The function f (x) = cos x is periodic with period

Options

A. `2 pi`
B. ` pi`
C. ` (pi)/(2)`
D. ` (pi)/(4)`

Answer

To determine the period of the function \( f(x) = \cos x \), we need to check the condition for periodicity. A function \( f(x) \) is periodic with period \( T \) if:

\[ f(x + T) = f(x) \quad \text{for all } x \]

### Step 1: Identify the function
The function we are analyzing is:
\[ f(x) = \cos x \]

### Step 2: Test the options for periodicity
We need to check the given options to see which one satisfies the periodic condition \( f(x + T) = f(x) \).

#### Option 1: \( T = \frac{\pi}{4} \)
Calculate \( f\left(\frac{\pi}{4} + x\right) \):
\[ f\left(\frac{\pi}{4} + x\right) = \cos\left(\frac{\pi}{4} + x\right) \]
Using the cosine addition formula:
\[ \cos\left(\frac{\pi}{4} + x\right) = \cos\frac{\pi}{4} \cos x - \sin\frac{\pi}{4} \sin x = \frac{\sqrt{2}}{2} \cos x - \frac{\sqrt{2}}{2} \sin x \]
This is not equal to \( \cos x \), so \( T = \frac{\pi}{4} \) is not a period.

#### Option 2: \( T = \frac{\pi}{2} \)
Calculate \( f\left(\frac{\pi}{2} + x\right) \):
\[ f\left(\frac{\pi}{2} + x\right) = \cos\left(\frac{\pi}{2} + x\right) = -\sin x \]
This is not equal to \( \cos x \), so \( T = \frac{\pi}{2} \) is not a period.

#### Option 3: \( T = \pi \)
Calculate \( f(\pi + x) \):
\[ f(\pi + x) = \cos(\pi + x) = -\cos x \]
This is not equal to \( \cos x \), so \( T = \pi \) is not a period.

#### Option 4: \( T = 2\pi \)
Calculate \( f(2\pi + x) \):
\[ f(2\pi + x) = \cos(2\pi + x) = \cos x \]
This is equal to \( \cos x \), so \( T = 2\pi \) is a period.

### Conclusion
The period of the function \( f(x) = \cos x \) is:
\[ \boxed{2\pi} \]