Question

The period of the function $f\left( x \right)=4{{\sin }^{4}}\left( \dfrac{4x-3\pi }{6{{\pi }^{2}}} \right)+2\cos \left( \dfrac{4x-3\pi }{3{{\pi }^{2}}} \right)$
(A) $\dfrac{3{{\pi }^{2}}}{4}$
(B) $\dfrac{3{{\pi }^{3}}}{4}$
(C) $\dfrac{4{{\pi }^{2}}}{3}$
(D) $\dfrac{4{{\pi }^{3}}}{3}$

Answer 1

We start solving this problem by first assuming that $\dfrac{4x-3\pi }{3{{\pi }^{2}}}=2t$. Then we solve the expression by using the trigonometric identities $\cos 2\theta =1-2{{\sin }^{2}}\theta$ and $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and simplify the function until we transform it into a single trigonometric function. Then we use the property of period of any function $f\left( x \right)=\cos \left( ax+b \right)$ is $\dfrac{2\pi }{a}$ to find the period of our given function.

Complete step-by-step answer:
First let us consider $\dfrac{4x-3\pi }{6{{\pi }^{2}}}=t$.
Then we get $\dfrac{4x-3\pi }{3{{\pi }^{2}}}=2t$.
Then we can write the function as $f\left( x \right)=4{{\sin }^{4}}t+2\cos 2t$
Now, let us simplify the expression $4{{\sin }^{4}}t+2\cos 2t$.
Now let us consider the formula $\cos 2\theta =1-2{{\sin }^{2}}\theta \Rightarrow 2{{\sin }^{2}}\theta =1-\cos 2\theta$.
So, we can write our above expression as,
$\begin{aligned} & \Rightarrow 4{{\sin }^{4}}t+2\cos 2t={{\left( 1-\cos 2t \right)}^{2}}+2\cos 2t \\\ & \Rightarrow 4{{\sin }^{4}}t+2\cos 2t={{\cos }^{2}}2t-2\cos 2t+1+2\cos 2t \\\ & \Rightarrow 4{{\sin }^{4}}t+2\cos 2t={{\cos }^{2}}2t+1 \\\ \end{aligned}$
Now let us consider the formula $\cos 2\theta =2{{\cos }^{2}}\theta -1\Rightarrow {{\cos }^{2}}\theta =\dfrac{\cos 2\theta +1}{2}$.
Using the above formula, we can write above equation as,
$\begin{aligned} & \Rightarrow 4{{\sin }^{4}}t+2\cos 2t=\dfrac{\cos 4t+1}{2}+1 \\\ & \Rightarrow 4{{\sin }^{4}}t+2\cos 2t=\dfrac{\cos 4t+3}{2} \\\ \end{aligned}$
So, we can write it as
$\Rightarrow 4{{\sin }^{4}}t+2\cos 2t=\dfrac{1}{2}\cos 4t+\dfrac{3}{2}$
So, we get the function f as,
$f\left( x \right)=\dfrac{1}{2}\cos 4t+\dfrac{3}{2}$
Now, substituting the value of t, we get
$\begin{aligned} & \Rightarrow f\left( x \right)=\dfrac{1}{2}\cos 4\left( \dfrac{4x-3\pi }{6{{\pi }^{2}}} \right)+\dfrac{3}{2} \\\ & \Rightarrow f\left( x \right)=\dfrac{1}{2}\cos \dfrac{8x-6\pi }{3{{\pi }^{2}}}+\dfrac{3}{2} \\\ & \Rightarrow f\left( x \right)=\dfrac{1}{2}\cos \left( \dfrac{8x}{3{{\pi }^{2}}}-\dfrac{2}{\pi } \right)+\dfrac{3}{2} \\\ \end{aligned}$
Now we need to find the period of the above function f(x).
We know that the period of any function $f\left( x \right)=\cos x$ is $2\pi$.
Then the period of any function $f\left( x \right)=\cos \left( ax+b \right)$ is $\dfrac{2\pi }{a}$.
Using that we can say that the period of our function f(x) is given by,
$\dfrac{2\pi }{\dfrac{8}{3{{\pi }^{2}}}}=\dfrac{2\pi \times 3{{\pi }^{2}}}{8}=\dfrac{3{{\pi }^{3}}}{4}$
So, we get the period of the function f(x) as $\dfrac{3{{\pi }^{3}}}{4}$.

So, the correct answer is “Option B”.

Note: There is a possibility of making a mistake while solving this problem by taking the period of cosx as $\dfrac{\pi }{2}$. But it is wrong. For any function sinx or cosx, the period of the function is $\pi$. Another mistake possible while solving this question is one might take the formula for cos2x as $\cos 2\theta =2{{\sin }^{2}}\theta -1$. But the actual formula is $\cos 2\theta =1-2{{\sin }^{2}}\theta$.