Find a cosine expansion of the following function:
$$f(x)=\frac{1}{2}+\frac{\pi}{4}\sin x \space,\space\space\space x\in[0,\pi)$$
And then, using the result you got, compute the following sum:
$$ \sum_{k=1}^{\infty} \frac{1}{4k^2-1}$$
Ok, basically, this shouldn't be much of a problem, but one part of this question confuses me, at the beginning, it says:
"Find cosine expansion of the following function".
First thing that pops-up in my mind is Fourier expansion of the function, since it's formula is: $$f(x)=\frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n\cos nx +b_n \sin nx) $$
Now, whenever I have a function that is even I would only have cosine terms since $b_n=0$ every time we are dealing with even functions. However, this example requires to find a expansion of this function on $[0, \pi)$ and the period of this function is $2\pi$ so non of the coefficients shouldn't be equal to zero.
So, to sum up everything into two main questions:
What does it even means "to find cosine expansion of a function", is it really related to Fourier series?
If it is, then how to make series strictly cosine as it is required in the question?
Make $f$ even: $f(x)=\frac12+\frac\pi4\sin\bigl(|x|\bigr)$, with $x\in(-\pi,\pi)$. Then compute the Fourier series of this function. It will only have cosines in it.
Note that, if $n\in\mathbb N$, then$$\int_{-\pi}^\pi f(t)\cos(nt)\,\mathrm dt=\begin{cases}0&\text{ if $n$ is odd}\\-\frac\pi{4\left(\frac n2\right)^2-1}&\text{ otherwise.}\end{cases}$$