Problem 49 - Algebra 2

Question

Find the period and amplitude of each sine function. Then sketch each function from 0 to \(2 \pi\) $$ y=0.5 \sin \frac{\pi}{3} \theta $$

Step-by-Step Solution

Answer

The period of the function \(y = 0.5 \sin \left(\frac{\pi}{3} \theta\right)\) is 6 and the amplitude is 0.5. The function can be sketched starting at the origin, peaking at (1.5, 0.5), crossing x-axis at (3,0), reaching its minimum at (4.5, -0.5), and closing the period at (6,0)

Step 1: Identify B

On closer inspection of the function, you can spot \(B = \frac{\pi}{3}\).

Step 2: Calculate the Period

The period is calculated by \[T = \frac{2\pi}{B}\]. So, replacing B with \(\frac{\pi}{3}\) results in period \(T = 2\pi \times \frac{3}{\pi} = 6\). So the period is 6.

Step 3: Determine the Amplitude

Amplitude is the coefficient before \(\sin\), and in this function it is given as 0.5.

Step 4: Sketch the Function

Start by drawing a horizontal line (the x-axis). This will represent the possible values of \(\theta\). Mark points along the line corresponding to one period of the function, from 0 to 6. Draw a vertical line, the y-axis, through zero point on x-axis. This represents the output of the function. Mark points on it for the maximum and minimum values of y, which are the amplitude and its negative, 0.5 and -0.5 in this case. Now, sketch the sine curve starting at origin (0,0), peaking at (1.5, 0.5), crossing x-axis at (3,0), reaching its minimum at (4.5, -0.5), and finally closing the period at (6,0).

Key Concepts

Sine Function

The sine function is one of the fundamental trigonometric functions, frequently appearing in various mathematical contexts. It's defined as \[ y = A \sin(B\theta + C) + D \] where:

• A: Amplitude
• B: Frequency multiplier, which affects the period
• C: Horizontal shift
• D: Vertical shift

In our specific function, \[ y = 0.5 \sin \frac{\pi}{3} \theta \] there are no horizontal or vertical shifts, simplifying our equation. The sine function produces values that oscillate between its amplitude and its negative counterpart, creating a wave-like pattern. This wave-like pattern is periodic in nature, meaning it repeats itself at regular intervals, which leads us to understand the concepts of period and amplitude next.

Period of a Function

The period of a trigonometric function like the sine function defines how long it takes for the function to complete one full cycle. A complete cycle refers to the function starting at a point, rising to a maximum value, decreasing to a minimum, and returning back to the starting point.
In the general form of a sine function, the period \(T\) is calculated using the formula:\[ T = \frac{2\pi}{B} \]where \(B\) is the coefficient of \(\theta\) within the sine function.
For the function \( y = 0.5 \sin \frac{\pi}{3} \theta \), our frequency multiplier \(B\) is \(\frac{\pi}{3}\). Applying the formula gives us: \[T = \frac{2\pi}{\frac{\pi}{3}} = 6\].
This tells us that the function completes one cycle over an interval of 6 units on the x-axis.

Amplitude

Amplitude in a trigonometric function refers to the height of the wave from the center line to the maximum peak. It represents how far the wave varies above and below its average value. Mathematically, if the function is written as \( y = A \sin(B\theta + C) + D \), then amplitude is given by \(|A|\).
In the given function \( y = 0.5 \sin \frac{\pi}{3} \theta \), the coefficient 0.5 represents the amplitude. This means the wave oscillates 0.5 units above and 0.5 units below the center line, providing a total wave height of 1.
The amplitude's role is crucial in defining the range of the sine wave, creating the peaks and troughs characteristic of trigonometric curves.

Graphing Trigonometric Functions

Graphing sine functions involves understanding its key characteristics: period, amplitude, and possible shifts. First, establish the axes where the x-values represent the angle \(\theta\) and the y-values reflect the sine's output. 1. **Start with the axes:** Draw a horizontal x-axis and a vertical y-axis. Determine the range based on amplitude.2. **Calculate Critical Points:** For our function \( y = 0.5 \sin \frac{\pi}{3} \theta \), plot key points within one period, from 0 to 6.Let's plot based on the function's characteristics:- Start at (0,0), the beginning of the sine wave.- At quarter period \((1.5, 0.5)\), the wave reaches its peak due to the amplitude.- At half period \((3,0)\), it crosses back through the midline.- At three-quarters period \((4.5, -0.5)\), it reaches the negative peak.- Finally, at the end of the period \((6,0)\), the wave completes its cycle back at the midline.Drawing these points and connecting with a smooth curve gives us the classic sinusoidal shape, making it easy to visualize how the sine function alternates between maximum, minimum, and equilibrium points.