Question Video: Recognizing Trigonometric Functions from their Graphs Mathematics

Consider the following figures. Which function does the plot in the graph, figure (a), represent? [A] tangent [B] cosine [C] sine. 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

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

Looking at figure (a) then, we identify first that this graph has vertical asymptotes. These asymptotes occur when π‘₯ is equal to negative πœ‹ by two, πœ‹ by two, and three πœ‹ by two. In fact, if we were to continue the graph in both directions, we’d see that they occur at every interval of πœ‹ radians from these values. The graph also has no maximum and no minimum value. It is unbounded in the vertical direction.

Now, we should recall that neither the sine nor the cosine functions have vertical asymptotes. And they also oscillate between a minimum value of negative one and a maximum value of positive one. The tangent function, however, does have the properties we’ve identified. And so this tells us that the graph must represent the tangent function. Additionally, we can observe that the period of this graph is πœ‹ radians and that the roots of the graph occur at every integer multiple of πœ‹, which are also properties of the tangent function.

Let’s now consider the second part of the question, in which we’re asked to assign each region of the plot in figure (a) with the corresponding quadrant of the unit circle in figure (b). Looking at the graph then, region 𝐴 corresponds to angles that are between negative πœ‹ by two and zero radians. On the unit circle, an angle of zero radians would lie on the positive π‘₯-axis. Negative angles are measured clockwise from the positive π‘₯-axis. So an angle of negative πœ‹ by two represents a quarter revolution clockwise from the positive π‘₯-axis, which places this angle on the negative 𝑦-axis.

So the angles in region 𝐴 lie between the negative 𝑦-axis and the positive π‘₯-axis, which corresponds to the fourth quadrant. Next, we’ll consider region 𝐡, which represents angles between zero and πœ‹ by two radians. We’ve already said that an angle of zero radians would lie on the positive π‘₯-axis on the unit circle. An angle of πœ‹ by two radians, that’s a positive angle, would be a quarter turn in the counterclockwise direction from the positive π‘₯-axis, which brings us to the positive 𝑦-axis. Angles between these two values therefore lie in the first quadrant.

Region 𝐢 next represents angles between πœ‹ by two and πœ‹ radians. An angle of πœ‹ by two radians, we’ve already said, would take us to the positive 𝑦-axis. And an angle of πœ‹ radians represents a half turn in the counterclockwise direction from the positive π‘₯-axis, which takes us to the negative π‘₯-axis. The angles in region 𝐢 therefore lie in the second quadrant of the unit circle. Finally then, region 𝐷, which contains angles between πœ‹ and three πœ‹ by two radians, an angle of πœ‹ radians takes us to the negative π‘₯-axis, and an angle of three πœ‹ by two radians is three-quarters of a turn in the counterclockwise direction from the positive π‘₯-axis, taking us to the negative 𝑦-axis. So angles in region 𝐷 do indeed lie in the third quadrant of the unit circle.

So we found that the given graph represents the tangent function and that region 𝐴 corresponds to quadrant four, region 𝐡 corresponds to quadrant one, region 𝐢 corresponds to quadrant two, and region 𝐷 corresponds to quadrant three of the unit circle.

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