Question Video: Recognizing Trigonometric Functions from Their Graphs | Nagwa Question Video: Recognizing Trigonometric Functions from Their Graphs | Nagwa

# Question Video: Recognizing Trigonometric Functions from Their Graphs Mathematics • First Year of Secondary School

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

So, first, we need to determine whether the graph in figure (a) represents the sine or cosine function. We recall that the graphs of these two functions have the same basic shape and a lot of the same 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 this graph represents, letβs consider the roots. We can see that the values of this function are zero when the angle is two π and three π. Continuing along the π₯-axis, there will also be roots at four π, five π, and so on, essentially all the integer multiples of π.

Looking specifically at the value of two π then, we need to determine whether it is sin of two π or cos of two π that is equal to zero. And this will tell us whether it is the sine or cosine function that has been plotted in figure (a). To do this, weβre going to use the unit circle. We recall that points on the unit circle have the coordinates cos of π, sin of π, where π is the counterclockwise angle between the positive π₯-axis and the radius that connects the point to the origin. Positive angles are measured in the counterclockwise direction from the positive π₯-axis. An angle of two π radians corresponds to one full counterclockwise revolution, which brings us back onto the positive π₯-axis.

The point on the unit circle which lies on the positive π₯-axis has the coordinates one, zero. So this tells us that the cos of two π must be equal to one; thatβs the π₯-coordinate. And the sin of two π must be equal to zero; thatβs the π¦-coordinate. We found then that it is the sine function that is equal to zero when the angle is two π, and so this is the function represented in figure (a).

Letβs now consider the second part of the question, in which weβre asked to assign each region of the plot in figure (a) with the corresponding quadrant of the unit circle in figure (b). Considering region A first of all then, this contains all the angles between three π over two and two π radians. We already said that an angle of two π radians corresponds to one full counterclockwise revolution from the positive π₯-axis, bringing us back to the positive π₯-axis. An angle of three π over two radians will correspond to three-quarters of a counterclockwise revolution, which brings us to the negative π¦-axis. Angles between three π over two and two π radians therefore lie in the fourth quadrant of the unit circle. So region A is assigned to quadrant four.

Next, we consider region B, which contains all the angles between two π radians and five π over two radians. We already know that an angle of two π brings us back to the positive π₯-axis. An angle of five π over two radians is an angle of two π plus another π by two. Thatβs another quarter turn counterclockwise from the positive π₯-axis. We go all the way around once and then quarter of the way around again. This brings us to the positive π¦-axis. So angles between two π and five π over two radians correspond to the first quadrant of the unit circle.

We can then continue in the same way for the remaining two regions. Region C contains all the angles between five π by two and three π radians. An angle of three π radians is equal to two π plus π radians. So thatβs one full counterclockwise revolution and then another half a revolution, which brings us to the negative π₯-axis. So angles in region C correspond to quadrant two. Finally, for region D, an angle of three π, weβve already said, corresponds to the negative π₯-axis. For an angle of seven π over two, we need a further angle of π by two radians. So thatβs a further quarter turn counterclockwise, which brings us to the negative π¦-axis. The angles in region D therefore correspond to the third quadrant of the unit circle.

So, by using the unit circle to find the values of the sine and cosine functions for an angle of two π, we found that the plot in figure (a) represents the sine function. And weβve then found that region A corresponds to the fourth quadrant of the unit circle, region B corresponds to the first quadrant, region C corresponds to the second quadrant, and region D corresponds to the third quadrant.

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