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Chapter 15: Problem 6

Sketch \(y=3 \sin 2 A\) from \(A=0\) to \(A=2 \pi\) radians.

Short Answer

Expert verified
The sine wave oscillates between 3 and -3 with a period of \( \pi \).

Step by step solution

01

Understand the Function

The function given is in the form of \( y = a \sin(bA) \), where \( a = 3 \) and \( b = 2 \). This indicates that the amplitude of the function is 3 and the period is determined by dividing \( 2\pi \) by \( b \).
02

Determine the Amplitude

The amplitude of a sine function is the maximum value it reaches. In this function, \( a = 3 \), so the amplitude is 3. This means the sine wave will oscillate between 3 and -3.
03

Determine the Period

To find the period of \( y = 3 \sin(2A) \), use the formula for the period of a sine function: \( \text{Period} = \frac{2\pi}{b} = \frac{2\pi}{2} = \pi \). Thus, the sine wave completes one full cycle when \( A \) goes from 0 to \( \pi \).
04

Identify Key Points

The key points of one cycle of the sine wave \( y = 3 \sin(2A) \) occur at these values due to the period \( \pi \): \( A = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi \). They correspond to \( y = 0, 3, 0, -3, 0 \) respectively.
05

Sketch the Function

Using the amplitude and period determined, sketch two cycles of the sine wave: Starting at 0, rising to 3 at \( \frac{\pi}{4} \), returning to 0 at \( \frac{\pi}{2} \), descending to -3 at \( \frac{3\pi}{4} \), and returning to 0 at \( \pi \). Repeat the wave from \( \pi \) to \( 2\pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude of Sine Waves
The amplitude of a sine wave is a fundamental concept in trigonometry, representing the height of the wave's peaks and valleys. In the function given, \( y = 3 \sin(2A) \), the amplitude is determined by the coefficient \( a \), which is 3 in this case.
Therefore, the sine wave oscillates between the maximum value of 3 and the minimum value of -3.
This indicates a vertical stretch, making the wave taller compared to the standard sine function, which oscillates between 1 and -1.
  • A larger amplitude implies a stronger wave, meaning more significant oscillations.
  • The amplitude never changes over the domain; it consistently defines the vertical limits of the wave.
Understanding amplitude is crucial for interpreting real-world scenarios, such as vibration strengths or sound wave intensities, linking math with everyday applications.
Period of Trigonometric Functions
The period of a trigonometric function like sine defines the interval length needed for the wave to complete one full cycle. In the function \( y = 3 \sin(2A) \), the period is calculated using the formula \( \text{Period} = \frac{2\pi}{b} \). Here, \( b = 2 \) modifies the typical period of a sine function.Substituting the values, the period becomes \( \frac{2\pi}{2} = \pi \).
This means the sine function completes a full wave cycle as \( A \) moves from 0 to \( \pi \).
  • A smaller \( b \) value elongates the period, stretching the wave horizontally.
  • A larger \( b \) value shortens the period, compressing the wave horizontally.
Knowing the period helps in predicting the wave's behavior over specific intervals, which is valuable for timing signals or cyclical patterns.
Sketching Graphs of Sine Functions
Sketching graphs of sine functions involves plotting key points and interpreting amplitude and period. For \( y = 3 \sin(2A) \), the amplitude is 3 and the period is \( \pi \). Understanding these elements allows for accurate sketching.To sketch one cycle:
  • Start at \( A = 0 \) with \( y = 0 \).
  • At \( A = \frac{\pi}{4} \), the graph reaches its peak at 3.
  • The graph returns to \( y = 0 \) at \( A = \frac{\pi}{2} \).
  • At \( A = \frac{3\pi}{4} \), the graph dips to -3.
  • Finally, it returns to \( y = 0 \) at \( A = \pi \).
Repeating these steps leads to a full sketch from \( A = 0 \) to \( A = 2\pi \).
Visualizing these points sketch by hand or using graphing software helps in understanding the sine function's wave pattern.

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