71. The expression x2 - 2x cos +1 is the product of the two factors (1 - xeie) (1-xe-i), and therefore an expression having x- 2x cos 0+1 for its denominator may often be expanded in ascending powers of x by finding its equivalent partial fractions. =eia {1+xeïß +x22iß + ... } +e ̄ia {1+xe ̄iß +x2e-2iß + =eia+e ̄ia+x (eia+iß +ė¬ia¬¿ß)+x2 (¿2ía+2iß +e−2ia−) =2 cos a +x 2 cos (a +ẞ) +x2 2 cos 2 (a+ß) + ... [Art. 23.] } a−2iß)+. ... [Art. 23.] 1 EXAMPLE II. Expand in ascending powers of x. 1-2x cos a + x2 This expression may be written (1-x)−1 (1 − xe ̄ia)–1 = {1+x1a +x3 ̧2ia +x3¿sia +...} {1+xe ̄ia +x2e−2ia+x3e-3ia + ... } then EXAMPLE III. In any triangle c2=a2 - 2ab cos C+b2; let a be>b, .. c=2 oe-ic), = (1-10) (1-10). +10g (1-00-00) - 2 log e-2 log a+ log (1-0) + 108 (1-10) [Art. 71.] Given sin 0=x sin (0+a), expand 0 in a series of ascending powers 2i0+2irπ= x (e1a − e−ia) +4x2 (e2ia − e−2ia)+etc. [Art. 4.] 0+r=x sin a+ x2 sin 2a + 1x3 sin 3a + ... [Art. 23, 28.] If in the above x= put 20 for a. =-1, then sin 0 = − sin (0+a), so that we may Hence we obtain when @ is less than 0=sin 20-sin 40+ } sin 60 – etc. EXAMPLES. XXIX. 1 (1) Expand in a series of ascending powers 1- 2a cos +a2 of a; and prove that if P-1, P., Pn+1 be the coefficients of three consecutive terms 2p, cos &=Pn-1+Pn+1• Expand the following expressions in ascending powers of x. α (12) In any triangle sin 4= sin (4+ C), hence prove that (13) If tann tan 0, find a series for & in terms of 6. (15) Prove that cos na cos" a + i sin na cos" a= expand cos na cos" a in ascending powers of tan a. (16) Sum to infinity the series 1 (1 − i tan a)" (i) 4+9 cos 0+21 cos 20+51 cos 30+ etc. and CHAPTER VI. RESOLUTION OF sin @ AND COS INTO FACTORS. 73. To prove sin 0-0 (1-4) (1-) (1-*).. = By Arts. 38, 52 or 62 we have when n is a positive integer 2n x2 - 2x" cos 2na + 1 = (x2 - 2x cos 2a + - 1){a2 – 2x cos 2a + · n factors ... in this result let x=1, and let 2n=π, then 2 (1- cos 2na) = 2′′ (1 − cos 2a) {1 — cos (2a +44)}... Now 1- cos 2na = 2 sin3 na; hence taking the square root + 2 sin na = 2" sin a. sin (a + 2p) . sin (a + 46) ..... But sin (a + 2np − 24) = sin (a + π − 24) = sin (24 − a), hence, when n is odd, we have +2 sin na = 2" sin a sin (24+ a) sin (24-a). sin (4+ a) sin (44 - a) ... sin {(n − 1) p + a} sin {(n − 1) 4 — a}. But sin (24+ a) sin (24 − a) = sin3 24 – sin3a. + 2 sin na = 2" sin a (sin3 24-sin' a) (sin 44 - sin' a)... Next, divide both sides by sin a, and let a be diminished without limit, and we obtain 2n = 2" sin3 24. sin' 44 sin' 6p... sin' (n-1) . Divide the first of these last two results by the second, Write for na, and let n be increased while a is diminished Now, when lies between 0 and π, sin 0 is positive and every factor on the right-hand side is positive; when lies between π and 2π, sin 0 is negative and one factor only on the right-hand side is negative; and so on. Therefore the upper sign must be taken in the above result instead of the ambiguity and the proposition is established. *74. We can prove that the upper sign must be taken in each of the foregoing identities as follows: : 29 In the figure, let ROP1 = a; produce PO to Q, and divide the semicircumference PQ into n equal parts P. P1, P, P, etc. Then since n. 24π, each of the angles POP1, POP... is equal to 24 and ROP1 = a + 26, ROP1 = a + 44, etc. Now consider the first ambiguity on page 74. |