B)cos. (A-B)=cos.2B-sin.2A=(cos. B+sin. A) (cos.B―sin. A) 4. By division, (5) The right-hand members of these equations will assume other useful forms by dividing both numerator and denominator of each by certain expressions: thus, let the divisors for the first equation be cos. A cos. B, sin. A sin. B, sin. A cos. B; those for the second, cos. A sin. B, sin. A cos. B, cos. A cos. B; and those for the third the same as those for the first; we shall then have If A = 45°, then tan. A = cot. A = 1, therefore, Such are the most useful theorems respecting the sums.and differences of two unequal arcs, and they may be converted into other expressions, involving three or more arcs by simply substituting B+C+D+ &c. for B. We shall briefly consider the case of three arcs, or angles, because of a curious property belonging to them whenever they make up either 180° or 90°. Let A, B, C, be any three arcs, and consider A + B as one, then by equa. (1) sin. (A+B+C) = sin. (A + B) cos. C + cos. (A + B) sin. C = = (sin. A cos. B + cos. A sin. B) cos. C + (cos. A cos. B sin. A sin. B) sin. C cos. (A+B+ C) = cos. (A + B) cos. C — sin. (A + B) sin. C =(cos. A cos. B—sin. A sin. B) cos. C (sin. A cos. B + cos. A sin. B) sin. C. Let now the sum of the three arcs be 180°, or, indeed, any multiple of 180°, then the sine of this sum will be 0, so that the first of these equations gives sin. A cos. B cos. C + cos. A sin. B cos. C + cos. A cos. B sin. C=sin. A sin. B sin. C; a remarkable property of the angles of a plane triangle. Again, let the sum of the three arcs be 90°, or any multiple thereof, then the cosine of this sum will be 0, so that the second general equation above becomes cos. A cos. B cos. Cer A een Beos. C+ sin. A cos. B sin. C + cos. A sin. B sin. C; dividing both sides by sin. A sin. B sin. C, we have cot. A cot. B cot. C=cot. A + cot. B+ cot. C. (28.) To deduce formulas for multiple arcs we have only to put n A for A+B in the preceding expressions. We thus get from (1) sin. n Asin. A cos. (n - 1) A + sin. (n − 1) A cos. A cos. n A = cos. A cos. (n − 1) A — sin. A sin. (~ — 1 ) A ; so that putting for n, 1, 2, 3, &c. successively, we have We may put the general expressions for sin. nA, and cos. nA, under a different form, by making use of the second equation in (1) and (2), thus putting (n-1) A for A, and A for B, these become sin. (n-2) A= sin. (n − 1) A cos. A sin. A cos. (n − 1) A cos. (n-2) A= cos. (n - 1) A cos. A + sin. (n − 1) A sin. A; or, by transposing, 0 sin. A cos. (n − 1) A + sin. (n − 1) A cos. A — sin. (n − 2) A 0= cos. A cos. (n - 1) A + sin. A sin. (n—1) A—cos. (n − 2) A ; adding these two equations to those above, there results sin. n A2 sin. (n - 1) A cos. A — sin. (n—2) A (12); cos.n A2 cos. (n-1) A cos. A . — cos. (n — 2) A hence, (29.) The sines and cosines of multiple arcs may also be developed in terms of the powers of the sine and cosine of the simple arc, by help of a remarkable formula, known by the name of De Moivre's formula, which may be easily established, as follows. Multiply together the two expressions, (cos. A sin. A, + sin. A cos. A1) √—1; which, by the equations (1), (2), is the same as cos. (A + A1) + sin. (A + A1) √— 1; which is of the same form as the original factors, consequently, multiplying this by the new factor cos. A,+ sin. A, .√1, we must have for the product cos. (A+A, + A1) + sin. (A + A+ A2) √—I, and thus by continually introducing a new factor, we must have generally (cos. A + sin. A. √—1) (cos. A, + sin. A, . √ — 1) (cos. A2 + sin. A, . 1) &c. = cos. (A+A, +A, + &c.) + sin. (A + A, + A2+ &c.) √I Suppose now that A=A,A,&c. then this equation will become (cos. A + sin. A.√√=1)" = cos. n A + sin. n A . or, writing the radical with the double sign, (cos. A± sin. A. √=1)" = cos. n A ± sin. nA . √—1 |