Question

Period of $\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}+x \right)$ is
$\begin{aligned} & a)\dfrac{\pi }{2} \\\ & b)\pi \\\ & c)\dfrac{3\pi }{2} \\\ & d)2\pi \\\ \end{aligned}$

Answer 1

Now first we will convert the given expression with the help of formula $\cos \left( C-D \right)+\cos \left( C+D \right)=2\sin C\sin D$ . Now we know that $\cos \left( \dfrac{\pi }{2} \right)=0$ hence we will get a simplified equation in terms of cos. Now again we know that the period of functions of type $a\cos \left( bx \right)$ is given by $\dfrac{2\pi }{b}$ . Hence we can find the period of the given function.

Complete step-by-step answer:
First let us understand the term period. A periodic function is a function which repeats the same values at regular intervals. The distance between this repetition is called the period of function. For a periodic function we have $f\left( x+T \right)=f\left( x \right)$ where T is the period of function. Let us take an example to understand.
We know that $\sin \left( 2\pi +\theta \right)=\sin \left( \theta \right)$ hence we can say that the period of the function $\sin \theta$ is $2\pi$ .
Now consider the given function $\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}+x \right)$ .
We know that $\cos \left( C-D \right)+\cos \left( C+D \right)=2\sin C\sin D$
Hence we can say that $\dfrac{\cos \left( C-D \right)+\cos \left( C+D \right)}{2}=\sin C\sin D$
Now comparing the given function with RHS of above equation we get $C=\left( \dfrac{\pi }{4}-x \right)$ $D=\left( \dfrac{\pi }{4}+x \right)$ .
Hence we get,
$\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}+x \right)=\dfrac{\cos \left( \left( \dfrac{\pi }{4}-x \right)-\left( \dfrac{\pi }{4}+x \right) \right)+\cos \left( \left( \dfrac{\pi }{4}-x \right)-\left( \dfrac{\pi }{4}+x \right) \right)}{2}$
Now opening the brackets we get,
$\begin{aligned} & \sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}+x \right)=\dfrac{\cos \left( \dfrac{\pi }{4}-x+\dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x-\dfrac{\pi }{4}-x \right)}{2} \\\ & \Rightarrow \sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}+x \right)=\dfrac{\cos \left( \dfrac{\pi }{2} \right)+\cos \left( -2x \right)}{2} \\\ \end{aligned}$
Now we know that $\cos \dfrac{\pi }{2}=0$ , using this we get and $\cos \left( -x \right)=\cos \left( x \right)$
$\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}+x \right)=\dfrac{\cos \left( 2x \right)}{2}...................\left( 1 \right)$ .
Now for any function of the type $a\cos \left( bx \right)$ the period is given by $\dfrac{2\pi }{b}$ .
When we compare the expression $\dfrac{\cos \left( 2x \right)}{2}$ with $a\cos \left( bx \right)$ we get $a=\dfrac{1}{2}$ and b = 2.
Hence we get the period of $\dfrac{\cos \left( 2x \right)}{2}$ is $\dfrac{2\pi }{2}=\pi$ .
Hence from equation (1) we get the period of $\sin \left( \dfrac{\pi }{4}-x \right)\sin \left( \dfrac{\pi }{4}+x \right)$ is $\pi$ .

So, the correct answer is “Option b”.

Note: Now note that the period is the shortest distance after which the function repeats itself. Hence even though we have $\sin \left( 4\pi +x \right)=\sin x$ $4\pi$ is not the shortest distance and hence not the period of the function. Also note that for a function $a\cos \left( bx \right)$ the period does not depend on amplitude a.