Question

What is the equation for a sine function with a period of $\dfrac{3}{7}$, in radians?

Answer 1

First understand the meaning of period of a function. Assume T’ as the period of the required sine function. Further, assume the equation of the sine function as $\sin \left( ax \right)$ where ‘a’ is any constant that we have to determine. Now, use the fact that the period of the sine function represented as $\sin x$ is $T=2\pi$ and use the relation $T'=\dfrac{T}{\left| a \right|}$. Substitute the given values and find the value of ‘a’ to get the answer.

Complete step by step solution:
Here we have been provided with the period of a sine function as $\dfrac{3}{7}$ which is in radian and we are asked to determine the sine function. First we need to understand the meaning of ‘period’ of a function.
Now, in mathematics the period of a function $y=f\left( x \right)$ is defined as the interval of x after which the value of the function starts repeating itself. Such a function is known as a periodic function. Generally we denote the period of a function with T and mathematically we have $f\left( T+x \right)=f\left( x \right)$. T is a constant. There is a property of periodic function that if T is the period of $f\left( x \right)$ then the period of the function $f\left( ax \right)$ is given by $T'=\dfrac{T}{\left| a \right|}$. Here $\left( ax \right)$ is called linear in x.

Let us come to the question. We know that the period of sine function is $T=2\pi$ so assuming the required sine function as $\sin \left( ax \right)$ and its period as T’ we must have,
$\Rightarrow T'=\dfrac{2\pi }{\left| a \right|}$
Substituting the given value of $T'=\dfrac{3}{7}$ we get,
$\begin{aligned} & \Rightarrow \dfrac{3}{7}=\dfrac{2\pi }{\left| a \right|} \\\ & \Rightarrow \left| a \right|=\dfrac{7}{3}\times 2\pi \\\ & \Rightarrow \left| a \right|=\dfrac{14\pi }{3} \\\ \end{aligned}$
Removing the modulus function using the relation: - if $\left| a \right|=n$ then $a=\pm n$ we get,
$\Rightarrow a=\pm \dfrac{14\pi }{3}$
Therefore, the equation of the sine function can be given by substituting the above value of ‘a’ in $\sin \left( ax \right)$. So we get,
$\Rightarrow \sin \left( ax \right)=\sin \left( \pm \dfrac{14\pi x}{3} \right)$
Using the property $\sin \left( -x \right)=-\sin x$ we get,
$\therefore \sin \left( ax \right)=\pm \sin \left( \dfrac{14\pi x}{3} \right)$
Hence, the above relation is our answer.

Note: You must remember the period of all the trigonometric functions because they are used in the chapter like integration where the limits will be made very large and this property of the periodic function will be used. The period of sine, cosine, secant and cosecant function is $2\pi$ whereas the period of the tangent and co – tangent function is $\pi$.