Determine all functionf $f : \mathbb{R} \mapsto \mathbb{R}$ such that 

\[ f(x+y) = f(x)f(y) - c \sin{x} \sin{y}  \] for all reals $x,y$ where $c> 1$ is a given constant.
1. **Substitute \( y = 0 \) into the functional equation:**
   \[
   f(x + 0) = f(x)f(0) - c \sin{x} \sin{0}
   \]
   Simplifying, we get:
   \[
   f(x) = f(x)f(0)
   \]
   This implies:
   \[
   f(x)(1 - f(0)) = 0
   \]
   Therefore, either \( f(x) = 0 \) for all \( x \), or \( f(0) = 1 \). Since \( f(x) = 0 \) for all \( x \) is not a useful solution, we assume \( f(0) = 1 \).

2. **Substitute \( x = t \) and \( y = \pi \) into the functional equation:**
   \[
   f(t + \pi) = f(t)f(\pi) - c \sin{t} \sin{\pi}
   \]
   Simplifying, we get:
   \[
   f(t + \pi) = f(t)f(\pi)
   \]

3. **Substitute \( x = \frac{\pi}{2} + t \) and \( y = \frac{\pi}{2} \) into the functional equation:**
   \[
   f\left(\frac{\pi}{2} + t + \frac{\pi}{2}\right) = f\left(\frac{\pi}{2} + t\right)f\left(\frac{\pi}{2}\right) - c \sin\left(\frac{\pi}{2} + t\right) \sin\left(\frac{\pi}{2}\right)
   \]
   Simplifying, we get:
   \[
   f(\pi + t) = f\left(\frac{\pi}{2}\right)f\left(\frac{\pi}{2} + t\right) - c \cos{t}
   \]

4. **Substitute \( x = \frac{\pi}{2} \) and \( y = t \) into the functional equation:**
   \[
   f\left(\frac{\pi}{2} + t\right) = f\left(\frac{\pi}{2}\right)f(t) - c \sin\left(\frac{\pi}{2}\right) \sin{t}
   \]
   Simplifying, we get:
   \[
   f\left(\frac{\pi}{2} + t\right) = f\left(\frac{\pi}{2}\right)f(t) - c \sin{t}
   \]

5. **Combine the equations obtained in steps 3 and 4:**
   \[
   f(\pi + t) = f\left(\frac{\pi}{2}\right)f\left(\frac{\pi}{2} + t\right) - c \cos{t}
   \]
   and
   \[
   f\left(\frac{\pi}{2} + t\right) = f\left(\frac{\pi}{2}\right)f(t) - c \sin{t}
   \]

6. **Assume \( f\left(\frac{\pi}{2}\right) = a \) and solve for \( f(t) \):**
   \[
   f(\pi) = a^2 - c
   \]
   Using the form \( f(t) = a \sin{t} + \cos{t} \), substitute \( t = \pi \):
   \[
   f(\pi) = a \sin{\pi} + \cos{\pi} = -1
   \]
   Therefore:
   \[
   a^2 - c = -1 \implies a^2 = c - 1
   \]
   Thus, \( a = \sqrt{c - 1} \) or \( a = -\sqrt{c - 1} \).

7. **Verify the solution:**
   Substitute \( f(t) = \sqrt{c - 1} \sin{t} + \cos{t} \) or \( f(t) = -\sqrt{c - 1} \sin{t} + \cos{t} \) back into the original functional equation to verify that it satisfies the equation.

The final answer is \( \boxed{ f(t) = \sqrt{c - 1} \sin{t} + \cos{t} } \) or \( f(t) = -\sqrt{c - 1} \sin{t} + \cos{t} \).