For any real $ a$ define $ f_a : \mathbb{R} \rightarrow \mathbb{R}^2$ by the law $ f_a(t) \equal{} \left( \sin(t), \cos(at) \right)$.
a) Prove that $ f_{\pi}$ is not periodic.
b) Determine the values of the parameter $ a$ for which $ f_a$ is periodic.
[b]Remark[/b]. L. Euler proved in $ 1737$ that $ \pi$ is irrational.
### Part (a): Prove that \( f_{\pi} \) is not periodic.

1. **Assume \( f_{\pi} \) is periodic**: Suppose \( f_{\pi} \) is periodic with period \( T > 0 \). This means that for all \( t \in \mathbb{R} \), we have:
   \[
   f_{\pi}(t + T) = f_{\pi}(t)
   \]

2. **Evaluate at \( t = 0 \)**: Specifically, at \( t = 0 \), we get:
   \[
   f_{\pi}(T) = f_{\pi}(0)
   \]
   Given the definition of \( f_{\pi} \), this translates to:
   \[
   (\sin(T), \cos(\pi T)) = (\sin(0), \cos(0)) = (0, 1)
   \]

3. **Analyze the equations**: From the above, we have two conditions:
   \[
   \sin(T) = 0 \quad \text{and} \quad \cos(\pi T) = 1
   \]

4. **Solve \( \sin(T) = 0 \)**: The solutions to \( \sin(T) = 0 \) are:
   \[
   T = k\pi \quad \text{for some integer } k
   \]

5. **Solve \( \cos(\pi T) = 1 \)**: The solutions to \( \cos(\pi T) = 1 \) are:
   \[
   \pi T = 2m\pi \quad \text{for some integer } m
   \]
   Simplifying, we get:
   \[
   T = 2m
   \]

6. **Combine the conditions**: For \( T \) to satisfy both conditions, \( T \) must be an integer multiple of \( \pi \) and an even integer. Therefore, \( T = 2m \) must be an integer multiple of \( \pi \). This implies:
   \[
   2m = k\pi
   \]

7. **Contradiction**: Since \( \pi \) is irrational (as proved by Euler in 1737), \( k\pi \) cannot be an integer unless \( k = 0 \). Thus, there is no positive \( T \) that satisfies both conditions simultaneously. Therefore, \( f_{\pi} \) cannot be periodic.

\(\blacksquare\)

### Part (b): Determine the values of the parameter \( a \) for which \( f_a \) is periodic.

1. **Assume \( f_a \) is periodic**: Suppose \( f_a \) is periodic with period \( T > 0 \). This means:
   \[
   f_a(t + T) = f_a(t) \quad \text{for all } t \in \mathbb{R}
   \]

2. **Evaluate at \( t = 0 \)**: Specifically, at \( t = 0 \), we get:
   \[
   f_a(T) = f_a(0)
   \]
   Given the definition of \( f_a \), this translates to:
   \[
   (\sin(T), \cos(aT)) = (\sin(0), \cos(0)) = (0, 1)
   \]

3. **Analyze the equations**: From the above, we have two conditions:
   \[
   \sin(T) = 0 \quad \text{and} \quad \cos(aT) = 1
   \]

4. **Solve \( \sin(T) = 0 \)**: The solutions to \( \sin(T) = 0 \) are:
   \[
   T = k\pi \quad \text{for some integer } k
   \]

5. **Solve \( \cos(aT) = 1 \)**: The solutions to \( \cos(aT) = 1 \) are:
   \[
   aT = 2m\pi \quad \text{for some integer } m
   \]
   Simplifying, we get:
   \[
   a = \frac{2m\pi}{k\pi} = \frac{2m}{k}
   \]

6. **Conclusion**: For \( f_a \) to be periodic, \( a \) must be a rational number of the form \( \frac{2m}{k} \). Therefore, \( a \) must be rational.

\(\blacksquare\)

The final answer is \( \boxed{ a } \) must be rational.