Question

The period of ${{\sin} ^ {2}} x$ is $$$$
A.\dfrac{\pi} {2}$$$$$ B. \pi C. $\dfrac{3\pi }{2}
D. $2\pi$$$$$

Answer 1

We recall the definition of period and try find the minimum value positive value of $P$ such that ${{\sin }^{2}}\left( x+P \right)={{\sin }^{2}}x$. We use the formula ${{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$ , find all the solution of $\cos \theta =\cos \alpha$ that is $\theta =2n\pi +\alpha$and check for which value of $n$ we get minimum $P$, which will be the period.$$$$

Complete step-by-step solution:
We know from cosine double angle formula that for any acute angle $\theta$,

& \cos 2\theta =1-2{{\sin }^{2}}\theta \\\ & \Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}....\left( 1 \right) \\\ \end{aligned}$$ We know that the solutions of the equation $\cos \theta =\cos \alpha $ are given by $$\theta =2n\pi +\alpha ,n\in Z$$ We know that the period of a real valued function $f\left( x \right)$ is a nonzero positive constant $P$ for all $x$ in the domain of $f$ if $$f\left( x+P \right)=f\left( x \right)$$ $P$ is also called the fundamental period or basic period. All the functional values of $f\left( x \right)$ will repeat after the interval of length $P$. The plot of periodic function exhibits translational symmetry which means the graph of function $f\left( x \right)$ is invariant in $x-$ axis with respect to translation by a length $P$. We can also writ for some positive integer $n$ $$f\left( x+nP \right)=f\left( x \right)$$ Here the given function is $f\left( x \right)={{\sin }^{2}}x$ where $f:{{R}^{+}}\bigcup \left\\{ 0 \right\\}\to {{R}^{+}}\bigcup \left\\{ 0 \right\\}$. Let us assume the period of ${{\sin }^{2}}x$ is $P$. So we have $${{\sin }^{2}}\left( x+P \right)={{\sin }^{2}}x$$ We use relation (1) for $\theta =x+P,x$ and have $$\begin{aligned} & \dfrac{1-\cos \left( 2\left( x+P \right) \right)}{2}=\dfrac{1-\cos 2x}{2} \\\ & \Rightarrow \cos \left( 2x+2P \right)=\cos 2x \\\ \end{aligned}$$ The solutions of the above equation with some integer $n$ are $$\begin{aligned} & 2x+2P=2x+2n\pi \\\ & \Rightarrow x+P=x+n\pi \\\ & \Rightarrow P=n\pi \\\ \end{aligned}$$ The fundamental period is the minimum non-zero positive value of $P$ which we find with $n=1$ and hence the period of ${{\sin }^{2}}x$ is $P=1\times \pi =\pi $. So the correct option is C. If we plot $y={{\sin }^{2}}x$ then we can see the repeat of functional values after a period$\pi =3.14$.$$$$  **Note:** We can alternatively solve by first finding the period of $\cos 2x$ using the fact that the period of $f\left( ax \right)$ is $\dfrac{P}{\left| a \right|}$. We then use the fact that the period of functions $f\left( x \right)$ and $g\left( x \right)$ are equal if and only if $f\left( x \right)=a+bg\left( x \right)$for ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$. If a function foes not have a period it is called aperiodic and if $f\left( x+T \right)=-f\left( x \right)$ then it is called anti-periodic.