Problem 92
Question
Determine whether the statement is true or false. Justify your answer. The graph of \(y=-\cos x\) is a reflection of the graph of \(y=\sin \left(x+\frac{\pi}{2}\right)\) in the \(x\) -axis.
Step-by-Step Solution
Verified Answer
False.
1Step 1: Understand trigonometric functions
The basic shapes of the sine and cosine functions need to be remembered. In a unit circle, the cosine function corresponds to the x-coordinate and the sine function to the y-coordinate of a point moving along the circle. The sine function starts at zero, then goes to 1, back to 0, down to -1, and then returns to 0 for a full cycle from 0 to \(2\pi\). The cosine function, however, starts at 1, goes to zero, then to -1, back to zero, and then returns to 1 for a full cycle from 0 to \(2\pi\) position on the circle.\n
2Step 2: Analyze given functions
The function \(y=-\cos x\) is a cosine function reflected along the x-axis. The function \(y=\sin (x+\frac{\pi}{2})\) is a sine function, shifted to the left by \(\frac{\pi}{2}\) units. The phase shift of \(\frac{\pi}{2}\) to the left actually converts the sine function into a cosine function.
3Step 3: Comparison
Comparing the functions, it can be seen that the graph of \(y=\sin (x+\frac{\pi}{2})\) matches exactly with the shape of the basic cosine function, which starts at the maximum point, and not with the reflected one. So, the statement 'The graph of \(y=-\cos x\) is a reflection of the graph of \(y=\sin (x+\frac{\pi}{2})\) in the x-axis' is false.
Key Concepts
Reflection of FunctionsPhase ShiftThe Unit Circle
Reflection of Functions
When it comes to transforming graphs, the reflection of a function is a fundamental concept in mathematics. A reflection flips a graph over a specific line, called the axis of reflection. For functions, the two primary axes of reflection are the x-axis and the y-axis.
Reflecting over the x-axis will multiply the y-values of the function by -1, which essentially inverts the graph. For example, if you have the function \(y=f(x)\), its reflection across the x-axis will be \(y=-f(x)\). Similarly, reflecting a graph over the y-axis will multiply the x-values by -1, producing \(y=f(-x)\).
In our exercise, the graph of \(y=-\text{cos} x\) is the reflection of the graph of \(y=\text{cos} x\) over the x-axis. It is important to recognize that reflecting a cosine function does not transform it into a sine function. This observation is key in understanding why the statement from the exercise is false.
Reflecting over the x-axis will multiply the y-values of the function by -1, which essentially inverts the graph. For example, if you have the function \(y=f(x)\), its reflection across the x-axis will be \(y=-f(x)\). Similarly, reflecting a graph over the y-axis will multiply the x-values by -1, producing \(y=f(-x)\).
In our exercise, the graph of \(y=-\text{cos} x\) is the reflection of the graph of \(y=\text{cos} x\) over the x-axis. It is important to recognize that reflecting a cosine function does not transform it into a sine function. This observation is key in understanding why the statement from the exercise is false.
Phase Shift
The term 'phase shift' refers to the horizontal displacement of a trigonometric function along the x-axis. In essence, phase shifting a function moves it left or right without altering its shape. When you apply a phase shift to the sine or cosine function, you essentially change where the function starts its cycle on the unit circle.
Mathematically, a function \(y = \text{sin}(x)\) with a phase shift can be written as \(y = \text{sin}(x + \text{phase shift})\) for a shift to the left, or \(y = \text{sin}(x - \text{phase shift})\) for a shift to the right. A phase shift of \(\frac{\rm{\text{\textpi}}}{2}\) to the left, as seen in the exercise \(y=\text{sin}(x + \frac{\rm{\text{\textpi}}}{2})\), shifts the start of the sine function to the same starting point as the cosine function, thus converting its appearance to that of a cosine function.
Understanding phase shifts is crucial for recognizing the transformation of trigonometric graphs and identifying whether two functions will have the same graph after a phase shift is applied.
Mathematically, a function \(y = \text{sin}(x)\) with a phase shift can be written as \(y = \text{sin}(x + \text{phase shift})\) for a shift to the left, or \(y = \text{sin}(x - \text{phase shift})\) for a shift to the right. A phase shift of \(\frac{\rm{\text{\textpi}}}{2}\) to the left, as seen in the exercise \(y=\text{sin}(x + \frac{\rm{\text{\textpi}}}{2})\), shifts the start of the sine function to the same starting point as the cosine function, thus converting its appearance to that of a cosine function.
Understanding phase shifts is crucial for recognizing the transformation of trigonometric graphs and identifying whether two functions will have the same graph after a phase shift is applied.
The Unit Circle
The unit circle is a tool essential to understanding trigonometry and its functions. It's a circle with a radius of one unit, centered at the origin of the coordinate system. The angle in a unit circle is measured from the positive x-axis, and as it increases, it represents the rotation from that axis.
Each point on the circumference of the unit circle corresponds to the endpoint of an arc whose length is equal to the angle's measure. These points have coordinates that are the values of the cosine and sine of the angle: \( (\text{cos} \theta, \text{sin} \theta) \), where \( \theta \) is the angle measured in radians. For any angle \( \theta \), the cosine function gives the x-coordinate, and the sine function gives the y-coordinate.
By using the unit circle, students can visualize how the sine and cosine functions oscillate and how their values are determined by the angle's position. This circle underpins the relationship between the trigonometric functions and the angles, hence playing a pivotal role in solving trigonometric problems.
Each point on the circumference of the unit circle corresponds to the endpoint of an arc whose length is equal to the angle's measure. These points have coordinates that are the values of the cosine and sine of the angle: \( (\text{cos} \theta, \text{sin} \theta) \), where \( \theta \) is the angle measured in radians. For any angle \( \theta \), the cosine function gives the x-coordinate, and the sine function gives the y-coordinate.
By using the unit circle, students can visualize how the sine and cosine functions oscillate and how their values are determined by the angle's position. This circle underpins the relationship between the trigonometric functions and the angles, hence playing a pivotal role in solving trigonometric problems.
Other exercises in this chapter
Problem 91
Use the given value and the trigonometric identities to find the remaining trigonometric functions of the angle. $$\tan \theta=-4, \cos \theta
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Find (if possible) the complement and supplement of the angle. $$\frac{3 \pi}{2}$$
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Write the function in terms of the sine function by using the identity \(A \cos \omega t+B \sin \omega t=\sqrt{A^{2}+B^{2}} \sin \left(\omega t+\arctan \frac{A}
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Use the given value and the trigonometric identities to find the remaining trigonometric functions of the angle. $$\cot \theta=5, \sin \theta > 0$$
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