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# 1.2: Circular Functions

Difficulty Level: At Grade Created by: CK-12

Most scientific and graphing calculators have a π\begin{align*}\pi\end{align*} key, primarily to make calculating angles in radians easier. Make sure your students know where this key is on their calculators.

When a question like example 4\begin{align*}4\end{align*} comes up, students may wonder how they are supposed to know that this angle measure is in radians and not degrees. After all, 3π4\begin{align*}\frac{3\pi}{4}\end{align*} is just a number like any other, so an angle of 3π4\begin{align*}\frac{3\pi}{4}\end{align*} degrees could exist too—which means we can’t just assume any angle measure with π in it is in radians. And as we’ve seen in the text, measures like 1 radian and 2\begin{align*}2\end{align*} radians are meaningful as well, so we can’t assume any angle measure without π\begin{align*}\pi\end{align*} in it is in degrees.

The solution to this conundrum is to assume all angle measures are in radians unless otherwise specified; that’s the convention used by mathematicians. So if you’re asked what the cosine of 5\begin{align*}5\end{align*} is, if the problem says just 5''\begin{align*}5\text{''}\end{align*} and not 5\begin{align*}5^\circ\end{align*},” assume it means 5\begin{align*}5\end{align*} radians. But beware of typographical errors! If an angle measure doesn’t include a degree sign, but is a suspiciously familiar round number like 60\begin{align*}60\end{align*} or 90\begin{align*}90\end{align*}, it may be worth looking over the problem for signs that the author might have really meant to specify degrees and just left out the degree sign by mistake.

Example 7 contains a new concept: the inverse sine function. You may need to help students find the inverse sine on their calculators (usually they’ll need to press the 2nd''\begin{align*}2^{nd}\text{''}\end{align*} key followed by the “sin” key), and you may also need to explain the inverse sine function itself.

First, a brief review of inverse functions may be needed: remind students that the inverse of a function is simply what you get when you apply the function “backwards,” so the input becomes the output and the output becomes the input. In the case of the sine function, normally the input is an angle measure (which can be any real number) and the output is a ratio of side lengths (which can be any real number between 1\begin{align*}1\end{align*} and 1\begin{align*}1\end{align*}). The inverse sine function, therefore, takes a number between 1\begin{align*}1\end{align*} and 1\begin{align*}1\end{align*} as its input, and its output is the measure of an angle whose sine is that number. For example, the inverse sine of 1\begin{align*}1\end{align*} is 90\begin{align*}90^\circ\end{align*} (or π2\begin{align*}\frac{\pi}{2}\end{align*} radians), because the sine of 90\begin{align*}90^\circ \end{align*} (or Π2\begin{align*}\frac{\Pi}{2}\end{align*} radians) is 1\begin{align*}1\end{align*}.

The notation used here also bears explaining. Inverse functions are written with what looks like an exponent: the inverse of f(x)\begin{align*}f(x)\end{align*} is written as f1(x)\begin{align*}f^{-1}(x)\end{align*}, and the inverse of sin(x)\begin{align*}\sin(x)\end{align*} is written as sin1(x)\begin{align*}\sin^{-1}(x)\end{align*}. Emphasize that sin1(x)\begin{align*}\sin^{-1}(x)\end{align*} does not mean sin(x)\begin{align*}\sin(x)\end{align*} raised to the power of 1\begin{align*}-1\end{align*}, even though it looks as if it does. (Normally, as we learned in the previous chapter, placing the exponent right after the f\begin{align*}f\end{align*} in f(x)\begin{align*}f(x)\end{align*} is the standard notation for raising a function to a power, and when raising a trig function like sin(x)\begin{align*}\sin(x)\end{align*} to a power, we put the exponent right after the “sin” part. But when we want to raise sin(x)\begin{align*}\sin(x)\end{align*} to the power of 1\begin{align*}-1\end{align*}, we must write (sin x)1\begin{align*}(\sin\ x)^{-1}\end{align*} instead, so that we can use sin1(x)\begin{align*}\sin^{-1}(x)\end{align*} to designate the inverse sine of x\begin{align*}x\end{align*}.)

Rotations

Example 1 contains a slight error: since the hour hand has rotated 13\begin{align*}\frac{1}{3}\end{align*} of the way from 11\begin{align*}11\end{align*} to 12\begin{align*}12\end{align*}, the distance between the hour hand and the 12\begin{align*}12\end{align*} is 23\begin{align*}\frac{2}{3}\end{align*}, not 13\begin{align*}\frac{1}{3}\end{align*}, of that twelfth of the circle; it is π9\begin{align*}\frac{\pi}{9}\end{align*} rather than π18\begin{align*}\frac{\pi}{18}\end{align*} radians.

Length of Arc

The illustration for example 2 shows 12feet\begin{align*}12 \;\mathrm{feet}\end{align*} as the diameter of the circle, but the solution worked out in the text is in fact based on the radius of the circle being 12feet\begin{align*}12 \;\mathrm{feet}\end{align*}. If you wish, you can have students solve the problem both ways just for practice, but make sure to keep everyone on the same page about which formulation of the problem you are using at which time.

Perceptive students may try to solve example 3 the short way, by noticing that since the radius of the larger circle is 74\begin{align*}\frac{7}{4}\end{align*} that of the smaller circle, the circumference is also 74\begin{align*}\frac{7}{4}\end{align*} as great, and so a complete rotation of the smaller circle would be \begin{align*}\frac{4}{7}\end{align*}of a rotation of the larger circle. This is certainly a legitimate way of solving the problem, but you might encourage them to do it over again the way the book describes, just so they can get some practice doing calculations with arc lengths.

Additionally, there is a more precise way to express the answer to this problem. Instead of finding a decimal approximation for \begin{align*}\theta\end{align*} and then multiplying it by \begin{align*}\frac{180}{\pi}\end{align*} to approximate the angle measure in degrees, it is better to simply leave the value of \begin{align*}\theta\end{align*} in fraction form as \begin{align*}\frac{8\pi}{7}\end{align*}, so that multiplying it by \begin{align*}\frac{180}{\pi}\end{align*} yields the answer \begin{align*}\frac{1440}{7}\;\mathrm{degrees}\end{align*}.

Area of a Sector

You may need to explain the setup of the equation for \begin{align*}1\end{align*} radian. Basically, we are starting with the equation \begin{align*}2\pi\ \mathrm{(radians)} = \pi r^2\end{align*} (area) and dividing by \begin{align*}2\pi\end{align*} to make the left side equal \begin{align*}1\end{align*} radian. Dividing the right side by \begin{align*}2\pi\end{align*} then gives us \begin{align*}\frac{1}{2} r^2\end{align*} as the area.

The diagram given for example 4 contains a slight error ; you might encourage students to find it for themselves. (It’s a tricky error to catch—just remember that \begin{align*}\frac{2\pi}{3}\end{align*} is not in fact \begin{align*}\frac{2}{3}\end{align*} of \begin{align*}2\pi\end{align*}.) This is a good opportunity to remind them not to make the same error themselves, as it’s quite a common one.

Length of a Chord

Example 5 presents a slightly convoluted solving method, as it has students find half the length of the chord and then double it to get the final answer. In future, students will find it easier to simply use the formula for the whole chord length: twice the radius of the circle times the sine of half the angle, or \begin{align*}2r\ \sin \left (\frac{\theta}{2} \right )\end{align*}.

Also, when applying the chord-length formula, it isn’t actually necessary for the angle measure to be in radians (as it is with the arc-length and area formulas), because the sine of the angle is the same whether the angle is in radians or degrees. However, it is still useful to have students practice converting from degrees to radians.

## Circular Functions of Real Numbers

\begin{align*}y = \sin(x)\end{align*}, the Sine Graph

Students should notice that the height of the point tracing out the sine graph at any given stage is exactly the same (in graph-units) as the height of the point moving around the circle. (This may not be immediately obvious because the sine graph and the circle graph are depicted on slightly different scales.)

\begin{align*}y = \cos(x)\end{align*}, the Cosine Graph

The relationship between graph height and location on the circle is harder to see for the cosine graph, because the height of the graph represents the horizontal rather than the vertical location of the point on the circle. Imagining the circle rotated a quarter-turn to the left may help make the connection more visible, as you can now see the heights of the two points matching the same way they did on the sine graph—but your students probably needn’t try this, as long as they understand the basic principle that the cosine represents the \begin{align*}x-\end{align*}value of the corresponding angle.

\begin{align*}y = \tan(x)\end{align*}, the Tangent Graph

It may seem strange that the tangent line can get infinitely long when the sine and cosine lines can’t. Remind students that the length of the tangent line represents the ratio between the sine and cosine, and so it gets infinitely big as the cosine gets infinitesimally small.

To explain this in terms of the similar triangles shown in the diagram: The sides marked \begin{align*}t\end{align*} and \begin{align*}1\end{align*} have the same ratio as the sides marked \begin{align*}y\end{align*} and \begin{align*}x\end{align*}. As side \begin{align*}y\end{align*} gets longer, side \begin{align*}t\end{align*} gets longer—but as side \begin{align*}x\end{align*} gets shorter, side \begin{align*}1\end{align*} can’t get shorter because its length is fixed at 1. So if that side can’t get shorter, side t has to get even longer to keep the ratio the same.

The Three Reciprocal Functions: \begin{align*}\cot(x), csc(x)\end{align*}, and \begin{align*}\sec(x)\end{align*}

You might stop and ask your students why it makes sense that \begin{align*}1\end{align*} and \begin{align*}-1\end{align*} are the only values for which a function and its reciprocal are the same. (Hint: What has to be true of \begin{align*}y\end{align*} in order for \begin{align*}y = \frac{1}{y}\end{align*} to be true? Further hint: What happens when you solve that equation for \begin{align*}y\end{align*}?)

The illustrations showing the cosecant segment for angles greater than \begin{align*}180^\circ\end{align*} may be a little confusing, as the segment looks the same as it did for angles less than \begin{align*}180^\circ\end{align*}, but its length is now being described as a negative number. The reason it is now negative is that the segment is now pointing in the opposite direction from the line segment that forms the terminal side of the angle, whereas it was pointing in the same direction for angles less than \begin{align*}180^\circ\end{align*}.

To make it extra clear that the graph of the secant function is not made up of parabolas, you can point out that the secant graph has vertical asymptotes, whereas parabolas have no domain restrictions and extend infinitely far in both the positive and negative \begin{align*}x-\end{align*}directions.

Lesson Summary

It’s possible to express the domain restrictions on the cotangent and cosecant functions in a way that makes clearer their relationship to the domains of the tangent and secant functions. Instead of {\begin{align*}x : x \neq n\pi\end{align*}, where n is any integer}, we can express the domain of the cotangent or cosecant as {\begin{align*}x : x \neq n \left (\frac{\pi}{2} \right )\end{align*}, where \begin{align*}n\end{align*} is any even integer}. When we compare this to the domain of the secant or tangent, {\begin{align*}x : x \neq n \left (\frac{\pi}{2} \right )\end{align*} , where \begin{align*}n\end{align*} is any odd integer}, we can see much more clearly that the cosecant and cotangent have the same pattern of asymptotes as the secant and tangent, just shifted by \begin{align*}\pi\end{align*} units.

## Linear and Angular Velocity

Linear Velocity \begin{align*}v = \frac{s}{t}\end{align*}

You may need to stress that s represents distance and does not stand for “speed.” Students will still be prone to forget this when plugging in values without thinking too hard, so they may need reminding when they slip up.

In example 2, students might need to be reminded that the bar over the \begin{align*}3\end{align*} in \begin{align*}21.\bar{3}\end{align*} signifies a repeating decimal; the quantity being expressed is \begin{align*}21.233333333\ldots\end{align*} with the 3’s continuing on forever.

Angular Velocity \begin{align*}\omega = \frac{\theta}{t}\end{align*}

After finding the angular velocity in example 4, you may want to pause and explain why this answer makes sense: if the girls rotate through \begin{align*}\frac{1}{6}\end{align*} of the circle each second, it will take them \begin{align*}6 \;\mathrm{seconds}\end{align*} to make a complete rotation, and 6 seconds is indeed the time we were given at the beginning of the problem.

Similarly, it is useful to check the linear velocities and see why they make sense. Since Lindsey is \begin{align*}7 \;\mathrm{feet}\end{align*} from the center while Megan is \begin{align*}2.5 \;\mathrm{feet}\end{align*} from the center, Lindsey should be traveling about \begin{align*}\frac{7}{2.5}\end{align*} or \begin{align*}\frac{14}{5}\end{align*} times as fast as Megan. Sure enough, their speeds are about \begin{align*}7.3\end{align*} and \begin{align*}2.6 \;\mathrm{feet}\end{align*} per second respectively, and \begin{align*}\frac{7.3}{2.6}\end{align*} is about the same as \begin{align*}\frac{14}{5}\end{align*}.

Another problem for students to consider: if your arms are \begin{align*}2\;\mathrm{feet}\end{align*} long, and you swing a baseball bat that is also \begin{align*}2\;\mathrm{feet}\end{align*} long, how much faster is the end of the bat traveling than the end of your arms? (You can make up an angular velocity for the swing and have students find the two linear velocities, but you don’t need to—it may be more educational to have them look at the problem more abstractly. The key insight here is that no matter what the actual speeds are, since the arc of the circle that the bat sweeps out has a radius twice as great as the arc of the circle swept out by the swinger’s arms, the end of the bat will be traveling twice as fast. Students can see this by looking at the formula for linear velocity in terms of angular velocity, \begin{align*}v = r \omega\end{align*}, and considering what happens if \begin{align*}r\end{align*} is doubled—or multiplied by any other constant.) What does this suggest about why we use baseball bats—or for that matter, tennis rackets, golf clubs, or croquet mallets?

You may also want to have students try solving the formulas backwards: for instance, if the girls’ angular velocity in Example 4 were \begin{align*}\frac{\pi}{2}\end{align*} radians per second, what would be their linear velocity if the merr\begin{align*}y-\end{align*}go-round were still the same size?

## Graphing Sine and Cosine Functions

Amplitude

You may need to stress that the amplitude is the greatest distance the wave gets from the center of the wave, so it is only half the distance between the minimum and maximum values.

Period and Frequency

You may want to clarify the exact relationship between period and frequency, perhaps by helping students to work it out on their own. Remind them that an inverse relationship means that one quantity decreases when the other increases, and that it also means that the two quantities will yield a constant result when multiplied together. The three examples given in the text are a function with a period of \begin{align*}2 \pi\end{align*} and a frequency of \begin{align*}1\end{align*}; a function with a period of \begin{align*}\frac{\pi}{4}\end{align*} and a frequency of \begin{align*}8\end{align*}; and a function with a period of \begin{align*}4 \pi\end{align*} and a frequency of \begin{align*}\frac{1}{2}\end{align*}. In each of these cases, what do we get when we multiply the period by the frequency? (Answer: \begin{align*}2 \pi\end{align*}.) So what does that suggest the relationship is between the two? (Answer: \begin{align*}\mathrm{period} * \mathrm{frequency} = 2 \pi\end{align*}, which is more usefully expressed as either \begin{align*}\mathrm{period} = \frac{2\pi}{\mathrm{frequency}}\end{align*} or \begin{align*}\mathrm{frequency} = \frac{2 \pi}{\mathrm{period}}\end{align*}; both expressions are useful in different situations.)

Students may find it a little counterintuitive that the period of the cosecant graph is \begin{align*}2\pi\end{align*}, because the graph divides up so neatly into \begin{align*}\pi-\end{align*}sized chunks (and also perhaps because they’ve just seen that the period of the tangent graph is \begin{align*}\pi \;\mathrm{units}\end{align*}). Stress that it takes two of those chunks, in this case, before the graph actually repeats itself, just as in Example 2 at the beginning of the lesson it took one “high” portion and one “low” portion of the graph together to make up one period.

Transformations of Sine and Cosine Graphs: Dilations

It’s a little hard to tell from the graphs of those linear functions that the line is being “stretched” vertically when the slope increases, because vertically stretching a straight line looks just like rotating it. You may want to also show graphs of the dilations of \begin{align*}x^2\end{align*} discussed in the text, because those graphs will make the stretching and shrinking a lot clearer.

Also, the statement “Constants greater than \begin{align*}1\end{align*} will stretch the graph out vertically and those less than \begin{align*}1\end{align*} will shrink it vertically” is really only part of the story. That is, it’s true for positive constants, but negative constants will first flip the graph upside down and then stretch or shrink it vertically (stretch it if their absolute value is greater than \begin{align*}1\end{align*}, shrink it if it’s less than \begin{align*}1\end{align*}.) And of course, a constant of 0 will shrink the graph all the way down to a straight horizontal line.

Review Questions

Encourage students to sketch graphs for question 2. Also, remind them if necessary that an amplitude of \begin{align*}A\;\mathrm{units}\end{align*} (where \begin{align*}A\end{align*} is the constant multiplier in front of the trig function) means that the graph goes both \begin{align*}A\;\mathrm{units}\end{align*} above the \begin{align*}x-\end{align*}axis and \begin{align*}A\;\mathrm{units}\end{align*} below it, so the maximum is \begin{align*}A\end{align*} and the minimum is \begin{align*}-A\end{align*}. (The exception, of course, is when \begin{align*}A\end{align*} is negative, as in part c; then \begin{align*}A\end{align*} is the minimum and \begin{align*}-A\end{align*} is the maximum.)

## Translating Sine and Cosine Functions

Vertical Translations

Another way to find the answer to example 1 is to find the minimum and maximum of the base function, \begin{align*}\cos(x)\end{align*}, and then subtract \begin{align*}6\end{align*} from both of them.

Horizontal Translations (Phase Shift)

Here is another way to explain the apparent “backwards-ness” of phase shift: when we look at a given number \begin{align*}x\end{align*}, an expression like \begin{align*}\cos(x-2)\end{align*} means “the cosine of the number that’s \begin{align*}2\;\mathrm{units}\end{align*} to the left of this one.” So it’s as though we’re taking the value of the cosine function from \begin{align*}2\;\mathrm{units}\end{align*} to the left of “here,” and moving it over to “here”—which means we’re moving it \begin{align*}2\;\mathrm{units}\end{align*} to the right of where it started out. And of course we’re doing the same thing with the whole function, so \begin{align*}\cos(x-2)\end{align*} describes the whole cosine function shifted \begin{align*}2\;\mathrm{units}\end{align*} to the right.

Yet another way to explain it is in terms of inverse functions. Students may remember from previous algebra classes that the graph of an inverse function is the graph of the original function reflected about the line \begin{align*}y = x\end{align*}; in other words, flipped diagonally. This means that a horizontal shift of the original graph would result in a vertical shift of the inverse graph, and vice versa. Specifically, shifting the original graph to the right corresponds to shifting the inverse graph up, and shifting the original graph left is the same as shifting the inverse graph down.

Now, one way to find the equation for an inverse function is to swap the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}values of the original equation and then solve for \begin{align*}y-\end{align*}so if the original equation was \begin{align*}y = 3x\end{align*}, the inverse equation would be \begin{align*}x = 3y\end{align*}, or in other words \begin{align*}y = \frac{x}{3}\end{align*}. But note that if we add a constant to the \begin{align*}x-\end{align*}value in the original equation, that constant ends up being subtracted from the inverse equation—if we start with \begin{align*}y = 3(x + 4)\end{align*} instead of just \begin{align*}y = 3x\end{align*}, the inverse is \begin{align*}x = 3(y + 4)\end{align*}, or \begin{align*}x = 3y + 12,\end{align*} or \begin{align*}x - 12 = 3y,\end{align*} or \begin{align*}y = \frac{x}{3} - 4\end{align*}. Because everything gets reversed in an inverse operation, increasing the \begin{align*}x-\end{align*}value of the original function means decreasing the \begin{align*}y-\end{align*}value of the inverse function. That means the inverse function gets shifted down (not up), and that must mean the original function was shifted left (not right) when the \begin{align*}x-\end{align*}value was increased. So, increasing the \begin{align*}x-\end{align*}value means shifting the graph left, and vice versa.

One way to explain why sin and cos are just phase-shifted versions of each other is to recall that they are based on the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates of a point moving around the unit circle. These coordinates behave in the same way as the point rotates—they both oscillate between \begin{align*}1\end{align*} and \begin{align*}-1\end{align*}—it’s just that they start out at different points in the cycle.

Another way to explain it is this: Recall that the sine of an angle is the cosine of the angle’s complement (draw a triangle to see why this is so—the sine of one acute angle is the cosine of the other acute angle, and the two angles add up to \begin{align*}90^\circ\end{align*} by the Triangle Sum Theorem). We can write this fact as \begin{align*}\sin(x) = \cos(90 - x)\end{align*}, and we can rewrite \begin{align*}\cos(90 - x)\end{align*} as \begin{align*}\cos(-x + 90)\end{align*}. But the graph of \begin{align*}\cos(-x + 90)\end{align*} is simply the graph of \begin{align*}\cos x\end{align*} flipped over vertically (since all the negative \begin{align*}x-\end{align*}values become positive and vice versa) and shifted horizontally by \begin{align*}90^\circ\end{align*} (or \begin{align*}\frac{\pi}{2}\;\mathrm{radians}\end{align*}), and that happens to be the same as the graph we get if we shift it \begin{align*}\frac{\pi}{2}\;\mathrm{units}\end{align*} in the other direction and don’t flip it.

## General Sinusoidal Graphs

Drawing Sketches/Identifying Transformations from the Equation

Some students may notice that translating a basic sine or cosine graph \begin{align*}\pi\;\mathrm{units}\end{align*} horizontally is essentially the same as flipping it upside down (i.e. multiplying it by \begin{align*}-1\end{align*}). This isn’t a necessary fact to know, but it can be useful. For particularly curious students, here’s how to explain why it is true:

When you add \begin{align*}\pi\end{align*} to an angle measure, you get another angle two quadrants away with the same reference angle. (Sketching a couple of angles will make this obvious.) That means all the trig functions for that new angle will be either the same or the opposite of the trig functions for the old angle. Now, in any two quadrants that are opposite each other, the sine function has opposite signs—it’s positive in I but negative in III, and positive in II but negative in IV. Similarly, the cosine function is positive in I but negative in III, and negative in II but positive in IV. So, whenever we add \begin{align*}\pi\;\mathrm{units}\end{align*} to an angle, the sine and cosine of the new angle are the negatives of the sine and cosine of the old angle—or, to put it more formally, for any angle \begin{align*}x\end{align*}, \begin{align*}\sin(x + \pi)\end{align*} equals \begin{align*}\sin(x)\end{align*} and \begin{align*}\cos(x+\pi)\end{align*} equals \begin{align*}-\cos(x)\end{align*}. In other words, applying a phase shift of \begin{align*}\pi\;\mathrm{units}\end{align*} to the sine or cosine graph is the same as multiplying it by \begin{align*}-1\end{align*}, or flipping the graph over.

(You may also notice that, since the sign of the tangent function alternates from quadrant to quadrant, the tangent function keeps the same sign when you add \begin{align*}\pi\end{align*} to the angle. This is why the tangent function simply repeats itself after \begin{align*}\pi\;\mathrm{units}\end{align*}, and a phase shift of \begin{align*}\pi \;\mathrm{units}\end{align*} is equivalent to no change at all.)

Examples 1 and 2 demonstrate two different ways to approach the problem of sketching a graph: starting with the horizontal and vertical translations, or starting with the amplitude and frequency. Students will probably find they prefer one method or the other, and there’s certainly no need to be strict about which one to use.

Also, some students may find it easier to sketch a complete curve at each step of the process until they end up with the final curve, while others may find it easier to simply sketch the key points of the graph, move them around as necessary, and not connect them with a curve until the final step. Again, either method should work fine; it may in fact be a good idea to point out that both methods exist.

Writing the Equation from a Sketch

A second way to find the amplitude after finding the phase shift is just to subtract the phase shift from the maximum value of the function. For instance, in the example given, we would subtract \begin{align*}20\end{align*} from \begin{align*}60\end{align*} to get \begin{align*}40\end{align*}.

Review Questions

Problems 6-10 should contain the instruction “Write an equation that describes the given graph.” Of course, either sine or cosine may be used.

## Date Created:

Feb 23, 2012

Nov 17, 2014
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