JUMP TO TOPIC

- What Is Local Extrema?
- How To Find the Local Maximum
- How To Find Local Minimum
- Finding Local Maxima and Minima for Transformations
- The y-value of Local Extrema
- The x-values of Local Extrema
- Local Extrema for Tangent and Cotangent
- Examples
- Example 1
- Solution
- Example 2
- Solution
- Example 3
- Solution
- Example 4
- Solution
- Example 5
- Solution
- Practice Questions

# Local Extrema – Explanation and Examples

*Local extrema in trigonometry are the maximum or minimum values a trig function takes before changing direction.*

Because they are periodic and repeat every $2\pi$ or $\pi$ radians, trig functions do not have absolute maxima and minima. Instead, their maximum and minimum values are localized.

Local extrema are important because the maximum and minimum value points of a function reveal aspects of physical applications of the functions. For example, the local extrema of sine functions modeling a real electromagnetic wave reveal the amplitude of that wave.

To get the most out of this article, review trig function graphs first.

## What Is Local Extrema?

Local extrema are maximum and minimum values in a function that change direction more than once. The term “local extrema” is the plural form of “local extremum,” referring to extreme values in a specific domain for a function.

Since trigonometric functions are periodic, they may change direction infinitely many times. In particular, sine, cosine, secant, and cosecant change direction twice over a domain of $2\pi$. Tangent and cotangent do not change direction at all in their period of $\pi$. Therefore, discussion of local extrema in relation to trigonometric functions focuses on the sine, cosine, secant, and cosecant functions.

Since sine and cosine have the same general shape with horizontal displacement, one can use similar strategies for finding the coordinates of local extrema for the two functions. Similarly, since secant and cosecant have the same general shape with only a horizontal displacement, similar strategies will reveal the local maxima and minima for these functions.

Note that there may be more than one strategy for finding local maxima and minima for the different functions.

### How To Find the Local Maximum

To find a local maximum of a trig function, find the local maxima on the interval $[0, 2\pi)$ radians. (Note that “maxima” is just the plural of “maximum.”) These will occur at points where the function changes direction from increasing to decreasing. It may sometimes also occur at the endpoints of a given interval.

Recall that the trigonometric functions with local extrema, namely sine, cosine, secant, and cosecant, are $2\pi$ periodic. This means that they repeat themselves every $2\pi$ units. Thus, every input corresponds to an input over the interval $[0, 2\pi)$ radians. Since the inputs are angles, another way of saying this is that each angle outside the interval $[0, 2\pi)$ radians is coterminal with an angle inside that interval.

Since finding the local maxima on these intervals requires a familiarity with the base graphs, it helps to consider the sine and cosine functions together and then consider the secant and cosecant functions together.

#### Local Maxima for Sine and Cosine

The local maxima of the sine and cosine functions are the crests of the waves. Recall that both sine and cosine have a wave-like shape with a midline of the x-axis. The crests are the top parts of these waves, and the distance from the midline to a crest is the amplitude of such a wave.

Sine and cosine have an amplitude of $1$, which means that the distance from the midline to a crest is $1$ unit. Since the midline is the x-axis, the tops of the crests have a y-value of $1$. In fact, since the range of values for sine and cosine is $[-1, 1]$, $1$ is the overall maximum for both functions. Because the functions are periodic, however, the functions achieve this maximum infinitely many times.

The goal, then, for finding local maxima for sine and cosine is to find the local x-value for which the function achieves a y-value of $1$. This requires knowing for which angles the functions achieve their maximum on the interval $[0, 2\pi)$ and finding a local value coterminal with these angles.

Specifically, the sine reaches its maximum of $1$ at $\frac{\pi}{2}$ radians. Therefore, finding local maxima for a given interval involves finding an angle (or angles) coterminal with this angle. That is, it involves finding an angle $\alpha$ such that $\alpha = \frac{\pi}{2}+2n\pi$ for some integer $n$.

Similarly, since cosine reaches its maximum of $1$ at $0$, local maxima are angles coterminal with the zero angle. These are $\beta$ such that $\beta = 2n\pi$ for some integer $n$.

#### Other Maxima for Sine and Cosine

In the event that a given interval does not contain an angle coterminal with $\frac{\pi}{2}$ radians for sine or $0$ radians for cosine, finding the local maximum or maxima requires testing the endpoints of the interval. That is, if the interval is smaller than $2\pi$ radians, the largest value will occur at one (or both) of the endpoints.

A sine or cosine graph may also have a local maximum at a left-side bound if the graph is decreasing at that point regardless of whether or not the interval contains an angle coterminal with the crests. Similarly, the graph will have a local maximum at the right-side bound if the graph is increasing at that point.

Knowing whether the function is increasing or decreasing at a point requires knowing whether its coterminal angle on the interval $[0, 2\pi)$ is increasing or decreasing at that point.

Sine is increasing over the subintervals $(0, \frac{\pi}{2})$ radians and $(\frac{3\pi}{2}, 2\pi)$ radians. Right-bounds coterminal with angles in these subintervals will be local maxima. Then, the sine is decreasing from $(\frac{\pi}{2}, \frac{3\pi}{2})$. Left-bounds coterminal with angles in these sub intervals will be local maxima as well.

Cosine increases from $(\pi, 2\pi)$ radians. Therefore, right-bounds with angles coterminal to angles in this subinterval will be local maxima. Since it decreases over the subinterval $(0, \pi)$ radians, left-bounds with angles coterminal to angles in this subinterval will be local maxima too.

#### Local Maxima for Secant and Cosecant

The local maxima for secant and cosecant functions can vary. For both functions, the range is $(-\inf, -1)$ and $(1, \inf)$.

Recall that secant and cosecant functions essentially consist of two parabolic shapes right next to each other. One is a parabola that points upwards and has a lowest point of $1$, and the other points downwards and has a highest point of $-1$.

The high point of $-1$ on the parabolic section that goes to negative infinity is considered a local maximum because the function changes direction there. As before, it makes the most sense to just consider the interval $[0, 2\pi)$ since both functions are $2\pi$ periodic. On this interval, the local maximum of $-1$ occurs at the angle $\frac{3\pi}{2}$ radians for cosecant. For secant, it occurs at $\pi$ radians.

Therefore, any angles coterminal with these angles would be local maxima of their respective functions. That is, an angle $\alpha$ is a local maximum of the cosecant function if there exists some integer $n$ such that $\alpha = 2n\pi+\frac{3\pi}{2}$ radians. Similarly, an angle $\beta$ is a local maximum of secant if there exists some integer $n$ such that $\beta = 2n\pi+\pi$ radians.

#### Other Maxima for Secant and Cosecant

An interval for secant or cosecant may also have one or more local maxima at the endpoints. This happens for left-side bounds if the graph is increasing at that point. It happens for right-side bounds if the graph is decreasing at that point.

Secant increases over the subintervals $(0, \frac{\pi}{2})$ and $\frac{\pi}{2}, \pi)$. Then cosecant increases over the subintervals $(\frac{\pi}{2}, \pi)$ and $(\pi, \frac{3\pi}{2})$. Any right-bounds that are coterminal with angles in these intervals will be local maxima for their respective functions.

Similarly, secant decreases over the subintervals $(\pi, \frac{3\pi}{2})$ and $)\frac{3\pi}{2}, 2\pi)$, while cosecant decreases over the subintervals $(0, \frac{\pi}{2})$ and $(\frac{3\pi}{2}, 2\pi)$. Therefore, any left-bound coterminal with angles in these intervals will also be local maxima for their respective functions.

It is also possible, however for the endpoints to have undefined values. These happen at points where the secant’s and cosecant’s respective reciprocal functions have ratios of $0$.

Theoretically, then, it is possible to have an interval for secant or cosecant with no local maxima. This would happen if the endpoints of an interval of length $\pi$ radians were the vertical asymptotes for the function.

### How To Find Local Minimum

Finding the local minimum of a trigonometric function is similar to finding the maximum of a trigonometric function. To find these, it helps to consider what happens to each of the trig functions of the interval $[0, 2\pi)$, since the functions are $2\pi$ periodic.

Like local maxima, local minima (note that “minima” is the plural of “minimum”) mainly occur in the trigonometric functions sine, cosine, secant, and cosecant. This is because a local minimum is a point at which the function changes from decreasing to increasing.Note that, like local maxima, local minima may also be the endpoints of an interval.

Since the overall shape of sine and cosine are the same, they can be considered together. Likewise, the local minima of secant and cosecant can be considered together.

#### Local Minimum for Sine and Cosine

Local minima for sine and cosine will be points where the function achieves the smallest possible ratio for sine and cosine, namely $1$, and/or endpoints of the interval.

Both sine and cosine have a wave shape. Over one period, they will decrease from $1$ to $-1$ and then increase from $-1$ back to $1$. Since the function changes direction from decreasing to increasing at $-1$, any angle with a sine or cosine ratio of $-1$ will be a local minimum.

Since $sinx=-1$ when $x=\frac{3\pi}{2}$ radians on the interval $[0, 2\pi)$, any angle coterminal with this angle will also have a ratio of $-1$. Thus, such angles will be a local minimum. These angles, $\alpha$ will all have the form $\alpha = \frac{3\pi}{2}+2n\pi$ for some integer $n$.

Similarly, $cosx=-1$ when $x=\pi$ radians on the interval $[0, 2\pi)$. Thus, any angle coterminal with this one will also have a ratio of $-1$ and also be a local minimum. Such angles, $\beta$, have the form $\beta = \pi+2n\pi$ for some integer $n$.

The sine and cosine may also have local minima at the endpoints. These endpoints will be local minima if they are left-side bounds and lie on a segment of the graph that is increasing. If they are right-side bounds and lie on a section of the graph that is decreasing, then they will be local minima.

#### Local Minimum for Secant and Cosecant

The local minima of secant and cosecant graphs will be the bottoms of the parabolic segments that point upward or endpoints of the graph.

A cosecant or secant function has a local minimum with a value of $1$ at the base of the upper section of the graph. On the interval $[0, 2\pi)$, cosecant achieves this value at $0$ radians. Consequently, all angles coterminal with the zero angle will also be local minima. These angles will have the form $2n\pi$ for some integer $n$.

Then, the secant has a value of $1$ at the point $\frac{\pi}{2}$ radians on the interval $[0, 2\pi)$. Thus, all coterminal angles of the form $2n\pi+\frac{\pi}{2}$ for some integer $n$ will also be local minima.

Similar to the sine and cosine local minima, left-bounds of the secant and cosecant graphs on intervals where the graphs are increasing will be local minima. Then, right-bounds on intervals where the graphs are decreasing will also be local minima.

### Finding Local Maxima and Minima for Transformations

Finding local maxima and minima for transformations requires familiarity with how different transformations change the trig functions.

Begin by considering a transformed function $y=afun(bx+c)+d$, where $fun$ stands in for the trigonometric functions sine, cosine, secant, and cosecant.

### The y-value of Local Extrema

The y-value of local extrema for a transformed function $y=afun(bx+c)+d$ depends on the coefficient $a$ and the vertical shift $d$.

When the function shifts $d$ units, the new midline of the trigonometric function is the horizontal line $y=d$. This midline is the line directly between the peaks and troughs in the wave functions sine and cosine. In secant and cosecant, it is the line directly between the “vertices” of the two parabolic segments. For all four functions, the original midline is the x-axis, $y=0$.

Then, the coefficient $a$ in front of the function changes the location of the local extrema. In particular, it changes the distance from the peaks and troughs to the midline in the wave functions sine and cosine. Then, it changes the distance from the “vertices” of the parabolic segments in secant and cosecant.

Specifically, the new location of peaks in the wave functions and the base of the upward-pointing parabolic segments for the reciprocal functions will be $d+a$. Then, the troughs of the wave functions and the high points of the downward-pointing parabolic segments of the reciprocal functions will be $d-a$.

### The x-values of Local Extrema

The x-values of local extrema can be found with a little bit of algebra. While what is inside the parenthesis of the function will still come out to the original angle of the local extrema, $b$ and $c$ may change the angle associated with this value.

That is, if a function achieves a local extremum at an angle $\gamma$, then the transformed function $afun(bx+c)+d$ will achieve a local extremum at the angle $\theta$ such that $\gamma = b\theta + c$. Solving for $\theta$ yields $\frac{\gamma-c}{b}$.

Clearly, there will be no local extrema if $b=0$.

### Local Extrema for Tangent and Cotangent

Local extrema for tangent and cotangent functions may or may not exist depending on the boundaries of the given interval.

If the boundaries of the given interval are vertical asymptotes of the tangent or cotangent function, there will be no local maxima or minima. This is because the periods of the tangent function increase from $-\inf$ to $\inf$, and the periods of the cotangent function decrease from $\inf$ to $-\inf$. Thus, there is no highest or lowest value.

But a left-boundary on the tangent graph that is not at a vertical asymptote will be a local minimum. Similarly, a right-boundary on the cotangent graph that is not at a vertical asymptote will be a local maximum.

This is reversed for cotangent. Left-bounds that are not at asymptotes will be local minima, and right-bounds that are not asymptotes will be local maxima.

### Examples

This section goes over common examples of problems involving local extrema in trig functions and their step-by-step solutions.

### Example 1

Find the local extrema values (maxima and minima) for the sine function on the interval $[5\pi, 8\pi]$.

### Solution

First, note that this interval has a length greater than $2\pi$ radians, so there may be more than one local maximum or minimum.

Start by finding the angles in the range of $[0, 2\pi)$ that correspond to the angles in the given interval. The interval $[5\pi, 6\pi)$ is coterminal with the interval $[\pi, 2\pi)$, and $[6\pi, 8\pi)$ is coterminal with $[0, 2\pi)$. The angle $8\pi$ radians is also coterminal with the angle $0$ radians.

#### Local Maxima

The local maxima in this interval will be angles coterminal with $\frac{\pi}{2}$. There are no such angles on the subinterval $[5\pi, 6\pi)$, but there is one on the subinterval $[6\pi, 8\pi)$. Since $6\pi$ is coterminal with $0$, the angle $6\pi+\frac{\pi}{2} = \frac{13\pi}{2}$ radians is coterminal with $\frac{\pi}{2}$. Thus, it is a local maximum with a y-value of $1$, the y-value at $\frac{\pi}{2}$ radians.

#### Local Minima

Similarly, the local minima on the given interval will be angles coterminal with $\frac{3\pi}{2}$. There will be one such angle on the subinterval $[5\pi, 6\pi)$ because it is coterminal with $[\pi, 2\pi)$. Since $\frac{3\pi}{2}$ radians is exactly between $\pi$ radians and $2\pi$ radians, the corresponding angle is $\frac{11\pi}{2}$. This is a local minimum with a y-value of $-1$, the y-value at $\frac{3\pi}{2}$ radians.

There is another local minimum on the subinterval $[6\pi, 8\pi)$. Since $6\pi$ radians is coterminal with $0$ radians, the angle $6\pi+\frac{3\pi}{2} = \frac{15\pi}{2}$ radians is coterminal with $\frac{3\pi}{2}$ radians. This angle is also a local minimum with a y-value of $-1$.

#### Endpoints

Finally, determine whether the function is increasing or decreasing at the endpoints of the given interval to determine if these points are local maxima, minima, or neither.

Since the sine function decreases from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$ radians, the function is decreasing at the left endpoint. This is because $5\pi$ is coterminal with $\pi$, which lies on that decreasing interval. Thus, this point will be a local maximum because decreasing left-bounds are local maxima.

Then the angle $8\pi$ is coterminal with $0$. At $0$, the sine function is increasing, so this endpoint is a local maximum. This is because all increasing right-bounds are local maxima.

### Example 2

Find the local maxima and minima for the secant over the interval $[-\frac{13\pi}{2}, -5\pi]$.

### Solution

This interval is shorter than $2\pi$ in length. Specifically, it has a length of $\frac{3\pi}{2}$ radians.

As before, find the coterminal angles on the interval $[0, 2\pi)$. The endpoint $-5\pi$ is coterminal with $\pi$ radians, and the endpoint $-\frac{13\pi}{2}$ is coterminal with $\frac{3\pi}{2}$. Therefore, the subinterval $[-\frac{13\pi}{2}, -6\pi)$ is coterminal with the interval $[\frac{3\pi}{2}, 2\pi)$. Then, since $2\pi$ is coterminal with $0$ and $-6\pi$, the subinterval $[-6\pi, -5\pi]$ is coterminal with the interval $[0, \pi]$.

Recall that the local maximum of the secant function is at $\pi$ radians. Therefore, since $-5\pi$ is coterminal with $\pi$ radians, it is a local maximum. Since the function has no transformations, its y-value is $-1$.

The local minimum of a secant function is at $0$ radians. Since $-6\pi$ radians is coterminal with $0$, the local minimum of the function is there. Its y-value is $1$.

One of the endpoints, $-5\pi$ has already been shown to be a local extremum. The other endpoint, $-\frac{13\pi}{2}$, is coterminal with $\frac{3\pi}{2}$. At this point, the secant function has an undefined value. Therefore, this point is not a local extremum.

### Example 3

Find the local extrema for the cosine function $y=8cos(\frac{1}{2}x-\pi)+3$ over the interval $[0, 4\pi]$. Make sure to include x-values and y-values for these points.

### Solution

Begin by finding the transformed function’s new maximum and minimum values.

This function has the form $y=acos(bx+c)+d$. In this case, $a=8$, $b=\frac{1}{2}$, $c=-\pi$, and $d=3$. Therefore, the new midline of the function is the horizontal line $y=3$. Then, because $a=8$, the amplitude of this function is also $8$. This means that the local maximum values will have a y-value of $8+3=11$. Similarly, the local minima will have y-values of $3-8=-5$.

Next, recall that the local minima and maxima will occur at angles $bx+c$ such that $bx+c$ equals the extrema points on the interval $[0, 2\pi)$.

That is, since the local maxima of the basic cosine function on the interval $[0, 2\pi)$ are at $0$ radians and $2\pi$ radians, the local maxima of this function will be at points $x$ such that $bx+c=0$ and $bx+c=2\pi$. Then, since the local minimum of the basic cosine function on the interval $[0, 2\pi)$ is at $\pi$ radians, the local minima of this function will occur at solutions to the equation $bx+c=\pi$.

Use the values of $b$ and $c$ from the given transformation to solve these equations.

#### Local Maxima

The local maxima will be at points $x$ such that $\frac{1}{2}x-\pi=0$ and $\frac{1}{2}x-\pi=2\pi$. Solving for $x$ yields $x=2\pi$ and $x=6\pi$. Only the maximum $x=2\pi$ will be on the given interval $[0, 4\pi)$.

#### Local Minima

Then, the local minimum will be at the point $x$ such that $\frac{1}{2}x-\pi=\pi$. Solving for $x$ yields $x=4\pi$. Note, however, that the period of this function is $\frac{2\pi}{\frac{1}{2}} = 4\pi$. This means that the angles $4\pi$ radians and $4\pi-4\pi = 0$ radians will be coterminal. Thus, $0$ radians will also be a local minimum.

Both endpoints have already been shown to be local minima, so there are no more points to consider. Thus, the local maximum of the function $y=8cos(\frac{1}{2}x-\pi)+3$ over the interval $[0, 4\pi]$ is at the point $(2\pi, 11)$, and the local minima will be at the points $(0, -5)$ and $(4\pi, -5)$.

### Example 4

Find the local extrema for a tangent function over the interval $[\pi, 2\pi]$.

### Solution

First, recall that tangent has a period of $\pi$ radians, not $2\pi$ radians. The tangent function is also not like the other trigonometric functions because it does not have local maxima and minima on some periods such as $[-\frac{\pi}{2}, \frac{\pi}{2}]$. At these endpoints, the tangent function is undefined and extends to positive or negative infinity.

Thus, the only thing to check regarding local maxima and minima for a tangent function is the endpoints. If the endpoints are at undefined locations, which include all angles coterminal with $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, then there are no extrema. Otherwise, a defined left-bound will be a local minimum and a defined right-bound will be a local maximum since tangent is always increasing in defined intervals.

In the given interval, $\pi$ and $2\pi$ are coterminal, since the period is just $\pi$ radians. Since $\pi}$ is on the standard interval and not coterminal with $-\frac{\pi}{2}$ or $\\frac{\pi}{2}$, the tangent is defined at this point. Therefore, both $\pi$ radians and $2\pi$ radians will be local extrema. Specifically, as a left-bound of an increasing function, $\pi$ will be a local minimum. Similarly, $2\pi$ will be a local maximum.

At these points, the y-values will be the same. Recall that the sine value at $\pi$ and $2\pi$ is $0$. Since tangent is equal to sine divided by cosine, and cosine is not equal to $0$ at these points, the tangent at these points is also $0$.

Thus, the local minimum will be $(\pi, 0)$, and the local maximum will be $(2\pi, 0)$.

### Example 5

A transformation of the cosecant function has a local maximum of $1$ at $\frac{\pi}{2}$. It also has a period of $\pi$ radians. Find at least two possible equations with the form $acsc(bx+c)+d$ for this function.

In this case, it is assumed the local maximum is due to the graph shape and not due to the location of an interval’s endpoints.

### Solution

There are a couple of ways to do this problem. Begin by finding the transformation necessary to create a period of $\pi$ radians.

Since the basic period of a cosecant function is $2\pi$, and the transformed period is $\frac{2\pi}{b}$, the new period only depends on $b$. Because of the formula, there is only one possible value for $b$.

Thus, solve the equation $\pi=\frac{2\pi}{b}$ to find that $b=2$. Therefore, the equation will have the form $y=acsc(2x+c)+d$.

#### Maximum Value

It is required that the local maximum of the transformed function is $1$. Recall that in a regular cosecant function, the local maximum is the peak of the lower parabolic shape, which is $-1$.

The location of this point can change in two ways. First, changing the location of the midline, the horizontal line directly between the local maxima and minima in a trig function, changes the location of these points. Then, a vertical dilation moves the points further apart or closer together.

Since the original midline for a cosecant function is $y=0$, the x-axis, there is no way to move the top of the bottom parabolic shape to $1$ without moving the midline. Vertical dilations can bring the point farther from or closer to the midline, but it can’t move it to the other side.

The simplest thing to do in this case is to move the midline up two units. This will similarly move the local maximum up two units. It is, however, also possible to combine a vertical dilation with a midline shift upwards by more than one unit.

#### Horizontal Shift

There is also a horizontal shift in this transformation because the angle of the local maximum is different. In the basic cosecant function, the maximum occurs at $\frac{3\pi}{2}$ radians. In this function, it occurs at $\frac{\pi}{2}$.

This means that what is inside the parenthesis of the transformed function must be equal to $\frac{\pi}{2}$ when $x=\frac{3\pi}{2}$. That is, $b(\frac{3\pi}{2})+c=\frac{\pi}{2}$. Since $b=2$, this is $3\pi+c=\frac{\pi}{2}$.

Then solving for $c$ yields $-\frac{5\pi}{2}$. Any angle coterminal with $-\frac{5\pi}{2}$ would also work.

#### The Transformations

Therefore, the simplest option for the equation is $csc(2x-\frac{5\pi}{2})+2$. Another option is $csc(2x-\frac{\pi}{2})+2$, since $-\frac{5\pi}{2}$ and $-\frac{pi}{2}$ are coterminal.

This equation could also have a different vertical shift combined with a vertical dilation. A dilation factor of $2$, for example, means that the local maximum’s y-value will be two units below the midline. Therefore, if the local maximum value is $1$ with a dilation factor of $2$, the midline must be at $y=1+2$, or $y=3$. Since $y=d$ is the midline of a transformed function, another option is $2csc(2b-\frac{\pi}{2})+3$.

There are infinitely many options for changing the function by changing the vertical dilation factor and vertical shift together or by changing the $c$ value to a different angle coterminal with $-\frac{5\pi}{2}$, of which there are infinitely many.

### Practice Questions

Images/mathematical drawings are created using Geogebra.