Question

Determine the period of the following trigonometric function.
y, equals, one quarter, sine, left parenthesis, start fraction, 8, pi, x, divided by, 3, end fraction, plus, 3, right parenthesis
y=
4
1
​
sin(
3
8πx
​
+3)

Answer 1

To determine the period of the given trigonometric function \( y = \frac{1}{4} \sin\left( \frac{8\pi}{3} x + 3 \right) \), we will focus on the argument of the sine function, which is \( \frac{8\pi}{3} x + 3 \).

The standard form of a sine function is given by:

\[
y = A \sin(Bx + C) + D
\]

where \( B \) affects the period of the sine function. The period \( T \) of the sine function is calculated using the formula:

\[
T = \frac{2\pi}{|B|}
\]

In this case, we identify \( B \) as \( \frac{8\pi}{3} \). Now, we can calculate the period:

\[
T = \frac{2\pi}{\left|\frac{8\pi}{3}\right|} = \frac{2\pi}{\frac{8\pi}{3}} = 2\pi \cdot \frac{3}{8\pi} = \frac{6}{8} = \frac{3}{4}
\]

Thus, the period of the function \( y = \frac{1}{4} \sin\left( \frac{8\pi}{3} x + 3 \right) \) is:

\[
\boxed{\frac{3}{4}}
\]