CRASH COURSE IN PRECALCULUS

Shiah-Sen Wang

The graphs are prepared by Chien-Lun Lai

Based on : Precalculus: Mathematics for Calculusby J. Stuwart, L. Redin & S. Watson,6th edition, 2012, Brooks/ColeChapter 5-7.

LECTURE 6. TRIGONOMETRY: PART I

This lecture is the first part of reviewing high schooltrigonometry: Pythagorean theorem, the definitions oftrigonometric and inverse trigonometric functions and some oftheir properties.

Trigonometry of Right TriangleAngle Measure: If a circle of radius 1 (so its circumference is2π) with vertex of an angle at its center, then the measure ofthis angle in radians (abbreviated rad) is the length of the arcthat subtends the angle. See

where the second graph shows that the length s of an arc thatsubtends a central angle of θ radians in a circle of radius r is

s = rθ

Trigonometry of Right TriangleRELATIONSHIP BETWEEN DEGREES AND RADIANS

180○ = π rad, 1 rad = (180π

)

○

1○ =π

180rad

Examples

1. Express 30○ in radians. 30○ = 30(π

180) rad =

π

6rad.

2. Express 45○ in radians. 45○ = 45(π

180) rad =

π

4rad.

3. Expressπ

3rad in degrees.

π

3rad = (

π

3)(

180π

) = 60○.

A note on terminology: We often use a phrase such as “a 30○

degree angle” to mean an angle whose measure is 30○. Alsofor an angle θ, we write θ = 30○ or θ = π/6 to mean the measureof θ is 30○ or π/6rad. When no unit is given, the angle isassume to be in radians.

Area of a Circular SectorIn a disc of radius r , the area of a sector with central angle of θradians

isA = πr2

×θ

2π=

12πr2.

Trigonometry of Right TrianglePythagorean Theorem: In any right triangle, the square of thelength of the hypotenuse equals the sum of the squares of thelengths of the two other sides. If the hypotenuse has length c,and the legs have lengths a and b, then

c2= a2

+ b2.

Trigonometric Function: Right Triangle Approach

Trigonometric Ratios for 0 < θ <π

2

sin θ =bc

cos θ =ac

tan θ =ba

csc θ =cb

sec θ =ca

cot θ =ab

Examples1. Suppose 0 < θ <

π

2and cos θ =

13

. Find the values ofsin θ, tan θ, cot θ,sec θ and csc θ.By

we see that the length of the opposite side is¿ÁÁÀ1 − (

13)

2=

2√

23

= sin θ, tan θ =√

24,cot θ = 2

√2,

csc θ =3√

24

and sec θ = 3

Examples

2. Suppose 0 < θ <π

2and tan θ = 2. Find the values of

sin θ,cos θ, cot θ,sec θ and csc θ.Use

We see the length of the hypotenuse side is√

5. Hence

sin θ =2

√5,cos θ =

1√

5,cot θ =

12,sec θ =

√5 and csc θ =

√5

2

Trigonometric Function: Unit Circle Approach

The unit circle is the circle of radius 1 centered at the origin inthe xy -plane. Its equation is

x2+ y2

= 1.

Suppose θ is a real number. Mark off a distance θ along theunit circle, starting at the point (1,0) and moving in acounterclockwise direction if θ > 0 or in a clockwise direction ifθ < 0. In this way we arrive at a point P(x ,y) on the unit circle.The point P(x ,y) obtained in this way is called the terminalpoint determined by the real number θ.

Trigonometric Function: Unit Circle Approach

,

Trigonometric Function: Unit Circle Approach

Trigonometric Functions for θ ∈ R

sin θ = y cos θ = x tan θ =yx(x ≠ 0)

csc θ =1y(y ≠ 0) sec θ =

1x(x ≠ 0) cot θ =

xy(y ≠ 0)

Note that, when 0 < θ <π

2,a = x ,b = y and c = 1, the unit circle

approach is identical with the right triangle approach in thedefining these trigonometric functions, but the domains of thesetrigonometric functions are larger using the unit circle approach.

Trigonometric Function: Special Triangles

Special Triangles

θ in ○ θ in rad sin θ cos θ tan θ csc θ sec θ cot θ0○ 0 0 1 0 − 1 −

30○π

612

√3

2

√3

32

2√

33

√3

45○π

4

√2

2

√2

21

√2

√2 1

60○π

3

√3

212

√3

2√

33

2√

33

90○π

21 0 − 1 − 0

Trigonometric Function: Special Triangles

Fundamental Identities of Trigonometric Function

Reciprocal Identities

csc θ =1

sin θsec θ =

1cos θ

cot θ =1

tan θ

tan θ =sin θcos θ

cot θ =cos θsin θ

Pythagorean Identities

sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ 1 + cot2 θ = csc2 θ

Fundamental Identities of Trigonometric Function

Even-Odd Identities

sin(−θ) = −sin θ cos(−θ) = cos θ tan(−θ) = − tan θ

csc(−θ) = −csc θ sec(−θ) = sec θ cot(−θ) = −cot θ

Fundamental Identities of Trigonometric Function

Cofunction Identities

sin(π

2− θ) = cos θ tan(

π

2− θ) = cot θ sec(

π

2− θ) = csc θ

cos(π

2− θ) = sin θ cot(

π

2− θ) = tan θ csc(

π

2− θ) = secθ

Examples on Applications of Trigonometric Identities

1. Simplify tan θ +cos θ

1 + sin θ.

tan θ +cos θ

1 + sin θ=

sin θcos θ

+cos θ

1 + sin θ=

sin θ(1 + sin θ) + cos2 θ

cos θ(1 + sin θ)

=sin θ + sin2 θ + cos2 θ

cos θ(1 + sin θ)=

1 + sin θcos θ(1 + sin θ)

=1

cos θ= sec θ

2. Prove1 − sin θ1 + sin θ

= (sec θ − tan θ)2.

1 − sin θ1 + sin θ

=1 − sin θ1 + sin θ

⋅1 − sin θ1 − sin θ

=(1 − sin θ)2

1 − sin2 θ= (

1 − sin θcos θ

)

2

= (1

cos θ−

sin θcos θ

)

2= (sec θ − tan θ)2.

Trigonometric GraphsA function f is periodic if there is a positive p such thatf (θ + p) = f (θ) for every θ. The least such positive number is theperiod of f . If f has period p, then the graph of f on any intervalof length p is called one complete period of f .

Periodic Properties of Sine and CosineThe functions sine and cosine have period 2π:

sin(θ + 2π) = sin θ cos(θ + 2π) = cos θ

Trigonometric Graphs

Periodic Properties of tangent and CotangentThe functions tangent and cotangent have period π:

tan(θ + π) = tan θ cot(θ + π) = cot θ

Trigonometric Graphs

Periodic Properties of Secant and CosecantThe functions secant and cosecant have period 2π:

sec(θ + 2π) = sec θ csc(θ + 2π) = csc θ

Examples

1. sin2π3

= sin(π

2+π

6) = cos(−

π

6) = cos

π

6=

√3

2.

2. cosπ = cos(π

2+π

2) = sin(−

π

2) = −sin

π

2= −1.

3. cot5π4

= cot [π

2+ (

π

2+π

4)] = tan [−(

π

2+π

4)]

= − tan (π2 +

π4 ) = −cot (−π

4 ) = cot π4 = 1.

4. tan3π2

= tan(2π −π

2) = tan(−

π

2) = − tan

π

2= 0.

5. sec5π6

= sec(2π −π

6) = sec(−

π

6) = sec

π

6=

2√

33

.

6. csc31π

6= csc(5π +

π

6) = csc(π +

π

6) = sec [−(

π

2+π

6)]

= sec (π2 +

π6 ) = csc (−π

6 ) = −csc π6 = − 1

sin π6= − 1

12= −2.

Domains and Ranges of Trigonometric Functions

sin ∶ RÐ→ [−1,1]

cos ∶ RÐ→ [−1,1]

tan ∶ R ∖ {kπ +π

2∣ k ∈ Z}Ð→ R

cot ∶ R ∖ {kπ ∣ k ∈ Z}Ð→ R

sec ∶ R ∖ {kπ ∣ k ∈ Z}Ð→ (−∞,−1] ∪ [1,∞)

csc ∶ R ∖ {kπ +π

2∣ k ∈ Z}Ð→ (−∞,−1] ∪ [1,∞)

Signs of Trigonometric Functions

Quadrant Positive Functions Negative FunctionsI all noneII sin,csc cos,sec, tan,cotIII tan,cot sin,csc,cos,secIV cos,sec sin,csc, tan,cot

Inverse Trigonometric Functions● The Inverse Sine Function is the function

sin−1∶ [−1,1]Ð→ [−

π

2,π

2]

defined bysin−1 x = y ⇐⇒ sin y = x

The inverse sine function is also called arcsine, denotedby arcsin.

Inverse Trigonometric Functions

Examples

i sin−1 12=π

6.

ii arcsin(−

√3

2) = −

π

3.

iii sin−1 2 =? Because 2 > 1, 2 is not in the domain of arcsin,so sin−1 2 is not defined.

iv arcsin(sinπ

4) =

π

4.

v sin−1(sin

7π6

) = sin−1(−

12) = −

π

6.

Inverse Trigonometric Functions● The Inverse Cosine Function is the function

cos−1∶ [−1,1]Ð→ [0, π]

defined bycos−1 x = y ⇐⇒ cos y = x

The inverse cosine function is also called arccosine,denoted by arccos.

Inverse Trigonometric Functions● The Inverse Tangent Function is the function

tan−1∶ RÐ→ (−

π

2,π

2)

defined bytan−1 x = y ⇐⇒ tan y = x

The inverse tangent function is also called arctangent,denoted by arctan.

Inverse Trigonometric Functions● The Inverse Cotangent Function is the function

cot−1∶ RÐ→ (0, π)

defined bycot−1 x = y ⇐⇒ cot y = x

The inverse cotangent function is also calledarccotangent, denoted by arccot.

Inverse Trigonometric Functions● The Inverse Secant Function is the function

sec−1∶ {x ∈ R ∣ ∣x ∣ ≥ 1}Ð→ [0,

π

2) ∪ (

π

2, π]

defined bysec−1 x = y ⇐⇒ sec y = x

The inverse cosecant function is also called arcsecant,denoted by arcsec.

Inverse Trigonometric Functions● The Inverse Cosecant Function is the function

csc−1∶ {x ∈ R ∣ ∣x ∣ ≥ 1}Ð→ [−

π

2,0) ∪ (0,

π

2]

defined bycsc−1 x = y ⇐⇒ csc y = x

The inverse cosecant function is also called arccosecant,denoted by arccsc.

Inverse Trigonometric Functions

Examples

1. sin−1 cos2π3

= sin−1 12=π

6.

2. cos−1 tan(−π

4) = cos−1

(−1) = π.

3. tan arcsin12= tan

π

6=

√3

3.

4. csc cos−1√

22

== cscπ

4=√

2.