Find all functions $f:[-1,1] \rightarrow \mathbb{R},$ which satisfy
$$f(\sin{x})+f(\cos{x})=2020$$
for any real number $x.$
1. **Given the functional equation:**
   \[
   f(\sin{x}) + f(\cos{x}) = 2020
   \]
   for any real number \( x \).

2. **Symmetry and Even Function:**
   Note that:
   \[
   f(\sin{x}) + f(\cos{x}) = 2020 = f(\sin(-x)) + f(\cos(-x))
   \]
   Since \(\sin(-x) = -\sin(x)\) and \(\cos(-x) = \cos(x)\), we have:
   \[
   f(-\sin{x}) + f(\cos{x}) = 2020
   \]
   This implies that \( f \) is an even function, i.e., \( f(-x) = f(x) \).

3. **Expressing \( f \) in terms of another function \( g \):**
   Let us write \( f(x) \) in the form:
   \[
   f(x) = g(1 - 2x^2) + 1010
   \]
   where \( g: [-1, 1] \rightarrow \mathbb{R} \).

4. **Substituting into the original equation:**
   Substitute \( f(x) = g(1 - 2x^2) + 1010 \) into the original equation:
   \[
   f(\sin{x}) + f(\cos{x}) = g(1 - 2\sin^2{x}) + 1010 + g(1 - 2\cos^2{x}) + 1010
   \]
   Since \(\sin^2{x} + \cos^2{x} = 1\), we have:
   \[
   1 - 2\sin^2{x} = -\cos(2x) \quad \text{and} \quad 1 - 2\cos^2{x} = \cos(2x)
   \]
   Therefore, the equation becomes:
   \[
   g(-\cos(2x)) + 1010 + g(\cos(2x)) + 1010 = 2020
   \]
   Simplifying, we get:
   \[
   g(-\cos(2x)) + g(\cos(2x)) = 0
   \]

5. **Odd Function \( g \):**
   The equation \( g(-\cos(2x)) + g(\cos(2x)) = 0 \) implies that \( g \) is an odd function, i.e., \( g(-y) = -g(y) \) for all \( y \in [-1, 1] \).

6. **Conclusion:**
   We have shown that all functions \( f \) of the form:
   \[
   f(x) = g(1 - 2x^2) + 1010
   \]
   where \( g \) is an odd function, satisfy the given condition. Hence, the general solution is:
   \[
   f(x) = g(1 - 2x^2) + 1010
   \]
   for any odd function \( g: [-1, 1] \rightarrow \mathbb{R} \).

The final answer is \( \boxed{ f(x) = g(1 - 2x^2) + 1010 } \) for any odd function \( g: [-1, 1] \rightarrow \mathbb{R} \).