Question

How do you find the period of $y = 2\tan (3\pi x + 4)$

Answer 1

Compare the given equation with the general form of the tangent function and use the period of tangent function formula. Here $y = 2\tan \left( {3\pi x + 4} \right)$ must be compared to the general form of the tangent function i.e. $y = A\tan \left( {Bx - c} \right)$. Here the coefficients of the given tangent 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 $\dfrac{\pi }{B}$ to find the period of the given tangent equation. We divide the numerator i.e. $\pi$ 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 $y = 2\tan \left( {3\pi x + 4} \right)$
The tangent function equation is of the general form$y = A\tan \left( {Bx - c} \right)$
The formula to calculate the period of the tangent function is $\dfrac{\pi }{B}$
In the given tangent equation, $B = 3\pi$
Therefore, substituting B in the formula of finding the period of tangent function we get, $\dfrac{\pi }{B} = \dfrac{\pi }{{3\pi }}$
By cancelling out the common terms we get the period as $\dfrac{1}{3}$

Hence, the period of the given tangent equation $y = 2\tan \left( {3\pi x + 4} \right)$is $\dfrac{1}{3}$

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 tangent and different for other trigonometric functions. The formula for finding the period of a tangent function is $\dfrac{\pi }{B}$, whereas the for sine and cosine function it is $\dfrac{{2\pi }}{B}$. So make sure you use the correct formulas every time.