Is the function f(x)`=sqrt(sinx)` periodic?

Answer

To determine if the function \( f(x) = \sqrt{\sin x} \) is periodic, we need to check if there exists a positive number \( T \) such that \( f(x + T) = f(x) \) for all \( x \).

### Step-by-Step Solution:

1. **Define the Function**:
We start with the function \( f(x) = \sqrt{\sin x} \).

2. **Check the Condition for Periodicity**:
A function \( f(x) \) is periodic if there exists a period \( T \) such that:
\[ f(x + T) = f(x) \quad \text{for all } x. \]

3. **Assume a Period \( T \)**:
Let's assume \( T \) is the period. Then we have:
\[ f(x + T) = \sqrt{\sin(x + T)}. \]

4. **Use the Sine Addition Formula**:
We can use the sine addition formula:
\[ \sin(x + T) = \sin x \cos T + \cos x \sin T. \]
Therefore,
\[ f(x + T) = \sqrt{\sin x \cos T + \cos x \sin T}. \]

5. **Set Up the Equation**:
For \( f(x + T) \) to equal \( f(x) \), we need:
\[ \sqrt{\sin x \cos T + \cos x \sin T} = \sqrt{\sin x}. \]

6. **Square Both Sides**:
Squaring both sides gives:
\[ \sin x \cos T + \cos x \sin T = \sin x. \]

7. **Rearranging the Equation**:
Rearranging this equation leads to:
\[ \sin x \cos T + \cos x \sin T - \sin x = 0. \]
This simplifies to:
\[ \sin x (\cos T - 1) + \cos x \sin T = 0. \]

8. **Analyzing the Equation**:
For this equation to hold for all \( x \), both coefficients must be zero:
- \( \cos T - 1 = 0 \) implies \( \cos T = 1 \) (which occurs at \( T = 2n\pi \) for \( n \in \mathbb{Z} \)).
- \( \sin T = 0 \) implies \( T = n\pi \) for \( n \in \mathbb{Z} \).

9. **Finding the Minimum Period**:
The smallest positive \( T \) that satisfies both conditions is \( T = 2\pi \).

10. **Conclusion**:
Since we found a positive \( T \) such that \( f(x + T) = f(x) \), we conclude that the function \( f(x) = \sqrt{\sin x} \) is periodic with a period of \( 2\pi \).

### Final Answer:
The function \( f(x) = \sqrt{\sin x} \) is periodic with a period of \( 2\pi \).