Consider the following function: f(x) = { sin x , if 0 ≤ x ≤ π 4 cos x , if π 4 ≤ x ≤ π 2 . Expand f(x) in a Fourier series of sine terms.

The Fourier series of a function can be represented as a sum of sine and cosine terms. However, the function you provided is piecewise and defined differently in two intervals.

The Fourier series of a function f(x) is given by:

f(x) = a0/2 + Σ [an cos(nx) + bn sin(nx)]

where the coefficients an and bn are given by:

an = (1/π) ∫ from -π to π [f(x) cos(nx) dx] for n = 0,1,2,3,...

bn = (1/π) ∫ from -π to π [f(x) sin(nx) dx] for n = 1,2,3,...

In your case, you want to expand f(x) in a Fourier series of sine terms only. This means that we will only calculate the bn coefficients and ignore the an coefficients.

However, the function you provided is defined from 0 to π/2, not from -π to π. Therefore, we need to extend the function to the interval from -π to π in a way that makes it odd, because only odd functions have sine Fourier series.

We can extend the function as follows:

f(x) = {

sin x , if 0 ≤ x ≤ π/4

cos x , if π/4 ≤ x ≤ π/2

-sin x , if -π/4 ≤ x < 0

-cos x , if -π/2 ≤ x < -π/4

Now, we can calculate the bn coefficients:

bn = (1/π) ∫ from -π to π [f(x) sin(nx) dx]

This integral will be split into four integrals according to the four intervals that we defined for f(x).

After calculating these integrals, you will get the bn coefficients, and therefore, the Fourier series of f(x) in sine terms.