1. Key Concepts and Formulas

• Periodicity: A function $f(x)$ is periodic with period $T > 0$ if $f(x+T) = f(x)$ for all $x$ in its domain. The smallest such $T$ is the fundamental period.
• Period of basic trigonometric functions: The period of $\sin x$ and $\cos x$ is $2\pi$. The period of $\tan x$ is $\pi$.
• Period transformation: If the period of $g(x)$ is $T$, then the period of $g(ax+b)$ is $\frac{T}{|a|}$.
• Trigonometric Identity: The power-reducing identity for $\sin^2\theta$ is $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$.

2. Step-by-Step Solution

Step 1: Express ${\sin ^2}\theta$ in a simpler form.

• What: We will use the power-reducing identity to rewrite ${\sin ^2}\theta$ in terms of a linear cosine function.
• Why: ${\sin ^2}\theta$ involves a squared term, which is generally harder to analyze for periodicity directly. Expressing it as a linear combination of simpler trigonometric functions will make the period determination straightforward.
• Using the identity $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$, we have: $f(\theta) = \sin^2\theta = \frac{1 - \cos 2\theta}{2}$

Step 2: Analyze the periodicity of the transformed function.

• What: We need to find the period of $f(\theta) = \frac{1}{2} - \frac{1}{2}\cos 2\theta$.
• Why: The period of a sum or difference of functions is determined by the period of the term with the smallest period. Adding or multiplying by a constant does not change the period of a function. Therefore, the period of $f(\theta)$ will be determined by the period of $\cos 2\theta$.

Step 3: Determine the period of the core trigonometric component.

• What: We find the period of $\cos 2\theta$.
• Why: We apply the rule for the period of transformed trigonometric functions: if the period of $g(x)$ is $T$, then the period of $g(ax+b)$ is $\frac{T}{|a|}$.
• The period of $\cos x$ is $T = 2\pi$.
• For $\cos 2\theta$, we have $a=2$.
• Therefore, the period of $\cos 2\theta$ is $\frac{2\pi}{|2|} = \frac{2\pi}{2} = \pi$.

Step 4: Conclude the period of ${\sin ^2}\theta$.

• What: Since the period of $\cos 2\theta$ is $\pi$, and the constants $\frac{1}{2}$ and $-\frac{1}{2}$ do not affect the period, the period of $f(\theta) = \frac{1 - \cos 2\theta}{2}$ is also $\pi$.
• Why: As established in Step 2, transformations involving addition and multiplication by constants do not change the period.
• Thus, the period of ${\sin ^2}\theta$ is $\pi$.

3. Common Mistakes & Tips

• Mistake: Assuming that the period of $\sin^2\theta$ is half the period of $\sin\theta$ without proper justification. While this often holds for even powers, it's crucial to use identities to confirm.
• Tip: For powers of sine and cosine, remember the general rules: If $n$ is odd, the period of $\sin^n\theta$ and $\cos^n\theta$ is $2\pi$. If $n$ is even, the period is $\pi$. In this case, $n=2$ is even, so the period is $\pi$.
• Tip: Always simplify trigonometric expressions involving powers using identities before determining periodicity. This usually leads to linear forms of basic trigonometric functions, making period calculation easier.

4. Summary

To find the period of ${\sin ^2}\theta$, we used the power-reducing trigonometric identity $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$. This transformed the expression into a linear combination of a constant and $\cos 2\theta$. The period of a constant is undefined or considered infinite, and adding/subtracting constants or multiplying by non-zero constants does not alter the period of a function. Therefore, the period of ${\sin ^2}\theta$ is determined by the period of $\cos 2\theta$. Since the period of $\cos x$ is $2\pi$, the period of $\cos 2\theta$ is $\frac{2\pi}{|2|} = \pi$. Thus, the period of ${\sin ^2}\theta$ is $\pi$.

The final answer is $\boxed{\pi}$, which corresponds to option (D).