2=(2-2 cos 4) (2-2 cos 3p)...up to {2-2 cos (2n − 1) p}...(1), n = (2-2 cos 24) (2 - 2 cos 46)...up to {2-2 cos (2n−2) p}...(2). Take the square root. Thus √2 = 2′′ sin § 6 . sin § 34.......up to sin † (2n − 1) p..... ... (3), √n=2"-1 sin 24. sin 44...up to sin (2n − 2) p...(4). Now sinλ = cos(2n-λ) p. Thus √2=2" cos. cos 3p...up to cos (2n-1) .......(5), √n = 2n−1 cos 24. cos § 44...up to cos § (2n − 2) p.......(6). 1 Hence, multiplying (3) by (5); and (4) by (6); 1 = 2′′-1 sin & sin 34...up to sin (2n − 1) p.............(7), n = 2n-1 sin 24 sin 44...up to sin (2n − 2) p.........................(8). Again, sin λ = sin (2n − λ) 4. Hence, taking the square root, 1/2"-1. sin o sin 34 sin 5p... &c. = √n= √2n-1. sin 24 sin 44 sin 6p...&c. (9), (10), where the coefficients of have every odd or even value less than n. For the unpaired factor sin no (when it occurs) = 1. 427. To apply geometrically the factorisation of Take any circle with centre 0. Draw the radius OI. Let A be any point on the circumference, and angle IOA: =α. Divide the whole circumference into n equal arcs, AB, BC, CD... beginning at A: so that, if L is any one of the points of division, Also angle IOL = a + 2λπ/n. OA = OB = = OL= . = OI. I. Let be any point on the circumference, such that angle IOQ=0: and let Qq, Aa, Bb,... be perpendiculars upon OI. Then will OI" (cos no cos na) = 2n-1. aq.bq. cq ... to n factors..(1). For OQ cos - OL cos (a + 2λπ/n) = Oq — Ol = lq. .., multiplying cos no - cos na by OI", we have (1). II. Let P be any point on OI, such that OP= x. OI; and join PA, PB, PC .... Then will OP2 –20P .01" cos na+(1TM=PA2. PB2. PC2...to n factors... (2). For PLOP2 + OL2 – 20P. OL cos IOL -n .., multiplying x2 + x−1 2 cos na by x". Oľ, we have (2). 428. Particular cases of the two theorems, proved above, should be noted. Thus (1) Let and P coincide with I. (3) Let the arcs AB, BC, CD ... in (2) be bisected in A', B', Then C'.... - 01. sin2 n6 = 22n-2. aq. a'q. bq. b'q to 2n factors. ... OI2n = AP. A'P.BP.B'P ... to 2n factors. 429. The result (2) of Art. 427 is called De Moivre's property of the circle: and the particular cases of it in Art. 428 are called Cotes's properties of the circle. 430. We have shown that xnxn is the same function of x+x-1 as 2 cos no is of 2 cos 0. It follows that (2 cos 0)" can be expanded in cosines of multiples of according to the same formula by which (x+x1)" can be expanded in functions of the form x2 + x ̄". Thus, by the binomial theorem, (x + x−1)n = x2 + nx2-1 . x−1 + §n (n − 1)xn−2, x2 + + 1n (n − 1) x2 . x¬n+2 + nx . x¬n+1 +x-n. Writing the series again in the reverse order; and adding: we have 2(x+x-1)"= -2 (x2 + x-n) + n (xn−2 + x−n+2) + ‡n (n − 1) (xn−4 + x−n+4) + ... Hence ... (2 cos 0)n = cos n✪ + n cos (n 2)0+ + n cos (−n + 2) 0+cos(-n0). This is another mode of proving the result of Art. 362. 1. EXAMPLES XVI. Show that, if f(x) is an integral function of degree not where (x) is some integral function of x, and ɑ ɑ ɑз...ɑn are any n quantities, all different. 2. Show that the sum of the coefficients of ƒ (a), ƒ (a2)...... f(an) in the above question is unity. 3. If a = one nth of half a right angle, sin a sin 5a sin 9a...sin (4n - 3) a = √2.2−". 4. The product of all the lines that can be drawn from one of the angles of a regular polygon inscribed in a circle whose radius is a to all the other angular points is nan-1. 5. If P1, P2P2n-1, Pan be the perpendiculars drawn from any point in the circumference of a circle of radius a on the sides of a regular circumscribing polygon of 2n sides, then P1P3P5P2n-1 + P2P 4· · ·P2n = an. 2−n+2 ̧ 6. A polygon is described about a circle touching it at the angular points of an inscribed polygon: the product of the perpendiculars drawn to the several sides of the inscribed polygon from any point in the circumference of the circle is equal to the product of the perpendiculars drawn from the same point to the several sides of the circumscribed polygon. 7. AB is the diameter of a circle of centre O, and Q, any point on the circumference; Q1, Q2 Q...are the points of bisection of the arcs AQ0, AQ1, AQ2..., prove that 29 9. Find the equations whose roots shall be (a) 2 cosπ, 2 cosπ, 2 cosπ. (b) sin2, sin2 & π, sin2 & π. tan a tan[(a + 4) tan (a + 24)...tan {a + (n − 1) p} − (− 1)”. 13. Prove that COS = 15 15 15 15 15 15 14. From Ex. 26, p. 278, show that COS = 15 27 CHAPTER XVII. SUMMATION OF SERIES. The difference method. 431. To find the sum of the sines or cosines of angles in Arithmetical Progression. 1. Let S sin a + sin (a + 8) + + sin {a + (n − 1) 8}. Multiply each term by 2 sin (1⁄2 the common difference). Thus 2 sin a sin 8= cos (a − 18) — cos (a + 18), 2 sin (a + d) sin 18 = cos (a + 18) – cos (a + 1⁄2 38) and so on, up to 2 sin {a + (n − 1)8} sin 18=cos {a + (2n−3) 8}-cos {a + † (2n−1) 8)} As before, multiply each term by 2 sin (the common difference). Thus we shall find 432. To find the sum of the sines or cosines, alternately negatived, of angles in arithmetical progression. 1. Let S sin a sin (a + d) + sin (a + 28)...+(-1)-1 sin {a + (n − 1) 8}. |