Question

The time period of \[{\sin ^3}x + {\cos ^3}x\] is
(A) \[\dfrac{\pi }{3}\]
(B) \[\pi \]
(C) \[2\pi \]
(D) \[\dfrac{{2\pi }}{3}\]

Answer 1

Hint: First, we will determine the time period \[{\sin ^3}x\] and the time period of \[{\cos ^3}x\]. Then, by using this we will determine the time period of the required function \[{\sin ^3}x + {\cos ^3}x\] by taking L.C.M of the time period of \[{\sin ^3}x\] and \[{\cos ^3}x\] and dividing it by H.C.F. of 1 and 1.

Complete step-by-step answer:
Given the function is \[{\sin ^3}x + {\cos ^3}x\].
The time period of \[{\sin ^a}\theta \] (when a is odd) is \[2\pi \].
Therefore, the time period of \[{\sin ^3}x\] is \[2\pi \].
The time period of \[{\cos ^a}\theta \] (when a is odd) is \[2\pi \].
Therefore, the time period of \[{\cos ^3}x\] is \[2\pi \].
The time period of \[{\sin ^3}x + {\cos ^3}x\] is given by taking L.C.M of the time period of both the given functions \[{\sin ^3}x\] and \[{\cos ^3}x\] is given by
 \[ \Rightarrow \dfrac{{L.C.M.(2\pi ,2\pi )}}{{H.C.F.(1,1)}}\]
Since, the L.C.M. of \[2\pi \] and \[2\pi \] is \[2\pi \]. And the H.C.F. of 1 and 1 is 1. Therefore,
 \[ \Rightarrow \dfrac{{L.C.M.(2\pi ,2\pi )}}{{H.C.F.(1,1)}} = \dfrac{{2\pi }}{1} = 2\pi \]
Hence, the time period of \[{\sin ^3}x + {\cos ^3}x\] is \[2\pi \].

Note: The time period of \[{\sin ^a}\theta \] ( when a is even) is \[\pi \] and the time period of \[{\cos ^a}\theta \] ( when a is even) is $\pi $. The time period of \[{\sin ^a}\theta \] ( when a is odd) is \[2\pi \] and the time period of \[{\cos ^a}\theta \] ( when a is odd) is \[2\pi \].