Chapter 12

Radians and Graphs

Radians

Thousands of years ago, ancient Mesopotamians divided a full rotation into 360 parts. Why 360 parts? No one knows. $ ^{*} $ This tradition of measuring angles in degrees passed eventually to India and Greece, from whence it ultimately spread throughout the civilized world. The degree as a unit of measurement is a historical accident, devoid of mathematical significance. In fact, its arbitrariness renders it unsuitable for calculus, where it yields unnecessarily messy formulas. Fortunately, the mess vanishes if we adopt a more intrinsically meaningful unit of angle measurement: the radian.

Definition. If we lay a circle’s radius along its circumference, the corresponding angle at the circle’s center is one radian.

A radian is a large unit. Even a full rotation (i.e. $ 360^{} $ ) encompasses just over 6 radians. More precisely, since a circle’s full circumference is exactly $ 2$ radii long, there are exactly $ 2$ radians in a full rotation. Consequently, there are $ $ radians in a half-turn, a relationship that merits its own boldface line:

π radians = 180°.

Bearing this relationship in mind, we can easily convert back and forth between degrees and radians.

Example 1. Convert $ 27.4^{} $ to radians.

Solution. Since $ 180^{} = $ radians, one degree is $ /180 $ radians.

Hence, $ 27.4^{} $ is $ 27.4(/180) $ radians.

Example 2. Convert 2.32 radians to degrees.

Solution. Since \(\pi\) radians = \(180^{\circ}\), one radian is \(180/\pi\) degrees.

Hence 2.32 radians is $ 2.32(180/) ^{} $ .

Since the special angles $ 30^{}, 45^{}, 60^{} $ , and $ 90^{} $ are simple fractions of $ 180^{} $ , their radian measures are simple fractions of $ $ . For example, $ 30^{} $ is a sixth of $ 180^{} $ , so it is $ /6 $ radians. Similarly, a right angle is half of $ 180^{} $ , so it is $ /2 $ radians. When you encounter an angle of, say, $ 5/3 $ radians, you should think, without having to calculate on paper, “Well, $ /3 $ radians is $ 60^{} $ , so $ 5/3 $ radians is $ 5(60^{}) = 300^{} $ .”

Radians are ubiquitous in higher mathematics, so by convention, angles with unspecified units are always understood to be in radians. Thus, $ (1) $ is the sine of 1 radian, which is quite different from $ (1^{}) $ , the sine of 1 degree. (Numerically, $ (1) $ , $ (1^{}) $ .)

Example 3. Find the exact value of $ (5/3) $ .

Solution. $ (5/3)=(300^{})=(60{})=1/2. $

Exercises

1. In your head, convert the following radian measurements to degrees.

1. $ $ b) $ /2 $ c) $ /3 $ d) $ /4 $ e) $ /6 $ f) $ 2/3 $ g) $ 5/6 $ h) $ 3/4 $

2. $ 2$ j) $ 3/2 $ k) $ 4/3 $ l) $ 5/3 $ m) $ 5/18 $ n) $ 7/6 $ o) $ 11/6 $ p) $ 3/20 $

2. In your head, convert the following degree measurements to radians.

1. $ 30^{} $ b) $ 45^{} $ c) $ 60^{} $ d) $ 90^{} $ e) $ 120^{} $ f) $ 150^{} $ g) $ 210^{} $ h) $ 240^{} $

2. $ 300^{} $ j) $ 330^{} $ k) $ 180^{} $ l) $ 360^{} $ m) $ 270^{} $ n) $ 135^{} $ o) $ 225^{} $ p) $ 315^{} $

3. With pencil and paper (or in your head, when you can), convert these radian measurements to degrees.

1. 1

2. 3

3. $ /180 $

4. $ 1/$

5. 3.14

6. 5.39

4. With pencil and paper (or in your head, when you can), convert these degree measurements to radians.

1. $ 15^{} $ b) $ 10^{} $ c) $ 1^{} $ d) $ 32^{} $ e) $ 212^{} $ f) $ ^{} $ g) $ 5.39^{} $

5. Find the exact values of the following expressions, without a calculator.

1. \(\sin(\pi/4)\) b) \(\sin(\pi/3)\) c) \(\cos(\pi/3)\) d) \(\tan(\pi/4)\) e) \(\sin(\pi/6)\) f) \(\cos(\pi/6)\)

2. \(\cos(\pi/4)\) h) \(\sin(\pi/2)\) i) \(\cos(\pi/2)\) j) \(\cos(-\pi)\) k) \(\tan(\pi/6)\) l) \(\tan(\pi/3)\)

3. \(\cos(2\pi/3)\) n) \(\sin(5\pi/3)\) o) \(\sin(5\pi/6)\) p) \(\cos(11\pi/6)\) q) \(\tan(3\pi/4)\) r) \(\sec(-\pi/3)\)

4. \(\csc(2\pi/3)\) t) \(\cot(5\pi/3)\) u) \(\csc(7\pi/4)\) v) \(\cos(7\pi/3)\) w) \(\sin(7\pi/2)\) x) \(\sec(-11\pi/6)\)

6. As I mentioned above, radians simplify some important calculus formulas. In this exercise, you’ll see that they also simplify some geometric formulas.

Consider an angle \(\theta\) at a circle’s center, and the circular arc that subtends it, as in the figure. The angle takes up a certain fraction of a full rotation. Clearly, the arc takes up that same fraction of the full circumference. (For instance, a right angle is 1/4 of a full rotation, so the corresponding arc is exactly 1/4 of the circumference.) Rephrased algebraically, the following must be true:

\[ \mathrm{a r c}=\left(\frac{\theta}{\mathrm{a~f u l l~r o t a t i o n}}\right)2\pi r. \]

To complete this formula, we must supply the full rotation, but this depends on our choice of unit $ (360^{} $ or $ 2) $ , so the formula will look different depending on which unit we use for angle measurements.

1. If we measure angles in degrees, what is the circular arc length formula?

2. If we measure angles in radians, what is the circular arc length formula? You should commit this very simple formula to memory – and be able to explain to others (and yourself) where it comes from.

7. The shaded region in the figure at right is a circular sector (or more informally, a “pie slice”) Use the previous exercise’s ideas to find a formula for the area of a circular sector in terms of $ $ and r…

1. when $ $ is measured in degrees.

2. when $ $ is measured in radians.

3. when \(\theta\) is measured in gradians (a gradian is a hundredth of a right angle).\(^{*}\)

You should memorize the radian version in Part B – and be able to derive it.

8. The circles at right are tangent circles (each pair touch at only one point). If their radii are 1, 2, and 4 units, find the area of the region trapped between them.

[Hints: The line joining the centers of any two tangent circles passes through their point of tangency. Apply what you learned in the previous exercise.]

9. The previous problem’s circles are now rearranged. Each pair still touch at only one point. Find the areas of the two regions that lie inside the largest circle, but outside the others.

[Hint: The first hint on the previous problem holds even when one of the two tangent circles lies inside the other. When you’ve found an isosceles triangle, you are on the right track.]

The Graphs of the Trigonometric Functions

We briefly met the sine wave two chapters ago. If you’ve forgotten how we generated it from sine’s unit-circle definition, please go back and review that (or better yet, work it out for yourself). Here it is again, but now with $ $ measured in radians.

I’ve emphasized one full cycle of the sine wave (corresponding to one full lap around the unit circle). Notice that the graph completely repeats itself after it runs through $ 2$ units in the domain. To describe this mathematically, we say that sine is periodic and that its period – the length of its basic cycle – is $ 2$ . All six trigonometric functions are periodic, which is hardly surprising since they are all ultimately defined in terms of a point orbiting the origin.

Cosine’s graph is nearly identical to sine’s graph, except that it begins at 1, since $ = 1 $ .

Like sine’s graph, cosine’s graph repeats itself every $ 2$ units. That is, cosine’s period is also $ 2$ .

The graph of tangent is an altogether different sort of beast. The easiest way to produce tangent’s graph is to use the alternate characterization of $ $ that we found in Chapter 10, Exercise 35. Namely, $ $ is the slope of the radius in the figure at right. By thinking about this alternate characterization of tangent and keeping track of how the radius’s slope varies as $ $ runs through a full rotation (and beyond), we easily obtain the graph of tangent:

Observe that tangent’s period – the length of its basic cycle – is only $ $ , in contrast to sine and cosine’s $ 2$ . Tangent’s vertical asymptotes leave it with fewer opportunities to model periodic physical phenomena, but they do exist, if you know where to look for them. (See Exercise 12).

The graphs of the three reciprocal trig functions are relatively unimportant, but thinking about what they look like is a good exercise in coordinate geometry. (It also happens to be exercise 16 below.)

Exercises

10. Tangent’s period is $ $ , so $ (+ ) = $ for all values of $ $ in tangent’s domain.

1. Explain why tangent’s characterization as a radius’s slope makes this fact obvious.

2. Explain why, still more generally, $ (+ k) = $ for all integers k.

11. You can also understand why tangent’s graph looks as it does by thinking about tangent’s unit-circle definition. Do so. (It’s nice to have two ways to understand the tangent function on the unit circle, isn’t it?)

12. Late one night, Xu Fu approaches the Great Wall of China (which in this problem is infinitely long and perfectly straight), wearing a helmet to which he has strapped a curious flashlight that emits laser-like beams from both ends. Standing a foot away from the wall, he beholds the illuminated spot directly in front of him, marks its center with a bit of chalk, and then, apparently satisfied, begins slowly rotating clockwise. He continues this strange behavior all night long. Find a function describing the illuminated spot’s position after he has rotated through the radians. [Note: Positive and negative positions should correspond to distances right and left (respectively) of the chalk mark. Verify that your function holds for all positive values of $ $ : not just acute ones.]

13. True or false. (If true, explain why. If false, provide a counterexample.) For all $ $ …

\[ \sin(\theta+\pi)=\sin\theta \]

2. sin(θ + 2π) = sin θ

3. sin(θ + 3π) = sin θ

4. sin(θ + 4π) = sin θ

5. cos(θ + π) = cos θ

6. cos(θ − 6π) = cos θ

7. $ (+ 23) = $ h) $ (+ 23) = (+ ) $ i) $ (/2 - ) = $

14. Write identities for sine and cosine analogous to the one that you explained for tangent in exercise 10B.

15. Recall that the variables that appear in any function’s formula are just “dummy variables”, mere placeholders. For instance, $ f(x) = x^{2} $ , $ y = t^{2} $ , and $ h(z) = z^{2} $ all represent the exact same function – the squaring function. Consequently, if we change a formula’s variables, we do not change the function’s graph at all, apart (obviously) from the trivial change of the axes’ labels.

This being so, sketch the graphs of $ y = x $ , $ y = x $ , and $ y = x $ .

[Yes, this exercise is as easy as it sounds. Still, go ahead and draw the graphs. We’ll often write sine as a function of x rather than $ $ ; it’s important that you realize that this changes nothing.]

16. If you understand the graphs of sine, cosine, and tangent, you shouldn’t have much trouble graphing their reciprocals. After all, if sine sends a certain input value to 1/10, cosecant will send that same input to 10; if sine sends some other input to -2/3, cosecant sends that same input to -3/2, and so forth. Consequently, to graph cosecant, you need only look at sine’s graph, and think about where its points will go if you change all of their y-coordinates to their reciprocals. Think about this idea until it is clear, then use it to produce graphs of…

1. $ y = $ b) $ y = $ c) $ y = $ . d) $ y = 1 / (x^{2} + 1) $

17. 1. Explain why, in the upper picture of the unit circle at right, $ OT = $ .

2. The Latin verb secure means “to cut,” while tangere means “to touch”. Ponder the line segments TO and TQ, and their relationship to the circle.

3. Think of some English cognates of secure and tangere.

4. If you divide both sides of the Pythagorean identity by $ ^{2}$ , you’ll obtain another useful identity. Find it. Then explain how you can see this identity directly by looking at the upper figure in the right way.

5. If θ is in quadrant two (see the lower figure at right), sec θ = -OT. Explain why.

6. Draw figures for the cases where $ $ lies in quadrants three and four, and locate sec $ $ on these figures. (It will be OT or -OT in each case. Explain why.)

7. Now that you can visualize secant on the unit circle, produce the graph of the secant function without recourse to the method you used in exercise 16b.

8. Find $ $ and $ $ on the unit circle.

[Hint: Draw the line tangent to the unit circle at the point $ (0, 1) $ .]

Transformed Sine Waves

Nature’s wavelike functions never behave exactly like sine, but they often resemble it. For example, in the graph at right, y represents the population, in thousands, of angler fish in some region of the ocean; x represents time, measured in years since 2000. The graph is certainly “siney” (“sinusoidal” is the more respectable word), but it differs from the standard sine wave in obvious ways: It has a shorter period (3 instead of $ 2$ ), a higher “midline” (y = 15 instead of y = 0), and a greater amplitude (the distance from its midline to its peaks: here, 5 instead of 1).

We can bridge these differences with transformations. By shifting and stretching the familiar graph of $ y = x $ , we can transform the standard sine wave into the sinusoidal graph above. These geometric transformations cast algebraic shadows; studying them helps us find the sinusoidal graph’s equation.

Example 1. Find an equation whose graph matches the one in the figure above.

Solution. The following sequence of geometric transformations will turn the graph of $ y = x $ into the graph above (as you should verify by drawing some sketches of the intermediate stages):

1. Stretch vertically by a factor of 5.

2. “Stretch” horizontally by a factor of $ 3/2$ .

3. Shift up by 15 units.

(This changes its amplitude from 1 to 5),

(This changes its period from $ 2$ to 3),

(This makes it oscillate about y = 15).

In chapter 6, you learned that the corresponding sequence of algebraic transformations is:

1. Multiply the RHS by $ 5.^{} $

2. Substitute $ (x) $ for x.

3. Add 15 to the RHS.

[This turns $ y = x $ into $ y = 5 x $ .]

[This yields $ y = 5 (x) $ .]

[This gives us $ y = 5 (x) + 15 $ .]

Hence, the graph’s equation is $ y = 5 ( x ) + 15 $ .

We’ve just used trigonometry to model a periodic relationship between time and fish. Neither triangles nor angles appeared anywhere in this problem, yet we solved it with trigonometry. $ ^{4} $ We’ve now witnessed sine’s full metamorphosis: It began as a simple triangle-solving tool, but it has now been reborn as the very archetype of periodicity, with the unit circle serving as the great god SOH CAH TOA’s cocoon.

Sine often models phenomena in which angles play no role, so we tend to use the neutral symbol x (rather than $ $ , with its specifically angular connotations) as the symbol for its independent variable.

The moral of the story: From a higher perspective, sine’s essence is… periodicity itself.

As a reminder of some Chapter 6 material, here is a condensed transformation table summarizing the correspondences between the basic geometric and algebraic transformations of any function $ y = f(x) $ . (The abbreviation RHS stands for “right-hand side”.)

	Horizontal	Vertical
Stretch by a factor of $ k $ (with a reflection if $ k < 0 $ )	Substitute $ ( x) $ for each $ x $	Multiply the RHS by $ k $
Shift by $ k $ units	Substitute $ (x - k) $ for each $ x $	Add $ k $ to the RHS

Let us consider another example like the previous one.

Example 2. Find a function with the graph shown at right.

Solution. It begins at the crest of a wave, so this resembles cosine. It differs from cosine by oscillating around y = 4 with a period of 8 and an amplitude of 10, but we can turn cosine’s graph into the one at right by applying an ordered sequence of geometric transformations:

1. Stretch vertically by a factor of 10.

2. Stretch horizontally by a factor of $ 4/$ .

3. Shift up by 4 units.

(This changes the amplitude from 1 to 10),

(This changes its period from $ 2$ to 8),

(This makes it oscillate about y = 4).

According to our transformation table above, these geometric transformations correspond to this ordered sequence of algebraic transformations:

1. Multiply the RHS by 10.

2. Substitute $ (x) $ for x.

3. Add 4 to the RHS.

[This turns $ y = x $ into $ y = 10 x $ .]

[This yields $ y = 10 ( x) $ .]

[This gives us $ y = 10 ( x) + 4 $ .]

Hence, the graph can be modeled by $ y = 10 ( x ) + 4 $ .

Naturally, there are variations on this basic theme, some of which you’ll explore in the exercises.

Exercises

18. In the preceding example, we could have obtained the given graph by subjecting the sine wave to the following transformations (in this order): vertical stretch by a factor of 10, horizontal stretch by a factor of $ 4/$ , shift left by 2 units, shift up by 4 units.

1. Convince yourself that this is so.

2. Carrying out the corresponding algebraic transformations yields $ y = 10 ((/4)x + (/2)) + 4 $ . Verify that this is so.

3. To reconcile the equation in Part B with the equation that we found in example 2, you’ll need an identity for $ (+ /2) $ . First, discover this identity by thinking about the unit circle and sketching some pictures. (You did this sort of thing in Chapter 10, Exercises 41-43.) Then use your identity to demonstrate that the equation in Part B is equivalent to the equation we found in Example 2.

19. Find formulas for functions with the following graphs:

a)

b)

c)

d)

20. Draw the sine and cosine functions on the same set of axes.

1. A single geometric transformation will transform cosine’s graph into sine’s. Which one?

Verify that the corresponding algebraic transformation turns $ y = $ into $ y = $ .

[For the verification, you’ll need a trigonometric identity and an algebraic identity: $ a - b = -(b - a) $ .]

2. Similarly, a single geometric transformation will transform sine’s graph into cosine’s. Which one? Verify that the corresponding algebraic transformation turns $ y = $ into $ y = $ .

21. Repeat the previous exercise, but with tangent and cotangent. (Two transformations will be necessary.)

22. Repeat it again, but with secant and cosecant.

23. Answer the following questions about the strange graph in the figure at right, the graph of

\[ f(x)=x\sin\left(\frac{1}{x}\right). \]

1. Does the graph pass through the origin?

2. How many times does the portion of the graph shown at right cross the x-axis?

[Hint: look for exact solutions to $ f(x) = 0 $ .]

3. Will the graph continue to cross and re-cross the x-axis as we move away from the big bang at the origin? If so, how do you know? If not, then what is the greatest solution of $ f(x) = 0 $ ?

Graphing Transformed Trig Functions

Functions of the form $ y = a (bx + c) + d $ are common in models of periodic phenomena. Graphing some by hand will help you hone your transformation skills.

Example. Graph the function $ y = -2 (4x - ) + 1 $ .

Solution. The game is to think of a sequence of algebraic transformations that will turn $ y = x $ into the given function. It’s often best to stretch before shifting (your old gym coach would agree), so let’s bring the horizontal shift to the foreground by factoring out a $ 4$ inside the parentheses:

\[ y=-2\sin\left(4\pi\left(\boldsymbol{x}-\frac{\mathbf{1}}{\mathbf{8}}\right)\right)+1. \]

We can obtain this equation from $ y = x $ by applying this sequence of algebraic transformations:

1. Substitute $ 4x $ for x

[This transforms $ y = x $ into $ y = (4x) $ .]

2. Multiply RHS by -2

[This gives us $ y = -2 (4x) $ .]

3. Substitute $ (x-) $ for x

[This yields $ y = -2(4(x - )) $ .]

4. Add 1 to the RHS.

[This puts the cherry on top.]

The corresponding sequence of geometric transformations is:

1. Stretch horizontally by a factor of $ 1/(4) $ .

2. Stretch vertically by a factor of 2, and reflect the graph vertically (that is, over the x-axis).

3. Shift right by 1/8 unit.

4. Shift up by 1 unit.

The first two geometric transformations (the two stretches) shrink the sine wave’s period from $ 2$ to $ 2(1/4)=1/2 $ , while increasing its amplitude from 1 to 2 and flipping it over the x-axis. Thus, the net effect of the two stretches on the sine wave is to produce the graph at right. I’ve marked a few key points on the x-axis (peaks, valleys, midline intersections), because they will help us visualize the effects of the next transformations in

our sequence: the shifts.

The shifts merely change the graph’s position while preserving its shape. Shifting all the previous graph’s points 1/8 unit to the right and 1 unit up brings us, at last, to our final graph, which is shown at right.

One last thought: Identities, both algebraic and trigonometric, can simplify your work considerably. For instance, in the preceding example, we might have noted, before doing any graphing whatsoever, that

\[ \begin{aligned}y&=-2\sin\left(4\pi x-\frac{\pi}{2}\right)+1\\&=-2\sin\left(-\left(\frac{\pi}{2}-4\pi x\right)\right)+1\quad&(since(a-b)=-(b-a))\\&=2\sin\left(\frac{\pi}{2}-4\pi x\right)+1\quad&(sine is an odd function)\\&=2\cos\left(4\pi x\right)+1,\quad&(cofunction identity)\\ \end{aligned} \]

which would have been easier to graph since it doesn’t require a horizontal shift, that most troublesome of transformations. In the exercises that follow, use identities to simplify the given function when you can.

Exercise

24. Sketch graphs of the following functions; note key points and features within one full cycle of each.

\[ \begin{aligned}&a)y=5\sin(\pi x)\quad&b)y=-2\cos(x-\pi)\quad&c)y=4\sin(2x-(3\pi/2))-3\\&d)y=\cos(7x-10\pi)+\pi\quad&[Hint:Exercise14]\quad&e)y=\sin(x+5341\pi)\quad&[Hint:Exercise13h.]\\&f)y=2-3\sin(\pi-6x)\quad&g)y=\pi\cos(\pi x+\pi)+\pi\quad&h)y=\tan(x+\pi/4)+2\\&i)y=-2\tan(2x)\quad&j)y=3\tan(2\pi x-\pi/4)\end{aligned} \]

Calculators are Big Fakers

Press a few buttons, and your calculator will inform you that $ (22^{}) $ . This value isn’t stored in the calculator’s memory, so where did it come from?

Calculators are masters of arithmetic – and of nothing else. $ ^{*} $ Evaluating a polynomial is a purely arithmetic task, so a calculator can do it at lightning speed. The sine function, however, is no polynomial; it is defined geometrically. Consequently, your calculator does not know, and cannot know, what $ (22^{}) $ means. Still, it is an excellent faker. It sports a “sin” button, and its programmer has taught it to bluff whenever someone hits that button. If, for instance, we ask a calculator to evaluate $ (22^{}) $ , the calculator, as instructed by its programmer, might evaluate the following $ 15^{th} $ -degree polynomial at 22:

\[ \begin{align*}y=-\left(\frac{\pi^{15}}{180^{15}\cdot15!}\right)x^{15}+\left(\frac{\pi^{13}}{180^{13}\cdot13!}\right)x^{13}-\left(\frac{\pi^{11}}{180^{11}\cdot11!}\right)x^{11}+\left(\frac{\pi^{9}}{180^{9}\cdot9!}\right)x^{9}\\-\left(\frac{\pi^{7}}{180^{7}\cdot7!}\right)x^{7}+\left(\frac{\pi^{5}}{180^{5}\cdot5!}\right)x^{5}-\left(\frac{\pi^{3}}{180^{3}\cdot3!}\right)x^{3}+\left(\frac{\pi}{180}\right)x.\end{align*} \]

The calculator, always happy to evaluate a polynomial, quickly churns out the sum and gives us a value. And we, for our part, are happy to accept that value as an honest approximation for $ (22^{}) $ . Why?

We accept the approximation because the $ 15^{th} $ -degree polynomial’s graph (the dashed curve below) is virtually indistinguishable from the sine wave for small values of x, such as 22. In fact, the two graphs are so close that for input values between -90 and 90, the polynomial’s output matches the sine’s output in its first nine decimal places, which is precisely what the calculator shows us on its screen. The calculator is a very good faker indeed.

But where did that crazy polynomial come from? An excellent question, but its answer requires calculus. When you study calculus, you’ll learn about so-called Taylor polynomials, which are marvelous creatures capable of mimicking all sorts of functions. The polynomial discussed above is one such example. Incidentally, those ugly powers of $ $ and 180 that the Taylor polynomial contains are consequences of my decision to use degrees in this example. Had I used radians instead, they would all have vanished, and the Taylor polynomial would have looked much cleaner. Radians may seem awkward at first, but they do in fact simplify life.

Exercises

25. The Taylor polynomial above gives a terrible approximation for $ (460^{}) $ . Explain how a calculator could, nonetheless, approximate $ (460^{}) $ by doing some preliminary arithmetic on the argument, $ 460^{} $ . Then explain further how one might program the calculator to use the Taylor polynomial to approximate the sine of any number outside the sweet spot between $ -90^{} $ and $ 90^{} $ .

26. Scientists and engineers often use the approximation $ x x $ for small values of x, which applies only if we use a radian measure. Convince yourself of this first by playing around with your calculator, and then by thinking about the graphs of $ y = x $ and y = x.