Problem 2
Question
Find the amplitude, if it exists, and period of each function. Then graph each function. $$ y=2 \sin \theta $$
Step-by-Step Solution
Verified Answer
Amplitude is 2 and period is \( 2\pi \).
1Step 1: Understand the Sine Function
The standard form of a sine function is \( y = a \sin(b\theta + c) + d \). The amplitude is given by the absolute value of \(a\), and the period is calculated by \(\frac{2\pi}{b}\). In the given function \( y = 2 \sin \theta \), we have \( a = 2 \) and \( b = 1 \).
2Step 2: Calculate the Amplitude
The amplitude is the coefficient \( a \) of the \( \sin \) function. Here, \( a = 2 \). Therefore, the amplitude is \( |2| = 2 \).
3Step 3: Determine the Period
The period of a sine function is calculated using the formula \( \frac{2\pi}{b} \). In this function, \( b = 1 \), so the period is \( \frac{2\pi}{1} = 2\pi \).
4Step 4: Graph the Function
To graph \( y = 2 \sin \theta \), start by plotting key points of the sine function cycle from 0 to \( 2\pi \), which is one full period. The points \((0,0)\), \((\frac{\pi}{2}, 2)\), \((\pi, 0)\), \((\frac{3\pi}{2}, -2)\), and \((2\pi, 0)\) represent the peaks, zero crossings, and troughs. Use these points to plot the smooth, wave-like curve of the sine function across the x-axis.
Key Concepts
Understanding AmplitudePeriod of Trigonometric FunctionsGraphing Trigonometric Functions
Understanding Amplitude
The amplitude of a sine function determines how "tall" or "short" the waves of the graph appear. Think of it as the height from the centerline of the wave to the peak or trough. In mathematical terms, the amplitude is derived from the coefficient in front of the sine function, represented by \( a \) in the equation \( y = a \sin(b\theta + c) + d \). In our example, \( y = 2 \sin \theta \), the coefficient \( a \) is 2. Thus, the amplitude is \(|2| = 2\). This means that the graph of our sine function will reach a maximum height of 2 units above the centerline and a minimum of 2 units below the centerline.
Amplitude does not alter the length of the wave; it only changes how high and low the wave can go. This feature is vital for understanding the behavior and appearance of trigonometric graphs.
Amplitude does not alter the length of the wave; it only changes how high and low the wave can go. This feature is vital for understanding the behavior and appearance of trigonometric graphs.
Period of Trigonometric Functions
The period of a trigonometric function describes how long it takes for the function to repeat its wave pattern. In essence, it tells you the "length" of one full cycle of the wave on the graph. For a standard sine function \( y = a \sin(b\theta + c) + d \), the period is calculated by the formula \( \frac{2\pi}{b} \), where \( b \) is the coefficient of \( \theta \).
In the provided function \( y = 2 \sin \theta \), the coefficient \( b \) is 1. Substituting into the formula, the period becomes \( \frac{2\pi}{1} = 2\pi \). This means that every \( 2\pi \) units along the x-axis, the sine wave starts to repeat its cycle. Understanding the period is essential as it helps in accurately plotting the graph over the specified interval and predicting future behavior of the wave.
In the provided function \( y = 2 \sin \theta \), the coefficient \( b \) is 1. Substituting into the formula, the period becomes \( \frac{2\pi}{1} = 2\pi \). This means that every \( 2\pi \) units along the x-axis, the sine wave starts to repeat its cycle. Understanding the period is essential as it helps in accurately plotting the graph over the specified interval and predicting future behavior of the wave.
- The period affects how "stretched" or "compressed" the graph appears along the x-axis.
- A larger \( b \) value results in a shorter period, leading to more waves being plotted within the same length on the axis.
- A smaller \( b \) value yields a longer period, resulting in fewer waves over the same span.
Graphing Trigonometric Functions
Graphing a sine function involves plotting key points over one complete cycle of the function and then extending this pattern throughout the graph. For \( y = 2 \sin \theta \), the cycle starts at 0 and completes at \( 2\pi \). We want to focus on positioning each part of the wave accurately to capture the sine's characteristic wave shape.
Firstly, identify and plot the following critical points:
This visual representation is significant as it not only reflects the defined amplitude and period but also provides insight into the wave's behavior over varying intervals. Mastery of this graphical approach makes handling more complex trigonometric functions easier, as seeing patterns aids in predicting and solving problems. Plus, it looks pretty cool too!
Firstly, identify and plot the following critical points:
- Start at \((0,0)\) - the wave begins at the origin.
- Reach the peak at \((\frac{\pi}{2}, 2)\) - this is the maximum point, corresponding to the amplitude.
- Cross back through the axis at \((\pi, 0)\).
- Hit the trough at \((\frac{3\pi}{2}, -2)\) - this is the minimum point.
- Complete the cycle at \((2\pi, 0)\).
This visual representation is significant as it not only reflects the defined amplitude and period but also provides insight into the wave's behavior over varying intervals. Mastery of this graphical approach makes handling more complex trigonometric functions easier, as seeing patterns aids in predicting and solving problems. Plus, it looks pretty cool too!
Other exercises in this chapter
Problem 2
Find the value of each expression. $$ \csc \theta, \text { if } \cos \theta=-\frac{3}{5} ; 180^{\circ} \leq \theta
View solution Problem 2
State the amplitude, period, and phase shift for each function. Then graph the function. $$ y=\tan \left(\theta+60^{\circ}\right) $$
View solution Problem 3
Find all solutions of each equation for the given interval. \(\sin 2 \theta=\cos \theta ; 0 \leq \theta
View solution Problem 3
Find the exact values of \(\sin 2 \theta, \cos 2 \theta, \sin \frac{\theta}{2},\) and \(\cos \frac{\theta}{2}\) for each of the following. $$ \sin \theta=\frac{
View solution