The period of `sin^(2) theta` , is

Options

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

Answer

To find the period of the function \( \sin^2 \theta \), we can follow these steps:

### Step 1: Rewrite the function
We start by rewriting \( \sin^2 \theta \) using a trigonometric identity. The identity states that:
\[ \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \]

### Step 2: Identify the periodic component
From the rewritten form, we can see that the function \( \sin^2 \theta \) is expressed in terms of \( \cos(2\theta) \). The cosine function has a known period.

### Step 3: Determine the period of \( \cos(2\theta) \)
The standard period of \( \cos \theta \) is \( 2\pi \). However, since we have \( \cos(2\theta) \), we need to adjust the period. The period of \( \cos(k\theta) \) is given by:
\[ \text{Period} = \frac{2\pi}{|k|} \]
where \( k = 2 \) in this case. Thus, the period of \( \cos(2\theta) \) is:
\[ \text{Period of } \cos(2\theta) = \frac{2\pi}{2} = \pi \]

### Step 4: Conclude the period of \( \sin^2 \theta \)
Since \( \sin^2 \theta \) is derived from \( \cos(2\theta) \), the period of \( \sin^2 \theta \) is the same as the period of \( \cos(2\theta) \):
\[ \text{Period of } \sin^2 \theta = \pi \]

### Final Answer
Thus, the period of \( \sin^2 \theta \) is \( \pi \).

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