Question

Find the period for \[{{\sin }^{3}}x\].

Answer 1

Hint: For solving this question you should know about the period of \[{{\sin }^{n}}x\]. In this problem first we will find the period for \[\sin nx\] and then we determine the period for \[{{\sin }^{n}}x\] and thus we will get the period for \[{{\sin }^{3}}x\].

Complete step by step solution:
According to the question it is asked to find the period for \[{{\sin }^{3}}x\].
As we know that in order to find the period of the function, one just needs to find out the interval when the values of the function starts repeating itself.
Since, the value of \[\sin \left( nx \right)\] can only lie in between (-1 and 1), and there are no such values in between \[\sin \left( -1 \right)\] and \[\sin \left( 1 \right)\] that repeat itself therefore our solution lies in the period of \[\sin \left( nx \right)\].
So, when the value of \[\sin \left( nx \right)\] repeats itself, so will our value of \[\sin x=\sin \left( \sin \left( nx \right) \right)\].
And period of \[\sin x=2\pi \] because \[\sin \left( x+2n\pi \right)=\sin x\] and for our fundamental case, to find the smallest value, we get our period as \[2\pi \].
However, here we have \[{{\sin }^{n}}x\], therefore its period will also be \[2\pi \]. Similar to our n when I was explaining about the periodicity of \[\sin x\].
\[\sin \left( n\left( x+2\pi \right) \right)=\sin \left( nx+2n\pi \right)={{\sin }^{n}}x\]
So, the period of \[{{\sin }^{n}}x\] will be \[2\pi \].
And this also remains the same for \[{{\sin }^{3}}x\].
So, the period of \[{{\sin }^{3}}x\] is \[2\pi \].

Note: While solving the period for any function of trigonometry first you have to ensure that the given function contains any finite value or not. Because it will be different for both the values of trigonometric functions. And then find the fundamental period of the function \[f\left( x \right)\].