How to find the side of the square ,by using trigonometry There is a square with diagonal length of 'a", the question is to find the length of the sides. it can be found by Pythagorean theorem. but I tried to do it with trigonometry, considering the properties of a square but it doesn't resemble the first answer, so I was wondering what did I forget/mistake?

Let the side length of the square equal $b$, and the hypotenuse $a$. By Pythagoras' Theorem, we have
\begin{align}
b^2+b^2&=a^2 \\
2b^2&=a^2 \\
b^2&=\frac{a^2}{2}\\
b&=\sqrt{\frac{a^2}{2}}=\frac{\sqrt{a^2}}{\sqrt 2}=\frac{a}{\sqrt 2}=\frac{\sqrt 2}{2}a
\end{align}
Alternatively, the angle between $a$ and $b$ is $45$ degrees. We know this because a square has $4$ right angles, which are bisected (cut in half) by the diagonal. Therefore,
\begin{align}
\sin(45)&=\frac{b}{a}\\
\end{align}
Now find the value of $\sin(45)$, and rearrange the equation to find $b$. Hopefully this answer is familiar.
It seems that part of your confusion stems from the fact that you are unsure about rationalising the denominator:
$$
\frac{1}{\sqrt 2}=\frac{\sqrt 2}{2}
$$
Consider the left-hand side of the above equality. Multiplying $\frac{1}{\sqrt 2}$ by $1$ does not change its value:
$$
\frac{1}{\sqrt 2}\times1=\frac{1}{\sqrt 2}
$$
We also know that
$$
1=\frac{\sqrt{2}}{\sqrt{2}}
$$
Therefore,
$$
\frac{1}{\sqrt 2}\times1=\frac{1}{\sqrt 2}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt 2}{2}
$$
