Question

Which of the following functions has a period of 8 π ?

Answer 1

Answer: Choice B

Explanation:

The general template of a sine function is

y = Asin(B(x-C))+D

The value of B is the only thing that determines the period T. We ignore every other term.

It turns out that T = 2pi/B which can be solved to B = 2pi/T

Plug in the given period of T = 8pi and compute to get

B = 2pi/T

B = (2pi)/(8pi)

B = 2/8

B = 1/4

This means that the term out front of the x is 1/4 and the equation must be $y = \sin\left(\frac{1}{4}x\right)$ which matches with answer choice B.