Fourier cosine series expansion of $f(x)=1$

Question

Fourier cosine series expansion of $$f(x)=1,~~~ x\in (0,\pi)$$

Hint is "thought is better than calculation".

Answer 1

So, do I just let the Fourier series for cosine equal $1$?

Yes, the cosine expansion of this function has just one nonzero term: $1$. The rest of coefficients are therefore zero.

This is something that always happens when you expand a vector in a basis which contains that vector. If your basis of $\mathbb{R}^2$ is $\{(2,3), (1,-5)\}$, and you want to expand the vector $v=(1,-5)$ in this basis, the expansion is just that: $$(1,-5) = 0\cdot (2,3)+1\cdot (1,-5)$$

Indeed, this particular choice of coefficients evidently works, and since the coefficients are unique, any other method would produce same ones.