Question

The period of ${\sin }^2\theta$ is

• A. ${\pi }^2$
• B. $\pi$
• C. $2\pi$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$\pi$

Step 1: Form a function for given trigonometric expression

Given, ${\sin }^2\theta$

We know that,

$\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }$

$\Rightarrow$ $2si{n}^2\theta =1-cos2\theta$

$\Rightarrow$ $si{n}^2\theta =\frac{1-\cos 2\theta }{2}$

$\Rightarrow$ $si{n}^2\theta =\frac{1}{2}-\frac{\cos 2\theta }{2}$ $...\left(\mathrm{i}\right)$

Step 2: Calculate the period

Now, as we know that the period of $\cos \theta$ is $2\pi$.

i.e., $T\left(\cos \theta \right)=2\pi$

$\Rightarrow$ $T\left(\cos 2\theta \right)=\frac{2\pi }{2}$

$\Rightarrow$ $T\left(\cos 2\theta \right)=\pi$

$\Rightarrow$ $T\left(\frac{1}{2}-\frac{\cos 2\theta }{2}\right)=\pi$

$\Rightarrow$ $T\left({\sin }^2\theta \right)=\pi$ [using equation $\left(\mathrm{i}\right)$]

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

Hence, option (B) is the correct option.