Module 2: Trigonometry - Open School...

Module 2: Trigonometry

Section 1 Trigonometric Functions 3

Lesson 1 The Trigonometric Values for θ, 0°≤ θ ≤ 360° 5

Lesson 2 Solving Trigonometric Equations, 0° ≤ θ ≤ 360° 9

Lesson 3 Special Trigonometric Values 17

Lesson 4 The Unit Circle 25

Lesson 5 The Circular Functions 33

Lesson 6 Radian Measures 45

Summary of Section 1 55Review 57Section Assignment 2.1 59

Section 2: Graphs of Trigonometric Functions 65

Lesson 1 Sine and Cosine Graphs 67

Lesson 2 Transformations of the Sine and Cosine Functions 75

Lesson 3 Graphs of the Remaining Circular Functions 85


Section 3: Trigonometric Identities 105

Lesson 1 Elementary Identities 107

Lesson 2 Using Elementary Identities 117

Lesson 3 Sum and Difference Identities 123

Lesson 4 Double Angle Identities 131


Principles of Mathematics 12 Contents 1

Module 2

2 Contents Principles of Mathematics 12

Module 2

Section 4: Problem Solving 151

Lesson 1 Modelling with Trigonometric Functions 153

Lesson 2 Solving Trigonometric Equations UsingYour Graphing Calculator 161

Lesson 3 Solving Trigonometric Equations :Finding the General Solution 161

Summary of Section 4 165

Review 167Section Assignment 2.4 171

Module 2: Answer Key 177

Module 2

Module 2, Section 1

Trigonometric Functions

IntroductionYou have studied trigonometric ratios since Grade 9 Mathematics. Inthis module you will study those ratios as functions. You will look attables of values of these functions, generate their graphs, and listtheir properties. It is important for you to do all the questions in theassignments because some new concepts are developed within theassignments. Using techniques that you developed in Module 1, youwill study transformations of these circular functions. In order tofind the zeros of the functions, you will develop skills in circularfunction equation solving and in relationships amongst thesefunctions which are called identities.

Section 1 — Outline

Lesson 1 The Trigonometric Values for θ, 0° ≤ θ ≤ 360°

Lesson 2 Solving Trigonometric Equations for 0° ≤ θ ≤ 360°

Lesson 3 Special Trigonometric Values

Lesson 4 The Unit Circle

Lesson 5 The Circular Functions

Lesson 6 Radian Measures

Lesson 7 Sine and Cosine Graphs

Review

Principles of Mathematics 12 Section 1, Introduction 3

Notes

Module 2

4 Section 1, Introduction Principles of Mathematics 12

Module 2

Lesson 1The Trigonometric Values for θθ, 0° ≤≤ θθ ≤≤ 360°

To this point, you have determined the values of the trigonometricratios for angles between 0° and 180°. In this short introductorylesson, we will explore the trigonometric ratios for any angle θbetween 0° and 360°. First we will re-define the primary trig ratiossine, cosine, and tangent.

Suppose θ is any angle in standard position (its initial arm lies onthe positive x-axis). Further, let P(x,y) be any point on θ’s terminalarm and r be the distance from P to the origin (0,0). Since r is adistance, it will always be considered positive.

Now, using Pythagorean Theorem, we can determine the value of r:

These definitions can be applied to determine the trigonometricratios for any angle, θ, in standard position if the coordinates of apoint on its terminal arm are known.

Example 1Determine the three primary trig ratios for the angle θ in standardposition, with the point P(–4,5) on its terminal arm.

x

�

y

P(x,y)

y(opposite)

x(adjacent)

(hyp

oten

use)

r

θ

terminal arm

initial arm

2 2 and we define:

sin

cos

tan

θ

θ

θ

= +

= =

= =

= =

r x yy oppositer hypotenusex adjacentr hypotenusey oppositex adjacent

Principles of Mathematics 12 Section 1, Lesson 1 5

Solution

θ terminates in Quadrant II and we will be using our new definitions tosolve this problem, but first we need to determine r.

As you can see, for anyangle θ terminating in Quadrant II, as in Example 1, the only one ofthe three primary trig ratios which is positive is sin θ. We will comeback to this result shortly.

Example 2An angle θ in standard position has the point Q(–12,–5) on its terminalarm. Determine its primary trigonometric ratios.

For θ in Quadrant III:

y

x

3

–5–10

–5P(–12,–5)�

r = 13

θ

2 2( 12) ( 5) 135sin 0.385

1312cos 0.923

135tan 0.417

12

ryrxryx

θ

θ

θ

= − + − =−= = ≈ −

−= = ≈ −

−= = = +−

y

x

5

–3

3

�

–3

P(–4,5)

θr = √41

2 2( 4) (5) 415sin 0.781414cos 0.62541

5tan –1.2504

ryrxryx

θ

θ

θ

= − + =

= = ≈

−= = ≈ −

= = =−

6 Section 1, Lesson 1 Principles of Mathematics 12

Module 2

Note that the only positive one of the primary trigonometric ratios foran angle θ terminating in Quadrant III is tan θ.

If we examine the signs of our x and y-coordinates in the other twoquadrants, we see the following.

Quadrant IVIn Quadrant IV, x is positive and y is negative.

So for θ terminating in Quadrant IV only cos θ is positive.

Quadrant IIn Quadrant I, x is positive and y is positive.

So for θ terminating in Quadrant I, all three of the primarytrigonometric ratios are positive.

x

�

y

P(+,+)

θ

r

(always positive)

( )( ) ( )

( )( ) ( )

( )( ) ( )

So sin

cos

tan

yr

xr

yx

θ

θ

θ

+= = = +

+

+= = = +

+

+= = = +

+

x

�

y

P(+,–)

θ

r

(always positive)

( )( ) ( )

( )( ) ( )

( )( ) ( )

θ

θ

θ

−= = = −

+

+= = = +

+

−= = = −

+

So sin

cos

tan

yr

xr

yx


Module 2

Now we are able to summarize what we know about the trig ratios ofany angle in standard position.

From the first letter of each sentence, this is called the “CAST” diagramand it will prove very helpful to you in the future!

Now for a little practice.

Guided Practice

Using the new definitions for sin θ, cos θ, and tan θ, determine the exactvalues for these trigonometric ratios. Please leave your answers infractional and/or radical form if they are not integer values.

1. P(1,11) is on the terminal arm.2. Q(–6,8) is on the terminal arm.3. R(5,–12) is on the terminal arm.4. S(–6,–3) is on the terminal arm.5. M(5,0) is on the terminal arm.6. N(0,4) is on the terminal arm.7. L(–3,0) is on the terminal arm.8. K(0,–2) is on the terminal arm.

Check your answers in the Module 2 Answer Key.

Sine only is positive All trig ratios are positive

Tangent only is positive Cosine only is positive

Quadrant III

Quadrant II Quadrant I

Quadrant IV


Module 2

Lesson 2

Solving Trigonometric Equations, 0° ≤≤ θθ ≤≤ 360°OutcomesUpon completing this lesson, you will be able to:• determine the three primary trigonometric ratios for any

angle θ, in standard position 0° ≤ θ ≤ 360°• solve trigonometric equations for 0° ≤ θ < 360°

OverviewIn Lesson 1 we learned new definitions for sin θ, cos θ, and tan θ. Using these new definitions, we were able to determine thetrigonometric ratios for angles terminating in any one of the fourquadrants and make observations about their signs (the “CAST”diagram below).

In this lesson we will be using the CAST diagram along with a givenangle’s reference angle (defined below) to solve simple trigonometricequations.

Definition:The reference angle for any angle θ in standard position is theacute angle between its terminal arm and the nearest part of the x-axis.

Example 1Determine the quadrant where each of the following terminate andthe reference angle.

a) 55° b) 169° c) 228° d) 310°

Solutiona) Angles which terminate in QI

(Quadrant 1) are identical inmeasure to their reference angle.

x

y

55º

QI

Reference Angle = 55º

S A

T C


Module 2

b) The reference angle for θterminating in QII is given by: 180° – θ. This is because thenearest part of the x-axis to theterminal arm of θ is the negative x-axis, where the angle is 180° from the initial arm.

c) The reference angle for θterminating in QIII is given by: θ –180°. θ comes before 180 to makethe reference angle positive.

d) The reference angle for θterminating in QIV is given by:360° – θ.Note: our abbreviation for“reference angle” is θR.

Angles having the same referenceangles have very similartrigonometric ratios. Thedifferences are in the signs(positive or negative) of each.

Definition: Coterminal angles are angles with the same terminalarm. Coterminal angles have the same reference angle

x

y

•

θ = 310º

θR = 360º – 310º = 50º

sin 310º = –0.766

cos 310º = 0.643

tan 310º = –1.192

•

QIV

x

y

= 220º – 180º

= 40ºQIII

Reference Angle

•

θ = 220º

x

y

= 180º – 169º

= 11º

QII

Reference Angle

•

θ = 169º


Module 2

Example 2Find an angle coterminal with:

a) 30° b) 220°

Solutiona)

30° and 390°are coterminalbecause theyboth end up inthe sameposition

or

30° and –330° are coterminalbecause they bothend up in thesame position

b)

220° and 580°are coterminalbecause theyboth end up inthe sameposition

or

220° and–140° arecoterminalbecause theyboth end upin the sameposition

x

y

140°

x

y

220º

x

y

580º

x

y

220º

x

y

–330º

x

y

30º

x

y

390º

x

y

30º


Example 3Use your calculator to determine the three primary trigonometric ratiosfor each of the following:

a) 50° b) 130° c) 230° d) 310°

Solutiona) b)

c) d)

You can check these results against the CAST diagram, too. InQuadrant I(a) all trig ratios are positive, in Quadrant II (b) only sinis positive, in Quadrant III (c) only tan is positive, and in Quadrant IV (d) only cos is positive.

Now we will turn things around a little and solve some equations for θ,0° ≤ θ < 360°.

x

y

•

θ = 310º

θR = 360º – 310º = 50º

sin 310º = –0.766

cos 310º = 0.643

tan 310º = –1.192

•

x

y

•

230ºθR = 230º – 180º = 50º

sin 230º = –0.766

cos 230º = –0.643

tan 230º = 1.192

•

x

y

•

130º

θR = 180º – 130º = 50º

sin 130º = 0.766

cos 130º = –0.643

tan 130º = –1.192

•

x

y

•

θR = 50º

sin 50º = 0.766

cos 50º = 0.643

tan 50º = +1.192

•

50º


Module 2

Example 4Solve for θ if sin θ – 0.528 = 0. Express θ to the nearest whole degree.

0° ≤ θ < 360°

SolutionIf sin θ – 0.528 = 0

sin θ = 0.528 (positive)

Now using the “CAST” diagram below . . .

we see that sine is positive in both Quadrants I and II, so there are twosolutions; θ, in Quadrant I and θ2 in Quadrant II. Both θ1 and θ2 willhave the same reference angle which we will determine first.

Example 5Solve for θ if 12 cos θ + 5 = 2 cos θ, 0° ≤ θ < 360°

SolutionRearrange the equation first.

( )θ

=

= =

10 cos –5–5cos –0.500 negative10

x

y

•

θ2= 148º

x

y

•

32º = θ 1

( )–1

1 2

sin 0.528 [Calculator]

32º. This is the reference angle.32º and 180º 32º 148º

Rθ

θ θ

=

∴ = = − =

S A

T C


Module 2

If cosine is negative, then θ must terminate in QuadrantII or Quadrant III. (Using CAST we see that cosine ispositive in the other two Quadrants.)

Example 6

SolutionRearrange the equation first.

tan θ positive means that θ terminates in Quadrant I orQuadrant III. (Using CAST.)

( )θ

θ θ

• =

= = + =

–1

1 3

tan 2.000

63°63° and 180° 63° 243°

R

S A

T C

( )

θ

θθθ

−

− ===

3tan 4=1

23tan 4 2

3tan 6

tan 2.000 positive

θθ θ− ≤ <3tan 4Solve for : =1 0° 360°. Answer to nearest degree.

2

x

y

•

θ3 = 240º

x

y

•

θ2= 120º

( )–1

2 3

Now cos 0.500

60180 – 60 120 180 60 240

R

R

θθθ θ

=

• = °∴ = ° ° = ° = ° + ° = °

S A

T C


Module 2

[We ignore signs when determining thereference angle because reference anglesalways lie between 0° and 90°.]

Example 7

Solution

( )( )θ θ

θ θ

θ θ

θ θ

+ − =− + =

∴ = =

= =

2If 2cos 3cos 2 0

2cos 1 cos 2 0 [Factoring the Trinomial]

1cos or cos –2

2cos 0.500 or cos –2

θ θ θθ

+ − =≤ <

2Solve for if 2cos 3cos 2 00° 360°

x

y

•

θ3 = 243º

x

y

•

θ1 = 60º


Module 2

Case 1:

If cos θ = 0.500 (positive)

then θ terminates in Quadrant I or Quadrant IV.

Case 2:

cos θ = –2

This is because sin θ and cos θ range between –1 and 1.

So the solution set in Example 5 is {60°, 300°}.

Now it’s your turn to try a few.

( )–1cos 2 No solutionRθ• = −

x

y

•

θ4 = 300º

x

y

•

θ1 = 60º

( )θθ θ

• = =

∴ = = =

–1

1 4

cos 0.500 60°

60° or 360° – 60° 300°R

S A

T C


Module 2

Guided Practice

Solve for θ, 0° ≤ θ < 360°, in each of the following. Answer to thenearest degree, or state an exact solution where possible.

1. a) Find the reference angles for 116°, 265°, 289°, 323°.

b) Evaluate the three trigonometric ratios for the angles above.

c) Find a coterminal angle with the angles above.

2. 3 sin θ = –2

3. 5 tan θ + 4 = –2 tan θ

4. 3 sin θ + 1 = 0

5. 3 cos2θ + 14 cos θ = 5

6. 7 cos θ – 8 = 4

7.

8. 4 tan2x – 13 tan x = 12


22 sin 3 sinθ θ=


Module 2

Lesson 3Special Trigonometric Values

OutcomesUpon completing this lesson, you will be able to:• state the trigonometric ratios for any “Quadrantal” angle θ,

0° ≤ θ ≤ 360°• state the trigonometric ratios of any multiple of 30°, 45°, or 60°

without the use of a calculator or trigonometric table• solve linear and quadratic trigonometric equations with

solutions which are multiples of 30°, 45°, or 60° (theseequations will be solved without the use of any aids)

OverviewIn Lesson 2 you determined the trigonometric ratios for any θ,where 0° ≤ θ ≤ 360°. You also solved various trigonometricequations for θ, 0° ≤ θ < 360°, using a calculator and the CASTdiagram. In this lesson you will be solving the same type ofproblems except without the use of a calculator. This will requiresome recall of Lesson 1 and the establishment of some geometricfacts.

In the Lesson 1, Guided Practice 5–8, you determined the threeprimary trigonometric ratios for four very special angles which wecall Quadrantal. They are angles which terminate on thepositive or negative x or y-axis, thereby dividing the coordinateplane into the four quadrants. Let’s look at them once more sothat we can summarize the results.


Module 2

0°, 360°

90°

4sin 90° 140cos90° 044tan 90° undefined0

yrxryx

= = =

= = =

= =

The point (0,4) is on theterminal arm of the angleθ, which represents arotation of 90° from itsinitial arm. Therefore:

y

x

5� N(0,4)

θ=90º

r = 4{ 4

0sin 0° sin 360° 055cos0° cos360° 150tan 0° tan 360° 05

yrxryx

= = =

= = =

= = =

The point (5,0) is on theterminal arm of the angleθ, which represents arotation of either 0° or360° from its initial arm.Therefore:

y

x

4

–4

�

M(5,0)

θ = 360º

r = 5 }θ = 0º

or


Module 2

180°

270°

The following table summarizes the trigonometric values for theQuadrantal angles.

θ 0°/360° 90° 180° 270°

sin θ 0 1 0 –1

cos θ 1 0 –1 0

tan θ 0 ∞ 0 ∞

–2sin 270 12

0cos 270 12–2tan 270 undefined0

yrxryx

° = = = −

° = = =

° = =

The point (0,–2) is on theterminal arm of the angleθ, which represents arotation of 270° from itsinitial arm. Therefore:

y

x

–4

�K(0,–2)

θ=270º

r = 2}

° = = =

° = = = −

° = = =

0sin180 0

3–3

cos180 130

tan180 0–3

yrxryx

The point (–3,0) is on theterminal arm of the angleθ, which represents arotation of 180° from itsinitial arm. Therefore:

y

x3

�

L(–3,0)

θ=180º

r = 3 {


Module 2

Now for some geometric facts.

Fact 1: 45°- 45°- 90° Triangle

Since ∆ ABC is isosceles it follows that AB = AC. Let these two equallengths be x and use the Pythagorean Theorem to find the length ofside BC.

Therefore, the ratios of a 45°- 45°- 90° triangle are alwaysrespectively. This fact enables you to find the

trigonometric ratios of a 45° angle.

Therefore,

opposite side 1sin45hypotenuse 2

adjacent side 1cos45

hypotenuse 2

opposite side 1tan45 1adjacent side 1

° = =

° = =

° = = =

4545B C

A

11

2

: : 2, or 1:1: 2,x x x

BC AB AC

BC

BC

BC

b g b g b gb g b g b gb g b g

2 2 2

2 2 2

2 22

2

= +

= +

=

=

x x

x

x

4545B C

A


Module 2

You should memorize these ratios. When you are answering aquestion based on the 45°-45°-90° triangle, you may find thatsketching the above reference triangle, with its ratios, is useful.

Example 1Without the use of a calculator:

Solution

Fact 2: 30°-60°-90° TriangleBeginning with an equilateral ∆ ABC each side of length 2x, youform two 30°-60°-90° triangles by dropping a perpendicular, AD,from vertex A.

Triangles ABDand ACD arecongruent by ASA.Hence,

Note: (ASA: Angle-Side-Angle: the two triangles have equal 30°angles, equal 60° angles, and sides of equal length 2x which liebetween the two angle(s).

BD BC= =12

x.

60

A

30

DB

2x

30º

60º

C

2x

2x

( )θ θ

θ θ

θ θ

θ θθ

− = − = − − =

− =

∴ = =

= ° ° ° °

2

1 1 1 1a) 1 12 22 2

b) sin 2 sin 0

sin 1 2 sin 0 (factor)

1 Either sin 0 or sin2

sin is positive in both cases, so lies in Quadrants 1 and 2 only.0 , 180 , 45 , 135

( )2

a) simplify sin45 cos45 tan45

b) solve for , if 0 360 : sin 2 sinθ θ θ θ

° ° − °

°≤ ≤ ° =


Module 2

You use the Pythagorean Theorem to find the third side AD.

Therefore, the ratios of a 30°-60°-90° triangle are alwaysrespectively. This fact enables you to find

the trigonometric ratios of a 30° or a 60° angle.

You should memorize these values! When you are answering aquestion based on a 30° - 60° - 90° triangle, you may find thatsketching the above reference triangle, with its ratios, is useful.Observe the length of the longest side—it’s 2, not √3. Many studentsoverlook that and put √3 on the wrong side of the triangle.

Example 2Without the use of a calculator:

a) find the value of 2 cos 30° sin60°tan30°

b) solve for if 0° 360° : sin tan 3 sin 0

+

≤ < ⋅ − =x x x x x

opposite side 1 3sin30 sin60hypotenuse 2 2

adjacent side 3 1cos30 cos60hypotenuse 2 2

opposite side 1 3tan30 tan60 3adjacent side 13

° = = ° =

° = = ° =

° = = ° = =

60

A

30

DB1

2 3

1 3 2: : ,: 3:2 , orx x x

( ) ( ) ( )( ) ( ) ( )( )

2 2 2

2 2 2

2 2

AD AB BD

AD 2

AD 3

AD 3

x x

x

x

= −

= −

=

=


Module 2

Solution

The following activity should serve as a review of your previoustrigonometry experiences. The only new work is that you are to try todo this work without using a calculator.

Guided Practice1. Complete the following table with exact values. Do as much as

you can without a calculator. You will be required to be able tostate these values without a calculator, whenever the questionasks for exact values.

2. Determine the exact value for each expression.

( )( )x x 22 2 2

a) cos 45 sin 45 sin 60b) tan 30 tan 60 sin 90

c) sin 120 cos 120 Recall that sin sin .

° °− °° °+ °

°+ ° =

210° 225° 240° 270° 300° 315° 330° 360°

sin θ

cos θ

tan θ

0° 30° 45° 60° 90° 120° 135° 150° 180°

sin θ

cos θ

tan θ

( )

( )

3 3 1 1 2 3 1a) 2 32 2 2 23

b) sin tan 3 0

sin 0 or tan 3 0

sin 0 or tan 3

0 , 180 , 60 , 180 60

0 , 180 , 60 , 240

x x

x x

x xxx

+ + = + =

− =

= − =

= == ° ° ° + °= ° ° ° °


Module 2


3. Find the value(s) of θ, where 0° < θ < 360°, that satisfy thefollowing equations:


2 2

2

2

1a) sin b) tan 323 1c) cos d) cos and tan 0

2 23e) sin and cos < 0 f) 2 cos 2

2g) 2 sin sin 0 h) 4 cos 1

i) 2 tan 2 3 0 j) 2 sin sin 1

k) cos cos 0

θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θθ θ

= =

= = − >

= =

+ = =

+ = − =− =

2

2

2 2

tan 60d) cos 135

sin 60

e) sin 210 cos 210 tan 210

f) cos 150 sin 150 cos 300

g) sin 150 cos 60 cos 150 sin 60 sin 210

−

− ++ −


Module 2

Lesson 4

The Unit Circle

OutcomesUpon completing this lesson, you will be able to:• locate the position of the terminal point P(θ) on the unit circle for

special values of the real number θ• state a value of θ which would produce a given position of a special

circular point P(θ)

OverviewThis lesson is a lead-in to the study of circular functions. However,your ability to locate special points on the unit circle is extremelyimportant for the rest of this module.

Definition: A unit circle is the circle with centre at the origin andwith a one-unit radius.

The circumference of this circle is 2πr = 2π(1) = 2π.

We now set up a circular function game in which you move certaindistances around this circle governed by the following rules:1. The point with coordinates (1,0) is the starting point for all of

your trips around the circle. It corresponds to the initial armdefined in Lesson 1. The arc is then said to be in standardposition.

2. If the distance to be moved is positive you will move in acounterclockwise direction along the unit circle.

3. If the distance to be moved is negative you will move in aclockwise direction along the unit circle.

x

y

�

(1, 0)

+

–

4. The notation P(θ) will be used to indicate where your trip hasterminated after you have moved a distance of θ units. P(θ) willbe called the circular, or terminal, point.

We use θ to represent a distance here rather than an angle, but do notbe concerned. We will connect this distance to an angle soon!

The following diagram will be used to label some special points onthe unit circle so that you can try some trips:

Where is P(0)? This notation asks, “What is your terminal point ifyou do not move from the starting point?”

Answer S, since you always begin your trips from the point with coordinates(1,0).

Where is P(π)? This asks for the circular point of a counterclockwisetrip halfway around the circle.

Answer B

Note: Since the circumference of the unit circle is 2π it follows thatthe distance:

Therefore, the distance halfway around the circle must be π.

Many students feel that going once around the circle should be π.Well it is not! You may wish to remember the following fact:

halfway around the unit circle is π

once around the unit circle is 2π

x

y

A

S

C

B

Figure A


Module 2

Exercise 1Use Figure A (previous page) to fill in the table.

Solution

To make the game more interesting, four additional points areinserted at the midpoints of each quarter arc.

The midpoints subdivide a semi-circle of arc length π into four equal arcs, each of length

x

y

llll

llll

π4

.

x

y

B

A

S

C

KL� �

��NM

Figure B

θ –π

P(θ) B A C C A S A A S B S

π2

− π2

3

2

π− 32

π −2π 5

2

π − 72

π101π2π 100π

θ –π

P(θ)

π2

− π2

3

2

π− 32

π −2π 5

2

π − 72

π101π2π 100π


Module 2

Exercise 2Use Figure B to fill in the table.

Solution

To make the game still more interesting, we subdivide the semicircle into six arcs of equal length by trisecting each of thequarter circles.

2Each small arc has a length of . Since reduces to ,6 6 3

it follows that you will have multiples of and multiples of 6 3

as you travel around the circle.

π π π

π π

x

y

llllllll

ll ll

θ

P(θ) K L N M L N M K L N L

3

4

ππ4

− π4

− 34

π − 54

π 7

4

π 5

4

π − 74

π − 574

π11

4

π 83

4

π

θ

P(θ)

3

4

ππ4

− π4

− 34

π − 54

π 7

4

π 5

4

π − 74

π − 574

π11

4

π 83

4

π


Module 2

Exercise 3Use the diagram below to fill in the table. (Assume that all thelittle arcs are equal.)

Solution

θ

P(θ) A B J H G G J E K B K

π3

π6

− π3

− 23

π − 56

π 7

6

π 5

3

π − 76

π 7

3

π11

6

π−13

6

π

θ

P(θ)

π3

π6

− π3

− 23

π − 56

π 7

6

π 5

3

π − 76

π 7

3

π11

6

π−13

6

π

x

y

S

C B�

�

�

�

��

�

�

�

�

��

A

K

JIH

G

F

E

D


Module 2

Guided PracticeIf we include the midpoints and the trisection points on one unitcircle, the following diagram results. Use this diagram to answer thequestions in the assignment.

1. Find the letter corresponding to the following value of θ:

2 7a) b)

3 6

5c) 27 d)

4

e) f) 146

5g) h) 53

2

11 17i) j)

6 3

k) l)2

4 3m) n)

3 2

5 15o) p)

6 4

7q) r) 4

3

π π

ππ

π π

π π

π π

π π

π π

π π

π π

−

−

−

−

−

x

y

S

DC

�

�

�

�

��

�

�

�

��

�

A

Q

ML

JI

H

FE

�B

�G

�K

�N


Module 2

2. Find the letter(s) corresponding to the following value of θ:a) kπ if k is an even integer (e.g., 2π, –2π, etc. . . .)b) kπ if k is an odd integer (e.g., –1π, 1π, 3π, etc. . . .)

c) if k is an integer

3. Find the letter(s) corresponding to the following value of θ:

a) if k is an odd integer

b) if k is an even integer (e.g., , etc. . . .)

4. State the smallest positive value of θ which will terminate at thepoint with letter:a) B b) Hc) M d) Le) A f) Cg) S h) Fi) I j) D

5. State the smallest negative value of θ which will terminate atthe point with letter:

a) B b) Hc) M d) Le) A f) Kg) S h) Fi) I j) N


44π π=2

4 2π π=

kπ4

– 3(e.g., , , , etc....)4 4 4π π πkπ

4

– 3(e.g., , , , etc....)2 2 2π π πkπ

2


Module 2


Module 2

Lesson 5

The Circular Functions

OutcomesUpon completing this lesson, you will be able to:• state the coordinates of special values of the circular points P(θ)• define the trigonometric functions as circular functions• define the three reciprocal functions in terms of their reciprocals• solve circular function equations, which have real number

solutions

OverviewFor every arc length θ on the unit circle there is only one point, P(θ),determined by this arc. Hence, you have a function which mapsevery arc into a unique point. Now your emphasis will switch fromthe point to the coordinates of the point. We redefine P(θ) as theordered pair of numbers rather than just the location.

For example, P(0) = (1,0), our starting point; = (0,1), the

coordinates at the top of the unit circle; and so on. For each realnumber θ there corresponds an ordered pair of numbers (x, y), whichare the coordinates of P(θ).

The above two examples were rather easy because the circular pointterminated on the axis. What happens when the point is not on anaxis? We will fill in the coordinates of the special terminal points inthe first quadrant and assign the rest as an exercise.

P2π

Since M is the midpoint of the arc it follows

since the radius of the unit

circle is one unit.

Furthermore, you know from Lesson 2 of this module that

Additionally, ∆ MON is isosceles, so

If points H and K are the trisection points of arc AS, it follows that ∠ KOS = ∠ HOK = ∠ AOH = 30°. Therefore, ∠ HOS = 60°.

x

y

A

S

ll

ll

Oll

303030

H

K

( )o o

1 1Hence, P , or coincidentally4 2 2

P cos 45 , sin 45 .4

π

π

= =

x = 12

.

y =o 1 1sin 45 = . Therefore, .2 2

oppositesin MONhypotenuse OM 1

y y∠ = = =

∠ = =MON o o12

90 45d i .

x

y

M

A

Sx

y

ll

ll

NO


Module 2

Let’s go after the coordinates of K and H!

You can use the figure below to conclude that:

At this point, by using the symmetry of a circle, you are capable ofcompleting the coordinates of every special point on the unit circle.

x

y

A

SO 60

H

y

x

1 3P , or P (cos60°, sin60 )

3 2 2 3π π = = °

3 1Hence, P , or P (cos30°, sin30 ).

6 2 2 6π π = = °

circlesin KOS since the radius of the unit is 1.OK 1

1And, since KOS 30° it follows sin30 (from Lesson 2).2

3Similarly, cos 30°= . Therefore, cos30 .

OK 2

∠ = =

∠ = = ° =

= ° =

y y

y

x x

x

y

A

SO 30

K

x

y


Module 2


Module 2

Exercise 1Use the stated coordinates to complete the coordinates of every otherlabeled special circular point.

Solution

Note: You should know the coordinates of these points by heart, or atleast those in Quadrant I—then you can use that information alongwith the CAST diagram to quickly determine the other points.

A(0, 1)

S(1, 0)E(–1, 0)

L(0, –1)

H1

2

3

2,FHG

IKJ

M1

2

1

2,

FHG

IKJ

K3

2

1

2,

FHG

IKJ

B−FHG

IKJ

1

2

3

2,

C−FHG

IKJ

1

2

1

2,

D−FHG

IKJ

3

2

1

2,

F− −FHG

IKJ

3

2

1

2, Q

3

2

1

2,−F

HGIKJ

R1

2

1

2,−F

HGIKJ

N1

2

3

2,−F

HGIKJ

J− −FHG

IKJ

1

2

3

2,

G1

2

1

2−

−−F

HGIKJ,

x

y

�

�

��

�

�

�

�

�

�

�

�

�

�

�

�

x

y

S(1,0)

A�

�

�

�

��

�

�

�

��

�

Q

NL

GF

E

CB

�

�D

�J

�R

H1

2

32,F

HG

IKJ

M

12

12,F

HG

IKJ

K

32

12,F

HG

IKJ


As mentioned previously, we have a function which matches a realnumber θ with an ordered pair of numbers (x, y). Mathe-maticiansprefer to match each real number with only one real number. This isaccomplished by taking a different approach to the trigonometricratios, an approach which was suggested by realizing that thecoordinates of the points on the unit circle are generated bytrigonometric function values.

Alternate definitions of trigonometric ratios (These are alsocalled circular functions.)

If θ is an arc in standard position (this means on a circle of radius 1)and P(θ) = (x, y) is the terminal point for the real number θ, then:

and

Example 1Use the coordinates on the unit circle and the definitions of thecircular functions to find:

Solution

This is a good sequence of steps to reason through:1. Locate where the point terminates on the unit circle.

(K in this case)

2. Think of the coordinates of this point.

3. Since sin θ = y, select the second coordinate as your

answer: 12

3 1,

2 2

a)12

2

a) sin b) cos c) tan6 3

d) sin e) cos f) sin 2cos4 6 4 6

π ππ

π π π π

−

+

y xx

θθ θθ

= ≠ = sin

tan , where 0 Since tancos

cos1

x x xr

θ = = =sin1

y y yr

θ = = =

This is a good sequence of steps to reason through:1. Locate where the point terminates on the unit circle: E2. Think of the coordinates of this point: (–1,0) 3. Since cos θ = x, select the first coordinate as your answer.

This is a good sequence of steps to reason through:1. Locate where the point terminates on the unit circle.2. Think of the coordinates of this point.

3. tan θ =

1. arc terminates at M

2. coordinates of M

3. Since

1. arc terminates at K

2. coordinates of K

3. Since

22 1 3 1 2 3 2 3 1f) sin 2cos 2

4 6 2 2 2 22π π + + = + = + =

3cos2

3cos6 2

xθ

π

= =

=

3 1,2 2

6π

e) cos :

1sin2

1sin4 2

yθ

π

= =

=

1 1,2 2

d) sin4π

3sin 2= 3

1cos2

yx

θθ

−

= = −

c) − 3

b) −1


Module 2

Now we’ll go backwards.

Example 2Solve the following over the interval 0 ≤ θ < 2π. Give solutions asexact values.

Solution

( )( )

2

2

b) If 2sin – sin 1then 2sin – sin –1 02sin 1 sin –1 0

2sin 1 0 or sin –1 0–1sin or sin 12

–1sin . The -coordinate is equal to or 12

7 – 3 –1 11 3 –1P = F , and P = Q ,6 2 2 6 2 2

P2

θ θθ θ

θθ θ

θ θ

θ

π π

π

==

+ =+ = =

= =

=

y y

A(0,1)

7 11: , ,6 6 2π π π

=

∴ Solutions

– 3a) cos2

– 3cos . The -coordinate is equal to 2

– 3 1 – 3 –1in two locations: D , and F ,2 2 2 2

– 3 1 5 – 3 –1 7D , P and F , P2 2 6 2 2 6

5 7: ,6 6

x x

Solutions

θ

θ

π π

π π

=

=

= =

∴

θ

θ θ

=

=2

– 3a) cos

2b) 2sin – sin 1


Module 2

There are three more circular functions defined as reciprocals of the basic three. They are called the secant (written as sec), cosecant (written as csc), and cotangent (written as cot) functions.

Definition:If θ is an arc length in standard position, with terminal circular point P(θ), then:

provided the denominators are not equal to zero.

Example 3Determine the exact value of

Solution

( )

22

2

1 1 2 2 4sec sec6 6 33 3 3cos

6 21 1cot 1 3cot 3 1 3

4 1 4tan4

4 4 9 –5sec – 3cot – 3 –6 4 3 3 3 3

π ππ

π ππ

π π

= = = → = =

= = = → = =

∴ = = =

π π2sec – 3cot6 4

1seccos

1cscsin

cos 1cot or sin tan

θθ

θθ

θθθ θ

=

=

=


Module 2

Example 4Solve for θ (exactly) for 0 ≤ θ < 2π:

Solution

The Link Between Angles in S.P. and Coordinates on Unit CircleThe following diagram shows the link between the angle in standardposition (or the length) in the unit circle and the coordinates of thecorresponding point on the unit circle.

Given θ in standard position,

and because r = 1 in the unit circle, it is simple to use thisdiagram to find exact trig. function values of any angle whichterminates on any of the points shown.

Examples

Also

π π

π π

= =

= =

= = =

2 33 2

2 3sec cot

3 21 1

cos tan

0–2 0

–1

r xx y

3 1sin4 2

11 3cos6 24 – 3 –1tan

2 233

yr

xryx

π

π

π

= =

= =

= =

=

θ θ θ= = =sin , cos , tany x yr r x

( )2If csc csc 0

csc csc 1 0csc 0 or csc 1 0

1sin csc –10

1(no solution) sin –1–1

3 (0, –1)2

32

y

P L

θ θθ θθ θ

θ θ

θ

π

πθ

+ =+ =

∴ = + =

= =

= = =

=

∴ =

θ θ+ =2csc csc 0


Module 2


For angles outside the interval (0, 2π), we can still locate thecorresponding point on the unit circle.

Example:Find the exact value of

Solution

is the arc of the unit circle obtained by starting at the

positive x-axis and travelling CLOCKWISE round the unit

circle for a distance of

This is the same point on the unit circle wewould have obtained by travelling COUNTERCLOCKWISE from the positve x-axis for the rest of the circumference of the circle, i.e.,

Thus π

ππ

π

=

=

–5csc6

5csc 2 –6

7csc6

ππ

π π

π

=

=

52 –

612 5

–6 6

76

�

–5π6

π–5.

6

π–56

–5csc .

6π

(0, 1)

(1, 0)(–1, 0)

(0, –1)

1

2

3

2,FHG

IKJ

1

2

1

2,

FHG

IKJ

3

2

1

2,

FHG

IKJ

−FHG

IKJ

1

2

3

2,

−FHG

IKJ

1

2

1

2,

−FHG

IKJ

3

2

1

2,

− −FHG

IKJ

3

2

1

2,

3

2

1

2,−F

HGIKJ

1

2

1

2,−F

HGIKJ

1

2

3

2,−F

HGIKJ

− −FHG

IKJ

1

2

3

2,

1

2

1

2−

−−F

HGIKJ,

x

y

�

�

��

�

�

�

�

�

�

�

�

�

�

�

�

π2

π

π

π

π

π

π

π

π

π

π

π

π

π ππ

3

4

6

(2

11

7

5

3

4

5

7

5

3

2

0)

6

4

3

2

3

4

6

6

4

3


Module 2

Guided PracticeThis activity is almost identical to the assignment at the end of theprevious two lessons except you will be using arc lengths instead ofangle measures.

1. Find sin θ, cos θ, and tan θ for each of the following values of θ. Ineach question, take the extra step to write the coordinates of P(θ).

2. Determine the exact value of each expression.Note: Although each expression involves functions and shouldcontain notation such as sin (π), mathematicians always likeshortcuts and as long as there is no ambiguity, they write thesecircular function statements as sin π, etc. This convention is usedthroughout this course.

( )

2

a) 2 cos tan3 4

15 13b) sin cos 20 tan2 6

c) tan6

d) cos sin

2 5 3 5e) tan cos sin tan3 4 2 6

f) cos2 6

π π

π ππ

π

π

π π π π

π π

+

−

− + −

π π π

π π π

π ππ

π π π

π π π

π π π

−

− −

−

−

2 7a) b) c) 273 6

5d) e) f) 144 65 11g) h) 53 i)2 6

17j) k) l)3 2

4 3 5m) n) o)3 2 6

15 7p) q) r) 44 3

3. Solve the following equations over the interval 0 ≤ θ < 2π. Dothese questions without the use of a calculator. All answers mustbe exact special values.

4. Determine the exact value of each expression.

5. Find the exact solution of the following equations over theinterval 0 ≤ θ < 2π.


2

a) sec 2 b) csc 2

c) cot 1, and sin 0 d) cot cot sin 0

e) tan 1 0 f) csc sin 0.

θ θ

θ θ θ θ θ

θ θ θ

= =

= − < − =

− = − =

2 2

2

2 2

a) sec sin cot b) sec csc3 6 4 6 6

5 2 2c) cot d) tan cot

6 3 3

7e) cos sec f) 4 csc

4 4 6

π π π π π

π π π

π π π

− +

−

2 2

2

2

1a) sin b) tan 32

3 1c) cos d) cos and tan 02 2

3e) sin and cos < 0 f) 2 cos 22

g) 2 sin sin 0 h) 4 cos 1

i) 2 tan 2 3 0 j) 2 sin sin 1

k) cos cos 0

θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ

= =

= = − >

= =

+ = =

+ = − =

− =

( )

2 7 3g) sin cos tan3 6 4

47h) sin cos 472

π π π

ππ

−

− −


Module 2

Lesson 6

Radian Measures

OutcomesUpon completing this lesson, you will be able to:• convert an angle from degrees to radians and vice versa• convert an angle from degrees to revolutions and vice versa• convert an angle from revolutions to radians and vice versa• solve angle measure problems, using any of the three units of

measure• relate arc length and central angles using the formula s = θr

OverviewYou should have noticed in the last lesson that a relationship seemsto exist between the arcs on the unit circle and the trigonometricvalues of the central angles (angles with vertices at the centre). Youwill develop this relationship in this lesson.

You are familiar with degree measure where:

A second method of measuring angles is in terms of radians.Definition: When the arc of a circle has the same length as theradius of the circle, then the measure of the angle, with vertex at thecentre, subtended by this arc is 1 radian.

Circle centre O

∠ AOB = 1 radian

x

y

�

(1, 0)

+

–

one revolution = 360°(once around the circle)


Module 2

Thus, for an arc of length s units, the radian measure of thecorresponding central angle is given by the number of radii needed tomake up the length of the arc s.

For example, an arc oflength 10 cm in a circle ofradius 4 cm has a centralangle whose measure isgiven by

Exercise 1Use the definition of radian measure to fill in the following table.

Mathematical agreement: If no symbol appears after an anglemeasure, the measure of the angle is in radians. To remind peoplethat the measure is in radians you can also use a superscript r todenote radian measure. Hence, the following are all equivalent:

x = x radians = xr

Arc Length Radius Central Angle (Radians)

2 2

6 3

20 5

2 4

16 4

=

=

=

104

2.5 radians

sO

r

= radianss

Or

x

y

llll

llll


Module 2


Module 2

Solution

The relationship among these three quantities is apparent. Thecentral angle is equal to the ratio of the arc length to the radius.Stated in a product form:

arc length = central angle in radian measure × radius

Many physics books use the symbol s for distance and θ for a angle,so the following formula for this relationship has endured:

s = θr, where s is the arc lengthθ is the measure of the central angle in radiansr is radius

Example 1a) Find the length of the arc subtending by a central angle of 3 if the

radius of the circle is 20 cm.

b) Find the length of the radius of a circle if a two-radian centralangle is subtended by a 15 m arc.

Solutiona) Since s = θr, it follows that s = 3(20) = 60 cm.

b) Since s = θr, it follows that 15 = 2(r) so that r = 7.5 m.

Furthermore, since the circumference of a circle is 2πr, it followsthat:

Using the formula

22

So one revolution is 2 radians.

s rr r

θπ θπ θ

π

===

Arc Length Radius Central Angle (Radians)

2 2 1

6 3 2

20 5 4

2 4 0.5

16 4 4

Exercise 2a) Convert 1° into radians.

b) Convert 1 radian into degrees.

This exercise provides you with a method to easily convert degrees toradians and radians to degrees.

oSince 360 2 radians, by dividing by 2 ,360 1 radian2

180 1 radian, so 1 radian is about 57.3 degr ees.

180Since 1 , to convert radians to degrees

180multiply by .

π π

π

π

π

π

=° =

° = ° =

x

x

o

o

Since 360 = 2 radians, by dividing by 36 0,360 2360 360 180

1 radians (about 0.0175 radians if you do the division)180

Since 1 radians, to convert degrees to radians180

multiply by .180

ππ π

π

π

π

°° = =

=

=

Method

x

x

One revolution = 2πr and since

one revolution = 360° it follows that

360° = 2π radians.


Module 2

Example 2a) Convert 45° into radians.

b) Convert π radians into degrees.

Solution

You are now in a position to link trigonometric ratios and circularfunction values.

Let’s look at the circular function values for central angles given inradians.

Example 3Determine the value (exact if possible, otherwise to three decimalplaces) of:

Solution

Example 4Find the exact value, if possible; otherwise use a calculator to findthe approximate value.

Solutiona) π is a special value so the memorized answer is cos π = –1.

oa) cos b) sin 20

3c) tan d) tan 14

π

π

°

° °

(45° is one of our 180 1a) sin sin sin45° 4 4 "special" angles)2

180 540b) cos3 cos3 cos –0.990 (calculator)

r r

r

π ππ

π π

= = =

= = ≈

radiansa) sin (Small superscript "r" means " .")4

b) cos3

r

r

π

π π

π ππ

= = = =

o

o

a) 45 45 radians180 4

180b) 180


Module 2

b) Since 20° is not a special value, use a calculator, in degree mode,to find sin 20° = 0.34202.

c) Special value, hence

d) Not a special value. Use the calculator, in radian mode, to findtan 1 = 1.55741.

Note: it was not 1º so it must mean 1 radian.

Unless stated otherwise, you should always attempt to find thetrigonometric, or circular function, value without a calculator. If themeasure is not a special value, or special value multiple, use acalculator—use radian mode for arcs or angles in radian measure, anddegree mode for angles stated in degree measure. To check if yourgraphing calculator is in radian mode or in degree mode, press MODE.To change the mode, move the cursor to the preferred mode (Radian orDegree) and press ENTER.

Example 5Convert each angle measure into revolutions.

Solution

When we solve an equation such as

we get not two but four solutions. Here’s why.

3cos2 for 0 2 ,

2θ θ π= ≤ <

a) If 1 revolution = 2 r

5and revolutions

61 2 5 5 1 5

then 2 revolutions6 6 2 125

6b) If 1 revolution = 360º

and revolutions 210º1 360 12 7

then 12 7 revolutions210 7 12

π

π

π π ππππ

=

= → = → = × =

=

= = → = → =

r

r

x

x xx

x

x xx

rπ o5a) b) 210

6

tan .34

1π = −


Module 2

Example 6

Using the CAST diagram, 2θ is positive in Quadrants I and IV.

When we divide by 2, we reduce the domain of the solutions from[0,2p) to [0,p). In order to obtain all solutions for 0 £ q < 2p, we needto travel around the unit circle twice.

We now have four solutions in the domain [0,2π).

In summary:

1. The equation would also have four solutions

over the domain [0,2π).

2. The equation would have six solutions over

the domain [0,2π).

3. If we solve trig equations in degrees instead of radians, the

equation would have four solutions over the

domain [0,360°).

3sin2

2θ =

3sin3

2θ =

3sin2

2θ =

11Instead of 2 or 6 6

11 11we need 2 or or 2 + or 2 +6 6 6 6

11 13 23, , , or

6 6 6 611 13 23Dividing by 2, , , or

12 12 12 12

π πθ

π π π πθ π π

π π π π

π π π πθ

=

=

=

=

2 or 2 2 –6 6

112 or 26 6

11or 12 12

π πθ θ π

π πθ θ

π πθ θ

= =

= =

= =

Solve cos2 for 0 26πθ θ π= ≤ <


Module 2

Guided Practice1. Convert each degree measure into radian measure.

a) 25° b) –125°c) 460° d) 330°

2. Convert each radian measure into degree measure. Round off toone decimal place where necessary.

3. Determine the complement of What is the supplement

of this angle? (Recall: Two angles are complementary if the sumof their measures is 90°. Two angles are supplementary if thesum of their measures is 180°.)

4. Express the supplement of 130° in radian measure, exactly.

5. An isosceles triangle has a base angle with measure

What are the measures of the other two angles? Express youranswer in radians.

6. Two parallel lines have interior alternate angles with measures24° and Find the value of x.

7. In what quadrant(s) are the following conditions satisfied?

a) sin θ < 0 and cos θ > 0b) tan θ > 0c) sin θ > 0 and cos θ < 0d) cos θ < 0

24º

πx

πx

.

27π

.

512π

.

a) b)

c) d)

−

−

76

1112

2 634 0 9825

π π

. .


Module 2


Module 2

Remember the CAST diagram.

8. An alien being, walking along the equator of her planet, travelleda distance of 12.3 clicks. If the diameter of this planet is 16.4clicks, through what central angle did this alien travel? Expressyour answer to the nearest tenth of a degree.

9. In the diagram AC = 6 cm, BC = 8, and AB is a diameter. Findthe circumference of this circle.

10. If a wheel having a circumference of 30 cm rolls 5 cm, how manyradians has it turned? How many degrees has it turned?

11. Convert each measure into revolutions:a) 3 radiansb) 75°

12. Use a calculator, using five digits after the decimal, in order tofind each value:

a) b)c) d)e) f)

g) h)

i) j)

k) l)

m) n)

o)

tan cossin . sec .cot sin

cos tan

csc sin

cot . sin

cos csc

tan

20 202 4 153 257

52

0 5 3

1 4

2

°

°

°

−

− °

−

ππ

πb g

b gb g

y

B

A

S

C

KL� �

��NM

S A

T C

13. Use a calculator to solve the following simple equations for x, 0 ≤x < 2π. Round your answer to two digits after the decimal. (Hint:Use sin–1 function on your calculator. Since x is in radians, haveyour calculator set to radian mode. Be sure to find all angles, notjust those in Quadrant I.)

The following questions are very important. Attempt to do as muchas you can without using a calculator. If you must use a calculator,be sure your setting is to the correct mode.

14. Solve each equation for x, where 0° ≤ x < 360°.

15. Solve each equation for x, where 0 ≤ x < 2π.


Review Section 1 before attempting the review questions followingthe summary on the next page. These questions should help youconsolidate your knowledge as you prepare Section Assignment 2.1.

( )

2a) cos 2 cos 0 b) tan 2 1

c) csc sec d) 2 sin 1 0

e) cos 0 f) sin tan sin 02

x x x

x x x

x x x x

π

π

− = =

= − − =

− = + =

2

2

a) 2 sin sin b) sin cos

c) sin 2 1 d) 2 cos 2 1

e) 2 csc 4 f) 2 tan 1 tan

x x x x

x x

x x x

= =

= =

= − =

a) b)c) d)e) f)g) h)

sin . cos .tan sin .cos . sec .cot csc

x xx xx xx x

= = −= = −= == − =

0123 01232 0 250 675 2 54 6


Module 2

Module 2

Summary of Section 1

The BasicsThe initial arm is the positive direction of the x-axis. A positive angleis made by moving in a counter-clockwise direction from the initialarm.

A reference angle is always made with the x-axis.

CAST DiagramThis diagram indicates the quadrants where thebasic trigonometric ratios are positive.

Exact ValuesSpecial trigonometric values can be calculated for Quadrant I andfrom there the values can be determined for the other quadrants.You should learn this table or learn how to derive these values fromtriangles (see Lesson 3, Facts 1 and 2).

The following table summarizes the trigonometric values for theQuadrantal angles:

undefined

Degrees 0 30 45 60 90

Arc Length 06 4 3 21 1 3

sin 0 12 22

3 1 1cos 1 0

2 221

tan 0 1 33

π π π π

θ

θ

θ

° ° ° ° °

S A

T C

y

x

x

y

referenceangle

x

y

θ

referenceangle

Principles of Mathematics 12 Section 1, Summary 55

Unit CircleDefinitions

For any point P(x,y) on the unit circle(radius = 1) with an arc length of s instandard position.

Solving EquationsThe equations y = sin θ, y = cos θ, y = tan θ have two solutions for thedomain 0 < θ < 2π, or the domain 0° ≤ θ < 360°.

The equations y = sin nθ, y = cos nθ, y = tan nθ, where n is aninteger, have 2n solutions for the domain 0° ≤ θ < 360° or the domain0° ≤ θ < 2π.

To solve the equations y = sec θ, y = csc θ, y = cot θ, use the

reciprocals:

Some more information can be found in Michelson’s book, Theoryand Problems for Senior High Math, on pages 47–49, #1–7, andfurther problems #1–5 on pages 52–53.

1 1 1cos , sin , tany y y

θ θ θ= = =

1 1sin cscsin

1 1cos seccos

1tan cottan

yy

xx

y xx y

θ θθ

θ θθ

θ θθ

= = =

= = =

= = =

x

y

θ

s

(x,y)P1

Module 2

56 Section 1, Summary Principles of Mathematics 12

Module 2, Section 1Review

1. Find the exact value for each expression.

2. Solve the following equations over the interval 0 ≤ θ < 2π. Dothese questions without the use of a calculator. All answers mustbe exact special values.

3. Convert each degree measure into radian measure.

a) 45° b) –150°

c) 500° d) 600°

2 2

2

2

1a) sin b) tan 1

2

1 1c) cos d) sin and tan 0

2 2

3e) cos and csc < 0 f) 2 cos 3

2

g) 2 cos cos 0 h) 4 cos 1

i) 2 cot 2 3 0 j) 2 cos cos 1

k) tan tan 0

θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ

= − = −

= = − >

= =

+ = =

+ = + =

− =

o o

o 2 2

2 2 o

tan4a) 12 sin30 cos 60 b)5sin6

3 3c) csc 210 2 cot d) sin cos4 2

5e) 2 sec cot 30 f) cos 1235 sin 2554

π

π

π π π

π π π

− − − −

− +

Principles of Mathematics 12 Section 1, Review 57

Module 2

4. Convert each radian measure into degree measure. Round off toone decimal place where necessary.

5. Solve each equation for x, where 0° ≤ x < 360°, or for θ, where0 ≤ θ < 2π.


Now do the Section Assignment which follows this section. When it iscomplete, send it in for marking.

( )

2

2

a) 2 sin sin b) sin cos

c) cos 2 1 d) 2 sin 2 1

e) 2 csc 4 f) 2 tan 1 tan

g) 2 cos 1 0 h) cos 02

i) cos tan cos 0

x x

x

x x

x x x

θ θ

θ

θ

πθ θ π

= =

=− =

=− − =−

− − = + =

+ =

a) b)

c) d)

56

1112

3 8 0 5432

π π−

−. .

58 Section 1, Review Principles of Mathematics 12

Module 2

Student's StudentName ____________________________ No. _______________________

Address ____________________________ School _______________________

____________________________Student's

____________________________ FAX No. _______________________

Instructor'sName ____________________________Date Sent _______________________

PRINCIPLES OF MATHEMATICS 12

Section Assignment 2.1

Your instructor will fill in this box with your percentage mark and lettergrade on the Section Assignment:

Comments/Questions:

Principles of Mathematics 12 Section Assignment 2.1 59

Version 05 Module 2


Address ____________________________ School _______________________

____________________________Student's

____________________________ FAX No. _______________________





Comments/Questions:


Version 05 Module 2


Address ____________________________ School _______________________

____________________________Student's

____________________________ FAX No. _______________________





Comments/Questions:


Version 05 Module 2

General Instructions for Section Assignments

These instructions apply to all the section assignments, but will notbe reprinted each time. Remember them for future sections.

(1) Treat this assignment as a test, so do not refer to your moduleor notes or other materials. A graphing calculator is required.

(2) Where questions require computations or have several steps,showing these can result in part marks for some exercises.Steps must be neat and well-organized, however, or theinstructor will only consider the answer.

(3) Always read the question carefully to ensure you answer whatis asked. Often unnecessary work is done because a questionhas not been read correctly.

(4) Always clearly underline your final answer so that it is notconfused with your work.

60 Section Assignment 2.1 Principles of Mathematics 12

Module 2


Trigonometric Functions

1. Fill in the blanks with the correct response.

a) If sin θ < 0 and sec θ > 0, then θ terminates in Quadrant: _____

b) The exact value of sec is: _____

c) A 2π cm wire is stretched along a circle of radius 4 cm. The degree measure of the central angle subtended by this wire arc is: _____

d) Find the smallest possible angle in standardposition of 4400°. _____

e) What is the reference angle for 4400°? _____

f) What is the reference angle for ? _____

g) If 0 < θ < π and cos θ = 0.5, then the value of cos (2θ) is: _____

h) The value of in degree measure, is: _____

i) The value of is: _____

j) The exact value of is: _____

k) What is a coterminal angle with 170°? _____

cot56πFHGIKJ

16 3cosπ

cos ,− −FHGIKJ

1 32

275π

76π


Module 2

Total Value: 48 marks(Mark values in margins)

(11 marks)

2. Find the value of

3. a) Evaluate tan 7.5.

b) Evaluate exactly

4. As the time changes from 4:10 p.m. to 6:45 p.m. on theanalog clock, determine the change in radian measure of theminute hand.

45cos – .

4π

sin cos .3 2

22

4−FHGIKJ +

π π


Module 2

(1)

(1)

(4)

(2)

5. Solve each equation for x, where 0 ≤ x < 2π.

a) 2 sin x + 1 = 0

b) cos x(2 sin x – 1) = 0

c) 2 cos 2x = 1

d) tan2x = 1


Module 2

(4)

(5)

(4)

(4)

6. Solve each equation or inequality for x:

b) Solve for 0° ≤ x < 360°3 tan 2x = 1

c) 2 sec x = 3, where 0° ≤ x < 360°

Send in this work as soon as you complete this section.

1a) sin where 0 2

2 2x x

π π + = ≤ <


Module 2

(4)

(4)

(4)

Total: 48 marks

Module 2: Trigonometry - Open School...

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