Chapter 6: Problem 23
a) Graph the curve \(y=4 \sin x-3 \cos x\) Notice that it resembles a sine
function.
b) What are the approximate values of \(a\) and \(c\) for the curve in the form
\(y=a \sin (x-c),\) where \(0
Short Answer
Expert verified
The graph resembles a sine wave with amplitude 5 and phase shift approximately 53.13 degrees. The function is \(y = 5 \sin (x - 53.13^{\circ})\).
Step by step solution
01
- Graph the Function
To graph the curve given by the function \(y = 4 \sin x - 3 \cos x\), plot the points for different values of \(x\) within a complete sine wave period, such as 0 to \(2\pi\). Notice its resemblance to a shifted and scaled sine function.
02
- Identify Values of a and c
To find the equivalent form \(y = a \sin (x - c)\), recognize that the coefficients 4 and -3 resemble amplitude and phase shift of a sine function. Use the equations: \(a = \sqrt{A^2 + B^2}\) and \(\tan(c) = \frac{B}{A}\), where \(A = 4\) and \(B = -3\).
03
- Calculate a
Find \(a\) using the equation: \[a = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\] The amplitude \(a\) is 5.
04
- Calculate c
Determine \(c\) using \(\tan(c) = \frac{-3}{4}\) and solve for \(c\): \[c = \arctan \left( \frac{-3}{4} \right)\approx -36.87^{\circ}\] Convert to a positive angle within the specified range by adding 360 degrees until in the range: \[c' = -36.87^{\circ} + 360^{\circ} \approx 323.13^{\circ}\] As it must be between 0 and 90 degrees, displace further: \[c'' = 323.13^{\circ} - 270^{\circ} = 53.13^{\circ}\] The phase shift \(c\) is approximately 53.13 degrees.
05
- Rewrite Function Using the Identity
Now use the difference identity for sine: \(\sin(A - B) = \sin A \cos B - \cos A \sin B\). Recognize that \(\sin(x - 53.13^{\circ})\) implies: \[4 \sin x - 3 \cos x = 5 \sin (x - 53.13^{\circ})\]. Therefore, the rewritten function is \(y = 5 \sin (x - 53.13^{\circ})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Trigonometric Functions
Graphing trigonometric functions is an important skill that helps visualize mathematical relationships.
For the function given in the exercise, we need to plot the points for different values of \(x\) over a full period of the sine wave, typically from 0 to \(2\pi\).
Start by choosing values of \(x\) at equal intervals such as 0, \( \frac{\pi}{6} \), \( \frac{\pi}{3} \), \( \frac{\pi}{2} \), etc.
Calculate the corresponding \(y\)-values using the function \(y = 4 \sin x - 3 \cos x\).
Once you have a set of points, plot them on a graph and draw a smooth curve through them.
The function \(y = 4 \sin x - 3 \cos x\) will look like a sine wave, but it is shifted and scaled.
By visualizing this function, you see how it mimics the sine function but with changes in amplitude and phase.
For the function given in the exercise, we need to plot the points for different values of \(x\) over a full period of the sine wave, typically from 0 to \(2\pi\).
Start by choosing values of \(x\) at equal intervals such as 0, \( \frac{\pi}{6} \), \( \frac{\pi}{3} \), \( \frac{\pi}{2} \), etc.
Calculate the corresponding \(y\)-values using the function \(y = 4 \sin x - 3 \cos x\).
Once you have a set of points, plot them on a graph and draw a smooth curve through them.
The function \(y = 4 \sin x - 3 \cos x\) will look like a sine wave, but it is shifted and scaled.
By visualizing this function, you see how it mimics the sine function but with changes in amplitude and phase.
Amplitude of Trigonometric Functions
The amplitude of a trigonometric function is the maximum value it reaches from its equilibrium position.
In the function \(y = a \sin (x - c)\), the amplitude is given by the coefficient \(a\).
For the initial function \(y = 4 \sin x - 3 \cos x\), we can find the amplitude by using the formula:
\[a = \sqrt{A^2 + B^2}\]
where \(A = 4\) and \(B = -3\).
Plugging in the values, we get:
\[a = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\]
Hence, the amplitude of the given function is 5.
This means that the wave oscillates between +5 and -5 from its mean position.
In the function \(y = a \sin (x - c)\), the amplitude is given by the coefficient \(a\).
For the initial function \(y = 4 \sin x - 3 \cos x\), we can find the amplitude by using the formula:
\[a = \sqrt{A^2 + B^2}\]
where \(A = 4\) and \(B = -3\).
Plugging in the values, we get:
\[a = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\]
Hence, the amplitude of the given function is 5.
This means that the wave oscillates between +5 and -5 from its mean position.
Difference Identity for Sine
The difference identity for sine allows us to rewrite trigonometric functions in different forms.
The general form for the difference identity for sine is:
\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]
Given the function \(y = 4 \sin x - 3 \cos x\), we want to rewrite it in the form \(y = a \sin(x - c)\).
From the calculation of amplitude, we found that \(a = 5\).
To find the phase shift \(c\), we use the relationship:
\[ \tan(c) = \frac{B}{A} = \frac{-3}{4} \]
Solving for \(c\), we get:
\[ c = \arctan \left( \frac{-3}{4} \right) \approx -36.87^{\circ} \]
Since angles need to be within the range from 0 to 90 degrees, adjust it:
\[ c = -36.87^{\circ} + 360^{\circ} \approx 323.13^{\circ} \]
Further adjust to fit within 0 to 90 degrees:
\[ c = 323.13^{\circ} - 270^{\circ} = 53.13^{\circ} \]
Therefore, you can rewrite the original function using the difference identity:
\[ y = 5 \sin(x - 53.13^{\circ}) \]
The general form for the difference identity for sine is:
\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]
Given the function \(y = 4 \sin x - 3 \cos x\), we want to rewrite it in the form \(y = a \sin(x - c)\).
From the calculation of amplitude, we found that \(a = 5\).
To find the phase shift \(c\), we use the relationship:
\[ \tan(c) = \frac{B}{A} = \frac{-3}{4} \]
Solving for \(c\), we get:
\[ c = \arctan \left( \frac{-3}{4} \right) \approx -36.87^{\circ} \]
Since angles need to be within the range from 0 to 90 degrees, adjust it:
\[ c = -36.87^{\circ} + 360^{\circ} \approx 323.13^{\circ} \]
Further adjust to fit within 0 to 90 degrees:
\[ c = 323.13^{\circ} - 270^{\circ} = 53.13^{\circ} \]
Therefore, you can rewrite the original function using the difference identity:
\[ y = 5 \sin(x - 53.13^{\circ}) \]
Converting Angles
Converting angles between degrees and radians is essential in trigonometry.
To convert degrees to radians, use the formula:
\[ \text{{Radians}} = \text{{Degrees}} \times \frac{\pi}{180} \]
Conversely, to convert radians to degrees, use:
\[ \text{{Degrees}} = \text{{Radians}} \times \frac{180}{\pi} \]
For example, to convert 53.13 degrees to radians, you'd do the following:
\[ 53.13 \times \frac{\pi}{180} \approx 0.927 \text{{ radians}} \]
Understanding these conversions helps in graphing and solving trigonometric problems since standard angles and functions are often given in radians.
It is also important for integrating trigonometric functions with other mathematical concepts.
Always keep these handy conversion formulas in mind for better comprehension of trigonometric identities and graphing.
To convert degrees to radians, use the formula:
\[ \text{{Radians}} = \text{{Degrees}} \times \frac{\pi}{180} \]
Conversely, to convert radians to degrees, use:
\[ \text{{Degrees}} = \text{{Radians}} \times \frac{180}{\pi} \]
For example, to convert 53.13 degrees to radians, you'd do the following:
\[ 53.13 \times \frac{\pi}{180} \approx 0.927 \text{{ radians}} \]
Understanding these conversions helps in graphing and solving trigonometric problems since standard angles and functions are often given in radians.
It is also important for integrating trigonometric functions with other mathematical concepts.
Always keep these handy conversion formulas in mind for better comprehension of trigonometric identities and graphing.