I am trying to express $\delta(x-1)$ on the trigonometric basis
on $L^2([0,1], \mathbb{R})$, given by
$$1, \sqrt{2}\cos(2 \pi j x), \sqrt(2)\sin(2 \pi j x). $$
For example,
$$ \delta(x - 1/2) = \sum_{n=0}^{\infty} c_n \sqrt{2}\cos(2 \pi j x) $$ with $c_n = \sqrt{2}\cos(\pi n )$.
When I try to expand $\delta(x -1)$, I get $$ \delta(x - 1) = \sum_{n=0}^{\infty} d_n \sqrt{2}\cos(2 \pi j x) $$ with $d_n = \frac{\sqrt{2}}{2}\cos(2 \pi n ) = \frac{\sqrt{2}}{2}$.
The result is an even function,
meaning the reconstruction fails to reproduce $\delta(x-1)$.
I know delta functions are not elements of $L^2([0,1], \mathbb{R})$ anyway, but why does $\delta(x -1/2)$ work and $\delta(x - 1)$ does not ?
Your formulas are incorrect. The correct formulas are as follows:
$\delta(x-\frac{1}{2})=\underset{N\to\infty}{\text{lim}}\left(1+2\sum\limits_{n=1}^N\cos(\pi n)\ \cos(2 \pi n x)\right),\quad -\frac{1}{2}<x<\frac{3}{2}\tag{1}$
$\delta (x-1)=\underset{N\to\infty}{\text{lim}}\left(1+2\sum\limits_{n=1}^N\cos(2 \pi n x)\right),\quad 0<x<2\tag{2}$
Figures (1) and (2) below illustrate both functions are periodic with a period of $1$ which is why formula (2) above evaluates non-zero at $x=0$.
Figure (1): Illustration of formula (1) for $\delta(x-\frac{1}{2})$ evaluated at $N=100$
Figure (2): Illustration of formula (2) for $\delta(x-1)$ evaluated at $N=100$
If you want the formula for $\delta(x-1)$ to evaluate closer to $0$ at $x=0$ then you need to increase the period as illustrated in formula (3) below. Note formula (3) below is equivalent to dividing formula (1) above by $2$ and evaluating it at $\frac{x}{2}$.
$\delta(x-1)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{2}+\sum\limits_{n=1}^N\cos(\pi n)\ \cos (\pi n x)\right),\quad -1<x<3\tag{3}$
Figure (3) below illustrates formula (3) above for $\delta(x-1)$ is periodic with period $2$ and evaluates closer to $0$ at $x=0$. Formula (3) only converges in a distributional sense and at $x=0$ actually evaluates to $\pm\frac{1}{2}$ depending on the evaluation limit $N$, more specifically formula (3) evaluates to $\frac{1}{2}(-1)^N$ at $x=0$.
Figure (3): Illustration of formula (3) for $\delta(x-1)$ evaluated at $N=100$