Find the trigonometric Fourier series for the function f(x): [-T/2, π/2] → R given by the expression: f(x) = { O cos 2x if x = [-π/2, 0] 0 if x = (0, π/2] FS(x) = -2 cos (2x) + En=1 FS(x) = cos(2x) + Σ-2 FS (x) = sin(2x) + Σn-2 FS(x) = cos(2x) + Σ FS(x) = cos(2x) + Ex-1 n² cos² (n²-1)π ² ( =) 2 2n cos² n cos² n cos² (n²-1) T 2(n²-1)T n=0 (n²-1) JLTT (+) 2 2n cos² 72T 2 (=) TLTT 2 -sin(2nx). (=) (n²+1) T -sin(2nx). -sin(nx). -sin(2nx). -sin(2nx).

= 1) The function f(x) periodic on the interval [0, 2л] has complex Fourier series f(x): Σ(1/n²) einx where the sum over n goes from - infinity to infinity. Convert this to cosine and sine Fourier Series by finding the values of A's and B's in the expression Ao + ΣAn cos(nx) + Σ Bn sin(nx) where each sum goes from 1 to infinity. Hint: consider the n and -n term together in the complex Fourier Series or use Euler's identity.