We are tasked with constructing an even function from $f(x) = \sin(x)$ over the interval $0 \leq x \leq \pi$, extending it to have a period of $\pi$, and then finding the sine-cosine Fourier series for this new function.

Step 1: Construct an Even Function

An even function satisfies $f(-x) = f(x)$. The original function $f(x) = \sin(x)$ is not even since $\sin(-x) = -\sin(x)$. However, we can construct an even function from $\sin(x)$ by defining a new function $g(x)$ such that:

\begin{cases} \sin(x) & \text{if } 0 \leq x \leq \pi, \\ \sin(-x) = -\sin(x) & \text{if } -\pi \leq x \leq 0. \\ \end{cases}$$ Thus, for $$g(x)$$, we define: $$g(x) = \sin(|x|), \quad \text{for all } x.$$ This function is even because: $$g(-x) = \sin(|-x|) = \sin(|x|) = g(x).$$ ### Step 2: Extend $$g(x)$$ to a Periodic Function We now extend $$g(x)$$ to be periodic with period $$\pi$$. This means that the function repeats every $$\pi$$ units. The extended function can be written as: $$g(x + \pi) = g(x) \quad \text{for all } x.$$ Thus, we construct a periodic function that is even and satisfies: $$g(x) = \sin(x) \quad \text{for } 0 \leq x \leq \pi, \quad \text{and repeats every } \pi.$$ ### Step 3: Sketch the Function The function $$g(x)$$ over three periods (from $$-3\pi$$ to $$3\pi$$) looks like a series of "bumps" due to the periodic nature of the sine function being reflected for negative $$x$$. At each integer multiple of $$\pi$$, the function reaches zero, and between these points, it oscillates in a smooth, continuous manner. ### Step 4: Fourier Series for the Even Periodic Function Since $$g(x)$$ is even, the sine terms in its Fourier series vanish, and we only have cosine terms. The general form of a Fourier cosine series for an even periodic function with period $$\pi$$ is: $$g(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{2n\pi x}{\pi}\right).$$ To find the coefficients $$a_0$$ and $$a_n$$, we use the following formulas: $$a_0 = \frac{2}{\pi} \int_0^\pi \sin(x) \, dx,$$ $$a_n = \frac{2}{\pi} \int_0^\pi \sin(x) \cos\left(\frac{2n\pi x}{\pi}\right) \, dx \quad \text{for } n \geq 1.$$ #### Calculate $$a_0$$: \[ a_0 = \frac{2}{\pi} \int_0^\pi \sin(x) \, dx = \frac{2}{\pi} \left[-\cos(x)\right]_0^\pi = \frac{2}{\pi} [1 - (-1)] = \frac{4}{\pi}. \] #### Calculate $$a_n$$ for $$n \geq 1$$: Using integration by parts or a standard Fourier series table, we know that: $$a_n = 0 \quad \text{for all } n \geq 1.$$ Thus, the Fourier series reduces to just the constant term: $$g(x) = \frac{2}{\pi}.$$ ### Conclusion The even, periodic function constructed from $$f(x) = \sin(x)$$ is a constant function over each interval due to the cancellation of higher-order terms. The Fourier series for this function is simply: $$g(x) = \frac{2}{\pi}.$$ Let me know if you'd like further details on any of the steps. ### Related Questions: 1. What is the general formula for the Fourier series of an odd function? 2. How do you find the Fourier sine series for $$f(x) = \sin(x)$$ without symmetry extensions? 3. What is the significance of even and odd functions in Fourier series? 4. How do you compute Fourier series for non-periodic functions? 5. How do the boundary conditions of $$f(x)$$ affect the Fourier coefficients? ### Tip: When constructing an even function from an odd function, reflect the odd part symmetrically and ensure the function maintains periodicity by adjusting its intervals accordingly.