It is given the function $f:\mathbb{R} \to \mathbb{R}$ such that it holds $f(\sin x)=\sin (2011x)$. Find the value of $f(\cos x)$.
1. We start with the given function \( f: \mathbb{R} \to \mathbb{R} \) such that \( f(\sin x) = \sin(2011x) \).
2. We need to find the value of \( f(\cos x) \).

To do this, we will use the trigonometric identity for cosine in terms of sine:
\[ \cos x = \sin\left(\frac{\pi}{2} - x\right) \]

3. Substitute \( \cos x \) into the function \( f \):
\[ f(\cos x) = f\left(\sin\left(\frac{\pi}{2} - x\right)\right) \]

4. Using the given function definition \( f(\sin x) = \sin(2011x) \), we substitute \( \sin\left(\frac{\pi}{2} - x\right) \) into the function:
\[ f\left(\sin\left(\frac{\pi}{2} - x\right)\right) = \sin\left(2011 \left(\frac{\pi}{2} - x\right)\right) \]

5. Simplify the argument of the sine function:
\[ 2011 \left(\frac{\pi}{2} - x\right) = 2011 \cdot \frac{\pi}{2} - 2011x \]

6. Therefore, we have:
\[ f(\cos x) = \sin\left(2011 \cdot \frac{\pi}{2} - 2011x\right) \]

7. Using the sine subtraction identity:
\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]

8. Let \( a = 2011 \cdot \frac{\pi}{2} \) and \( b = 2011x \):
\[ \sin\left(2011 \cdot \frac{\pi}{2} - 2011x\right) = \sin\left(2011 \cdot \frac{\pi}{2}\right) \cos(2011x) - \cos\left(2011 \cdot \frac{\pi}{2}\right) \sin(2011x) \]

9. Since \( 2011 \) is an odd number, we know:
\[ \sin\left(2011 \cdot \frac{\pi}{2}\right) = \sin\left(\frac{2011\pi}{2}\right) = \sin\left(\frac{\pi}{2} + 1005\pi\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
\[ \cos\left(2011 \cdot \frac{\pi}{2}\right) = \cos\left(\frac{2011\pi}{2}\right) = \cos\left(\frac{\pi}{2} + 1005\pi\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

10. Therefore:
\[ \sin\left(2011 \cdot \frac{\pi}{2} - 2011x\right) = 1 \cdot \cos(2011x) - 0 \cdot \sin(2011x) = \cos(2011x) \]

11. Thus, we have:
\[ f(\cos x) = \cos(2011x) \]

The final answer is \( \boxed{\cos(2011x)} \).