602. To find the cosine and sine of the sum of any number of angles in terms of the cosines and sines of the angles. = Multiply out the product on the left-hand side of the above equation : and equate the real and imaginary parts. Then; observing that q2 =-1; 23 =-; ; =+1; 38=i; and so on; ; we have cos (a + b + ) = cos a cos ß...cos & – Esin a sin ß cos y. cos Š + sin a sin ß sin y sin 8 cos e...cos $ -... n sin (a +B+...+É) = sin acosB...cos-sina sin ßsin ycosd...cos + ! a + ... where the symbol & denotes the sum of all terms of the form obtained by interchanging the letters a, b, y... & in the product to which it is attached.' This conclusion is the same as that obtained in Art. 350. The two methods of proof should be carefully compared. De Moivre's Theorem. 603. As in Art. 601 we have, for all real values m and ng (cos mo + i sin mo) (cos no + i sin në) = cos (m+n) 0+ i sin (m + n) 0. That is, calling cos mê + i sin mo, f (m), f(m) xf (n) = f (m+n). Hence, by the Index Theorem of Art. 463, m and n being real, f(n) is one value of {f(1)}", i.e. cos no + i sin nd is one value of (cos 6 + i sin 0), for all positive and negative values of n. This result is so important that we will give a proof of it without introducing the index theorem in its general form. It is usually known as De Moivri's theorem. 604. Whatever positive or negative value n may have, cos no + i sin nő is one of the values of (cos 0 + i sin 6)". We have cos (0+0+4 + ...) + i sin (0+$+4 + ...) =(cos @ + i sin 8) (cos $ + i sin $) (cos y + i sin y).... Now let 0 = $=y= ... and let there be n of these angles. Thus, if n is a positive integer, cos no + i sin no=(cos 0 + i sin O)n ...(1). Next let n be a negative integer m say. Then cosa mo - ** sin mo by (1) (2). Lastly, let n be a fraction LP say, where p and q are integral. a P P 9 (cos( + i sin 6)P by (1) and (2) ..cos no + i sin no is one of the values of $(cos 0 + i sin o)?: or cos no + i sin no is one of the values of (cos 6 + i sin 0)”. = m -m a Then (cos nø + i sin nøye = (cos Lo + i sin , 0)* 605. To find q values of (cos 0 + isin 6)ā where p and q are prime to one another. We have cos 0 + i sin 0 = cos (6 + 217) + i sin (@+ 217) where i has any integral value. .. any one of the values of cos. (0 + 217) + i sin P P (0 + 227) is 9 9 a value of (cos @ + i sin 6). 6 M 9 P 9 = and P a P would be equal, only if cos? (2 + 2um) = cos? (0 + 2ye'r) 0 9 9 and sin 2 (0+2 p.7) = sin ? (0+ 2u'r); 9 i.e. if the angles 2 (0+2um) and ? (6 + 2u'r) were equi-cosinal and 9 equi-sinal; i.e. if 2 (0+ 2mm) ~2 (0+2u'r) were a multiple of 27 ; P 9 P and q are prime to one another, this condition becomes .if u ~ pe' is a multiple of q.' Hence, if we give to i the q values 0, 1, 2... ..n-1, we shall different values of cos (6 + 2da) +i sin 2 (0 + 2Aa): and, р obtain q P a 9 if we give to a higher values, the values of this expression will simply recur. Hence 9 values of (cos 0 + i sin 6)? are given by the q values of cos? (0+217) + i sin (@+ 217). р P 9 P 606. To find q values of the oth root of any complex quantity. Let a + bi be any complex quantity : and let p denote the positive value of J(a + 62). Then p is the modulus of a + bi. р Since a/p and bp are each numerically less than unity, and are such that the sum of their squares is equal to unity, therefore, there is one and only one angle between 0 and 21 whose cosine is alp and whose sine is bp. Let this angle be 0. Thus a + bi =p (cos 0 + i sin 6). Now, since is positive, there cannot be more than one positive oth root of p. And there is one positive qth root of P, р whose value can be approximately determined by the Binomial Theorem. 1 + COS + i sin + pi(co ). Let pi denote this positive value of 3/p. Then, by the proposition of the last article, q values of the qth root of cos 0 + i sin 0 are given by the q values of θ + 2λπ θ + 2λπ 9 Hence q values of the oth root of a + bi are given by putting i successively equal to 0, 1, 2,...q-1 in the expression θ + 2λπ θ + 2λπ + i sin 9 9 607. There is one point in the above result that requires special attention. In the theory of indices, we use the symbol ağ as the equivalent of 3/(ap) or (a). Now, by the theory of fractions, a = a +7. Is it then legitimate to infer that y(ar)="/(a*P)? By the above proposition (ap) has q values; and, by Art. 599, it cannot have more than q values. Similarly "/(ap) has ro rq values and no more. Hence 3/(ap) and “/(arr) have not the same number of values: and, therefore, cannot be said to have the same meaning. The q values of Ja”) are, however, all contained amongst the rq values of "/(a?): as the student may show for himself. P = said The roots of +1 -1, +i, and - i. = 608. In De Moivre's theorem, Art. 605, let p=1. Now put 0=0. Thus coss A = 1, and sin 0 = 0. Hence the q qth roots of unity are given by the q values of 2λπ) 2λπ 9 COS COS Next put A = Thus cos 0 =-1 and sin 0 = 0. Hence the q qth roots of minus unity are given by the a values of 22+1 21+1 9 42+1 9 2. 9 values of 42 + 3 9 2 T COS T COS . ai (cos а = 609. Again, in Art. 606, put 0=0. Then a + bi reduces to a, i.e. p, a real positive quantity. Hence the qth roots of a are 2λπ 2λπ + i sin 9 9 where the expression in the bracket represents the qth roots of unity. Again put Q=T. Then a + bi reduces to a, i.e. - p, a real negative quantity. Hence the qth roots of a are 22+1 22 +1 9 9 where the expression in the bracket represents the qth roots of minus unity. We may illustrate the important result of Art. 606 by showing that the ratio of any two qth roots of a quantity is a qth root of unity. Thus giving to 1 any two values j and u', the ratio of the two corresponding qth roots of a +bi is cos (0+247)/q+i sin (0+ - 2um)/q. cos (0+2u'r)/q+i sin (0+2p's)/q (-a) (cos COS |