Let's break down how to find the amplitude and period of the function [tex]\( y = \frac{1}{2} \sin(3x) \)[/tex].
### Step-by-Step Solution
1. Given Function:
[tex]\[
y = \frac{1}{2} \sin(3x)
\][/tex]
2. Amplitude:
The amplitude of a sine function [tex]\( y = A \sin(Bx) \)[/tex] is given by the coefficient [tex]\( A \)[/tex] in front of the sine term.
In our function, [tex]\( A = \frac{1}{2} \)[/tex].
Therefore, the amplitude is
[tex]\[
A = \frac{1}{2}
\][/tex]
3. Period:
The period of a sine function [tex]\( y = A \sin(Bx) \)[/tex] is determined by the coefficient [tex]\( B \)[/tex] in front of [tex]\( x \)[/tex]. The formula for the period [tex]\( T \)[/tex] is
[tex]\[
T = \frac{2\pi}{B}
\][/tex]
In our function, [tex]\( B = 3 \)[/tex].
Therefore, the period is
[tex]\[
T = \frac{2\pi}{3}
\][/tex]
### Summary
- Amplitude:
[tex]\[
\frac{1}{2}
\][/tex]
- Period:
[tex]\[
\frac{2\pi}{3}
\][/tex]
### Final Answer
Therefore, the exact values are:
- Amplitude: [tex]\(\frac{1}{2}\)[/tex]
- Period: [tex]\(\frac{2\pi}{3}\)[/tex]
