Prove sin x is periodic and find its period.

Answer

To prove that \( \sin x \) is periodic and to find its period, we can follow these steps:

### Step 1: Definition of Periodic Function
A function \( f(x) \) is said to be periodic if there exists a positive constant \( T \) such that:
\[ f(x + T) = f(x) \quad \text{for all } x \]
In this case, we want to show that \( \sin x \) satisfies this condition.

### Step 2: Let \( T \) be the period of \( \sin x \)
Assume that \( T \) is the period of \( \sin x \). Therefore, we have:
\[ \sin(x + T) = \sin x \]

### Step 3: Use the Sine Addition Formula
Using the sine addition formula, we can express \( \sin(x + T) \):
\[ \sin(x + T) = \sin x \cos T + \cos x \sin T \]
Setting this equal to \( \sin x \), we have:
\[ \sin x \cos T + \cos x \sin T = \sin x \]

### Step 4: Rearranging the Equation
Rearranging gives us:
\[ \sin x \cos T - \sin x = -\cos x \sin T \]
Factoring out \( \sin x \) from the left side:
\[ \sin x (\cos T - 1) = -\cos x \sin T \]

### Step 5: Analyzing the Equation
For this equation to hold for all \( x \), both sides must equal zero. This leads us to two cases:
1. \( \sin T = 0 \)
2. \( \cos T - 1 = 0 \)

### Step 6: Finding Values of \( T \)
From \( \sin T = 0 \), we find:
\[ T = n\pi \quad \text{where } n \text{ is an integer} \]
From \( \cos T - 1 = 0 \), we find:
\[ T = 2k\pi \quad \text{where } k \text{ is an integer} \]

### Step 7: Determining the Fundamental Period
The smallest positive value of \( T \) that satisfies both conditions is:
\[ T = 2\pi \]
Thus, the period of \( \sin x \) is \( 2\pi \).

### Conclusion
We have shown that \( \sin x \) is periodic with a period of \( 2\pi \).

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