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2.2: Circular Functions

Difficulty Level: At Grade Created by: CK-12
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Radian Measure

In-Text Examples

One very common mistake to make when working in radians is to forget from time to time that a complete rotation is \begin{align*}2 \pi \;\mathrm{radians}\end{align*}2πradians, not \begin{align*}\pi\end{align*}π radians. This doesn’t happen often when one is thinking of a whole circle, but it is very easy to think of a quarter of a circle as being \begin{align*}\frac{\pi}{4}\;\mathrm{radians}\end{align*}π4radians instead of \begin{align*}\frac{\pi}{2}\end{align*}π2, a sixth of a circle as being \begin{align*}\frac{\pi}{6}\end{align*}π6 instead of \begin{align*}\frac{\pi}{3}\end{align*}π3, and so on. Students may make this type of error frequently, and may continue making it for quite some time, so remind them more than once to be particularly careful about checking their angle measures when working in radians.

(As evidence that not only students are prone to this particular error, see the illustration to example 4 in the lesson immediately following this one.)

Another common error is to get the degrees-to-radians formula and the radians-to-degrees formula mixed up. The note at the bottom of page 101 should help students with this, though: basically, they should multiply by \begin{align*}\frac{\pi}{180}\end{align*}π180 when they want \begin{align*}\pi\end{align*}π in their answer (that is, when converting to radians), and should multiply by \begin{align*}\frac{180}{\pi}\end{align*}180π when they have a \begin{align*}\pi\end{align*}π to get rid of (that is, when converting from radians to degrees). Of course, there won’t always be a \begin{align*}\pi\end{align*}π involved when working in radians (see problem 5 below, for example), but the mnemonic will remind them which way the conversion goes if they don’t take it too literally.

Review Questions

6) This problem should read “sine” instead of “cosine”; students may therefore think that Gina’s typing in “sin” instead of “cos” is the problem, instead of noticing that the calculator is in the wrong mode. They may also get \begin{align*}\frac{-\sqrt{3}}{2}\end{align*}32 instead of \begin{align*}\frac{-1}{2}\end{align*}12 as the actual value, and shouldn’t be penalized for this.

Additional Problems

1) What is \begin{align*}\sqrt{2}\;\mathrm{radians}\end{align*}2radians in degrees? (Round to the nearest tenth.)

2) What is \begin{align*}90\;\mathrm{radians}\end{align*}90radians in degrees, and what is its reference angle?

3) What is \begin{align*}\pi \;\mathrm{degrees}\end{align*}πdegrees in radians? (Round to four decimal places.)

Answers to Additional Problems

1) \begin{align*}81.03^\circ\end{align*}81.03

2) \begin{align*}5156.6^\circ\end{align*}5156.6, which is coterminal with \begin{align*}116.6^\circ\end{align*}116.6, so its reference angle is \begin{align*}63.4^\circ\end{align*}63.4.

3) \begin{align*}0.0548.\end{align*}0.0548.

Applications of Radian Measure

In-Text Examples

1) The most common mistake to make here is to forget that the hour hand is not right on the \begin{align*}11\end{align*}11, but a third of the way between \begin{align*}11\end{align*}11 and \begin{align*}12\end{align*}12. Students who forget this may jump to the conclusion that the hands are \begin{align*}\frac{5}{12}\end{align*}512 of the circle apart, and then may also forget (as mentioned above) that \begin{align*}\frac{5}{12}\end{align*}512 of the circle does not equal \begin{align*}\frac{5 \pi}{12}\;\mathrm{radians.}\end{align*}5π12radians.

2) A common error has actually been made in the text here: the numbers \begin{align*}12\end{align*}12 and \begin{align*}11.81\end{align*}11.81 have been substituted for \begin{align*}r\end{align*}r in the arc-length formula when each of those numbers is in fact a diameter and not a radius. Students should be cautioned against making the same error on future problems, but should probably be forgiven for making it here.

4) As mentioned earlier, another common error appears in the in-text illustration here: \begin{align*}\frac{2 \pi}{3}\end{align*}2π3 would of course be \begin{align*}\frac{1}{3}\end{align*}13 of the circle, not \begin{align*}\frac{2}{3}\end{align*}23. This will only lead to student error, though, if students try finding the area of the whole circle first and then taking \begin{align*}\frac{2}{3}\end{align*}23 of it instead of \begin{align*}\frac{1}{3}\end{align*}13 of it. If they apply the formula in the text instead, they will not go wrong, since the formula is not based on the illustration.

Review Questions

1) On part a.iii, students should not try to convert the approximate angle measure they found in part a.ii to degrees (as they may try to do with their calculators, especially since they have just used them to find that approximation). Instead they should start with the exact angle measure from part a.i, and convert it by hand using the formula they learned earlier.

3) Simply miscounting the dots may lead to error here. (There are 32.) Also, in finding the distance between the two dots selected in part b, students may include the dots at both the beginning and the end and conclude that the distance between them is \begin{align*}\frac{14}{32}\end{align*}1432 of the circle when it is really \begin{align*}\frac{13}{32}\end{align*}1332.

4) Students may forget to take into account both the radius of the outer circle and the radius of the inner circle when calculating the area of each section.

5) This problem is a particularly easy place to make the routine error of plugging in the diameter in place of the radius.

Circular Functions of Real Numbers

Review Questions

1) A quick hint to use similar triangles should help students who get stuck on this problem.

3) Students may get a little confused about where to do the labeling; point out if necessary that the largest circle in the diagram is the unit circle. The other circles are just there to give them a convenient place to write the angle measures without having to write them right on top of the coordinates of the points; if they get mixed up and write the radian measures on the inner circle and the degree measures on the middle circle, there’s no need to penalize them as long as they’ve matched the correct degree and radian measures with the correct angles.

4) Perceptive students may draw the cosine segment in a different place than the book suggests; they may draw it as a horizontal segment extending from the \begin{align*}y-\end{align*}yaxis to the point where the sine segment meets the unit circle. This isn’t an error; in fact, drawing the segment this way makes it easier to see that the sine and cosine segments have a relationship similar to the relationship between the tangent and cotangent or secant and cosecant segments.

(However, drawing it the standard way is fine too, and probably easier for most.)

5) The correct answer to this question is actually any combination of \begin{align*}\sin, \tan,\end{align*}sin,tan, and \begin{align*}\sec\end{align*}sec.

6) Some students may get the idea that answers b and d must both be true if either one of them is true—that the tangent must get infinitely large when the cotangent gets infinitely small, because they are “opposites” in a sense. The ambiguity of the phrase “infinitely small” doesn’t help; it can be taken to mean “approaching zero,” but in this case it really means “approaching negative infinity.” When the tangent approaches infinity, the cotangent approaches zero, not negative infinity; when the cotangent approaches negative infinity, the tangent approaches zero, and this latter case is what happens when \begin{align*}x\end{align*}x increases from \begin{align*}\frac{3 \pi}{2}\end{align*}3π2 to \begin{align*}2 \pi\end{align*}. (Referring to the graphs earlier in the lesson will confirm this.)

Additional Problems

1) Why does it make sense that the ranges of the secant and cosecant functions include all numbers except those between \begin{align*}1\end{align*} and \begin{align*}-1\end{align*}? (Hint: think in terms of sine and cosine.)

Answers to Additional Problems

1) The sine and cosine functions only take values between \begin{align*}1\end{align*} and \begin{align*}-1\end{align*}—in other words, only numbers whose absolute value is less than or equal to \begin{align*}1\end{align*}. The secant and cosecant are the reciprocals of the sine and cosine, and the reciprocal of a number whose absolute value is less than or equal to \begin{align*}1\end{align*} will have an absolute value greater than or equal to \begin{align*}1\end{align*}. (For example, think of a fraction between and \begin{align*}1\end{align*}: its numerator is less than its denominator, so its reciprocal will have a numerator greater than its denominator and so will be greater than \begin{align*}1\end{align*}.) So, the secant and cosecant functions can only take values whose absolute value is greater than or equal to \begin{align*}1\end{align*}.

Linear and Angular Velocity

In-Text Examples

1) A possible overcomplication of this problem is to think that \begin{align*}15\;\mathrm{feet}\end{align*} is the “length” of the oval shape formed by the track (i.e. the major axis of an ellipse), rather than simply the track’s circumference and thus the distance the car travels per circuit.

2) Students may forget to convert their final answer from miles per minute to miles per hour. If they do remember, they are likely to try to divide the miles per minute result by \begin{align*}60\end{align*}, instead of multiplying it by \begin{align*}60\end{align*}, to get the answer in miles per hour. Encourage them to slow down and think about what the answer really means: if Lois goes \begin{align*}.047\;\mathrm{miles}\end{align*} in one minute, then how far will she go in \begin{align*}60\;\mathrm{times}\end{align*} one minute?

Review Questions

1) Careless reading may lead students to jump to the answer \begin{align*}\frac{7}{9}\;\mathrm{cm/sec}\end{align*}. Remind them that the speed of the dial is the circumference, not the radius, divided by \begin{align*}9\;\mathrm{seconds.}\end{align*}

3) Note that Doris’ horse is not \begin{align*}7\;\mathrm{m}\end{align*} from the center, but rather \begin{align*}7\;\mathrm{m}\end{align*} farther from the center than Lois’, meaning it is \begin{align*}10\;\mathrm{m}\end{align*} total from the center.

4) We haven’t covered scientific notation in a while, so students may be prone to make mistakes with it. Watch for answers that are off by one or two orders of magnitude.

Additional Problems

1) Two gears mesh with each other so that they rotate in opposite directions, with both their outer edges moving at the same linear velocity. The radius of the larger gear is \begin{align*}10\;\mathrm{cm}\end{align*} and the radius of the smaller gear is \begin{align*}6\;\mathrm{cm}\end{align*}. The smaller gear makes two revolutions per second.

a) What is the angular velocity of the smaller gear?

b) What is the linear velocity of a point on its outside edge?

c) What is the angular velocity of the larger gear?

d) How many revolutions does the larger gear make per second?

e) A peg is attached to the larger gear at a point \begin{align*}2\;\mathrm{cm}\end{align*} from its outer edge. What is the peg’s linear velocity?

Answers to Additional Problems

1) a) \begin{align*}4 \pi\;\mathrm{radians/sec}\end{align*}, or approximately \begin{align*}12.57\;\mathrm{radians/sec}\end{align*}.

b) \begin{align*}24 \pi\;\mathrm{cm/sec}\end{align*}, or approximately \begin{align*}75.40\;\mathrm{cm/sec.}\end{align*}

c) The linear velocity of a point on its outside edge is the same as that of the smaller gear, or \begin{align*}24 \pi\;\mathrm{cm/sec}\end{align*}, so its angular velocity is \begin{align*}2.4\pi\;\mathrm{radians/sec}\end{align*}, or approximately \begin{align*}7.54\;\mathrm{radians/sec}\end{align*}.

d) \begin{align*}1.2\end{align*} revolutions per second.

e) The peg is \begin{align*}8\;\mathrm{cm}\end{align*} from the center of the gear, so its linear velocity is \begin{align*}19.2 \pi \;\mathrm{cm/sec}\end{align*}, or approximately \begin{align*}60.32 \;\mathrm{cm/sec}\end{align*}.

Graphing Sine and Cosine Functions

In-Text Examples

2) Students may think the period is \begin{align*}2\;\mathrm{units}\end{align*}, because each “portion” of the graph is \begin{align*}2\;\mathrm{units}\end{align*} wide. Explain that it takes one “high” portion and one “low” portion to make up one complete cycle of the graph.

3-6) Despite the explanations in the text, it is likely that some students will still habitually mix up the period with the frequency. Repeated drilling may be the only way to fix this.

Review Questions

2) A few students may be tripped up by part d; since there are no minimum and maximum values, they may get the idea they’re supposed to be looking for something else, and supply an answer like “\begin{align*}\frac{\pi}{2}\end{align*} and \begin{align*}\frac{-\pi}{2}\end{align*}” because those are the beginning and ending \begin{align*}x-\end{align*}values of one cycle of the graph.

The error they are most likely to make on problems like \begin{align*}e\end{align*} and \begin{align*}f\end{align*} is to think that the number within the parentheses affects the maximum and minimum \begin{align*}y-\end{align*}values. However, solving problems a through \begin{align*}d\end{align*} first should help reinforce that it is only the multiplier out front that matters.

3) Students may try to simply divide both sides by \begin{align*}\sin(x)\end{align*}; they will then end up with the equation \begin{align*}4 = 1\end{align*}, for which there is no solution. However, this technique is incorrect because \begin{align*}\sin(x)\end{align*} sometimes equals zero on the interval of \begin{align*}x-\end{align*}values given for the problem, and dividing by an expression that can equal zero often eliminates possible solutions.

Graphing the two functions instead (or simply thinking about them in the right way) will show that whenever \begin{align*}\sin(x)\end{align*} equals zero, \begin{align*}4 \ \sin(x)\end{align*} also equals zero and so the two expressions are equal. This happens at three places on the given interval, including the interval’s endpoints.

A more appropriate application of algebra will yield this solution as well: instead of dividing both sides by \begin{align*}\sin(x)\end{align*}, subtract \begin{align*}\sin(x)\end{align*} from both sides to yield \begin{align*}3 \ \sin(x) = 0\end{align*}, and then divide by \begin{align*}3\end{align*} to get \begin{align*}\sin(x) = 0\end{align*}.

4) Getting the period and frequency mixed up is the most likely error here; reviewing the definitions may help.

5) Students may get twice the correct value for the amplitude here, forgetting that it’s the distance from the minimum or maximum to the middle of the graph rather than to the maximum or minimum. Also, when writing out the equation, they may again get mixed up about which number goes where.

6) At this point, students may still be stretching the graph when they should be shrinking it, or vice versa.

Translating Sine and Cosine Functions

In-Text Examples

1) A few students may think we’re still dealing with amplitude here, and that the maximum and minimum are \begin{align*}6\end{align*} and \begin{align*}-6\end{align*}. Remind them that we are now shifting the graph rather than stretching it; the maximum and minimum remain the same distance apart, but have both moved together \begin{align*}6\;\mathrm{units}\end{align*} from where they started out.

Review Questions

1-5) Starting with the questions and trying to match them to the functions right away is tempting, but almost certainly the wrong way to approach this set of problems; it’s much better to sketch out graphs of the functions and then match them to the appropriate descriptions.

Also, the questions about \begin{align*}y-\end{align*}intercepts may be confusing, as we haven’t directly discussed those with regard to trig functions specifically. The \begin{align*}y-\end{align*}intercepts can’t be derived directly from the amplitude, frequency, period, phase shift, or vertical shift, as students may be tempted to try, but they can often be read off the graph fairly easily. The best way to find them, however, is simply to plug in for \begin{align*}x\end{align*} and then calculate the value of \begin{align*}y\end{align*}. (Remember, the \begin{align*}y-\end{align*}intercept is simply the value of \begin{align*}y\end{align*} when the graph crosses the \begin{align*}y-\end{align*}axis, which is to say when \begin{align*}x\end{align*} equals .)

6) “Express the equation as both a sine and cosine function” may be misconstrued to mean “write one equation that involves both sine and cosine” rather than “write one equation that involves sine and another that involves cosine.”

Also, watch out for students getting the sine and cosine graphs mixed up—not just on this problem, but on any problem that involves phase-shifted sinusoids (like the next four problems!).

7-10) Students may unfortunately be thrown by the fact that graphs A and C are incorrectly drawn: they are shifted up \begin{align*}2\;\mathrm{units}\end{align*} when they should be shifted 1 unit. Explain this to avoid confusing them.

Students may also still be getting mixed up about whether a shift to the left (or to the right) should be described with a negative or positive number, and this is a particularly bad time to make that error because it may be compounded with the error of mixing up sine and cosine graphs, resulting in students mistaking a sine graph shifted one way for a cosine graph shifted the other way (or even shifted the same way).

11) It may be hard to tell here that the tick marks on the \begin{align*}x-\end{align*}axis represent \begin{align*}1\;\mathrm{unit}\end{align*} each, rather than \begin{align*}\pi\end{align*} or \begin{align*}\frac{\pi}{2}\;\mathrm{units}\end{align*}. (The multiples of \begin{align*}\pi\end{align*} are indicated in approximately the right positions, but without tick marks.)

General Sinusoidal Graphs

In-Text Examples

1) On problems like this, students may still be getting mixed up about which transformations correspond to which parts of the equation. In particular, they may confuse the amplitude with the vertical shift and the frequency with the phase shift, or get the frequency and the period mixed up.

3) One knee-jerk error here is to assume that the amplitude is the same as the maximum value, or in this case \begin{align*}60\end{align*}. (In reality, the amplitude is indeed the same as the maximum value whenever the vertical shift is , but otherwise the maximum value is equal to the amplitude plus the vertical shift—in this case, the vertical shift is \begin{align*}20\end{align*}, so the amplitude is \begin{align*}40\end{align*}.)

Another common error is, once again, to forget that the amplitude is only half the total height of the graph.

Review Questions

1-5) Trying to find the maximum and minimum after finding the amplitude but before finding the vertical shift may lead students to get the wrong values, as they may default to treating the graph as if it were centered at .

Additional Problems

1) The graph of \begin{align*}y = 2 + \cos (3(x - \pi))\end{align*} is translated an additional \begin{align*}\frac{\pi}{2}\;\mathrm{units}\end{align*} to the left and \begin{align*}3\;\mathrm{units}\end{align*} down. What is the equation of the new graph?

2) The graph of \begin{align*}y = 3 + 2 \sin \left (6 \left (x - \frac{\pi}{3}\right )\right )\end{align*} is stretched so that its period is twice as long.

What is the frequency of the new sine wave?

3) Which of the following yields the same graph as \begin{align*}y = 2 + \cos \left (2 \left (x + \frac{\pi}{2}\right )\right )\end{align*}?

a) \begin{align*}y = 2 + \cos \left (2 \left (x - \frac{\pi}{2}\right )\right )\end{align*}

b) \begin{align*}y = 2 + \sin \left (2 \left (x - \frac{\pi}{2}\right )\right )\end{align*}

c) \begin{align*}y = 2 + \sin \left (2 \left (x + \frac{\pi}{2}\right )\right )\end{align*}

d) \begin{align*}y = 2 - \cos \left (2 \left (x + \frac{\pi}{2}\right )\right )\end{align*}

Answers to Additional Problems

1) \begin{align*}y = -1 + \cos \left (3 \left (x - \frac{\pi}{2}\right )\right )\end{align*}

2) If the period is doubled, the frequency is halved. The old frequency is \begin{align*}6\end{align*}, so the new frequency is \begin{align*}3\end{align*}.

3) a

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