Writing Equations of Cosine Functions

Question

Find an equation for a sinusoidal function that has period 3​𝜋, amplitude 2, and contains the point:(-3pi/4, -5)

Answer= 2cos(2/3x)+5

I am easily able to obtain A, B, and C, but am having trouble arriving at D.

I am told to use:

-5 = 2cos (-pi/2) +D How do I arrive at (-pi/2) from the given information?

-5= 2 (0)+D And how is (-pi/2) equal to 0?

-5= D

I have a working understanding of the unit circle, if that pertains.

.

Answer 1

Go to desmos.com and graph 2cosx and 2cos(2x/3). You might understand from the unit circle, but it might help to see it here as well. The two functions will have the same y values at the same percentage of the period. The useful information is that for both functions with an amplitude of 2:

y = 2 when x = 0

y = 0 when x = 1/4 of the period

y = -2 when x = 1/2 of the period

y = 0 when x = 3/4 of the period

y = 2 when x = the period

In your problem, since the period of the function is 3pi, the given x coordinate, -3pi/4, lands at 1/4 of the period to the left of the y axis. The answer to your first question is: they picked 2cos(-pi/2) because it is 1/4 of the period to the left of the y axis. From this, we can get the y value at that point, which is 0 and we know that the midline of the function (so far) is at y = 0.

To answer your second question, -pi/2 is not 0. cos(-pi/2) = 0.

Since we know that the given x coordinate is at 1/4 the period and the point is on the midline, in order for the graph of the function to go through point (-3pi/4, -5), the midline needs to move down 5 units.

You solved for D correctly, which is -5.

Note that either the answer they gave is wrong (It should be 2cos(2/3x) - 5) or the question should have given the point (-3/4pi/4, 5) in order to match the answer.

One more note: Although comparing the function to cosx in stead of 2cosx would have worked in this case (because y = 0 for both functions at 1/4 of the period (and 3/4 the period), the y values would be different at all other similar percentages of the period. Therefore, it would be a better habit to compare the problem function with a basic function with the same amplitude.