Question

How do you find the period of $y = \left( {\dfrac{1}{2}} \right)\sin \left( {\left( {\dfrac{x}{3}} \right) - \pi } \right)$

Answer 1

Hint: Try to compare the given equation with the general form of the sine function and use the period of sine function formula to yield the answer.
Here $y = \left( {\dfrac{1}{2}} \right)\sin \left( {\left( {\dfrac{x}{3}} \right) - \pi } \right)$ must be compared to the general form of the sine function i.e. $y = A\sin (Bx - C)$$2\pi $. Here the coefficients of the given sine form of the equation must be compared to the general form so that they can be substituted in the formula for finding the period. After comparing the values of the coefficients from the given equation we use the formula to find the period of the given sine equation. We divide the numerator i.e. $2\pi $with denominator i.e. B and then cancel out the common terms which will finally yield our answer.

Complete step by step solution:
Here the given equation is $y = \left( {\dfrac{1}{2}} \right)\sin \left( {\left( {\dfrac{x}{3}} \right) - \pi } \right)$
The sine function equation is of the general form $y = A\sin (Bx - C)$
The formula to calculate the period of the sine function is $\dfrac{{2\pi }}{B}$
In the given sine equation, $B = \dfrac{1}{3}$
Therefore, substituting B in the formula of finding the period of sine function we get
$\dfrac{{2\pi }}{B} = \dfrac{{2\pi }}{{\dfrac{1}{3}}}$
By simplifying the above equation we get the period as, $6\pi $

Hence, the period of the given sine equation $y = \left( {\dfrac{1}{2}} \right)\sin \left( {\left( {\dfrac{x}{3}} \right) - \pi } \right)$ is $6\pi $.

Note: While using the formula for finding the period of a given equation one must carefully look at the equation since the formula for finding the period is different for sine and different for tangent function, the formula for finding the period is $\dfrac{{2\pi }}{B}$, whereas the for other tangent function it is $\dfrac{\pi }{B}$.