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- Problem Statement
- Find the Fourier sine series of cos 2x on [0, π]

- Relevant Equations
- $$b_n = \dfrac{2}{l} \int^\pi _0 f(x) sin(\dfrac{n\pi x}{l})dx$$

I am attempting to find the sine representation of cos 2x by letting

$$f(x) = \cos2x, x>0$$ and $$-\cos2x, x<0$$

Which is an odd function. Hence using $$b_n = \dfrac{2}{l} \int^\pi _0 f(x) \sin(\dfrac{n\pi x}{l})dx$$ I obtain $$b_n = \dfrac{2n}{\pi} \left( \dfrac{(-1)^n - 1}{4-n^2} \right)$$

which when I plot the sine series

$$\sum^\infty _{n=3} \dfrac{2n}{\pi} \left( \dfrac{(-1)^n - 1}{4-n^2} \right) \sin(n x)$$

I don't get the fn of cos 2x on [0, π]. I don't understand where I went wrong any help would be great.

$$f(x) = \cos2x, x>0$$ and $$-\cos2x, x<0$$

Which is an odd function. Hence using $$b_n = \dfrac{2}{l} \int^\pi _0 f(x) \sin(\dfrac{n\pi x}{l})dx$$ I obtain $$b_n = \dfrac{2n}{\pi} \left( \dfrac{(-1)^n - 1}{4-n^2} \right)$$

which when I plot the sine series

$$\sum^\infty _{n=3} \dfrac{2n}{\pi} \left( \dfrac{(-1)^n - 1}{4-n^2} \right) \sin(n x)$$

I don't get the fn of cos 2x on [0, π]. I don't understand where I went wrong any help would be great.