This is a simpler "cousin" question to Would sine trigonometric series $f(x)$ for approximating $g(x) = x$ always be $f(x) \leq x$ for $0 \leq x \leq \pi$? . I am asking this as a separate question, since this may be relatively easier to answer than the linked question.
I am trying to use set of $c_k$ with $k \in \mathbb{N}$ such that $f(x) = \sum_{k=1}^{M} c_k \sin kx \approx x$.
$c_k$ is determined by setting $$f^{(2k+1)}(0) =0,\,\, k=1,..,M-1$$ $$f'(0) =1$$
These conditions provide $M$ linear equations to solve for $c_k$.
Question is, is it guaranteed that $f(x) \leq x$ for $0 \leq x \leq \pi$ for any $M\geq 2$?
The matrix of your system is $$ \begin{bmatrix} 1&1&1&\cdots&1\\ 1&2^3&2^5&\cdots&2^{2M-1}\\ 1&3^3&3^5&\cdots&2^{2M-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&M^3&M^5&\cdots&M^{2M-1}. \end{bmatrix} $$ It is not hard to check that this determinant is never zero, so your system has unique solution. That means that $c_1=1$, $c_2=\cdots=c_M=0$ is the only solution, and thus $$ f(x)=\sin x $$ for all $M$.