Question

The function sin²(2?x/L) is a periodic function of x (L is some fixed length). What is the period? Show explicitly using a trig identity that this can be written as a Fourier cosine series. In this case there are only two nonzero terms in the series.

Answer 1

Sin (pi + x) = -sin x sin^2 (pi + x) = sin^2 x period of sin^2 x in pi Given that, f(x) = sin^2 (2 pi.x/L) Its period is given by: 2 pi x/L = pi x = L/2 is the reg.period. further y = 2 pi x/L , then sin^2 y = 1 - cos^2 y [because sin^2 y + cos^2 y = 1] = 1 - (cos^2 y + sin62 y) [because cos2y = cos62 y - sin^2 y] sin^2 y = 1 - cos 2y - sin^2 y 2 sin^2 y = 1 - cos 2y sin^2 y = 1/2 - 1/2 cos 2y sin^2 (2 pi x/L) = 1/2 - 1/2 cos (2.2 pi.x/L) sin^2 (2 pi x/L) = 1/2 - 1/2 cos (4 pi x/L) comparing with fourier series, f(x) = a_0/2