University of Strathclyde Department of Mathematics & Statistics ...

University of Strathclyde Department of Mathematics & Statistics Circular Functions — Trigonometry Summer School Lecture Notes 1

Basic Definitions

OAB is a triangle with a right angle at A. b =θ AOB AB opposite = OB hypotenuse

B

OA adjacent = OB hypotenuse ( ) AB opposite sin θ tan θ := = = OA adjacent cos θ

use

n ote hyp

cos θ :=

cot θ :=

O

1 tan θ

REMINDER: SOH CAH TOA

1

θ adjacent

opposite

sin θ :=

A

2

Summer School: Circular Functions

2

Circular Functions Y circle, radius r

From the basic definitions, sin θ =

opp y = hyp r

B(x,y) ϕ

adj x cos θ = = hyp r ( ) opp y sin θ tan θ = = = adj x cos θ

r

y

θ x

O

A

x y , cos φ = so we have r r x sin(90◦ − θ) = sin φ = = cos θ r

Now φ = 90◦ − θ and sin φ =

cos(90◦ − θ) = cos φ =

y = sin θ. r

These are Complementary Angles. Also, tan(90◦ − θ) = tan φ =

3

x = cot θ y

Elementary Angles

When θ = 0◦ : y = 0, x = r sin(0◦ ) =

y = 0, r

cos(0◦ ) =

x y = 1, tan(0◦ ) = = 0 r x

and using complementary angles sin(90◦ ) = sin(90◦ − 0◦ ) = cos(0◦ ) = 1 cos(90◦ ) = cos(90◦ − 0◦ ) = sin(0◦ ) = 0 tan(90◦ ) = tan(90◦ − 0◦ ) =

1 = undefined 0

X

3


3.1

Isosceles Triangle B b + OBA b = 90◦ AOB

45

b = OBA b AOB

o

2a

a

a

A

b = OBA b = 45◦ ⇒ AOB OB 2 = OA2 + AB 2 = 2a2 √ ⇒ OB = 2 a

45

o

O

a 1 cos 45◦ = sin 45◦ = √ =√ 2a 2 tan 45◦ = 1

3.2

Equilateral Triangle

B

OB = BC = CO = 2l

o

30

OA = AC = l Then OB 2 = OA2 + AB 2 ⇒ 4l = l + AB ⇒ AB = 2

2

2

√

2l

3l

3l

b = OBC b = B CO b = 60◦ C OB o

b = ABC b = 30◦ OBA

60 O

l

Hence 1 = cos 60◦ 2 √ 3 = sin 60◦ cos 30◦ = 2 sin 30◦ =

1 tan 30◦ = √ 3 √ tan 60◦ = 3

A

C

4


3.3

Table of Values (IMPORTANT)

0◦ 30◦ 45◦ 60◦ 90◦ √ 1 1 3 √ sin θ 0 1 2 2 2 √ 3 1 1 √ cos θ 1 0 2 2 2 √ 1 tan θ 0 √ 1 3 − 3 θ

3.4 3.4.1

Extension to other angles 90◦ < θ ≤ 180◦

Now x < 0, y > 0 and r > 0.

Y

We define sin θ =

cos θ =

y >0 r

B

x 0

sin θ =

y 0 r

tan θ =

y 0 in the 1st and 2nd quadrants

sin

all

cos θ > 0 in the 1st and 4th quadrants

X tan

tan θ > 0 in the 1st and 3rd quadrants

3.6

cos

rd

th

3 Quad

4 Quad

Negative Angles

Angles are positive when measured anti-clockwise and negative

when

Y

measured

clockwise. y sin(−θ) = − = − sin θ r x = cos θ r y tan(−θ) = − = − tan θ x cos(−θ) =

r θ −θ

y X

x r

−y

7


4

Degrees and Radians

The angle θ in radians is θ=

length of arc AB r

B

(This does not depend on r)

r θ

The circumference is 2πr. Hence a

A

r

◦

complete rotation (360 ) is 2πr = 2π radians r

( ◦

2π radians ≡ 360 ⇒ 1 radian ≡

180 π

)◦

π radians ≡ 1◦ 180

4.1

Table of degrees and radians Degrees

0◦

30◦

45◦

60◦

90◦

120◦

135◦

150◦

Radians

0

π/6

π/4

π/3

π/2

2π/3 3π/4

5π/6

Degrees

180◦

210◦

225◦

240◦

270◦

300◦

330◦

Radians

π

315◦

7π/6 5π/4 4π/3 3π/2 5π/3 7π/4 11π/6

Now look back at the table of values and make sure you know exact values for radians!!! NOTE: Be careful to include

◦

when referring to degrees.

8


5

Pythagoras’ Theorem

Pythagoras’ Theorem says x2 + y 2 = r 2 Dividing by r2 , we get

r 2

y

2

x y + 2 =1 2 r r

θ

and using the definition of sin θ and cos θ:

x

cos2 θ + sin2 θ = 1

6

Graphs

The graph of sin θ has a maxi-

sin(θ)

mum value of 1 when θ takes the 1

values ···

−3π π 5π , , ··· 2 2 2

0.5

It takes the minimum value of −1 when θ takes the values ···

−π 3π , ,··· 2 2

0

−0.5

and is equal to 0 when θ takes −1

the values −6

0, ±π, ±2π, · · · The graph of sin θ has a period of 2π.

−4

−2

0 θ

2

4

6

9

Summer School: Circular Functions The graph of cos θ has a maxicos(θ)

mum value of 1 when θ takes the values

1

0, ±2π, ±4π · · · 0.5

It takes the minimum value of −1 when θ takes the values ±π, ±3π, · · ·

0

−0.5

and is equal to 0 when θ takes the values

−1 −6.2832

π 3π 5π ± ,± ,± ,··· 2 2 2

−3.1416

0 θ

3.1416

The graph of cos θ has a period of 2π.

Note the similarity between the graphs of sin θ and cos θ.

tan(θ) 6

The graph of tan θ takes the value 0 when θ takes the values 0, ±π, ±2π, ±3π · · · It is undefined when θ takes the values π 3π 5π ± ,± ,± ,··· 2 2 2 The graph of tan θ has a period of π.

4

2

0

−2

−4

−6 −4.7124

−1.5708

θ

1.5708

4.7124

6.2832

10


7

Compound Angles P α

L

α

N

Q

β α O

K

M

b is a right angle, so LN bQ = α Triangles OQK, P QN are similar, so QPbN = α. N LP PK P L + LK LK + P L NM + P L = = = PO PO PO PO N M ON PL PN = · + · ON P O P N P O = sin α cos β + cos α sin β

sin(α + β) =

Hence sin(α + β) = sin α cos β + cos α sin β It can also be shown that cos(α + β) = cos α cos β − sin α sin β sin(α − β) = sin α cos β − cos α sin β cos(α − β) = cos α cos β + sin α sin β tan(α + β) =

tan α + tan β 1 − tan α tan β

tan(α − β) =

tan α − tan β 1 + tan α tan β

Summer School: Circular Functions Example 7.1 Find the value of sin(75◦ ) sin(75◦ ) = sin(45◦ + 30◦ ) = sin 45◦ cos 30◦ + cos 45◦ sin 30◦ √ √ 1 1 1+ 3 1 3 = √ +√ = √ 2 2 22 2 2 Example 7.2 Find the value of cos(15◦ ) cos(15◦ ) = cos(45◦ − 30◦ ) = cos(45◦ ) cos(30◦ ) + sin(45◦ ) sin(30◦ ) √ √ 1 3 1 1 1+ 3 = √ +√ = √ 2 2 22 2 2 Example 7.3 Find the value of tan(105◦ ) tan(105◦ ) = tan(60◦ + 45◦ ) tan(60◦ ) + tan(45◦ ) = 1 − tan(60◦ ) tan(45◦ ) √ √ √ 3+1 3+1 1+ 3 √ = √ × √ = 1− 3 1− 3 1+ 3 √ √ 3+3+1+ 3 √ √ = 1+ 3− 3−3 √ √ 2 3+4 = =− 3−2 −2

8

Double Angles

From Section 7, we have sin(2a) = 2 sin a cos a cos(2a) = cos2 a − sin2 a We can use cos2 a + sin2 a = 1 to obtain alternative formulae for cos(2a): cos(2a) = cos2 a − sin2 a = cos2 a − (1 − cos2 a) = 2 cos2 a − 1 cos(2a) = cos2 a − sin2 a = (1 − sin2 a) − sin2 a = 1 − 2 sin2 a

11

12

Summer School: Circular Functions (

) 1◦ Example 8.1 Find the value of sin 22 . 2 Let a = 22

9

1◦ then consider cos(2a). 2 1 √ = cos(45◦ ) 2 ( ) 1◦ 2 ⇒ 2 sin 22 2 ( ) 1◦ 2 ⇒ sin 22 2 ( ) 1◦ ⇒ sin 22 2

(

) 1◦ = 1 − 2 sin 22 2 √ 1 2−1 =1− √ = √ 2 2 √ 2−1 = √ 2 2 √√ 2−1 √ = 2 2 2

Secant, Cosecant

We define sec θ =

1 cos θ

and

csc θ =

1 sin θ

Recall that cos2 θ + sin2 θ = 1. Divide by cos2 θ to get 1 + tan2 θ = sec2 θ Divide by sin2 θ to get cot2 θ + 1 = csc2 θ

10

Identities

Example 10.1 Show that

cos θ 1 − sin θ = 1 + sin θ cos θ

cos θ(1 − sin θ) cos θ(1 − sin θ) cos θ(1 − sin θ) 1 − sin θ cos θ = = = = 2 2 1 + sin θ (1 + sin θ)(1 − sin θ) cos θ cos θ 1 − sin θ Example 10.2 Show that 1 + sin(2θ) = (cos θ + sin θ)2 1 + sin(2θ) = cos2 θ + sin2 θ + 2 sin θ cos θ = cos2 θ + 2 cos θ sin θ + sin2 θ = (cos θ + sin θ)2

13

Summer School: Circular Functions Example 10.3 Show that

sin(2θ) − cos(2θ) + 1 = tan θ sin(2θ) + cos(2θ) + 1

sin(2θ) − cos(2θ) + 1 2 sin θ cos θ − (1 − 2 sin2 θ) + 1 = sin(2θ) + cos(2θ) + 1 2 sin θ cos θ + (2 cos2 θ − 1) + 1 2 sin θ cos θ + 2 sin2 θ = 2 sin θ cos θ + 2 cos2 θ sin θ(cos θ + sin θ) = cos θ(sin θ + cos θ) sin θ = = tan θ. cos θ

11

Pythagorean Triples

A set of three positive integers x, y, z such that x2 + y 2 = z 2 is called a Pythagorean Triple. Examples are {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {8, 15, 17}, {9, 40, 41}, {11, 60, 61}, {12, 35, 37}, {13, 84, 85}, {15, 112, 113}, {16, 63, 65}, {17, 144, 145}, {19, 180, 181}, {20, 21, 29}, {20, 99, 101}.

Example 11.1 If sin θ =

15 , what is 17

B

cos θ? Hint: draw a triangle! From diagram BC = 15, AB = 17

17

15

Hence AC 2 = 172 −152 = 289−225 = 64 = 82 AC 8 cos θ = = AB 17

A

θ 8

C

14

Summer School: Circular Functions Example 11.2 If sin α =

5 24 and cos β = , find tan(α − β) 13 25

13

25

5

α

7

β 24

12

tan α − tan β 1 + tan α tan β 7 5 − 12 24 = 5 × 24 − 7 × 12 = 5 7 12 × 24 + 5 × 7 1+ · 12 24 120 − 84 36 = = 288 + 35 323

tan(α − β) =

12

Product Formulae

Recall that sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β ⇒ sin(α + β) + sin(α − β) = 2 sin α cos β p+q p−q Put α + β = p, α − β = q ⇒ α = ,β = to get 2 2 ( ) ( ) p+q p−q sin p + sin q = 2 sin cos 2 2 ( cos p + cos q = 2 cos ( cos p − cos q = 2 sin Example 12.1

p+q 2 p+q 2 (

◦

◦

sin 105 + sin 15

)

( cos

)

( sin

p−q 2

q−p 2

) )

) ( ◦) 120◦ 90 = 2 sin cos 2 2 ◦ ◦ = 2 sin 60 cos√ 45 √ 3 1 3 √ = = 2 2 2 2

15

Summer School: Circular Functions Example 12.2 sin 105◦ − sin 15◦ = sin 105◦ + sin(−15◦ ) ( ◦) ( ) 120◦ 90 = 2 sin cos 2 2 ◦ ◦ = 2 sin 45 cos 60 1 1 1 = 2√ =√ 22 2 Example 12.3 (

◦

◦

cos 15 − cos 105

) ( ◦) 120◦ 90 = 2 sin sin 2 2 ◦ ◦ = 2 sin 60 sin 45 √ √ 3 1 3 √ =√ = 2 2 2 2

Example 12.4 (

) ( ) 5x + x 5x − x 2 sin cos sin(5x) + sin x 2 2 ( ) ( ) = 5x + x 5x − x cos(5x) + cos x 2 cos cos 2 2 sin(3x) = = tan(3x) cos(3x) Product formulae can also be re-written in the form 1 sin α cos β = {sin(α + β) + sin(α − β)} 2 1 cos α cos β = {cos(α + β) + cos(α − β)} 2 1 sin α sin β = {cos(α − β) − cos(α + β)} 2 Example 12.5 ( ) ( ◦) ) ( )] [ ( 1◦ 1 1 1◦ 1◦ 1◦ 1◦ cos 37 cos 7 = + cos 37 − 7 cos 37 + 7 2 2 2 2 2 2 2 1 = {cos(45◦ ) + cos(30◦ )} 2[ √ ] √ √ 1 1 3 2+ 3 √ + = = 2 2 4 2

16

Summer School: Circular Functions Example 12.6 ) ) ) ( ◦) ( ( 1◦ 1◦ 1◦ 1 1◦ 1◦ 2 sin 37 cos 7 = sin 37 + 7 + sin 37 − 7 2 2 2 2 2 2 ◦ ◦ = sin(45 ) + sin(30 ) √ 1 1 2+1 = √ + = 2 2 2 (

13

Trigonometric Equations

Question: Given a value c of sin(θ), what is θ? 1 Example 13.1 Suppose sin(θ) = : 2 sin(θ)

1

0.5

0

−0.5

−1 −6.2832

−3.1416

0

3.1416 θ

6.2832

9.4248

12.5664

1 There are infinitely many values of θ for which sin(θ) = , for example 2 θ=

π 5π π 5π , , 2π + , 2π + ,... 6 6 6 6

In general θ=

π + 2kπ, 6

θ=

5π + 2kπ, 6

k = 0, ±1, ±2, ±3, . . .

Because of these endless answers we are usually asked to find values of θ within a certain range.

17


Solving for θ ∈ [0, 2π]

13.1

Given we want to solve, for θ ∈ [0, 2π], trig f unction(θ) = c we first solve ˆ = |c| trig f unction(θ) i.e. for positive c. Here trig f unction is sin, cos or tan.

Y

nd

2

st

Quad

1 Quad sin

all

π−θ

rd

3

π ). We 2 then refer to our ’Rules of Signs’ from This gives an acute angle (θˆ