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

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 $\sin 3\theta$.

& 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{aligned}$$ Since we know that, $$\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{aligned} & \Rightarrow 4-\sin \left( 3\theta +2\pi \right) \\\ & \Rightarrow 4-\sin 3\theta \\\ \end{aligned}$$ So we can conclude that $$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.