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# 2.3: Circular Functions of Real Numbers

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Graph the six trigonometric ratios as functions on the Cartesian plane.
• Identify the domain and range of these six trigonometric functions.
• Identify the radian and degree measure, as well as the coordinates of points on the unit circle and graph for the critical angles.

## The Sine Graph

By now, you have become very familiar with the specific values of sine, cosine, and tangents for certain angles of rotation around the coordinate grid. In mathematics, we can often learn a lot by looking at how one quantity changes as we consistently vary another. We will be looking at the sine value as a function of the angle of rotation around the coordinate plane. We refer to any such function as a circular function, because they can be defined using the unit circle. Recall from earlier sections that the sine of an angle in standard position is the ratio of \begin{align*}\frac{y}{r}\end{align*}, where \begin{align*}y\end{align*} is the \begin{align*}y-\end{align*}coordinate of any point on the circle and \begin{align*}r\end{align*} is the distance from the origin to that point.

Because the ratios are the same for a given angle, regardless of the length of the radius \begin{align*}r\end{align*}, we can use the unit circle as a basis for all calculations.

The denominator is now 1, so we have the simpler expression, \begin{align*}\sin x=y\end{align*}. The advantage to this is that we can use the \begin{align*}y-\end{align*}coordinate of the point on the unit circle to trace the value of \begin{align*}\sin \theta\end{align*} through a complete rotation. Imagine if we start at 0 and then rotate counter-clockwise through gradually increasing angles. Since the \begin{align*}y-\end{align*}coordinate is the sine value, watch the height of the point as you rotate.

Through Quadrant I that height gets larger, starting at 0, increasing quickly at first, then slower until the angle reaches \begin{align*}90^\circ\end{align*}, at which point, the height is at its maximum value, 1.

As you rotate into the second quadrant, the height starts to decrease towards zero.

When you start to rotate into the third and fourth quadrants, the length of the segment increases, but this time in a negative direction, growing to -1 at \begin{align*}270^\circ\end{align*} and heading back toward 0 at \begin{align*}360^\circ\end{align*}.

After one complete rotation, even though the angle continues to increase, the sine values will repeat themselves. The same would have been true if we chose to rotate clockwise to investigate negative angles, and this is why the sine function is a periodic function. The period is \begin{align*}2\pi\end{align*} because that is the angle measure before the sine of the angle will repeat its values.

Let’s translate this circular motion into a graph of the sine value vs. the angle of rotation. The following sequence of pictures demonstrates the connection. These pictures plot \begin{align*}(\theta, \sin \theta)\end{align*} on the coordinate plane as \begin{align*}(x, y)\end{align*}.

After we rotate around the circle once, the values start repeating. Therefore, the sine curve, or “wave,” also continues to repeat. The easiest way to sketch a sine curve is to plot the points for the quadrant angles. The value of \begin{align*}\sin \theta\end{align*} goes from 0 to 1 to 0 to -1 and back to 0. Graphed along a horizontal axis, it would look like this:

Filling in the gaps in between and allowing for multiple rotations as well as negative angles results in the graph of \begin{align*}y = \sin x\end{align*} where \begin{align*}x\end{align*} is any angle of rotation, in radians.

As we have already mentioned, \begin{align*}\sin x\end{align*} has a period of \begin{align*}2\pi\end{align*}. You should also note that the \begin{align*}y-\end{align*}values never go above 1 or below -1, so the range of a sine curve is \begin{align*}\left \{-1 \le y \le 1 \right \}\end{align*}. Because angles can be any value and will continue to rotate around the circle infinitely, there is no restriction on the angle \begin{align*}x\end{align*}, so the domain of \begin{align*}\sin x\end{align*} is all reals.

## The Cosine Graph

In chapter 1, you learned that sine and cosine are very closely related. The cosine of an angle is the same as the sine of its complementary angle. So, it should not be a surprise that sine and cosine waves are very similar in that they are both periodic with a period of \begin{align*}2\pi\end{align*}, a range from -1 to 1, and a domain of all real angles.

The cosine of an angle is the ratio of \begin{align*}\frac{x}{r}\end{align*}, so in the unit circle, the cosine is the \begin{align*}x-\end{align*}coordinate of the point of rotation. If we trace the \begin{align*}x-\end{align*}coordinate through a rotation, notice the change in the distance is similar to \begin{align*}\sin x\end{align*}, but \begin{align*}\cos x\end{align*} starts at one instead of zero. The \begin{align*}x-\end{align*}coordinate at \begin{align*}0^\circ\end{align*} is 1 and the \begin{align*}x-\end{align*}coordinate for \begin{align*}90^\circ\end{align*} is 0, so the cosine value is decreasing from 1 to 0 through the \begin{align*}1^{st}\end{align*} quadrant.

Here is a similar sequence of rotations to the one used for sine. This time compare the \begin{align*}x-\end{align*} coordinate of the point of rotation with the height of the point as it traces along the horizontal. These pictures plot \begin{align*}(\theta, \cos \theta)\end{align*} on the coordinate plane as \begin{align*}(x, y)\end{align*}.

Plotting the quadrant angles and filling in the in-between values shows the graph of \begin{align*}y = \cos x\end{align*}

The graph of \begin{align*}y = \cos x\end{align*} has a period of \begin{align*}2\pi\end{align*}. Just like \begin{align*}\sin x\end{align*}, the range of a cosine curve is \begin{align*}\left \{-1 \le y \le 1 \right \}\end{align*} and the domain of \begin{align*}\cos x\end{align*} is all reals. Notice that the shape of the curve is exactly the same, but shifted by \begin{align*}\frac{\pi}{2}\end{align*}.

## The Tangent Graph

The name of the tangent function comes from the tangent line of a circle. This is a line that is perpendicular to the radius at a point on the circle so that the line touches the circle at exactly one point.

If we extend angle \begin{align*}\theta\end{align*} through the unit circle so that it intersects with the tangent line, the tangent function is defined as the length of the red segment.

The dashed segment is 1 because it is the radius of the unit circle. Recall that the \begin{align*}\tan \theta = \frac{y}{x}\end{align*}, and it can be verified that this segment is the tangent by using similar triangles.

\begin{align*}\tan \theta &=\frac{y}{x}=\frac{t}{1}=t\\ \tan \theta &=t\end{align*}

So, as we increase the angle of rotation, think about how this segment changes. When the angle is zero, the segment has no length. As we rotate through the first quadrant, it will increase very slowly at first and then quickly get very close to one, but never actually touch it.

As we get very close to the \begin{align*}y-\end{align*}axis the segment gets infinitely large, until when the angle really hits \begin{align*}90^\circ\end{align*}, at which point the extension of the angle and the tangent line will actually be parallel and therefore never intersect.

This means there is no finite length of the tangent segment, or the tangent segment is infinitely large.

Let’s translate this portion of the graph onto the coordinate plane. Plot \begin{align*}(\theta, \tan \theta)\end{align*} as \begin{align*}(x, y)\end{align*}.

In fact as we get infinitely close to \begin{align*}90^\circ\end{align*}, the tangent value increases without bound, until when we actually reach \begin{align*}90^\circ\end{align*}, at which point the tangent is undefined. Recall there are some angles (\begin{align*}90^\circ\end{align*} and \begin{align*}270^\circ\end{align*}, for example) for which the tangent is not defined. Therefore, at these points, there are going to be vertical asymptotes.

Rotating past \begin{align*}90^\circ\end{align*}, the intersection of the extension of the angle and the tangent line is actually below the \begin{align*}x-\end{align*}axis. This fits nicely with what we know about the tangent for a \begin{align*}2^{nd}\end{align*} quadrant angle being negative. At first, it will have very large negative values, but as the angle rotates, the segment gets shorter, reaches 0, then crosses back into the positive numbers as the angle enters the \begin{align*}3^{rd}\end{align*} quadrant. The segment will again get infinitely large as it approaches \begin{align*}270^\circ\end{align*}. After being undefined at \begin{align*}270^\circ\end{align*}, the angle crosses into the \begin{align*}4^{th}\end{align*} quadrant and once again changes from being infinitely negative, to approaching zero as we complete a full rotation.

The graph \begin{align*}y=\tan x\end{align*} over several rotations would look like this:

Notice the \begin{align*}x-\end{align*}axis is measured in radians. Our asymptotes occur every \begin{align*}\pi\end{align*} radians, starting at \begin{align*}\frac{\pi}{2}\end{align*}. The period of the graph is therefore \begin{align*}\pi\end{align*} radians. The domain is all reals except for the asymptotes at \begin{align*}\frac{\pi}{2}, \frac{3\pi}{2}, -\frac{\pi}{2}, etc.\end{align*} and the range is all real numbers.

## The Three Reciprocal Functions

For the three reciprocal functions, it gets increasingly difficult to show the segment representation on the unit circle. Instead of going through all of this, we will show the \begin{align*}\cot x\end{align*}, \begin{align*}\csc x\end{align*}, and \begin{align*}\sec x\end{align*} through the graphs of their reciprocal functions, \begin{align*}\tan x\end{align*}, \begin{align*}\sin x\end{align*}, and \begin{align*}\cos x\end{align*}.

## Cotangent

Cotangent is the reciprocal of tangent, \begin{align*}\frac{x}{y}\end{align*}, so it would make sense that where ever the tangent had an asymptote, now the cotangent will be zero. The opposite of this is also true. When the tangent is zero, now the cotangent will have an asymptote. The shape of the curve is generally the same, so the graph looks like this:

When you overlap the two functions, notice that the graphs consistently intersect at 1 and -1. These are the angles that have \begin{align*}45^\circ\end{align*} as reference angles, which always have tangents and cotangents equal to 1 or -1. It makes sense that 1 and -1 are the only values for which a function and it’s reciprocal are the same. Keep this in mind as we look at cosecant and secant compared to their reciprocals of sine and cosine.

The cotangent function has a domain of all real angles except multiples of \begin{align*}\pi \left \{\ldots -2\pi, -\pi, 0, \pi, 2\pi \ldots \right \}\end{align*} The range is all real numbers.

## Cosecant

Cosecant is the reciprocal of sine, or \begin{align*}\frac{1}{y}\end{align*}. Therefore, whenever the sine is zero, the cosecant is going to have a vertical asymptote because it will be undefined. It also has the same sign as the sine function in the same quadrants. Here is the graph.

The period of the function is \begin{align*}2\pi\end{align*}, just like sine. The domain of the function is all real numbers, except multiples of \begin{align*}\pi \left \{ \ldots -2\pi, -\pi, 0, \pi, 2\pi \ldots \right \}\end{align*}. The range is all real numbers greater than or equal to 1, as well as all real numbers less than or equal to -1. Notice that the range is everything except where sine is defined (other than the points at the top and bottom of the sine curve).

Notice again the reciprocal relationships at 0 and the asymptotes. Also look at the intersection points of the graphs at 1 and -1. Many students are reminded of parabolas when they look at the half-period of the cosecant graph. While they are similar in that they each have a local minimum or maximum and they have the same beginning and ending behavior, the comparisons end there. Parabolas are not restricted by asymptotes, whereas the cosecant curve is.

## Secant

Secant is the reciprocal of cosine, or \begin{align*}\frac{1}{x}\end{align*}. Therefore, whenever the cosine is zero, the secant is going to have a vertical asymptote because it will be undefined. It also has the same sign as the cosine function in the same quadrants. Here is the graph.

The period of the function is \begin{align*}2\pi\end{align*}, just like cosine. The domain of the function is all real numbers, except multiples of \begin{align*}\pi\end{align*} starting at \begin{align*}\frac{\pi}{2}. \left \{ \ldots \frac{-\pi}{2}, \frac{\pi}{2}, 0, \frac{3\pi}{2}, \frac{5\pi}{2} \ldots \right \}\end{align*}. The range is all real numbers greater than or equal to 1 as well as all real numbers less than or equal to -1. Notice that the range is everything except where cosine is defined (other than the tops and bottoms of the cosine curve).

Notice again the reciprocal relationships at 0 and the asymptotes. Also look at the intersection points of the graphs at 1 and -1. Again, this graph looks parabolic, but it is not.

The table below summarizes the functions with their domains and ranges:

Function Domain Range
\begin{align*}\sin x\end{align*} all reals \begin{align*}\left \{y : -1 \le y \le 1\right \}\end{align*}
\begin{align*}\cos x\end{align*} all reals \begin{align*}\left \{y : -1 \le y \le 1\right \}\end{align*}
\begin{align*}\tan x\end{align*} \begin{align*}\left \{x : x \neq n \times \frac{\pi}{2}, \ \text{where n is any odd integer} \right \}\end{align*} all reals
\begin{align*}\csc x\end{align*} \begin{align*}\left \{x : x \neq n\pi, \ \text{where n is any integer} \right \}\end{align*} \begin{align*}\left \{y : y \ge 1 \ \text{or} \ y \le -1 \right \}\end{align*}
\begin{align*}\sec x\end{align*} \begin{align*}\left \{x : x \neq n \times \frac{\pi}{2}, \text{where n is any odd interger} \right \}\end{align*} \begin{align*}\left \{y : y \ge 1 \ \text{or} \ y \le -1 \right \}\end{align*}
\begin{align*}\cot x\end{align*} \begin{align*}\left \{x : x \neq n\pi, \text{where n is any integer} \right \}\end{align*} all reals

## Points to Consider

• How are all the reciprocal functions’ graphs related to sine, cosine and tangent?
• What would the inverse function of \begin{align*}y=\sin x\end{align*} look like?

## Review Questions

1. Show that side \begin{align*}A\end{align*} (in orange) in this drawing is equal to \begin{align*}\sec \theta\end{align*}. Use similar triangles in your proof.
2. In Chapter 1, you learned that \begin{align*}\tan^2 \theta + 1 = \sec^2 \theta\end{align*}. Use the drawing and results from question 1 to demonstrate this identity.
3. This diagram shows a unit circle with all the angles that have reference angles of \begin{align*}30^\circ\end{align*}, \begin{align*}45^\circ\end{align*}, and \begin{align*}60^\circ\end{align*}, as well as the quadrant angles. Label the coordinates of all points on the unit circle. On the smallest circle, label the angles in degrees, and on the middle circle, label the angles in radians.
4. Which of the following shows functions that are both increasing as \begin{align*}x\end{align*} increases from 0 to \begin{align*}\frac{\pi}{2}\end{align*}?
1. \begin{align*}\sin x\end{align*} and \begin{align*}\cos x\end{align*}
2. \begin{align*}\tan x\end{align*} and \begin{align*}\sec x\end{align*}
3. \begin{align*}\sec x\end{align*} and \begin{align*}\cot x\end{align*}
4. \begin{align*}\csc x\end{align*} and \begin{align*}\sec x\end{align*}
5. Which of the following statements are true as \begin{align*}x\end{align*} increases from \begin{align*}\frac{3\pi}{2}\end{align*} to \begin{align*}2\pi\end{align*}?
1. \begin{align*}\cos x\end{align*} approaches 0
2. \begin{align*}\tan x\end{align*} approaches positive infinity
3. \begin{align*}\cos x < \sin x\end{align*}
4. \begin{align*}\cot x\end{align*} approaches negative infinity

1. Use similar triangles: So: \begin{align*}& \frac{x}{1}=\frac{1}{A} \rightarrow Ax = 1 \rightarrow A = \frac{1}{x}\\ & \cos \theta=x \rightarrow \frac{1}{\cos \theta}=\frac{1}{x} \rightarrow \frac{1}{\cos \theta}=\sec \theta=\frac{1}{x}\\ & \therefore \sec \theta=A\end{align*}
2. Using the Pythagorean theorem, \begin{align*}\tan^2 \theta + 1 = \sec^2 \theta\end{align*}.
3. b
4. d

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