Provide a trigonometric equation. Considering only the space between 𝑥 = 0 and 2𝜋, the equation must only have solutions at 𝑥 = 1 and 𝑥 = 2. Explain your thought process and the work you did to create the equation. You may round decimal values to 3 places.
I was thinking y=cos(x / 50)(x-1)(x-2), however I would like, if possible, some help to create one in a y= a sin,cos [k (x-d) ] + c format. Thanks.
To create an equation in the form y = a sin [k (x - d)] + c, where the solutions are at x = 1 and x = 2, we need to choose appropriate values for a, k, d, and c.
Considering that the solutions are at x = 1 and x = 2, we can deduce that the midline of the graph (the horizontal line y = c) will be the average of these two values, which is 1.5. Therefore, we can set c = 1.5.
Now let's focus on the amplitude of the graph. Since the sine function has an amplitude of 1, we can choose a value for a to adjust the amplitude accordingly. Let's set a = 0.5 to make the amplitude of the graph smaller.
Next, we need to determine the period of the graph, which corresponds to the value of k in the equation. Since we want the solutions to occur at x = 1 and x = 2, the period of the graph needs to be 1 unit. Thus, we set k = 2π.
Finally, we need to decide on the phase shift of the graph, represented by d in the equation. The phase shift indicates the displacement of the graph along the x-axis. Since we want the first solution at x = 1, we can set d = 1.
Putting all this together, the equation can be written as:
y = 0.5 sin[2π (x - 1)] + 1.5.
This equation satisfies the given conditions and has solutions only at x = 1 and x = 2.
To create a trigonometric equation in the form y = a sin(k(x - d)) + c that satisfies the given conditions, we can start by considering the periodic nature of trigonometric functions. Since we only want solutions at x = 1 and x = 2, we can set the period of the function to be 2.
Given that cosine has a maximum value of 1 and a minimum value of -1, we can choose the cosine function to create the desired behavior. Let's start with the equation y = cos(x).
However, cosine has a maximum value at x = 0, and we want it to have a maximum at x = 1. To achieve this shift, we can add a phase shift of π/2 to the cosine function, resulting in y = cos(x + π/2).
Now, we want the maximum value to be 1, but the cosine function has a maximum value of 0 at x = 1. To adjust this, we can multiply the function by a scaling factor of 2, resulting in y = 2cos(x + π/2).
Next, we want the function to have a minimum value of -1 at x = 2. Since a phase shift of π/2 would shift the maximum value to x = 2, we need a shift of π instead to create a minimum value at x = 2. So, we modify the equation to be y = 2cos(x + π).
Lastly, we need to adjust the vertical translation of the function to match the given condition y = 1 at x = 1. We can achieve this by adding 3 to the function, resulting in the final equation:
y = 2cos(x + π) + 3
This equation satisfies the given conditions and can be written in the desired form y = a sin(k(x - d)) + c.
To create a trigonometric equation in the form y = a sin(k(x - d)) + c, where the equation has solutions at x = 1 and x = 2, we can follow these steps:
Step 1: Determine the amplitude (a)
The amplitude represents the maximum absolute value of the function. Since our equation should only have solutions at x = 1 and x = 2, the amplitude should be nonzero. Let's select any nonzero value for simplicity, like a = 1.
Step 2: Determine the period (T)
The period represents the distance between two consecutive points with the same value. In our case, the only two solutions are at x = 1 and x = 2, which have a distance of 1. Therefore, the period T should be 1.
Step 3: Determine the horizontal shift (d)
The horizontal shift represents the displacement of the graph along the x-axis. Since we want solutions at x = 1 and x = 2, we need to shift the graph such that these two points lie on the same period. The midpoint between 1 and 2 is 1.5, which indicates a shift of d = 1.5.
Step 4: Determine the vertical shift (c)
The vertical shift represents the displacement of the graph along the y-axis. Since we want solutions at x = 1 and x = 2, we need the graph to pass through these points. From the given equation y = cos(x/50)(x-1)(x-2), we can observe that it is centered around y = 0 at x = 1.5. Therefore, the vertical shift c should be 0.
Step 5: Determine the angular frequency (k)
The angular frequency represents how quickly the graph oscillates. A higher value of k will make the graph oscillate more rapidly. Since we want the equation to have solutions only at x = 1 and x = 2, we need the graph to complete exactly one full oscillation between these two points. The distance between x = 1 and x = 2 is 1, so we need one complete oscillation in that interval. Therefore, the angular frequency can be calculated using the formula k = 2π/T, where T is the period. In our case, k = 2π/1 = 2π.
Using the calculated values of a = 1, T = 1, d = 1.5, c = 0, and k = 2π, we can write the trigonometric equation in the desired format:
y = 1 sin(2π(x - 1.5)) + 0
Simplifying further, we get:
y = sin(2πx - 3π)