(2) When n is odd. The only real root is other n 1 roots are found by putting r = 0, in (1), in succession. 1, 2, .. æ”+1= (x+1) | ∞2 − 2 x cos+1)(x2 - 2 x cos 3 п n ... 1; the n-3 2 +1)... 2 x cos 172 x + 1) (4) n EXAMPLES. 1. Find the roots of the equation 25 – 1 = 0. Ans. 1, cos (2 rπ) + i sin (2 rπ), where r = 1, 2, 3, 4. 2. Find the quadratic factors of 28 - 1. Ans. (x2 − 1) (x2 −√2 x + 1) (x2 + 1) (x2 + √2 x + 1). 3. Find the roots of the equation x + 1 = 0, and write down the quadratic factors of +1. Ans. ± +V-1 :; (x2 − x√2+1) (x2 + x√2 + 1). ... xn cos 0-1 sin 0 = ± i sin 0. since cos is unaltered if for 0 we put 0+2 rπ. If we put r = 0, 1, 2, ...n-1, successively in (1), we find 2n different roots, since each value of r gives two roots. х Cor. Change x into 2 in (3) and clear of fractions, and a we get x2n-2 anxn cos 0+ a2n. x2-2 ax cos Ꮎ = + a2 Find the quadratic factors of the following: 1. 28-2x+ cos 60° + 1 = 0. Ans. (x2 - 2 x cos 15° + 1)(x2 − 2 x cos 105° + 1) (x2 - 2 x cos 195° + 1) (x2 − 2 x cos 285° + 1) = 0. 2. x10-25 cos 10° + 1 = 0. - Ans. (x2 - 2 x cos 2o + 1) (x2 − 2 x cos 74° + 1) × (x2 - 2 x cos 146° + 1) (x2 − 2 x cos 218° +1) (x2 - 2 x cos 290° + 1) = 0. 170. De Moivre's Property of the Circle. centre of a circle, P any point in its plane. Divide the circumference into n equal parts BC, CD, DE, ..., beginning at any point B; and join O and P with the points of division B, C, D,... Let POB = 0; then will OB2n_2OB". OP" cos no+OP2 =PB2.PC2. PD2... ton terms. (1) Multiplying (1), (2), (3),..... together, we have 171. Cote's Properties of the Circle. These are particular cases of De Moivre's property of we have from (3) of Art. 170, after taking the square root of both members, OB" — OP" = PB. PC. PD... to n factors. ... I. (2) Let the arcs AB, BC, be bisected in the points a, b, ...; then we have, by (1), 2n 2n OB - OP2 Pa. PB. Pb. PC. Pc... to 2n factors. = Hence, by division, OB" + OP" = Pa. Pb. Pc... to n factors. Cor. If the arcs AB, BC,... be trisected in the points a1, a2, b1, b2, ..., then we have OB2+OB". OP”+OP2=Ра1 · Pа2 · Pb1· Pb2... to 2n factors. 172. Resolve sin 0 into Factors. (1) Put x=1; then we get from (3) of Art. 169 then extracting the square root, we have sin no 2"-1sin sin (+2a) sin (+4α) × = ... x sin (+2na-2α). . (1) (2) But sin (+2na −2 a) = sin (4+π−2 α) = sin (2 α — 4), sin (+2 na − 4 α) = sin (4 α — $), and so on. Hence, when n is odd, multiplying together the second factor and the last, the third and the last but one, and so on, we have sin no =2"-1sin & sin(2x+4) sin (2 a−4) sin(4a+) sin (4α-6) ... x sin [(n − 1) a + ] sin [ (n − 1)α — $]. - But sin (2x+4) sin (2 α— 4) = sin12 α- sin2, and so on. .. sin no 2-1 sin (sin 2a - sin2 ) (sin24 a-sin') X ... × [sin2 (n - 1)a — sin2] (3) get Divide both members of (3) by sin o, and then diminish indefinitely. Since the limit of sin no ÷ sin is n, we n = 2n-1 sin22 a sin24 a sin2 6 α × x sin2(n - 1) a (4) Divide (3) by (4); thus ... = 0, and let n be increased while is diminished remaining unchanged; then since 2nα = π, Put no without limit, sing (黒) 379 sin2 π n 0; and so on. that of n sin NOTE. The same result will be obtained if we suppose n even. |