Question

The period of the function \[f\left( \theta \right)=4+4\sin ^3\theta -3\sin \theta \]?
1.\[\dfrac{2\pi }{3}\]
2.\[\dfrac{\pi }{3}\]
3.\[\dfrac{\pi }{2}\]
4.\[\pi \]

Answer 1

Hint: In order to find the period of the given function \[f\left( \theta \right)=4+4\sin 3\theta -3\sin \theta \], firstly we will be trying to express the function in terms of \[\sin 3\theta \]. Then we will be checking the period of the function by substituting \[\theta +\dfrac{2\pi }{3}\] instead of \[\theta \]. Then we will be checking if the result obtained for \[\theta +\dfrac{2\pi }{3}\] is the same as when \[\theta \] was calculated. If they are equal, then we will be concluding with the period of the function.

Complete step by step answer:
Now let us learn about the period of a function. The distance between the repetition of any function is called the period of the function. In the case of the trigonometric function, the length of one complete cycle is called a period. Each function will have its own period. To any function, the reciprocal of the period is called the frequency of the function. We can find a period when it is represented as \[f\left( x \right)=f\left( x+p \right)\], \[p\] is the period of the function.
Now let us start finding the period of the given function \[f\left( \theta \right)=4+4\sin ^3\theta -3\sin \theta \].
Firstly, we will be expressing in terms of \[\sin 3\theta \].
\[\begin{align}
  & f\left( \theta \right)=4+4{{\sin }^{3}}\theta -3\sin \theta \\
 & \Rightarrow 4-\left( 3\sin \theta -4{{\sin }^{3}}\theta \right) \\
 & \Rightarrow 4-\sin 3\theta \\
\end{align}\]
Since we know that, \[\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta \]
So \[f\left( \theta \right)=4-\sin 3\theta \]
Now, \[f\left( \theta +\dfrac{2\pi }{3} \right)\]
Let us check for it,
\[f\left( \theta +\dfrac{2\pi }{3} \right)=4-\sin \left[ 3\left( \theta +\dfrac{2\pi }{3} \right) \right]\]
Upon solving this further, we get
\[\begin{align}
  & \Rightarrow 4-\sin \left( 3\theta +2\pi \right) \\
 & \Rightarrow 4-\sin 3\theta \\
\end{align}\]
So we can conclude that \[f\left( \theta \right)\] is periodic with \[\dfrac{2\pi }{3}\].

So, the correct answer is “Option 1”.

Note: While computing for the period, we must consider the given conditions and substitute into the function. We must also note that the period repeats at a particular interval of time. This would be termed as periodicity. We can represent the period of a trigonometric function in a graph. We must also note that the cotangent and the tangent functions are periodic but they do have breaks in the graph.