Find the period of the function
`f(x) = (sin x + sin 3x)/(cos x + cos 3x)`

Answer

To find the period of the function \( f(x) = \frac{\sin x + \sin 3x}{\cos x + \cos 3x} \), we can follow these steps:

### Step 1: Use the sum-to-product identities
We can apply the sum-to-product identities for sine and cosine. The identities are:
- \( \sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \)
- \( \cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \)

For our function, we have:
- \( a = x \) and \( b = 3x \)

### Step 2: Apply the identities
Using the identities, we can rewrite the numerator and denominator:
- **Numerator:**
\[ \sin x + \sin 3x = 2 \sin\left(\frac{x + 3x}{2}\right) \cos\left(\frac{x - 3x}{2}\right) = 2 \sin(2x) \cos(x) \]
- **Denominator:**
\[ \cos x + \cos 3x = 2 \cos\left(\frac{x + 3x}{2}\right) \cos\left(\frac{x - 3x}{2}\right) = 2 \cos(2x) \cos(x) \]

### Step 3: Simplify the function
Now substituting back into the function:
\[ f(x) = \frac{2 \sin(2x) \cos(x)}{2 \cos(2x) \cos(x)} \]
We can cancel \( 2 \cos(x) \) from the numerator and denominator (assuming \( \cos(x) \neq 0 \)):
\[ f(x) = \frac{\sin(2x)}{\cos(2x)} = \tan(2x) \]

### Step 4: Determine the period of \( \tan(2x) \)
The period of \( \tan(x) \) is \( \pi \). Therefore, the period of \( \tan(2x) \) is:
\[ \text{Period of } \tan(2x) = \frac{\pi}{2} \]

### Step 5: Verify the period
To confirm, we can check if \( f(x + \frac{\pi}{2}) = f(x) \):
\[ f\left(x + \frac{\pi}{2}\right) = \tan\left(2\left(x + \frac{\pi}{2}\right)\right) = \tan(2x + \pi) = \tan(2x) \]
This shows that the function is periodic with period \( \frac{\pi}{2} \).

### Conclusion
Thus, the period of the function \( f(x) = \frac{\sin x + \sin 3x}{\cos x + \cos 3x} \) is:
\[ \boxed{\frac{\pi}{2}} \]