91Ó°ÊÓ

The Fourier cosine series of a function f(x) can be represented as: $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos{(nx)} $$ Where the Fourier coefficients (aâ‚™) are given by: $$ a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx)dx $$ In this case, f(x) = sin(kx) and we need to find the coefficients aâ‚™ for n = 0, 1, 2, ...

First, let's calculate a₀, which is the average value of the function: $$ a_0 = \frac{1}{\pi}\int_0^\pi{\sin(kx)dx} $$ Integrate by substitution: Let u = kx, so du = kdx, and dx = (1/k)du. Change the limits of integration: When x = 0, u = k(0) = 0. When x = π, u = k(π) = kπ. Now, the integral becomes: $$ a_0 = \frac{1}{\pi k} \int_0^{k\pi}{\sin(u)du} $$ Now, integrate with respect to u: $$ a_0 = \frac{1}{\pi k}[-\cos(u)]_0^{k\pi} = \frac{-1}{\pi k}[\cos(k\pi) - 1] $$ Since k is not an integer, cos(kπ) ≠ 1, and a₀ ≠ 0.

Now, let's find the coefficients aₙ for n ≥ 1: $$ a_n = \frac{2}{\pi}\int_0^{\pi}\sin(kx)\cos(nx)dx $$ Integrate by parts: Let: u = sin(kx), dv = cos(nx)dx. du = kcos(kx)dx, v = (1/n)sin(nx). Now, using integration by parts: $$ \int u\,dv = uv - \int v\,du $$ Here, we have: $$ a_n = \frac{2}{\pi}\left[\frac{1}{n}\sin(kx)\sin(nx) \bigg|_0^\pi - \int_0^\pi \frac{k}{n}\cos(kx)\sin(nx)dx\right] $$ The first term of the expression will be 0 because sin(0) = sin(nπ) = 0. Now we are left with the second term: $$ a_n = \frac{-2k}{n\pi}\int_0^\pi \cos(kx)\sin(nx)dx $$ Now, to compute this integral, we can use the product-to-sum formula: $$ \cos(kx)\sin(nx) = \frac{1}{2}[\sin((n+k)x) - \sin((n-k)x)] $$ Substitute this into the integral: $$ a_n = \frac{-k}{n\pi}\int_0^\pi [\sin((n+k)x) - \sin((n-k)x)]dx $$ Now, integrate with respect to x: $$ a_n = \frac{-k}{n\pi}\left[\left(\frac{-1}{n+k}\cos((n+k)x)\right) - \left(\frac{1}{n-k}\cos((n-k)x)\right)\right]_0^\pi = \frac{-1}{n\pi}\left[\frac{\cos((n+k)\pi) - 1}{n+k} + \frac{\cos((n-k)\pi) - 1}{n-k}\right] $$ Since k is not an integer, aₙ ≠ 0.

Finally, put the calculated values of a₀ and aₙ into the Fourier cosine series formula: $$ f(x) = \frac{-1}{\pi k}[\cos(k\pi) - 1] + \sum_{n=1}^\infty\frac{-1}{n\pi}\left[\frac{\cos((n+k)\pi) - 1}{n+k} + \frac{\cos((n-k)\pi) - 1}{n-k}\right] \cos(nx) $$ This is the Fourier cosine series for the given function f(x) = sin(kx) where k is not an integer, in the interval [0, π].