]> Chapter 05 Notes (Sullivan & Sullivan 3e)

Chapter 05: Trigonometric Functions

§5.1: Angles and Their Measure

Definitions

We all know from Geometry class that an angle is formed by a pair of coterminal rays (or intersecting segments, or lines…). Now, though, we need to nail down a few more items.

When we measure an angle, we want to specify the direction of the angle…thus, one ray needs to be the initial side of the angle, and the other ray needs to be the terminal side of the angle. If the direction of the angle is counterclockwise, then the angle is said to be positive; an angle measured in the clockwise direction is called negative.

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We often measure angle that are in a coordinate plane. When this happens, we often want the angle to be in standard position —that is, with the initial side along the positive x-axis and the vertex at the origin.

Measuring Angles

It turns out that there are lots of ways to assign a number to the measure of an angle…

Degrees

The primary method for many real-world applications is the degree. An angle of one degree (also called one degree of arc) is one three-hundred-sixtieth of a full turn (circle). The symbol for degrees is a small circle in a superscript position—like this: $360 º$ . Without this symbol (or a word), the angle is NOT being measured in degrees!

If you’re really interested in why there are $360 º$ in a full circle…well, Google it; you’ll find out. There are several reasons, actually.

Degrees, Minutes, Seconds

It turns out that one degree isn’t sufficiently precise for some applications, and the idea of using decimals is still fairly new (I know you have trouble imagining this, but until recently, fractions were the only game in town!)…thus, one degree was subdivided into sixty minutes. One minute of arc is denoted $1'$ . This, in turn, was divided into sixty seconds. One second of arc is denoted $1 "$ . In this system (the DMS system), a really precise angle might look like $45 º 10' 45 "$ .

Gradians

A system that simplifies things a lot uses gradians. In this system, a right angle is 100 gradians. The benefit here is that the angle measure can be quickly interpreted as a part of a right angle—an angle of 50 gradians is half of a right angle.

The only place that I’ve ever seen gradians used is on an old Army plotting template…

Radians

Degrees are fine if you’re doing something real, but mathematicians never feel constrained by reality. Gradians are just weird. Mathematicians prefer to use radians to measure angle. One radian is an angle that subtends an arc with length equal to the radius of a circle.

Okay, get up off of the floor—that wasn’t a terrible explanation! Here, look—here’s a diagram.

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Note that a full circle in radians is $2 π$ ; thus, a half circle is $π$ radians. You can use this fact to set up a proportion to convert between radians and degrees: $\frac{ρ}{δ} = \frac{π}{180 º}$ , where $ρ$ is the angle measured in radians and $δ$ is the angle measured in degrees.

The big benefit of radians is that they don’t have any units—that makes the math work out (that’s the same reason why you can’t do anything useful with percentages until you switch them back to decimal form).

Arc Length

Radians are defined in terms of arc length, so finding arc length when the angle is measured in radians is easy! The length of an arc subtended by an angle of measure $θ$ in a circle of radius r is $A = r θ$ .

Arc length is harder when done in degrees—in that case, you have to set up a proportion with the angle measure and the arc length. In particular, the angle measure is to the measure of a full circle as the arc length is to the circumference of a circle: $\frac{θ}{360 º} = \frac{A}{2 π r}$ . If you solve that for A, then you’ll find that it’s just the radian version combined with a conversion from degrees to radians. Skip the middle man—just use radians in the first place!

The Area of a Sector

A sector is a “pie slice.” The ratio of the sector area to the circle area is the same as the ratio of the subtended angle to a full circle.

$\frac{A_{s}}{π r^{2}} = \frac{ρ}{2 π}$ , or $\frac{A_{s}}{π r^{2}} = \frac{δ}{360 º}$

Rotational and Linear Motion

Think about a circle rolling along…every time the circle makes one full turn, it travels its circumference in a linear direction. You can use this fact to convert between rotational speed (usually given as rotations per minute) into a linear speed.

Examples

[1.] Convert to radians: $72 º$ .

$72 º \cdot \frac{π}{180 º} = \frac{π}{5}$

[2.] Convert to degrees: 60

Note that this angle measure is in radians, since there is no degree symbol!

$60 \cdot \frac{180 º}{π} = \frac{10800 º}{π} \approx 3437.747 º$ .

[3.] Find the length of an arc subtended by an angle of measure $\frac{π}{8}$ in a circle of radius 3m.

$A = (3) (\frac{π}{8}) = \frac{3 π}{8} \approx 1.178$ meters.

[4.] A 12 inch diameter pizza is cut into eight slices. How much area is there for each slice?

A 12 inch diameter makes a radius of 6 inches, and eight slices makes the angle $\frac{2 π}{8} = \frac{π}{4}$ , so we have $\frac{A_{s}}{π {(6)}^{2}} = \frac{\frac{π}{4}}{2 π} \Rightarrow \frac{A_{s}}{36 π} = \frac{1}{8} \Rightarrow A_{s} = \frac{36 π}{8} = \frac{9 π}{2} \approx 14.137$ square inches.

[5.] Convert to a decimal: $15 º 12' 38 "$ .

That’s really $15 º + (\frac{12}{60}) º + (\frac{38}{3600}) º = (\frac{27379}{1800}) º \approx 15.211 º$ .

[6.] An 18 inch diameter wheel rotates at 2000rpm. What is the linear speed of the wheel?

Time to break out those unit conversion skills from science class!

$\frac{2000 rot}{1 min} \cdot \frac{18 π inches}{1 rot} = 3600 π \frac{inches}{min} \approx 11309.734 \frac{inches}{min}$ . That’s a perfectly good linear speed!

§5.2: The Unit Circle

Trigonometric Functions: The Right Triangle Approach

Been there; done that.

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Now we can say

Sine: $sin θ = \frac{opposite}{hypotenuse}$ , Cosine: $cos θ = \frac{adjacent}{hypotenuse}$ , and Tangent: $tan θ = \frac{opposite}{adjacent}$ . Also, Cosecant: $csc θ = \frac{hypotenuse}{opposite}$ , Secant: $sec θ = \frac{hypotenuse}{adjacent}$ and Cotangent: $\cot θ = \frac{adjacent}{opposite}$ .

Trigonometric Functions: The Coordinate Approach

Note that any point out in the plane can be used to create a triangle:

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Thus, for almost any point in the plane, we get $sin θ = \frac{y}{r}$ ; $cos θ = \frac{x}{r}$ ; $tan θ = \frac{y}{x}$ ; $sec θ = \frac{r}{x}$ ; $csc θ = \frac{r}{y}$ ; $\cot θ = \frac{x}{y}$ . We say almost because some points will have coordinate values of zero, and you can’t divide by zero! Thus, some of these functions fail to exist for certain angles.

Trigonometric Functions: The Unit Circle Approach

Every point in the plane makes a triangle; every point makes an angle to which sine, cosine, et. al., may be applied…however, there are a lot of points that make the same angle (and a lot of angles that pass through the same point)! Thus, we reduce and simplify: let's just consider the points that are one unit away from the origin. This creates the unit circle, and simplifies the trigonometric ratio definitions given above:

$sin θ = y$ ; $cos θ = x$ ; $tan θ = \frac{y}{x}$ ; $sec θ = \frac{1}{x}$ ; $csc θ = \frac{1}{y}$ ; $\cot θ = \frac{x}{y}$

The term unit circle is used by many people to simply refer to a diagram that helps one to memorize the sine and cosine values for certain special angles.

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I’ve only shown the first quadrant, since you can use symmetry to fill in the other quadrants!

Those points not on the axes correspond to the special right triangles (one of them can be turned two ways, remember?). Here is a list of the coordinates:

Angle (Degrees)	Angle (Radians)	Coordinates
0°	0	$(1, 0)$
30°	$\frac{π}{6}$	$(\frac{\sqrt{3}}{2}, \frac{1}{2})$
45°	$\frac{π}{4}$	$(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$
60°	$\frac{π}{3}$	$(\frac{1}{2}, \frac{\sqrt{3}}{2})$
90°	$\frac{π}{2}$	$(0, 1)$

OK $-$ here’s a full version I found out on the internet:

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Examples

[7.] Find the values of all six trigonometric functions for the angle in standard position that passes through the point $(- 2, 5)$ .

Note that this really makes a right triangle with legs -2 and 5, and hypotenuse ${(- 2)}^{2} + 5^{2} = h^{2} \Rightarrow h^{2} = 29 \Rightarrow h = \sqrt{29}$ . Thus, $sin (θ) = \frac{5}{\sqrt{29}}$ , $cos (θ) = - \frac{2}{\sqrt{29}}$ , $tan (θ) = - \frac{5}{2}$ , $sec (θ) = - \frac{\sqrt{29}}{2}$ , $csc (θ) = \frac{\sqrt{29}}{5}$ and $\cot (θ) = - \frac{2}{5}$ .

[8.] Give the exact value of $cos (\frac{7 π}{6})$ .

From the unit circle, that would be $- \frac{\sqrt{3}}{2}$ . I don’t care if you use special right triangles or the unit circle—you must be able to answer questions like this without a calculator. I can guarantee that such a question will be asked on the test!

§5.3: Properties of Trigonometric Functions

Domain and Range

If you look closely at the point-based definitions of the trigonometric functions, you’ll see that the only ones that can have domain issues are tangent, cotangent, secant and cosecant. Thus, the domain for sine and cosine is all real numbers. The domain for tangent and secant exclude any points with an x-coordinate of zero; that would be any angle which is an odd multiple of $\frac{π}{2}$ . The domain for cotangent and cosecant exclude any points with a y-coordinate of zero; that would be any angle which is an integer multiple of $π$ .

For range, let’s consider the unit circle for a moment. Since the value of the cosine function is the x-coordinate on the unit circle, it can be seen that the range of cosine is $[- 1, 1]$ . Similarly, since sine is the y-coordinate, it can be seen that range of sine is the same: $[- 1, 1]$ . For tangent and cotangent, since they are defined as the ratio of x and y, their values can be made as large or small as you like—thus, the range for those is all real numbers. For secant and cosecant, note that they are the reciprocals of since and cosine—thus, the range for these will be the reciprocal of some number between -1 and 1. The reciprocal of a number smaller than one (like, say, 0.2) is a number greater than one. Thus, the range of secant and cosecant is $(- \infty, - 1] \cup [1, \infty)$ .

Period

A function which repeats its values regularly is called periodic. It should not be hard to see that sine and cosine are periodic! The distance between inputs for the function values to repeat is called the period of the function. Again, it should not be hard to see that sine and cosine cycle through all of their values as an angle sweeps out a full circle; thus, their period must be $2 π$ .

Since secant and cosecant are the reciprocals of sine and cosine, their period must also be $2 π$ .

Because of symmetry in the unit circle (there are always two points on the circle that have the same $\frac{y}{x}$ ratio, for example), the period for tangent and cotangent must be half that of the others—thus, $π$ .

One way to express the periodic nature of sine is the equation $sin (θ + 2 k π) = sin (θ)$ for any integer value of k. The same sort of equation could be written for all of the trigonometric functions.

Identities

These identities come from clever replacements in the unit circle definitions of the trigonometric functions.

Reciprocal

$sec (θ) = \frac{1}{cos (θ)}$ , $csc (θ) = \frac{1}{sin (θ)}$ , $\cot (θ) = \frac{1}{tan (θ)}$

Quotient

$tan (θ) = \frac{sin (θ)}{cos (θ)}$ , $\cot (θ) = \frac{cos (θ)}{sin (θ)}$

Pythagorean

Add the Pythagorean theorem and you get this little gem!

${(sin (θ))}^{2} + {(cos (θ))}^{2} = 1$

Most people don’t write this as carefully as I have; you’d usually see that as ${sin}^{2} θ + {cos}^{2} θ = 1$ . Do a little division on this plain vanilla fellow and you get two more flavors of the Pythagorean Identity:

${tan}^{2} θ + 1 = {sec}^{2} θ$ , $1 + \cot^{2} θ = {csc}^{2} θ$

One use of these is to take the value of one trigonometric function and find the value of another trigonometric function (without knowing the measure of the angle).

Examples

[9.] Find the exact value of $sin (405 °)$ .

Since $sin (405 °) = sin (1 \cdot 360 ° + 45 °)$ , the answer is just $sin (45 °) = \frac{1}{\sqrt{2}}$

[10.] If $cos α = \frac{2}{3}$ , find the exact value of $\cot α$ .

Since $cos α = \frac{2}{3}$ , I can imagine a triangle with an adjacent side of length 2 and a hypotenuse of length 3…making the length of the opposite side $\pm \sqrt{5}$ . Note that I have to use the plus or minus there because there are multiple angles that have a cosine value of $\frac{2}{3}$ …some of them will have positive sine values (first quadrant) and some will have negative sine values (fourth quadrant).

Thus, $\cot α = \pm \frac{2}{\sqrt{5}}$ .

[11.] Evaluate $sin (\frac{14 π}{5}) \cdot csc (- \frac{6 π}{5})$ .

Note that $sin (\frac{14 π}{5}) = sin (\frac{14 π}{5} - 2 π) = sin (\frac{14 π}{5} - \frac{10 π}{5}) = sin (\frac{4 π}{5})$ , and $csc (- \frac{6 π}{5}) = \frac{1}{sin (- \frac{6 π}{5})} = \frac{1}{sin (- \frac{6 π}{5} + 2 π)} = \frac{1}{sin (\frac{4 π}{5})}$ . Thus, the value of the expression is just 1!

[12.] Officer Bob, atop a 5m observation tower, observes a car passing through his speed trap. He sights the car (as it enters the area) at an angle of depression of 50°. As the car leaves the area, the angle of depression is 65°. Write an expression for the length of the speed trap.

Let’s start with a picture:

Note that the length of the speed trap is labeled t in the diagram.

Clearly $tan 65 ° = \frac{5}{x}$ ; an expression for t is a little harder…in fact, you can’t write one directly. You can write an expression that involves t: $tan 50 ° = \frac{5}{x + t}$ . Let’s solve each of those for x…the first one comes out to $x = \frac{5}{tan 65 °}$ . The second needs some work: $x + t = \frac{5}{tan 50 °} \Rightarrow x = \frac{5}{tan 50 °} - t$ . Put those together to get $\frac{5}{tan 65 °} = \frac{5}{tan 50 °} - t \Rightarrow t = \frac{5}{tan 50 °} - \frac{5}{tan 65 °}$ . If you’ve got a calculator, you could even find that value…

§5.4: Graphs of Sine and Cosine Functions

The Sine Graph

Plot the angle in the horizontal axis and the value of sine for that angle in the vertical axis…here’s the graph that you get:

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The Cosine Graph

Plot the angle in the horizontal axis and the value of cosine for that angle in the vertical axis…here’s the graph that you get:

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Common Features

Amplitude

The amplitude of these graphs is half of the vertical distance from the highest point to the lowest point.

Period

Both of these graphs have a period of $2 π$ .

Transformations

Once you know the parent graph for sine, you can then graph any function of the form $y = a \cdot sin (b (x - c)) + d$ . The same is true of cosine!

The amplitude is $| a |$ and the period is $\frac{2 π}{b}$ for any transformation of these graphs.

Based on the graphs above, a little transformation gives you some new identities: $sin (x + \frac{π}{2}) = cos x$ , and $cos (x - \frac{π}{2}) = sin x$ .

Examples

[13.] Describe how the graph of $y = cos x$ can be transformed into the graph of $y = 2 - 3 cos (2 x)$ .

Flip the graph upside down and multiply the y-values by 3 (make the amplitude 3); divide the x-values by 2 (make the period $π$ ); shift up 2 units.

[14.] Determine the amplitude and period of $y = 5 sin (2 x + π) - 1$ .

The amplitude is 5 and the period is $\frac{2 π}{2} = π$ .

§5.5: Graphs of Other Trigonometric Functions

The Tangent Graph

Plot the angle in the horizontal axis and the value of tangent for that angle in the vertical axis…here’s the graph that you get:

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(you did remember that tangent has domain issues at every odd multiple of $\frac{π}{2}$ , didn’t you?)

The Cotangent Graph

Plot the angle in the horizontal axis and the value of cotangent for that angle in the vertical axis…here’s the graph that you get:

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Golly…that looks familiar! Make sure that you keep those two straight.

The Secant Graph

Plot the angle in the horizontal axis and the value of secant for that angle in the vertical axis…here’s the graph that you get:

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The Cosecant Graph

Plot the angle in the horizontal axis and the value of cosecant for that angle in the vertical axis…here’s the graph that you get:

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Again, that looks very familiar…be careful!

Transformations

Do I really have to write it out again?

Examples

[15.] Sketch the graph of $y = 3 csc (2 x - π)$ .

Take a regular cosecant graph, triple the y-values, halve the x-values, and shift right $\frac{π}{2}$ .

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[16.] Describe a sequence of transformations that turn the graph of $y = tan x$ into the graph of $y = tan (3 x - \frac{π}{4}) + 2$ .

Divide the x-values by 3 (making the period $\frac{π}{3}$ ), shift right $\frac{π}{12}$ and up 2.

§5.6: Sinusoids

I mentioned earlier that the sine and cosine graphs are horizontal shifts of one another. There are, in fact, many functions that can be written as some transformation of a sine graph. These functions are called sinusoids. One skill that you need to have is to be able to determine what transformations are needed to give you a sinusoidal graph.

Amplitude

The amplitude will be half of the vertical distance between the highest and lowest points on the graph. Note that amplitude is always positive! Amplitude takes the a spot in the transformation template.

Vertical Shift

The vertical shift will be the average of the highest and lowest y-values on the graph. This is put in the d spot in the template.

Period

The period is the horizontal distance between peaks, or half of the horizontal distance between the highest and lowest points on the graph. Note that the period of a transformed sine graph is $\frac{2 π}{b}$ …use this to determine the value of b that needs to be used in the template.

Horizontal Shift

This is probably the hardest value to just look at and see…note that the normal location of the first peak on a sine graph is at $(\frac{π}{2}, 1)$ . That means that the location of that first peak gets moved to $(\frac{π}{2 b} + c, a + d)$ after a transformation. Use the observed x-coordinate of the peak along with this to determine the value of c that needs to be used.

Example

[17.] Determine the equation of the graph shown below.

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The highest y-value is at 3 and the lowest is at -1. The amplitude will be half the difference: $\frac{3 - (- 1)}{2} = \frac{4}{2} = 2$ . The vertical shift will be the average height: $\frac{3 + (- 1)}{2} = \frac{2}{2} = 1$ . Thus, $a = 2$ and $d = 1$ .

The distance from peak to peak is $\frac{4 π}{8} = \frac{π}{2}$ ; $\frac{2 π}{b} = \frac{π}{2} \Rightarrow b = 4$ .

The peak is at $(\frac{3 π}{8}, 3)$ …so $\frac{π}{2 \cdot 4} + c = \frac{3 π}{8} \Rightarrow c = \frac{3 π}{8} - \frac{π}{8} = \frac{2 π}{8} = \frac{π}{4}$ and $2 + d = 3 \Rightarrow d = 1$ .

The equation is $y = 2 sin (4 (x - \frac{π}{4})) + 1$ .

Page last validated 2010-08-15