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

Given,

f(x) = sin²(2?x/L)

Period of sin²(x) is ?.

Period here is ,

f(x+L/2) = sin²(2?(x + L/2) / L) = sin²(2?x/L + ? )

So, period is L/2.

cos(2x) = cos²(x) - sin²(x) = 1 - 2sin²(x)

sin²(x) = 1/2 * ( 1 - cos(2x) )

Using the above identity ,

f(x) = sin²(2?x/L) = 1/2 * ( 1 - cos(4?x/L) )

      = 1/2 + ( (-1/2) * cos(4?x/L) )

which is indeed in the form of a cosine Fourier series and as given in the question has only two non-zero terms.