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Question Video: Recognizing Trigonometric Functions from their Graph Mathematics

Which function does the plot in the graph, figure (a), represent? [A] Sine [B] cosine [C] tangent. Assign each region of the plot in figure (a) with the corresponding quadrant of the unit circle in figure (b).

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Video Transcript

Which function does the plot in the graph, figure (a), represent? (A) Sine, (B) cosine, or (C) tangent. Assign each region of the plot in figure (a) with the corresponding quadrant of the unit circle in figure (b).

Let’s have a look at the graph in figure (a). We recall first that the graph of the tangent function has a vertical asymptote at πœ‹ by two radians. And then we can add and subtract πœ‹ to find the position of all other vertical asymptotes. If we keep subtracting πœ‹, we find that there are vertical asymptotes at negative seven πœ‹ by two, negative nine πœ‹ by two, and negative 11πœ‹ by two. But there are no vertical asymptotes on this graph. So we can rule the tangent function out. So it must be the graph of either the sine or cosine function.

Now, the graphs of the sine and cosine functions are very similar. They have the same basic shape and a lot of the same key features. For example, they both oscillate between a minimum value of negative one and a maximum value of positive one. To determine which of the two functions is represented here, we need to consider the values of the sine and cosine functions for certain key inputs. We can recall that the coordinates of points on the unit circle are given by cos of πœƒ, sin of πœƒ, where πœƒ is the counterclockwise angle between the positive π‘₯-axis and the radius that connects that point to the origin. In the given graph, we can see that the value of this function is zero when the angle is negative 11πœ‹ over two radians, negative nine πœ‹ over two radians, and negative seven πœ‹ over two radians.

Let’s think about the angle of negative nine πœ‹ over two specifically. Negative nine πœ‹ over two can be written as negative eight πœ‹ over two minus πœ‹ over two. That’s equal to negative four πœ‹ minus πœ‹ over two or two lots of negative two πœ‹ minus πœ‹ over two. Now, we know that on the unit circle, negative angles correspond to clockwise revolutions from the positive π‘₯-axis. An angle of negative two πœ‹ then would correspond to a full clockwise revolution from the positive π‘₯-axis. And two lots of negative two πœ‹ would correspond to two full clockwise revolutions from the positive π‘₯-axis. This would bring us right back to the positive π‘₯-axis.

We then need to perform a further revolution corresponding to an angle of negative πœ‹ over two. This corresponds to a quarter turn in the clockwise direction from the positive π‘₯-axis, which brings us to the negative 𝑦-axis. The point on the unit circle on the negative 𝑦-axis has the coordinates zero, negative one. So, recalling that the coordinates of this point will also be cos of negative nine πœ‹ over two, sin negative nine πœ‹ over two, we have that the cos of negative nine πœ‹ over two is equal to zero and the sin of negative nine πœ‹ over two is equal to negative one. Remember, the value of the function in the graph when π‘₯ is equal to negative nine πœ‹ over two is zero. And so this tells us that the graph represented here is the cosine graph.

So we’ve completed the first part of the question. The second part asked us to assign each region of the plot in figure (a) with the corresponding quadrant of the unit circle in figure (b). Now, we’ve just identified that an angle of negative nine πœ‹ over two radians is two full turns in the clockwise direction and another quarter turn, which brings us to the negative 𝑦-axis. Let’s consider region C, which contains all of the angles between negative nine πœ‹ over two and negative four πœ‹.

An angle of negative four πœ‹ will correspond to two full clockwise revolutions from the positive π‘₯-axis, which will bring us back to the positive π‘₯-axis. Angles between negative nine πœ‹ over two and negative four πœ‹ therefore lie in the fourth quadrant of the unit circle. So we can match up region C with quadrant four.

Next, let’s consider region B, which contains all the angles between negative five πœ‹ and negative nine πœ‹ over two. We know that an angle of negative nine πœ‹ over two brings us back to the negative 𝑦-axis. An angle of negative five πœ‹ is equal to negative four πœ‹ minus πœ‹. So that’s two full clockwise revolutions from the positive π‘₯-axis and then a further half revolution. Two full revolutions and another half brings us to the negative π‘₯-axis. Angles in region B then between negative five πœ‹ and negative nine πœ‹ over two correspond to the third quadrant of the unit circle.

We can assign regions A and D in the same way. Region A corresponds to angles between negative 11πœ‹ over two and negative five πœ‹. We know that an angle of negative five πœ‹ brings us to the negative π‘₯-axis. And then a further angle of negative πœ‹ by two is a further clockwise revolution of quarter of a turn, bringing us to the positive 𝑦-axis. So the angles in region A correspond to the second quadrant of the unit circle. Angles in the final region then, region D, which are angles between negative four πœ‹ and negative seven πœ‹ by two, correspond to the first quadrant.

We found then that the plot in figure (a) represents the cosine function and that region A corresponds to the second quadrant. Region B corresponds to the third quadrant. Region C corresponds to the fourth quadrant. And region D corresponds to the first quadrant of the unit circle.

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