and deduce from it that sin (2) AB is the diameter COS COS ad inf. 23 ... 22 is less than 0-103. of a circle and Qo any point on the circumference; Q1, 22, 23,... are the points of bisection of the arcs AQ0, AQ1, 4Q2,... prove that BQ1, BQ2, BQ3... BQn=OA”. AQo AQ (3) Find the limit of (cos )cote when 0=0. (5) Of what order is the error when for ? 3 sin 0 2+ cos 0 is substituted (6) Prove that 2 cos no-2 cos n = 2′′ (cos – cos 0) × (8) Expand cos1"+sin1" in a series of cosines of multiples CHAPTER V. ON THE SUMMATION OF TRIGONOMETRICAL SERIES. 66. There are two methods peculiarly applicable to the Summation of Trigonometrical Series. FIRST METHOD. 67. Sometimes each term of a series may be transformed into the difference of two quantities. EXAMPLE 1. To sum the series sin a+ sin (a + d) + sin (a +28) + + sin {a + (n-1) §}. ... We have 2 sin a. sin d=cos (a–d) - cos (a+18), 2 sin (a+8) . sin 18=cos (a+18) − cos (á +38), 2 sin (a+28). sin 18=cos (a + § d) − cos (a + §d), 2 sin {a + (n-1) d}. sin d=cos (a + 2n −38) 2n-3 δ - cos (a + 21-18). Therefore, if Sn stands for the sum of n terms, we obtain by Hence, proceeding as in Example 1, we obtain The results of these two examples are often useful. The student is advised to become familiar with them in words. The sum of n terms of a series of sines (or cosines) of angles in A. P. is equal to the sine (or cosine) of half the sum of the first and last angle, multiplied by the sine of n times half the difference, divided by the sine of half the difference. EXAMPLE 3. To prove that if n=2π, then sin a+ sin (a+4)+sin (a +24) + sin {a+ (n-1)ø}=0 for all values of a. ... In the result of Example 1, sin and occurs in the numerator, and sin nd=sinnp=sin=0, and the denominator sin is not=0. Therefore the sum of the series =0. Similarly cos a + cos(a+4)+cos(a+24)+ ... +cos {a+ (n-1)ø}=0. 68. The results of Example 3 may be stated geometrically: Let OR be the initial line and ROP, any angle, then if the whole circumference of a circle centre O and radius OR, be divided into n equal parts P ̧P1, P1 P ̧, etc. Then the sum 1 of the sines (or of the cosines) of all the ROP,... ROP 2-1 angles ROP, is zero t. This is an expression of the fact that the centre of gravity of equal particles placed at the points PoP... is at the centre of the circle. 2m sinm a=(-1) {cos ma-n cos (m − 2) a + etc.} and the required sum may be obtained from the known sum of the series {cos ma + cos m (a + d) + cos m (a +28) + etc.} + {cos (m − 2) a + cos (m − 2) (a + d) + cos (m − 2) (a +28) + etc. }. Similarly we may find the sum of the series cosm a+cosm (a + d) + cosm (a +28) + etc. to n terms. (3) sin a+sin 4a+sin 7a+...... (4) sin a. cos a + sin 2a. cos 2a + sin 3a. cos 3a +...... (5) cos2 a+ cos2 2a + cos2 3a + (8) sin 2a. cos a + sin 3a. cos 2a+sin 4a. cos 3a + ...... (9) sin a. sin 2a + sin 2a. sin 3a + sin 3a. sin 4a +...... (10) cos3 a + cos3 (a + d) + cos3 (a + 28) + ...... (11) sin1a+sin* (a + d) + sin1 (a +28) + ...... (12) Solve the equation sin0+ sin 20+ sin 30+ etc. to n terms=cos + cos 20+ cos 30+ etc. to n terms. (13) Write down the value of series (10) and (11) when n8=2π. (16) Deduce from Ex. (1) the sum of the series 1+2+3+ ...... +n. (17) Deduce from Ex. (6) the sum of the series 13+23+33 + etc. + n3. (18) Deduce from Ex. (9) the sum of the series 1.2+2.3+3.4+ etc. + n (n + 1). (19) Sum the series sin a-sin (a + d) + sin (a + 28) – etc. to n terms. (20) Sum the series cos a n terms. - cos (a+8) + cos (a +28) - etc. to (21) Prove that the series sin" a + sinm (a + ) + etc. to n terms, where no=2′′, is independent of a, provided m is less than n. (22) Prove that the series cosm a + cosm (a + ø) + etc. to n terms, where no=2, is independent of a if m is less than n. |