Answer :
To solve the problem of finding the correct graph for the function [tex]\( y = -\sin(x)\cos(x) - \cos(x)\sin(x) \)[/tex] over the interval [tex]\([-2\pi, 2\pi]\)[/tex], we start by simplifying the expression.
### Step-by-Step Solution:
1. Combine Like Terms:
The given function is
[tex]\[ y = -\sin(x)\cos(x) - \cos(x)\sin(x). \][/tex]
Notice that both terms are similar. Therefore, we can combine them to simplify the expression:
[tex]\[ y = -2\sin(x)\cos(x). \][/tex]
2. Use Trigonometric Identity:
We use the double-angle identity for sine, which states:
[tex]\[ \sin(2x) = 2\sin(x)\cos(x). \][/tex]
Applying this identity, we rewrite the function as:
[tex]\[ y = -\sin(2x). \][/tex]
3. Determine the Domain:
We are asked to plot the function over the interval [tex]\([-2\pi, 2\pi]\)[/tex].
4. Calculate Values at Key Points:
Let's identify a few key points and their function values:
- At [tex]\( x = -2\pi \)[/tex]:
[tex]\[ y = -\sin(-4\pi) = -\sin(0) = 0 \][/tex]
- At [tex]\( x = -\pi \)[/tex]:
[tex]\[ y = -\sin(-2\pi) = -\sin(0) = 0 \][/tex]
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -\sin(0) = 0 \][/tex]
- At [tex]\( x = \pi \)[/tex]:
[tex]\[ y = -\sin(2\pi) = -\sin(0) = 0 \][/tex]
- At [tex]\( x = 2\pi \)[/tex]:
[tex]\[ y = -\sin(4\pi) = -\sin(0) = 0 \][/tex]
5. Sketch the Graph:
The function [tex]\( y = -\sin(2x) \)[/tex] is a transformation of the sine function with double frequency and is reflected over the x-axis. A sine function with double frequency has twice as many oscillations in the same interval, thus making its period [tex]\( \pi \)[/tex] instead of [tex]\( 2\pi \)[/tex].
Plotting the function between [tex]\([-2\pi, 2\pi]\)[/tex], we get:
- The sine function starts at [tex]\(0\)[/tex] at [tex]\( x = -2\pi \)[/tex].
- Descends to [tex]\(0\)[/tex] at [tex]\(x = -\frac{3\pi}{2}\)[/tex].
- Reaches [tex]\(0\)[/tex] at [tex]\( x = -\pi \)[/tex].
- Descends to [tex]\(0\)[/tex] at [tex]\(x = -\frac{\pi}{2}\)[/tex].
- Reaches [tex]\(0\)[/tex] at [tex]\( x = 0 \)[/tex].
- Ascends to [tex]\(0\)[/tex] at [tex]\(x = \frac{\pi}{2}\)[/tex].
- Reaches [tex]\(0\)[/tex] at [tex]\( x = \pi \)[/tex].
- Ascends to [tex]\(0\)[/tex] at [tex]\(x = \frac{3\pi}{2}\)[/tex].
- Reaches [tex]\(0\)[/tex] at [tex]\( x = 2\pi \)[/tex].
6. Numerical Points:
The given points array for the x-values between [tex]\([-2\pi, 2\pi]\)[/tex] and their corresponding y-values could assist in plotting these points smoothly.
Try to visualize the graph based on these properties. It looks like a sine wave with a halved period, oscillating twice as often over [tex]\([-2\pi, 2\pi]\)[/tex] but reflected in the x-axis.
With the calculated [tex]\( (x, y) \)[/tex] values, you should now be able to match this with one of the given graph options properly. The graph appears like a sine wave with double the frequency and an amplitude inversion.
### Step-by-Step Solution:
1. Combine Like Terms:
The given function is
[tex]\[ y = -\sin(x)\cos(x) - \cos(x)\sin(x). \][/tex]
Notice that both terms are similar. Therefore, we can combine them to simplify the expression:
[tex]\[ y = -2\sin(x)\cos(x). \][/tex]
2. Use Trigonometric Identity:
We use the double-angle identity for sine, which states:
[tex]\[ \sin(2x) = 2\sin(x)\cos(x). \][/tex]
Applying this identity, we rewrite the function as:
[tex]\[ y = -\sin(2x). \][/tex]
3. Determine the Domain:
We are asked to plot the function over the interval [tex]\([-2\pi, 2\pi]\)[/tex].
4. Calculate Values at Key Points:
Let's identify a few key points and their function values:
- At [tex]\( x = -2\pi \)[/tex]:
[tex]\[ y = -\sin(-4\pi) = -\sin(0) = 0 \][/tex]
- At [tex]\( x = -\pi \)[/tex]:
[tex]\[ y = -\sin(-2\pi) = -\sin(0) = 0 \][/tex]
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -\sin(0) = 0 \][/tex]
- At [tex]\( x = \pi \)[/tex]:
[tex]\[ y = -\sin(2\pi) = -\sin(0) = 0 \][/tex]
- At [tex]\( x = 2\pi \)[/tex]:
[tex]\[ y = -\sin(4\pi) = -\sin(0) = 0 \][/tex]
5. Sketch the Graph:
The function [tex]\( y = -\sin(2x) \)[/tex] is a transformation of the sine function with double frequency and is reflected over the x-axis. A sine function with double frequency has twice as many oscillations in the same interval, thus making its period [tex]\( \pi \)[/tex] instead of [tex]\( 2\pi \)[/tex].
Plotting the function between [tex]\([-2\pi, 2\pi]\)[/tex], we get:
- The sine function starts at [tex]\(0\)[/tex] at [tex]\( x = -2\pi \)[/tex].
- Descends to [tex]\(0\)[/tex] at [tex]\(x = -\frac{3\pi}{2}\)[/tex].
- Reaches [tex]\(0\)[/tex] at [tex]\( x = -\pi \)[/tex].
- Descends to [tex]\(0\)[/tex] at [tex]\(x = -\frac{\pi}{2}\)[/tex].
- Reaches [tex]\(0\)[/tex] at [tex]\( x = 0 \)[/tex].
- Ascends to [tex]\(0\)[/tex] at [tex]\(x = \frac{\pi}{2}\)[/tex].
- Reaches [tex]\(0\)[/tex] at [tex]\( x = \pi \)[/tex].
- Ascends to [tex]\(0\)[/tex] at [tex]\(x = \frac{3\pi}{2}\)[/tex].
- Reaches [tex]\(0\)[/tex] at [tex]\( x = 2\pi \)[/tex].
6. Numerical Points:
The given points array for the x-values between [tex]\([-2\pi, 2\pi]\)[/tex] and their corresponding y-values could assist in plotting these points smoothly.
Try to visualize the graph based on these properties. It looks like a sine wave with a halved period, oscillating twice as often over [tex]\([-2\pi, 2\pi]\)[/tex] but reflected in the x-axis.
With the calculated [tex]\( (x, y) \)[/tex] values, you should now be able to match this with one of the given graph options properly. The graph appears like a sine wave with double the frequency and an amplitude inversion.
