Problem 31
Question
Use the graphs of the trigonometric functions to determine the number of solutions of the equation between 0 and \(2 \pi\) \(\sin t=3 / 5[\)Hint: How many points on the graph of \(f(t)=\sin t\) between \(t=0\) and \(t=2 \pi\) have second coordinate \(3 / 5 ?]\)
Step-by-Step Solution
Verified Answer
Answer: There are 2 solutions of the equation \(\sin t = \frac{3}{5}\) between \(0\) and \(2 \pi\).
1Step 1: Identify the domain and range
The domain of the sine function is all real numbers (\(\mathbb{R}\)) and its range is between -1 and 1 (\(-1 \leq \sin t \leq 1\)). The given equation is \(\sin t = \frac{3}{5}\) and it has a value between -1 and 1. So there is a possibility of having solutions in the given interval, \(0 \leq t \leq 2\pi\).
2Step 2: Sketch the graph
Sketch the graph of \(\sin t\) on the interval \(0 \leq t \leq 2\pi\). This will help to visualize how many times the graph touches the line \(y = \frac{3}{5}\). Remember that in the given interval, the sine function completes one full cycle and starts from 0 (origin) and reaches a maximum of 1, then comes back to 0, goes to a minimum of -1, and finally returns back to 0.
3Step 3: Determine the number of intersections
Count the number of times the graph of the sine function intersects with the line \(y = \frac{3}{5}\) in the interval \(0 \leq t \leq 2\pi\). It can be observed that the sine function intersects the line \(y = \frac{3}{5}\) twice in this interval.
4Step 4: Answer the question
There are two solutions of the equation \(\sin t = \frac{3}{5}\) between \(0\) and \(2 \pi\).
Key Concepts
Graphing Sine FunctionSolving Trigonometric EquationsDomain and Range of Sine Function
Graphing Sine Function
Visualizations greatly aid in understanding trigonometric functions. The sine function, denoted as \( \sin t \), oscillates in a smooth, wave-like pattern. This graph completes a full cycle over the interval of \(0\) to \(2\pi\).
- During this cycle, the sine wave begins at zero, peaks at one, returns through zero, dips to negative one, and finally rises back to zero.
- The characteristic wave is continuous and periodic, meaning it repeats its form indefinitely in both directions along the horizontal axis.
Drawing the sine function graph also helps to solve equations, such as determining when the sine of \( t \) equals a specific value, like \(\frac{3}{5}\). Sketching the function over a set range reveals how often the wave intersects a horizontal line, indicating potential solutions.
This visual method provides a straightforward way of counting intersections. Here, the graph of the sine function will touch the line \(y = \frac{3}{5}\) twice, leading to two solutions within the range of \(0\) to \(2\pi\). Working with graphs bridges the gap between abstract mathematical concepts and tangible, visual data.
- During this cycle, the sine wave begins at zero, peaks at one, returns through zero, dips to negative one, and finally rises back to zero.
- The characteristic wave is continuous and periodic, meaning it repeats its form indefinitely in both directions along the horizontal axis.
Drawing the sine function graph also helps to solve equations, such as determining when the sine of \( t \) equals a specific value, like \(\frac{3}{5}\). Sketching the function over a set range reveals how often the wave intersects a horizontal line, indicating potential solutions.
This visual method provides a straightforward way of counting intersections. Here, the graph of the sine function will touch the line \(y = \frac{3}{5}\) twice, leading to two solutions within the range of \(0\) to \(2\pi\). Working with graphs bridges the gap between abstract mathematical concepts and tangible, visual data.
Solving Trigonometric Equations
Solving trigonometric equations involves finding the values of the variable that will satisfy the equation. For \( \sin t = \frac{3}{5} \), the task is to determine those angles within a specific interval where this equality holds true.
- Since the sine function is periodic with a range from \(-1\) to \(1\), it is crucial to identify if the given value \(\frac{3}{5}\) lies within this spectrum.
- Given that it does, our focus shifts to finding the points of intersection on the graph of the sine function with the constant horizontal line of \(\frac{3}{5}\).
By visual inspection or by using algebraic methods, one can determine the points (angles) that make the equation true within the considered range. Here, the sine function reaches the value \(\frac{3}{5}\) twice, hence yielding two solutions for \(t\) within \(0\) to \(2\pi\). This demonstration underscores the importance of knowing the behavior and properties of trigonometric functions in solving similar equations.
- Since the sine function is periodic with a range from \(-1\) to \(1\), it is crucial to identify if the given value \(\frac{3}{5}\) lies within this spectrum.
- Given that it does, our focus shifts to finding the points of intersection on the graph of the sine function with the constant horizontal line of \(\frac{3}{5}\).
By visual inspection or by using algebraic methods, one can determine the points (angles) that make the equation true within the considered range. Here, the sine function reaches the value \(\frac{3}{5}\) twice, hence yielding two solutions for \(t\) within \(0\) to \(2\pi\). This demonstration underscores the importance of knowing the behavior and properties of trigonometric functions in solving similar equations.
Domain and Range of Sine Function
Understanding the domain and range is pivotal for working with trigonometric functions. Each function has its unique domain and range, outlining which input values are acceptable and what output values can occur.
- The domain of the sine function is all real numbers, written as \( \mathbb{R} \). This means sine can take any real value as its argument because its wave pattern extends horizontally ad infinitum.
- Its range, however, is restricted to values between \(-1\) and \(1\), demonstrated as \(-1 \leq \sin t \leq 1\). This is due to the wave's peaks and troughs hovering within this numerical boundary consistently.
Recognizing the domain allows one to evaluate sine at any angle, whereas knowing the range helps identify feasible output values. In equations like \( \sin t = \frac{3}{5} \), confirming that \(\frac{3}{5}\) falls within the function's range affirms that real solutions can exist. The domain and range inform the potential number and nature of these solutions, providing foundational knowledge necessary for analysis and problem-solving.
- The domain of the sine function is all real numbers, written as \( \mathbb{R} \). This means sine can take any real value as its argument because its wave pattern extends horizontally ad infinitum.
- Its range, however, is restricted to values between \(-1\) and \(1\), demonstrated as \(-1 \leq \sin t \leq 1\). This is due to the wave's peaks and troughs hovering within this numerical boundary consistently.
Recognizing the domain allows one to evaluate sine at any angle, whereas knowing the range helps identify feasible output values. In equations like \( \sin t = \frac{3}{5} \), confirming that \(\frac{3}{5}\) falls within the function's range affirms that real solutions can exist. The domain and range inform the potential number and nature of these solutions, providing foundational knowledge necessary for analysis and problem-solving.
Other exercises in this chapter
Problem 30
Convert the given degree measure to radians. $$36^{\circ}$$
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Sketch a complete graph of the function. $$h(t)=3 \sin (2 t+\pi / 2)$$
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In Exercises \(30-36,\) perform the indicated operations, then simplify your answers by using appropriate definitions and identities. $$\cos t \sin t(\csc t+\se
View solution Problem 31
In Exercises \(31-36\), write the expression as a single real mum. ber. Do not use decimal approximations. IHint: Exercises \(15-21 \text { may be helpful. }]\)
View solution