Expand f(x) = cos x in a Fourier sine series on the interval 0<= x <= pi Solutio

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b_n = (2/p) ?(x=0 to p) cos x sin(npx/p) dx = (2/p) ?(x=0 to p) cos x sin(nx) dx = (2/p) ?(x=0 to p) (1/2) * [sin(nx + x) + sin(nx - x)] dx, via trig identity = (1/p) ?(x=0 to p) [sin((n+1) x) + sin((n-1)x)] dx. If n = 1, then b_1 = (1/p) ?(x=0 to p) [sin(2x) + 0] dx = (1/p) * [-cos(2x)/2] {for x=0 to p} = 0. If n > 1, then b_n = (1/p) ?(x=0 to p) [sin((n+1) x) + sin((n-1)x)] dx = (-1/p) [cos((n+1) x)/(n+1) + cos((n-1)x)/(n-1)] {for x=0 to p} = (-1/p) [cos((n+1)p)/(n+1) + cos((n-1)p)/(n-1)] - (-1/p) [1/(n+1) + 1/(n-1)] = (-1/p) [(-1)^(n+1) / (n+1) + (-1)^(n-1) / (n-1)] - (-1/p) [1/(n+1) + 1/(n-1)] = (1/p) [(-1)^n / (n+1) + (-1)^n / (n-1)] + (1/p) [1/(n+1) + 1/(n-1)] = (1/p) * ((-1)^n + 1) (1/(n+1) + 1/(n-1)) = 0 if n is odd, and (1/p) * 4n/(n^2 - 1) if n is even. Therefore, the desired Fourier Sine Series is given by f(x) ~ S(n=1 to 8) b_n sin(nx) = S(even n) b_n sin(nx) + S(odd n) b_n sin(nx) = S(even n) [(1/p) * 4n/(n^2 - 1)] sin(nx) + S(odd n) 0 * sin(nx) = S(k=1 to 8) 8k sin(2kx) / [p(4k^2 - 1)] , with n = 2k

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