Lesson Explainer: Graphs of Trigonometric Functions Mathematics

In this explainer, we will learn how to graph trigonometric functions, such as sine, cosine, and tangent, and deduce their properties.

Let us begin by reviewing the special angles on the unit circle.

We know that the 𝑦-coordinates of these points represent sine values of the corresponding angles. Using degrees, we can construct the input–output table for the function sinπ‘₯.

π‘₯0∘30∘45∘60∘90∘120∘135∘150∘180∘210βˆ˜β‹―
sinπ‘₯012√22√321√32√22120βˆ’12β‹―

A key feature of sinπ‘₯, which is demonstrated in its graph, is that this function begins with the value 0 when π‘₯=0∘, and it increases to the maximum value 1 when π‘₯=90∘. By plotting the points from the input–output table above, we can approximate the graph of sinπ‘₯.

As noted earlier, the graph of sinπ‘₯ begins at zero when π‘₯=0∘, and it increases to the maximum value 1 when π‘₯=90∘.

Since π‘₯ represents the angle on the unit circle diagram, we know that each of these values repeats every 360∘, or 2πœ‹ radians. This leads to the fact that sinπ‘₯ is periodic, with a period of 360∘, or 2πœ‹ radians. The graph of sinπ‘₯ can be extended beyond the interval [0,360]∘∘ by creating copies of the graph of this interval. For example, the graph of sinπ‘₯ over [βˆ’1080,1080]∘∘ is shown below.

From this graph, we can see that sinπ‘₯ has roots at every 180∘ starting from 0∘. We also note that the sine function is odd, which means that its graph is rotationally symmetric about the origin. This property is algebraically expressed by sinsin(βˆ’π‘₯)=βˆ’π‘₯, for any real number π‘₯.

Properties: The Sine Function and Its Graph

The graph of the sine function demonstrates the following characteristics:

  • The 𝑦-intercept of sinπ‘₯ is 0, and it increases to the maximum value 1.
  • The roots of sinπ‘₯ are 180π‘›βˆ˜, or π‘›πœ‹, for any π‘›βˆˆβ„€.
  • The maximum value of the function is 1 and the minimum value is βˆ’1.
  • The function is periodic, with a period of 360∘, or 2πœ‹ radians.
  • sinπ‘₯ is an odd function; that is, sinsin(βˆ’π‘₯)=βˆ’π‘₯.

We can determine the graph of the cosine function using a similar process. We know that the π‘₯-coordinates of points on the unit circle represent cosine values of the corresponding angles. Then, we can obtain the input–output table.

π‘₯0∘30∘45∘60∘90∘120∘135∘150∘180∘210βˆ˜β‹―
cosπ‘₯1√32√22120βˆ’12βˆ’βˆš22βˆ’βˆš32βˆ’1βˆ’βˆš32β‹―

By plotting these points, we can approximate the graph of the cosine function.

Unlike the graph of sine, cosine begins at the maximum value 1 at π‘₯=0∘ and decreases to the minimum value βˆ’1 at π‘₯=180∘. Like sine, cosine is a periodic function with a period of 360∘, or 2πœ‹ radians, and we can extend this graph to a larger interval by making copies of the graph over [0,360]∘∘. The graph of cosπ‘₯ over [βˆ’1080,1080]∘∘ is shown below.

The roots of cosπ‘₯ begin at 90∘ and repeat every 180∘. Unlike sine, we can see that cosine is an even function, which means that it has reflective symmetry with respect to the 𝑦-axis. Algebraically, this means coscos(βˆ’π‘₯)=π‘₯.

Properties: The Cosine Function and Its Graph

The graph of the cosine function demonstrates the following characteristics:

  • The 𝑦-intercept of the function is 1, and it decreases to the minimum value βˆ’1.
  • The roots of cosπ‘₯ are (90+180𝑛)∘, or πœ‹2+π‘›πœ‹, for any π‘›βˆˆβ„€.
  • The maximum value of the function is 1 and the minimum value is βˆ’1.
  • The function is periodic, with a period of 360∘, or 2πœ‹ radians.
  • cosπ‘₯ is an even function; that is, coscos(βˆ’π‘₯)=π‘₯.

Finally, let us consider the graph of the tangent function. We know that tansincosπ‘₯=π‘₯π‘₯. This means that the tangent function is not defined when cosπ‘₯ is equal to zero, which is when π‘₯=(90+180𝑛),πœ‹2+π‘›πœ‹.∘orinradians

Using the input–output table for sine and cosine, we can construct one for tangent.

π‘₯0∘30∘45∘60∘90∘120∘135∘150∘180∘210βˆ˜β‹―
tanπ‘₯0√331√3Undefinedβˆ’βˆš3βˆ’1βˆ’βˆš330√33β‹―

By plotting these points, we can approximate the graph of the tangent function.

Again, it is worth highlighting that the function is undefined at 90∘ and 270∘, since cosπ‘₯=0 at these angles. Unlike sine and cosine, the graph of the tangent function is unbounded and includes vertical asymptotes. Similar to sine and cosine, we can extend this graph periodically to a larger interval, [βˆ’990,990]∘∘.

From this graph, we note that the period of the tangent function is 180∘, as opposed to the sine and cosine functions that have periods of 360∘. Also, like sine, the graph of the tangent function has rotational symmetry about the origin, meaning that the tangent function is odd; that is, tantan(βˆ’π‘₯)=βˆ’π‘₯.

Properties: The Tangent Function and Its Graph

The graph of the tangent function demonstrates the following characteristics:

  • The 𝑦-intercept of the function is 0, and it increases without bound until π‘₯=90∘.
  • The roots of tanπ‘₯ are the same as those of sinπ‘₯. These are 180π‘›βˆ˜, or π‘›πœ‹, for any π‘›βˆˆβ„€.
  • The graph of tanπ‘₯ has vertical asymptotes at the roots of cosπ‘₯, which are (90+180𝑛)∘, or πœ‹2+π‘›πœ‹, for any π‘›βˆˆβ„€.
  • The graph of tanπ‘₯ is unbounded.
  • The function is periodic, with a period of 180∘, or πœ‹ radians.
  • tanπ‘₯ is an odd function; that is, tantan(βˆ’π‘₯)=βˆ’π‘₯.

In the first three examples, we will determine which of the trigonometric functions correspond to the given graph, and consider which portion of the graph of a trigonometric function results from each quadrant in the unit circle diagram.

Example 1: Recognizing Trigonometric Functions from Their Graphs

Consider the following figures.

  1. Which function does the plot in the graph, figure (a), represent?
    1. Cosine
    2. Sine
    3. Tangent
  2. Assign each region of the plot in figure (a) to the corresponding quadrant of the unit circle in figure (b).

Answer

Part 1

Recall that the graph of the tangent function has vertical asymptotes at every interval of length πœ‹. Since this graph does not have any vertical asymptotes over an interval with length greater than πœ‹, this cannot be the graph of the tangent function. Hence, it must be the graph of either sine or cosine. Let us compare the values indicated by the graph with the values of sine and cosine functions.

The coordinates of points on the unit circle are given by (πœƒ,πœƒ)cossin, where πœƒ is the counterclockwise angle of the radius to the point with respect to the positive π‘₯-axis. In the given graph, we can see that the function value is equal to 0 when the angle is at 2πœ‹ radians. We know that 2πœ‹ is the angle of a full revolution, which brings the point back to the positive π‘₯-axis.

The coordinate of the point on the unit circle corresponding to the angle 2πœ‹ is (1,0), which tells us cossin2πœ‹=1,2πœ‹=0.

The given graph indicates that this function takes the value 0 at 2πœ‹, so this agrees with the sine function.

This is option B.

Part 2

We know that the values of the sine function are given by the 𝑦-coordinates of the points on the unit circle. To find the region of the unit circle corresponding to each part of the given graph, we draw these angles on the unit circle. Region A includes the angles between 3πœ‹2 and 2πœ‹. We know that 2πœ‹ is a full counterclockwise revolution. In the unit circle, these angles can be drawn as below.

Hence, the angles between these two values lie in the fourth quadrant, meaning that region A is assigned to quadrant IV.

Similarly, we can draw the angles in region B.

So, these angles lie on the first quadrant. Hence, region B is assigned to quadrant I.

Let us look at the remaining regions.

We can see that region C corresponds to quadrant II, and region D corresponds to quadrant III. In conclusion, we have the assignments A:IVB:IC:IIandD:III,,,.

Let us consider another example where we will determine the trigonometric function represented by a given graph and associate regions of the graph with parts of the unit circle.

Example 2: Recognizing Trigonometric Functions from Their Graphs

Consider the following figures.

  1. Which function does the plot in the graph, figure (a), represent?
    1. Cosine
    2. Sine
    3. Tangent
  2. Assign each region of the plot in figure (a) to the corresponding quadrant of the unit circle in figure (b).

Answer

Part 1

Recall that the graph of the tangent function has vertical asymptotes at every interval of length πœ‹. Since this graph does not have any vertical asymptotes over an interval with length greater than πœ‹, this cannot be the graph of the tangent function. Hence, it must be the graph of either sine or cosine. Let us compare the values indicated by the graph with the values of sine and cosine functions.

The coordinates of points on the unit circle are given by (πœƒ,πœƒ)cossin, where πœƒ is the counterclockwise angle of the radius to the point with respect to the positive π‘₯-axis. In the given graph, we can see that the function value is equal to 0 when the angle is at βˆ’7πœ‹2 radians. We know that negative angles are clockwise angles from the positive π‘₯-axis. Since βˆ’7πœ‹2=βˆ’3πœ‹βˆ’πœ‹2, this angle is obtained by rotating clockwise from the positive π‘₯-axis by 3πœ‹ (one revolution and a half) and going an additional quarter revolution clockwise.

The coordinates of the point on the unit circle corresponding to the angle βˆ’7πœ‹2 are (0,1), which tells us cossinο€Όβˆ’7πœ‹2=0,ο€Όβˆ’7πœ‹2=1.

The given graph indicates that this function takes the value 0 at βˆ’7πœ‹2, so this agrees with the cosine function.

This is option A.

Part 2

We know that the values of the cosine function are given by the π‘₯-coordinates of the points on the unit circle. To find the region of the unit circle corresponding to each part of the given graph, we draw these angles on the unit circle. Region A includes the angles between βˆ’11πœ‹2 and βˆ’5πœ‹. We know that negative angles are measured clockwise from the positive π‘₯-axis and that βˆ’2πœ‹ is a clockwise full revolution. We can write βˆ’11πœ‹2=βˆ’2πœ‹βˆ’2πœ‹βˆ’πœ‹βˆ’πœ‹2.

So, this angle represents two and a half clockwise revolutions followed by an additional quarter turn. Also βˆ’5πœ‹=βˆ’2πœ‹βˆ’2πœ‹βˆ’πœ‹, so this angle represents two and a half clockwise revolutions. These angles are drawn below.

Hence, the angles between these two values lie in the second quadrant, meaning that region A is assigned to quadrant II.

Similarly, we can draw the angles in other regions.

We can see that region B corresponds to quadrant III, region C corresponds to quadrant IV, and region D corresponds to quadrant I. In conclusion, we have the assignments A:IIB:IIIC:IVandD:I,,,.

Let us consider another example where the graph of a trigonometric function with vertical asymptotes is given.

Example 3: Recognizing Trigonometric Functions from Their Graphs

  1. Which function does the plot in the graph, figure (a), represent?
    1. Cosine
    2. Sine
    3. Tangent
  2. Assign each region of the plot in figure (a) to the corresponding quadrant of the unit circle in figure (b).

Answer

Part 1

Among sine, cosine, and tangent functions, the tangent function is the only one that grows without bound. In other words, the graph of a tangent function contains vertical asymptotes, while the graphs of the sine and cosine functions are bounded between βˆ’1 and 1. Hence, this graph represents the tangent function.

This is option C.

Part 2

Region A contains angles between βˆ’πœ‹2 and 0. Angle 0 represents the positive π‘₯-axis. We also know that negative angles are measured clockwise from the positive π‘₯-axis, so angle βˆ’πœ‹2 represents a quarter revolution clockwise from the positive π‘₯-axis, which places this angle on the negative 𝑦-axis. This tells us that angles in region A lie between the negative 𝑦-axis and the positive π‘₯-axis. Hence, region A corresponds to the fourth quadrant.

Angles in region B lie between 0 and πœ‹2. We know that 0 represents the positive π‘₯-axis, and πœ‹2 represents the positive 𝑦-axis since it is a counterclockwise quarter revolution from the positive π‘₯-axis. These angles lie in quadrant I.

Region C has angles between πœ‹2 and πœ‹. Since πœ‹ is a counterclockwise half-revolution, it represents the negative π‘₯-axis. Hence, angles in region C lie between the positive 𝑦-axis and the negative π‘₯-axis, which is quadrant II.

Finally, region C has angles between πœ‹ and 3πœ‹2. The latter angle is a three-quarter counterclockwise revolution, which lands on the negative 𝑦-axis. So, the quadrant between the negative π‘₯-axis and the negative 𝑦-axis is quadrant III.

In conclusion, we have the assignments A:IVB:IC:IIandD:III,,,.

In the next example, we will recognize the correct graph of a trigonometric function using its characteristics.

Example 4: Identifying the Graph of the Tangent Function

Which of the following is the graph of 𝑦=π‘₯tan?

Answer

Let us recall a few important characteristics of the tangent graph:

  • tanπ‘₯ is periodic, with a period of 180∘.
  • tanπ‘₯ has vertical asymptotes at the roots of cosπ‘₯, which are 90+180π‘›βˆ˜βˆ˜ for any π‘›βˆˆβ„€.
  • The roots of tanπ‘₯ are the same as those of sinπ‘₯, which are 180π‘›βˆ˜ for any π‘›βˆˆβ„€.

Let us begin by examining the first property, which tells us that the period of tanπ‘₯ is 180∘. We see that option A has a period of 90∘, so we can rule this out. All other graphs are periodic, with a period of 180∘.

The second characteristic tells us that we should see vertical asymptotes at 90+180π‘›βˆ˜βˆ˜. Since we already checked the periodicity of 180∘, it would be enough to see whether we can find a vertical asymptote at π‘₯=90∘. Option B does not have a vertical asymptote at π‘₯=90∘, so we can rule this one out. The remaining options, C, D, and E, all have a vertical asymptote at π‘₯=90∘.

Next, let us examine the roots. We know that the roots of the tangent function are 180π‘›βˆ˜. Once again, since we have already checked the periodicity, it is enough to check whether the graph demonstrates a root at π‘₯=0∘. Option C does not have a root at π‘₯=0∘, so we exclude this option. The remaining two options, D and E, both have a root at π‘₯=0∘.

Let us compare the remaining two graphs, D and E. We note that the function in D is negative over the interval π‘₯∈(0,90)∘∘, while the function in E is positive in that interval. We know that π‘₯∈(0,90)∘∘ represents acute angles, or angles in the first quadrant. Since the tangent function is positive for acute angles, we see that D is not the graph for the tangent function.

This leads to option E.

In our next example, we will identify a graph of the tangent function after a function transformation.

Example 5: Identifying the Graph of a Trigonometric Function after Transformation

Which of the following is the graph of 𝑦=βˆ’π‘₯tan?

Answer

We can begin by drawing the graph of tanπ‘₯. Let us recall a few important characteristics of the tangent graph:

  • tanπ‘₯ is periodic, with a period of 180∘.
  • tanπ‘₯ has vertical asymptotes at the roots of cosπ‘₯, which are 90+180π‘›βˆ˜βˆ˜ for any π‘›βˆˆβ„€.
  • The roots of tanπ‘₯ are the same as those of sinπ‘₯, which are 180π‘›βˆ˜ for any π‘›βˆˆβ„€.

Additionally, we know that tanπ‘₯ is positive for acute angles; that is, π‘₯∈(0,90)∘∘. This leads to the graph of the tangent function.

We know that multiplying a function by βˆ’1 results in reflection over the π‘₯-axis. Since we are multiplying tanπ‘₯ by βˆ’1 to obtain the given function, βˆ’π‘₯tan, we need to flip this graph over the π‘₯-axis.

In the graph above, the solid blue graph represents the original function, tanπ‘₯, and the dashed graph represents the reflected function, βˆ’π‘₯tan.

In particular, we note that both the roots and the vertical asymptotes remain the same after reflection. Hence, the graph of βˆ’π‘₯tan should have roots at 180π‘›βˆ˜ and vertical asymptotes at 90+180π‘›βˆ˜βˆ˜ for any π‘›βˆˆβ„€.

We can rule out options B, D, and E since these graphs do not demonstrate roots at 180π‘›βˆ˜, particularly at π‘₯=0∘.

We can see that the difference between the graphs in the two remaining options, A and C, is that the function is negative for π‘₯∈(0,90)∘∘ in A while it is positive over this interval in C. In the reflective graph of βˆ’π‘₯tan we constructed, we can see that our function takes negative values over (0,90)∘∘.

This leads to option A.

In our final example, we will apply function transformations to the sine function to obtain a new graph.

Example 6: Finding the Maximum Value of a Given Sine Function

Find the maximum value of the function 𝑓(πœƒ)=11πœƒsin.

Answer

We recall that the graph of sinπœƒ starts at 0 at πœƒ=0∘, and it oscillates between the maximum value 1 and the minimum value βˆ’1.

We know that multiplying a positive constant by π‘Ž>0 results in a vertical dilation (stretching or contracting) with the scale factor π‘Ž. Here, we are multiplying sinπœƒ by 11, so the new graph of this function can be obtained by stretching the graph above by a factor of 11.

Here, the solid blue graph represents the original function, sinπœƒ, and the dashed curve represents the function 11πœƒsin. The double-sided red arrows indicate vertical dilation. We can see from this graph that the function 11πœƒsin oscillates between βˆ’11 and 11.

We can also come to this conclusion algebraically. We know that βˆ’1β‰€πœƒβ‰€1.sin

Multiplying this inequality by 11, we have βˆ’11≀11πœƒβ‰€11.sin

This leads to the maximum value 11.

Hence, the maximum value of 𝑓(πœƒ)=11πœƒsin is 11.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • The graph of the sine function demonstrates the following characteristics:
    • The 𝑦-intercept of sinπ‘₯ is 0, and it increases to the maximum value 1.
    • The roots of sinπ‘₯ are 180π‘›βˆ˜, or π‘›πœ‹, for any π‘›βˆˆβ„€.
    • The maximum value of the function is 1 and the minimum value is βˆ’1.
    • The function is periodic, with a period of 360∘, or 2πœ‹ radians.
    • sinπ‘₯ is an odd function; that is, sinsin(βˆ’π‘₯)=βˆ’π‘₯.
  • The graph of the cosine function demonstrates the following characteristics:
    • The 𝑦-intercept of the function is 1, and it decreases to the minimum value βˆ’1.
    • The roots of cosπ‘₯ are (90+180𝑛)∘, or πœ‹2+π‘›πœ‹, for any π‘›βˆˆβ„€.
    • The maximum value of the function is 1 and the minimum value is βˆ’1.
    • The function is periodic, with a period of 360∘, or 2πœ‹ radians.
    • cosπ‘₯ is an even function; that is, coscos(βˆ’π‘₯)=π‘₯.
  • The graph of the tangent function demonstrates the following characteristics:
    • The 𝑦-intercept of the function is 0, and it increases without bound until π‘₯=90∘.
    • The roots of tanπ‘₯ are the same as those of sinπ‘₯. These are 180π‘›βˆ˜, or π‘›πœ‹, for any π‘›βˆˆβ„€.
    • The graph of tanπ‘₯ has vertical asymptotes at the roots of cosπ‘₯, which are (90+180𝑛)∘, or πœ‹2+π‘›πœ‹, for any π‘›βˆˆβ„€.
    • The graph of tanπ‘₯ is unbounded.
    • The function is periodic, with a period of 180∘, or πœ‹ radians.
    • tanπ‘₯ is an odd function; that is, tantan(βˆ’π‘₯)=βˆ’π‘₯.

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