Video: Graphing Trigonometric Functions Using Trigonometric Identities

Which of the following is the graph of 𝑦 = cos (π‘₯ βˆ’ 90)?

03:41

Video Transcript

Which of the following is the graph of 𝑦 equals cos of π‘₯ minus 90?

Let’s begin with a sketch of the graph of 𝑦 equals cos of π‘₯. The maximum and minimum values of cos of π‘₯ are one and negative one. So, these are the maximum and minimum values on our 𝑦 axis. Key points on the graph occur at multiples of 90 degrees on the horizontal scale. So, these are 90, 180, 270, 360, and so on, as well as, negative 90, negative 180, and so on. The graph begins at one on the 𝑦-axis because cos of zero is equal to one.

The graph is, then, a wave shape, falling to zero at 90 degrees, falling to negative one at 180 degrees, rising back to zero at 270 degrees, and rising all the way back to one at 360 degrees. The graph is periodic with a period of 360 degrees. So, this same shape then repeats to the right of 360 degrees and to the left of zero.

Now we haven’t actually been asked about the graph of 𝑦 equals cos of π‘₯. We’ve been asked about the graph of 𝑦 equals cos of π‘₯ minus 90. So, we need to determine what type of transformation has occurred. We recall that, in general, if we have the function 𝑓 of π‘₯, then the function 𝑓 of π‘₯ minus π‘Ž is a translation of the graph of 𝑓 of π‘₯ π‘Ž units to the right. It is a horizontal shift of the graph.

So, 𝑦 equals cos of π‘₯ minus 90 is a translation of the graph 𝑦 equals cos π‘₯ 90 degrees to the right. Each point moves 90 degrees to the right. So, the point zero, one will now be the point 90, one. The point 90, zero will now be the point 180, zero, and so on. We can sketch in the transformed graph in pink.

Now let’s look at the four graphs that we’ve been given as possibilities for the graph of 𝑦 equals cos of π‘₯ minus 90. We see that the second graph here has the correct shape and the correct features, the same as the graph we’ve sketched of 𝑦 equals cos of π‘₯ minus 90. So, this is the graph that we’re looking for.

We can also determine the equations of the other graphs we were given. The first graph is actually just the untranslated graph of 𝑦 equals cos of π‘₯. In the fourth graph, every point has been shifted upwards by one unit. So, this is a vertical translation of the graph of 𝑦 equals cos π‘₯ by one unit up. The equation of this graph is, therefore, 𝑦 equals one plus cos of π‘₯.

The third graph is a little trickier. But looking at the point 90, zero, which would originally have been the point 270, zero on our cos graph, we see that this graph represents a horizontal stretch with a scale factor of one third. And so, the equation of this graph is 𝑦 equals cos of three π‘₯.

You may notice that the graph we chose for 𝑦 equals cos of π‘₯ minus 90 is, in fact, identical to the graph of 𝑦 equals sin of π‘₯. And this is true, because the sine and cos graphs are just translations of one another. To get the graph of 𝑦 equals sin π‘₯, we translate the graph of 𝑦 equals cos π‘₯ 90 degrees to the right. We found then that the graph which represents 𝑦 equals cos of π‘₯ minus 90 is the second graph.

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