The period of sin² θ is

The correct option is π.

To find the period of the function $f\left(\theta \right)={sin}^2\theta$, we can use trigonometric identities to simplify the expression into a linear trigonometric function.

Recall the double-angle formula for cosine:
$cos\left(2\theta \right)=1-2{sin}^2\theta$

Rearranging this formula to express ${sin}^2\theta$ in terms of $cos\left(2\theta \right)$, we get:
${sin}^2\theta =\frac{1-cos\left(2\theta \right)}{2}$

This can be rewritten as:
$f\left(\theta \right)=\frac{1}{2}-\frac{1}{2}cos\left(2\theta \right)$

The constant term $\frac{1}{2}$ and the coefficient $-\frac{1}{2}$ shift and scale the function vertically but do not affect its period. The period of the function is entirely determined by the term $cos\left(2\theta \right)$.

The standard cosine function, $cos\left(\theta \right)$, has a fundamental period of $2\pi$. For a function of the form $cos\left(k\theta \right)$, the period $T$ is given by:
$T=\frac{2\pi }{\left|k\right|}$

In our function, $k=2$. Therefore, the period $T$ is:
$T=\frac{2\pi }{2}=\pi$

Thus, the period of ${sin}^2\theta$ is $\pi$.