3. Consider a pendulum of length L, released from rest at an angle a. It can be shown that such a pendulum has period T given by T = 4, xp: 1 – sin? (4) sin? x (a) Notice that the integrand has the form For simplicity, write A = sin? (9). Show that +Σ 1.3-5... (2n – 1) 2nn! 2n %3D V1- A sin? x n=1 (b) Calculate T/2 /2 */2 sin?: x dr, s sin“ r dr and sin® r x dr. (c) We can now determine the period T as a power series in A (which encodes the initial angle a) by integrating the power series from (a). Show that 1 1+-A+ 25 x² + 256 T = 27 64 (d) This is a great result, but the initial angle a is hidden from view. Let's turn this into a power series in a: i. We have the identity sin? t = (1 – cos(2t)). Use this to show that the Maclaurin series for sin? t is given for any t by (-1)n+12n-1 sin?t: (2n)! n=1 Write out the terms of this series up to degree 6. ii. Show that to degree 6, A is given in terms of a by 2 )"- 45 iii. Now show that to degree 6, the period T, in terms of a, is given by 11 173 T = 2n 1+ 16a +... 3072 737280 (e) You will often find the following statement: For small a, the period of a pendulum is given by T = 27/. Explain this based on your calculation.