Applying trigonometric functions

Contents

8.1 Degrees and radians

8.2 Trigonometric ratios and the unit circle

8.3 Trigonometric graphs

8.4 Trigonometric functions and applications Chapter summary Chapter review

Syllabus subject matter

Periodic functions and applications ■ Definition of a radian and its relationship with degrees ■ Definition of the trigonometric functions sin, cos and tan of any angle in degrees and in radians ■ Graphs of y= sin x, y= cos x and y= tan x for any angle in degrees (−360° x 360°) and in radians (−2πx 2π) ■ Significance of the constants A, B, C and D on the graphs of y=A sin (Bx+C) +D and y=A cos (Bx+C) +D ■ Applications of periodic functions Quantitative concepts and skills ■ Calculation and estimation with and without instruments ■ Basic algebraic manipulations ■ Plotting points using Cartesian coordinates

Syllabus Guide Chapter 8

Applying trigonometric functions

8.1

Degrees and radians

Degree angle measure is based on the ancient Babylonian number system. It is more convenient for many mathematical and scientific purposes to measure angles using circular angle measure.

It follows from the above definition that, for a circle of radius r, the radian measure of the central angle θ subtended by an arc of length l is found by calculating the multiple that l is of r.

For a central angle of 1 revolution (360°):

θ = = = 2π radians

so 2π= 360°

and π= 180°

But π≈ 3.14

so ≈

or 1 radian ≈ 57.3°

Have you ever wondered why there are 90° in a right angle? This is actually part of a sexagesimal system, based on the number 60. This number system originated over 4000 years ago with the ancient Sumerians and was transmitted to us through the Babylonian and Egyptian cultures. With lots of factors, a system based on 60 makes many fractions very simple. We still use this system for time and angles in navigation, astronomy and direction. However, there is another angle measure that is commonly used in mathematics and science—the radian. This is used because it makes calculations, functions and formulas simpler, although its does seem a little strange at first. In this chapter, you will be introduced to radian angle measure.

One radian (1c _{or 1 rad) is the measure of the angle} subtended at the centre of a circle by an arc equal in length to the radius of the circle.

It is not necessary to use the symbol ‘c_{’ or ‘rad’ after an} angle to show it is in radians. All angles are assumed to be in radians unless they are shown to be measured in degrees by the degree sign.

1

r

r r

!

Radian measure of a central angle Given that l and r are in the same units,

θ =

By rearranging this formula, the arc length can be calculated using the radius and central angle:

l = rθ

θ

r l

l r

--!

Extra Material

Length of an arc

θ

r l

l r -- 2πr

r

---π π

--- 180° 3.14

We can use the fact that π= 180° to assist with conversions between degrees and radians. Consider an angle θ measured in degrees (θd) and radians (θr).

θd =θr

so =

Rearranging gives θ_d =

or θ_r =

Radian facts

2π = 360°

π = 180° 1 radian ≈ 57.3°

The following multiples of π are often used and should also be remembered:

= 270° = 90° = 60° = 45° = 30°

3π 2

--- π 2

--- π

3

--- π

4

--- π

6

---!

θd

180° --- θr

π

---180°

π

---θ_r

π

180° ---θ_d

Radian–degree conversion formulas

Radians to degrees: θd =

Degrees to radians θ_r = 180°

π

---θ_r

π

180° ---θ_d

!

Change the following to degrees.

a b 2.07 rad

Solution

a Use the formula. = ×

= 324°

Alternative method

Use π= 180°. =

= 324°

b Use the formula. 2.07 rad = × 2.07

≈ 118.60°

Alternative method

Use 1 rad = . 2.07 rad = 2.07 ×

≈ 118.60° 9π

5

---9π 5

--- 180°

π

--- 9π 5

---9π 5

--- 9×180° 5

---180°

π

---180°

π

--- 180°

π

---Example

1

It is sometimes helpful to show angles at appropriate points on a unit circle, including the x-axis and y-axis on the diagram of a circle of radius 1. This is also known as locating the angle in

standard position. On a Cartesian coordinate plane, an angle

is said to be in standard position when the initial side of the angle is along the positive x-axis and the vertex of the angle is at the origin.

For an angle of θ, the point P(θ) on the circumference of the unit circle is sometimes called a trigonometric point corresponding to θ.

As shown on the right, ∠AOP is θ radians.

Angle θ subtends arc AP, so must be θ units long. Change the following to radians.

a 240° b 110.25°

Solution

a Use the formula. 240° = × 240°

Leave in exact form. =

b Use the formula. _{110.25° = × 110.25°}

Answer in exact form. = 0.6125π

Alternatively, evaluate and round off. ≈ 1.924

π

180°

---4π 3

---π

180°

---Example

2

Terminal side

Initial side

y

x

Vertex

θ

O y

x P(θ)

A (1, 0)

1

θ

AP)

Positive and negative angles

When drawn in standard position, angles are measured as positive in the anticlockwise direction and negative in the clockwise direction.

y

x

+θ

y

x

−θ

Two angles drawn in standard position and having a common terminal side are called coterminal

angles. Coterminal angles differ by an integer multiple of 2π radians or 360°.

Simplify the following and show the resulting angles on the unit circle.

a π+ b 2π−

Solution

The angle is halfway between π and .

The angle is between and 2π.

a Change to a common denominator. π+ = +

= (or 225°)

b Change to a common denominator. 2π− = −

= (or 300°)

π

4

--- π

3

---π

4 --- 4π

4 --- π

4

---5π 4

---O y

x

P

1

π

3π 2 ---5π

4

---⎝ ⎠ ⎛ ⎞ 3π

2

---π

3 --- 6π

3 --- π

3 ---5π

3 ---3π

2

---O y

x

1

2π

3π 2

---5π 3

---⎝ ⎠ ⎛ ⎞

P

Example

3

Coterminal angles

If α and β are coterminal angles,

α−β= 2πn = n360° where n = …, −4, −3, −2, −1, 0, 1, 2, 3, 4, …

Exercise 8.1

Degrees and radians

1 Find the value of each of the following angles drawn in standard position.

2 Express the following in radians, leaving your answers in terms of π.

a 60° b 45° c 90° d 135° e 120°

f 100° g 420° h 72° i 288°

3 Express the following in radians, correct to 4 decimal places where necessary.

a 90° b 56° c 140° d 185° e 276°

f 320° g 267.8° h 566.42° i 765.92°

4 Convert the following from radians to degrees, correct to 1 decimal place where necessary.

a b c d e

f g h i

Which of the following pairs of angles are coterminal?

a α = −135° and β = 585° b α = and β =

Solution

a Calculate the difference between the angles.

α− β =−135° − 585° = −720°

= (−2)360°

State the result. −135° and 585° are coterminal angles.

b Calculate the difference between the angles.

α − β = −

=

=

is not an integer. State the result. and are not coterminal angles.

π

3

--- 11π 3

---π

3 --- 11π

3 ---1–11

( )π

3 ---10π

3

---10 3

--- π

3

--- 11π 3

---Example

4

Additional Exercise 8 .1

y y

y

x

a b c

d e f

y

x

y

x

y

x x

x

θ

60°

θ

π

3

---θ

300°

120° θ

θ

3π 4

---π

4

---θ

2π 3

--- 5π 6

--- 7π

4

--- 6π

11

--- 11π 5

---13π 8

--- 15π 8

--- 17π 3

---5 Convert the following from radians to degrees, correct to 1 decimal place where necessary.

a 1 b 2.2 c 0.8 d 3.3 e 2.06

f 5.45 g 0.07 h 5.55 i 7.89

6 Which of the following pairs are coterminal angles?

a −225° and 135° b −420° and 120° c and

d and e 750° and 30° f and

7 Simplify the following and represent the resulting angles in standard location on the unit circle.

a π + b π − c π + d π − e π −

f 2π − g 2π + h 2π + i 2π −

8.2

Trigonometric ratios and the unit

circle

The sine (sin), cosine (cos) and tangent (tan) trigonometric (trig) ratios have already been defined using right-angled triangles as follows:

sin θ =

cos θ =

tan θ =

2π 3

---– π

3 ---7π

3 --- π

3

--- π

6

---– 25π

6 ---–

π

4

--- 2π

3

--- 2π

3

--- 5π

6

--- 3π 4

---2π 3

--- 3π

4

--- 2π

3

--- 5π

6

---The area of a sector or segment of a circle can be expressed as a rule involving the angle subtended at the centre of the circle. The rule is simplest when the angle is expressed in radians. Work in groups of three or four people to find the rules.

1 Write down the area of a circle.

2 Write the area of a sector as a fraction of the circle involving

the angle at the centre.

3 Write a rule for the area of the sector, using the angle at the

centre expressed in radians.

4 Write the area of a segment as the difference between the areas

of a sector and triangle.

5 Write the rule for the area of a segment.

6 Calculate the areas of some sectors and segments.

Sector

Segment

Investigation

Sectors and segments

Extra Material

Sectors and segments

Adjacent Hypotenuse

Opposite

θ

opposite hypotenuse

---adjacent hypotenuse

---You can use a calculator to find decimal approximations of the values of trig ratios. It is possible to find exact values of the trig ratios for the angles 30°, 45° and 60° by drawing special right-angled triangles known as standard triangles and using previous definitions, as shown below.

The definitions of sin, cos and tan as previously described are restricted to angles up to 90°. We can use the unit circle to extend the definitions to angles

of any size.

Suppose the point P(θ) on the unit circle has coordinates (x, y), so we can also write it as P(x, y).

Draw in a vertical line to make OBP as shown. Then OB = x, BP = y and OP = 1.

Using OBP in the unit circle, we can see that: sin θ = = y

cos θ = = x

tan θ = Angle = 30° = 45° = 60°

sin

cos

tan 1

45°

45°

1

1 2 3

60° 30°

1 2 π

6

-- π

4

-- π

3

--1 2

-- 1

2

--- 3 2

---3 2

--- 1 2

--- 1 2

--1 3

--- 3

O y

x P(θ) = P(x, y)

A (1, 0)

1

θ

x B

y

y 1 ---x 1 ---y x

--Defining trigonometric ratios using the unit circle If P(θ) = P(x, y) is a point on the unit circle found by rotating the point A(1, 0) anticlockwise through an angle of θ, then we define:

sin θ = y cos θ = x

tan θ = =

These definitions are sometimes extended further to a circle of radius r as follows: sin θ= cos θ = tan θ = =

O y

x P(θ) = P(x, y)

A (1, 0)

1

θ

y x

-- sinθ cosθ

---y r

-- x

r

-- y

x

-- sinθ cosθ

The previous definition can be used to assign values to the trig ratios for 0° (360°), 90°, 180° and 270°.

Angle 0 = 0°

2π = 360° = 90° π= 180° = 270°

sin 0 1 0 −1

cos 1 0 −1 0

tan 0 Not defined 0 Not defined

y

x

(1, 0)

(0, −1) (0,1)

(−1, 0) 90° =

180° =π 360° = 2π (0° = 0) π

2

270° =3π 2 π

2

-- 3π

2

---Use the unit circle to find the value of:

a sin (−π) b sin (−90°) c tan 5π

Solution

a Draw a sketch.

From the sketch it is clear that sin (−π) = sin π.

State the result. sin (−π) = 0

b Draw a sketch.

From the sketch it is clear that sin (−90°) = sin 270°.

State the result sin (−90°) =−1

c Draw a sketch.

5π= 2(2π) +π, so 5π is coterminal with π. From the sketch it is clear that tan 5π = tan π.

State the result. tan 5π = 0

y

x

(−1, 0) _−π

π

y

x

(0, −1) 270°

−90°

y

x

(−1, 0)

5π

π

Scientific and graphics calculators can both be used to find trigonometric ratios of angles expressed in degrees or radians. It is important to check that the calculator is in the correct mode for the angle measure before using it for evaluation. Scientific calculators normally have an indication on the display. It usually has deg _{for degrees or} rad _{for radians. Graphics calculators} normally show the angle measure in a SETUP or MODE menu. The values shown by calculators are actually approximations, but these are sufficient for most purposes. However, it is still important to be able to evaluate trigonometric ratios using only the acute values.

We can use the definitions of trig ratios with the unit circle to work out the signs of the sin, cos and tan of an angle in any quadrant of the circle:

■ sin θ= y, so sin θ will be positive in the quadrants where y is positive, i.e. the first and second quadrants.

■ cos θ= x, so cos θ will be positive in the quadrants where x is positive, i.e. the first and fourth quadrants.

■ tan θ= , so tan θ will be positive when x and y have the same sign, i.e. in the first and third quadrants.

This information can be summarised as follows.

We have previously seen how to use right-angled triangles to work out the values of trig ratios for acute angles. We have also seen how to determine the signs of the trig ratios for any angle in the unit circle. Now we will see how to use acute angles and the CAST diagram to find the values of the trig ratios for any angle.

Let’s begin with the second quadrant. For θ:

sin θ= b cos θ=−a tan θ=

Now draw in the acute angle between the terminal side of θ and the x-axis. This angle, β, is known as the reference angle. Because β is an acute angle, we can draw it in the first quadrant to determine the values of its trig ratios. In the first quadrant:

sin β= b cos β= a tan β = So in the second quadrant:

sin θ= sin β cos θ=−cos β tan θ=−tan β

y

x

(1, 0)

(0, −1) (0, 1)

(−1, 0)

1st 2nd

3rd 4th quadrant quadrant

quadrant quadrant y

x

--Signs of the trig ratios

1st quadrant: All ratios are positive (A) 2nd quadrant: Sin only is positive (S) 3rd quadrant: Tan only is positive (T) 4th quadrant: Cos only is positive (C)

The above can be remembered as the mnemonic ‘CAST’ or ‘All Science Teachers are Curious’.

S A

T C

!

P(−a, b) y

x

θ −b

a

---P(−a, b) y

x

θ

Q(a, b)

β β

---This means that the values of sin, cos and tan for θ and β differ only in their signs. The sign of the trig ratio can be determined using the CAST diagram.

Let’s see how this works with real values.

A similar approach can be used for the third quadrant. For θ: sin θ=−b cos θ=−a tan θ = For the reference angle, β:

sin β= b cos β= a tan β =

So in the third quadrant:

sin θ= −sin β cos θ= −cos β tan θ = tan β Calculate the value of cos 150° without the aid of a calculator.

Solution

Draw a diagram showing 150° drawn in standard position.

Mark in the reference angle (30°).

Using CAST, we know that cos 150° is negative.

Use CAST and the reference angle. cos 150° = −cos 30°

Use the standard triangle. =−

y

x

30°

S

150°

A

T C

3 2

---Example

6

y

x

β θ

Q(a, b)

P(−a, −b)

β

b a

---b a

---Find the value of tan 225° without using a calculator.

Solution

Draw a diagram showing 225° drawn in standard position.

Mark in the reference angle (45°).

Using CAST, we know that tan 225° is positive.

Use CAST and the reference angle. tan 225° = tan 45°

Use the standard triangle. = 1

y

x

225° 45°

S A

T C

In the fourth quadrant, for θ.

sin θ= −b cos θ= a tan θ= For the reference angle, β:

sin β = b cos β = a tan β =

So in the fourth quadrant:

sin θ= −sin β cos θ= cos β tan θ= −tan β

The procedures on the previous pages can be used to find the value of any trigonometric ratio in terms of an acute angle and are summarised below.

P(a, −b) y

x

β

Q(a, b)

β θ −b

a

---b a

---Find the value of sin 330° without using a calculator.

Solution

Draw a diagram showing 330° drawn in standard position.

Mark in the reference angle (30°).

Using CAST, we know that sin 330° is negative.

Use CAST and the reference angle. sin 330° = −sin 30°

Use the standard triangle. =−

30°

y

x

300°

1 2

---Example

8

Trigonometric ratios for an angle of any magnitude For any angle drawn in standard position on the Cartesian plane, the acute angle between the terminal side of the angle and the x-axis is known as the reference angle. The reference angle is always taken to be positive. In the diagram shown here, β is the reference angle for θ, which is in the third quadrant of the Cartesian plane.

For any angle θ, use the following steps to find the value of sin θ, cos θ or tan θ.

1 Draw θ in standard position on the Cartesian plane.

2 Calculate the reference angle, β, that corresponds to θ.

3 Use CAST to determine the sign (+ or −) of the value of the trig ratio.

4 Find the value of the required trig ratio for β and include the sign.

y

x

β θ

Find the exact value of:

a sin 225° b cos c tan

Solution

a Draw a sketch.

Calculate the reference angle. Using CAST, sin 225° is negative.

Write the trig ratio using the reference angle. sin 225° = −sin 45°

Use the standard triangle. = −

b Draw a sketch.

Calculate the reference angle.

Using CAST, cos is positive.

Write the trig ratio using the reference angle. cos = cos

Use the standard triangle. =

c Draw a sketch.

Calculate the reference angle.

Using CAST, tan is negative.

Use the reference angle to write the trig ratio. tan = −tan

Use the standard triangle. = −

11π 6

--- 2π

3

---y

x

45°

S

225°

A

T C

1 2

---y

x

S A

T C

π

6 11π

6

11π 6

---11π 6

--- π 6

---3 2

---y

x

S A

T C

π

3 23π

2π 3

---2π 3

--- π 3

---3

Find values of the following without using a calculator.

a cos 570° b sin − c tan

Solution

a Find θ (0°θ 360°) coterminal with 570°. cos 570° = cos (570°− 360°)

= cos 210° Draw a sketch.

Calculate the reference angle.

Using CAST, cos 570° (or cos 210°) is negative.

Write the trig ratio. cos 210° =−cos 30°

Use cos 570°= cos 210°. cos 570° =−

b Find θ (0 θ 2π) coterminal with − . sin − = sin

= sin

Draw a sketch.

Calculate the reference angle.

Using CAST, sin − is negative.

Write the trig ratio. sin =−sin

Use sin − = sin . sin − = −

c Find θ (0 θ 2π) coterminal with . tan = tan

= tan

Use the table of values. tan is not defined.

Use tan = tan . The value of tan is not defined.

25π 6

--- 11π 2

---y

x

30°

S

210°

A

T C

3 2

---25π 6

--- 25π 6

--- −25π 6

--- 36π 6 ---+

⎝ ⎠

⎛ ⎞

11π 6

---y

x

S A

T C

π

6 11π

6

25π 6

--- or sin 11π 6

---⎝ ⎠

⎛ ⎞

11π 6

--- π 6

---25π 6

--- 11π 6

--- 25π

6 --- 1

2

---11π 2

--- 11π

2

--- 11π 2 --- 8π

2 ---–

⎝ ⎠

⎛ ⎞

3π 2 ---3π

2 ---11π

2

--- 3π 2

--- 11π

2

The same procedure can used to simplify expressions involving trig ratios.

The right-angled triangle formed by the reference angle in the unit circle is known as the

reference triangle.

Simplify 2 sin x + sin (2π− x) where 0 x .

Solution

Draw a sketch for 2π− x.

The reference angle is x.

(2π− x) is in the 4th quadrant, so sin (2π− x) is negative.

Use the reference angle. sin (2π− x) = −sin x

Substitute into the given expression. 2 sin x + sin (2π − x)= 2 sin x − sin x

Simplify. = sin x

π

2

---y

x

S A

T C

x

2π− x

Example

11

Reference triangle

The terminal side of the reference angle β meets the unit circle at P(a, b), where a or b may be negative. The reference triangle is formed by drawing a perpendicular from P to the x-axis.

In the diagram at right, POM is the reference triangle for β.

Using Pythagoras’s theorem we can see that a2+ _b2= ₁

This definition can be extended beyond the unit circle reference angles.

For the reference triangle POM, a2+ _b2= _r2

So r =

y

x O

β θ

a M

b

P(a, b)

1

y

x

P(a, b) O

θ

βa M

b r

a2+b2

Exercise 8.2

Trigonometric ratios and the unit circle

1 Use the unit circle to find:

a cos 120° b sin 240° c tan 135°

d cos 300° e tan 210° f cos 330°

2 Use the unit circle to find:

a cos b sin c cos

d tan e sin f cos

3 Use the unit circle to find:

a sin 90° b cos π c tan 2π

d cos e tan 180° f tan 450°

4 Use the unit circle to find:

a sin −60° b cos − c tan −

d cos −π e tan −240° f cos −225°

5 If θ is drawn in standard position on the Cartesian plane, in which quadrants must the terminal side of θ lie for the following to be true?

a cos θ 0 b tan θ 0 c sin θ 0

d tan θ 0 e sin θ 0 f cos θ 0

6 a What is the greatest value of y in the unit circle? b What is the smallest value of y in the unit circle? c What is the range of sin θ?

7 a What is the greatest value of x in the unit circle? b What is the smallest value of x in the unit circle? c What is the range of cos θ?

The reference angle φ passes through P(5, −12). Use the reference triangle to find an exact value for sin φ.

Solution

Draw a diagram and mark in the reference triangle.

Use Pythagoras’s theorem. OP2 = ₅2+ ₍−₁₂₎2

OP is positive. = 13

Use the definition of sin. sin φ = −

y

x

O φ

M

P (5, −12)

−12 5

12 13

---Example

12

Additional Exercise 8.2

7π 4

--- 3π

4

--- 7π

6

---11π 6

--- 5π

3

--- 5π

6

---3π 2

---2π 3

--- 7π

---8 a Does have a greatest or least value when x and y are in the unit circle? b What is the range of tan θ?

c Are there any values of θ for which is not defined? d What is the domain of tan θ?

9 In each of the following cases, verify that P lies on the unit circle. If θ is an angle drawn in standard position and its terminal side passes through P, find the exact values of sin θ, cos θ and tan θ without actually calculating the value of θ.

a P b P c P

d P e P f P

10 In each of the following cases, θ is an angle drawn in standard position. Without actually calculating the value of θ, find the exact values of the other two trig ratios using the information supplied.

a sin θ= and θ is in quadrant 1 b cos θ= and θ is in quadrant 1

c cos θ=− and θ is in quadrant 2 d sin θ=− and θ is in quadrant 3

e sin θ=− and θ is in quadrant 4 f cos θ= and θ is in quadrant 1

g cos θ=− and θ is in quadrant 3 h sin θ= and θ is in quadrant 2

11 In each of the following cases, the reference angle β passes through point P. Use the reference triangle to find the exact values of sin β and cos β.

a P(3, 4) b P(−5, 12) c P(7, −24)

d P(−1, −1) e P(4, −3) f P(−12, 5)

12 Find the exact values of:

a sin 225° b cos 150° c tan 330°

d cos −135° e tan 315° f sin −210° g tan −240° h cos 300° i sin −120° 13 Find the exact values of:

a sin b cos c tan − d cos −

e tan f sin g sin −π h cos

i tan − j cos − k sin − l sin

14 Find the exact values of:

a sin −2π b cos c sin − d cos −

e cos f tan −π g sin − h cos −

i sin j tan k tan − l sin

y x --y x --3 5 --- 4 5 ---, ⎝ ⎠

⎛ ⎞ ₋12

13 --- 5 13 ---, ⎝ ⎠

⎛ ⎞ 1

2 --- − 1

2 ---, ⎝ ⎠ ⎛ ⎞ − 3 2 --- 1 2 ---, ⎝ ⎠

⎛ ⎞ 1

2 --- − 3

2

---,

⎝ ⎠

⎛ ⎞ ₋ 1

2 --- − 1

2 ---, ⎝ ⎠ ⎛ ⎞ 4 5 --- 5 13 ---7 25 --- 3 5 ---1 2 --- 1 2 ---24 25 --- 12 13 ---7π 6

--- 5π

6

--- 3π

4

--- 4π

3

---11π 6

--- 7π 4

--- 5π

3 ---4π 3 --- π 3 --- π 3

--- 5π

3

---7π 4

--- 3π

2

--- 3π

2

---7π 6

--- 3π

4

--- 2π

3

---5π 3

--- 7π

4

--- 7π

6

--- π

---15 Find the exact values of:

a sin b cos c tan −4π d cos −

e sin −540° f tan g tan − h sin

i cos 600° j cos −570° k tan −390° l sin 690°

Modelling and problem solving

16 Simplify the following.

a tan (π+ x) + 3 tan x b cos (π− x) + cos (2π− x) c sin (π+ 5y) − 5 sin 5y d cos (π+ x) + 2 cos (2π− x) e sin (y + 2π) − 3 sin y f cos (2π+ g) − 3 cos (π− g) g tan (q +π) + 3 tan (q −π) h 4 sin (m +π) + 3 sin (m − 2π) i 3 sin (π− c) + 2 sin (π+ c)

8.3

Trigonometric graphs

The graphs of the sine, cosine and tangent functions may be drawn by using their definitions and knowledge of the unit circle.

Graph of the sine function

1 Construct a unit circle on some graph paper, and on its circumference mark P , P ,

P , …, P by drawing diameters as shown in the diagram below.

2 Construct Cartesian axes with θ running from −π to 3π and y from −1 to 1 aligned with the unit circle. Mark the θ-axis in units of as shown.

3 Now, sin θ= y = y-coordinate of P. So the graph of y = sin θ can be drawn by transferring the y-coordinates of the points P(θ) on the unit circle to the Cartesian axes. This can be done physically as shown.

The graph above has been extended before 0 and past 2π using the symmetry of a circle or periodic nature of the sine function. The sign of the graph (whether it is above or below the axis) also matches the previous work on the sign of the trig ratios in each quadrant, as shown at the top of the following page.

13π 6

--- 17π 6

--- 5π

2

---19π 6

--- 11π 4

--- 7π 2

---π

12 ---⎝ ⎠

⎛ ⎞ π

6 ---⎝ ⎠ ⎛ ⎞

π

4 ---⎝ ⎠

⎛ ⎞ 23π 12 ---⎝ ⎠ ⎛ ⎞

π

12

---y

1

3π θ

−π π

−1

2π

y = sin θ

Graph of the cosine function

A similar method can be used to draw a graph of the cosine function, y = cos θ. However, the lengths of the x-coordinates must be used for cos, so it is necessary to transfer the lengths to the graph using a pair of dividers.

The resulting graph of y = cos θ shares many similar features with the graph of y = sin θ.

Graph of the tangent function

To draw the graph of the tangent function, we begin as for the sine and cosine except that the tangent, AT, is drawn to touch the unit circle at A. OP is extended to meet the tangent at T.

Now OMP ||| OAT

So = = AT

Thus tan θ = AT

So tan θ= length of tangent cut off by the extended radius. That is, tan θ= y-coordinate of T.

In the fourth quadrant, this is negative. In the second and third quadrants, the radius is extended backwards to meet the tangent.

y

1

2π θ

π

−1

π

2

--- 3π

2 ---+A

−C +S

−T

1

3π θ

−π π

−1

2π

y = cos θ

2

y

O

P T

A M

y x

θ

1 y

x -- AT

1

---T A

θ

T

A

θ

T A

For convenience, it is best to start plotting the graph y = tan θ between − and .

By looking ate the graphs of y = sin θ,y = cos θ and y = tan θ, you can see that these are all

periodic functions. Periodic functions were discussed in Chapter 4. You can also see that

y = cos θ and y = sin θ have the same period and amplitude.

A graphics calculator can also be used to draw graphs of the trigonometric functions, as already seen in Chapter 4.

Exercise 8.3

Trigonometric graphs

For questions 1 to 3, you may wish to print and use the blank grid on the CD-ROM. 1 Use a unit circle and graph paper to draw the graph of y = sin θ from θ=−2π to 2π. 2 Use a unit circle and graph paper to draw the graph of y = cos θ from θ=−2π to 2π. 3 Use a unit circle and graph paper to draw the graph of y = tan θ from θ=−π to π.

π

2 --- π

2

---y = tan θ

y

2

−2

2π θ

−π π

1

−1

Use a graphics calculator to draw the graph of y = sin θ from θ=−4π to 4π.

Solution

Use SETUP or to set the calculator to radian measure.

Put the function in as Y₁= .

Set the (or V-Window) so that X goes from −4π to 4π in units of , and so that

Y goes from −1 to 1 in units of 0.2. the function to obtain:

Casio fx-9860G AU Texas Instruments TI-84 Sharp EL-9900

MODE

SIN ( X,T,θ,n )

WINDOW π

4

---GRAPH

Example

13

Additional Exercise 8.3

4 Use the following graphs of y = sin θ and y = cos θ to answer the questions below.

a What are the periods and amplitudes of the sine and cosine functions? b Comment on the similarities and differences of the graphs.

5 a Use a graphics calculator to draw the graph of y = tan θ from θ=−2π to 2π. b What are the period and amplitude of the tangent function?

8.4

Trigonometric functions and

applications

To emphasise the fact that the variable is considered to be a real number instead of an angle, we use the symbol x instead of θ in most applications of trigonometric functions. The nature of the basic graphs is summarised below and on the next page.

1

0.5

−0.5

−1

y

x

π 2π 3π 4π

y = cos x

y = sin x

Basic trigonometric functions y= sin x

Domain = real numbers Range is −1 y 1. Period = 2π

Amplitude = 1

Zeros are at nπ: …, −2π, −π, 0, π, 2π, 3π, 4π, …

Maxima are at : …, − , , , …

Minima are at : …, − , , , … y= cos x

Domain = real numbers Range is −1 y 1. Period = 2π

Amplitude = 1

Zeros are at : …, − , − , , , , …

Maxima are at 2nπ: …, −2π, 0, 2π, 4π, … Minima are at (2n + 1)π: …, −π, π, 3π, …

−π π 3π 2π

2

---π

2

---π

2

---− x

y

1

−1

y = sin x

4n+1

( )π

2

--- 3π 2 --- π

2 --- 5π

2

---4n+3

( )π

2

--- π 2 --- 3π

2 --- 7π

2

---x y

1

−1

−π π 2π

y = cos x

3π 2

---π

2

---π

2

---−

2n+1

( )π

2

--- 3π 2 --- π

2 --- π

2 --- 3π

2 --- 5π

2 --- 7π

2

Some trigonometric function graphs have been shown in Chapter 4 as examples of periodic functions. In this section a more systematic study is made of the sine and cosine functions.

y= tan x

Domain = real numbers

except odd multiples of Range = real numbers

Period =π

Amplitude = infinite

Zeros are at nπ: …, −2π, −π, 0, π, 2π, 3π, 4π, …

The graph of tan x has asymptotes at : …, − , − , , , , , … At these values, the function is undefined, but near them the graph becomes very large or very small.

x y

4

−4

−π π 2π

y = tan x

3π 2

---π

2

---π

2

---3 2 1

−3

−2

−1

− π

2

---2n+1

( )π

2

--- 3π 2 --- π

2 --- π

2 --- 3π

2 --- 5π

2 --- 7π

2

---Make a table of values and hence sketch the graph of y = 3 sin x from −π to 2π. Comment on the relationship with the graph of y = sin x.

Solution

Set up the table with values at every . Round values of 3 sin x to 1 decimal place.

Sketch the graph, using knowledge of the basic shape of y = sin x to complete the curve.

The graph of y = 3 sin x is stretched vertically by a factor of 3 compared with y = sin x.

x −π − − − − − 0

3 sin x 0 −1.5 −2.6 −3 −2.6 −1.5 0 1.5 2.6 3

x π 2π

3 sin x 2.6 1.5 0 −1.5 −2.6 −3 −2.6 −1.5 0

π

6

---5π 6

--- 2π 3

--- π 2

-- π 3

-- π 6

-- π

6

-- π 3

-- π 2

--2π 3

--- 5π 6

--- 7π 6

--- 4π 3

--- 3π 2

--- 5π 3

--- 11π 6

---x

−π

y = sin x

y

3

−3 1 2

−2

−1 π

2

---y = 3 sin x

π 3π 2π

2

---−π

2

Use a graphics calculator to draw the graph of y = cos x + 2 from −π to 2π. Comment on the relationship with the graph of y = cos x.

Solution

Enter the function as Y1= 2.

Set the (or V-Window) so that: X_min =−π Y_min = 0 X_max = 2π Y_max = 4 Xscl =π 6 Yscl = 1 Then the function.

The display on your graphics calculator should look something like this. There will be minor variations depending on the brand of calculator, but the major features of the graph will be the same.

The graph of y = cos x + 2 is translated 2 units upwards compared with the graph of y = cos x.

COS ( X,T,θ,n ) +

WINDOW

GRAPH

Example

15

Make a table of values and hence sketch the graph of y = cos 3x from −π to 2π. Comment on its relationship with the graph of y = cos x.

Solution

π is about −3.1 and 2π is about 6.3. Make a table of values from −3.2 to 6.4.

Sketch the graph, using your knowledge of the basic shape of y = cos x.

The graph of y = cos 3x is compressed horizontally by a factor of 3 compared with the graph of y = cos x. The period of y = cos 3x is of the period of y = cos x.

x −3.2 −2.8 −2.4 −2 −1.6 −1.2 −0.8 −0.4 0 0.4 0.8 1.2 1.6

cos 3x −1.0 −0.5 0.6 1.0 0.1 −0.9 −0.7 0.4 1 0.4 −0.7 −0.9 0.1

x 2 2.4 2.8 3.2 3.6 4 4.4 4.8 5.2 5.6 6 6.4

cos 3x 1.0 0.6 −0.5 −1.0 −0.2 0.8 0.8 −0.3 −1.0 −0.5 0.7 0.9

x y

1

−1

−3.2 −2.4 −1.6 −0.8 0.8 1.6 2.4 3.2 4 4.8 5.6 6.4

y = cos 3x y = cos x

1 3

After working through the previous examples, you should be able to see that the following rules apply to general graphs of the sine and cosine functions.

Use a graphics calculator to compare the graphs of y = 2 sin 3x and y = 2 sin .

Solution

Enter the functions in Y₁ and Y₂ and set the (or V-Window) to show X from

− to with an X-scale (Xscl) of and Y from −2 to 2 with a Y-scale (Yscl) of 0.5.

Then the functions.

The display on your graphics calculator should look something like this. There will be minor variations depending on the brand of calculator, but the major features of the graph are the same.

The graph of y = 2 sin is translated to the right compared with the graph of y = 2 sin 3x.

There is a horizontal shift of ÷ 3 = to the right compared with y = 2 sin 3x. 3 x π

4 ---–

⎝ ⎠

⎛ ⎞

WINDOW

π

2 --- π

2

--- π

12

---GRAPH y = 2 sin 3x

y = 2 sin (3x − )π

4

---3 x π 4 ---–

⎝ ⎠

⎛ ⎞ π

12

---π

4

--- π 12

---Example

17

The graphs of the functions

y = A sin (Bx + C) + D and y = A cos (Bx + C) + D are related to graphs of the functions y = sin x and y = cos x by: Vertical translation of D

For D 0 it is upwards and for D 0 it is downwards.

Horizontal translation of

For 0 it is to the left and for 0 it is to the right.

x y

3

0 1

−1 2

π 2π 3π

x

π 2π 3π

y = sin x + 2

y = sin x y = cos x − 2

y = cos x

y

1

−2

−1

−3 0

C B

---C B

---- C

B

----y = sin (x − )

x

π 2π 3π

y = sin x

x y

1

0

−1

π 2π 3π

y

1

0

−1 π

2 y = cos (x + π₂ )

y = cos x

Horizontal change of scale of B

For B 1 it compresses by a factor of B and for 0 B 1 it stretches by a factor of . If B is negative, the graph is reflected in the y-axis, so the x values swap.

Vertical change of scale of A

For A 1 it stretches by a factor of A and for 0 A 1 it compresses by a factor of . If A is negative, the graph is inverted (reflected in the x-axis).

1 B

---x y

1

0

−1

π 2π 3π

x y

1

0

−1

π 2π 3π

y = sin

y = sin x

π

2 y = cos 3x

y = cos x

1 A

---x

π 2π 3π

y

3

1 2

−2

−1

−3 0

y = sin x

y = 3 sin x 1

0.5

−0.5

−1

y

y = cos x

x

π 2π 3π

0

y = cos 1 x

2

!

Summary of the features of sine and cosine functions

For the functions y= A sin (Bx + C) + D = A sin B + D

and y= A cos (Bx + C) + D = A cos B + D

■ the amplitude is the magnitude of A

■ the period is

■ the value is called the phase shift and is the horizontal translation

■ the average (mean) value is D.

x C

B ----+

⎝ ⎠

⎛ ⎞

x C

B ----+

⎝ ⎠

⎛ ⎞

2π B ---C B

----Sketching trigonometric graphs

Graphs of sine and cosine functions may be sketched from the equation by two methods:

Method 1: Identify the translations and changes of scale from the equation. Method 2: Identify the:

■ starting point

■ zeros

■ end of a cycle

■ maxima and minima.

Find the values of x that produce the sine or cosine of 0, , π, and 2π and the corresponding values of y.

π

2 --- 3π

2

Without using a calculator or a table of values, sketch the graph of y = 4 sin − 3.

Solution

Method 1

Method 2

The five points are shown on the ‘first’ cycle graph below in blue. Consider first the graph

of the simplified function y = 4 sin 2x.

It has the basic shape of the sine function, but is compressed by a horizontal factor of 2, so the period is

π instead of 2π. It is stretched vertically by a factor of 4, so the amplitude is 4.

The graph of y = 4 sin 2x is shown in green below. Now consider the translations

to obtain

y= 4 sin − 3

= 4 sin 2 − 3.

It is translated to the left by a phase shift of (compared with y = 4 sin 2x).

It is translated 3 downwards, so the average value is −3. The final graph is shown in blue below.

Find (x, y) when we get sin 0 = 0. = 0

x =−

y = 4 sin 0 − 3 =−3

The ‘first’ cycle starts at .

Find (x, y) so we get sin = 1. =

x = , so the point is .

Find (x, y) so we get sin π= 0. =π

x = , so the point is .

Find (x, y) so we get sin =−1. =

x = , so the point is . Find (x, y) so we get sin 2π= 0. = 2π

x = , so the point is . 2 x π

6 ---+

⎝ ⎠

⎛ ⎞

2 x π 6 ---+ ⎝ ⎠ ⎛ ⎞ x π 12 ---– ⎝ ⎠ ⎛ ⎞ π 12

---2 x π 6 ---+ π 12 ---π 12

---– ,–3

⎝ ⎠

⎛ ⎞

π

2

--- 2 x π

6 ---+ π 2 ---π 6 --- π 6

---,1

⎝ ⎠ ⎛ ⎞

2 x π 6 ---+

5π 12

--- 5π

12

---,–3

⎝ ⎠

⎛ ⎞

3π 2

--- 2 x π

6 ---+ 3π

2 ---2π 3

--- 2π

3

---,–7

⎝ ⎠

⎛ ⎞

2 x π 6 ---+

11π 12

--- 11π 12

---,–3

⎝ ⎠ ⎛ ⎞ π 2 --- π 2 --- x y 4 2 −4 −6

y = 4 sin 2x

y = 4 sin (2x + π) − 3 6

---−8

π

−π ₋ 2π

3 ---π 6 ---π 12

---− 11π

Without using a calculator or a table of values, sketch the graph of y = 1 − 2 cos . Check your sketch with the aid of a graphics calculator.

Solution

Method 1

Method 2

The five points are shown on the ‘first’ cycle graph below in blue. Consider first the graph

of the simplified function y =−2 cos 3x.

It has the basic shape of the cosine function, but is compressed by a horizontal factor of 3, so the period is instead of 2π.

It is stretched vertically by a factor of 2, so the amplitude is 2. The graph is inverted from the cosine graph because of the negative sign.

The graph of y =−2 cos 3x is shown in green below. Consider the translations to obtain

y= 1 − 2 cos

= 1 − 2 cos 3 .

It is translated to the right by a phase shift of (compared with y =−2 cos 3x).

It is translated 1 upwards, so the average value is 1. The final graph is shown in blue below.

Find (x, y) so we get cos 0 = 1. = 0

x = , so the ‘first’ point is .

Find (x, y) so we get cos = 0. =

x = , so the point is .

Find (x, y) so we get cos π=−1. =π

x = , so the point is .

Find (x, y) so we get cos = 0. =

x = , so the point is .

Find (x, y) so we get cos 2π= 1. = 2π

x = , so the point is . 3 x π

12 ---– ⎝ ⎠ ⎛ ⎞ 2π 3

---3 x π 12 ---– ⎝ ⎠ ⎛ ⎞ x π 36 ---– ⎝ ⎠ ⎛ ⎞ π 36

---3 x π 12 ---– ⎝ ⎠ ⎛ ⎞ π 36 --- π 36

---,–1

⎝ ⎠

⎛ ⎞

π

2

--- 3 x π

12 ---– ⎝ ⎠ ⎛ ⎞ π 2 ---7π 36

--- 7π

36

---,1

⎝ ⎠

⎛ ⎞

3 x π 12 ---– ⎝ ⎠ ⎛ ⎞ 13π 36

--- 13π 36

---,3

⎝ ⎠

⎛ ⎞

3π 2

--- 3 x π

12 ---–

⎝ ⎠

⎛ ⎞ 3π

2 ---19π

36

--- 19π 36

---,1

⎝ ⎠

⎛ ⎞

3 x π 12 ---– ⎝ ⎠ ⎛ ⎞ 25π 36

--- 25π 36

---,–1

⎝ ⎠ ⎛ ⎞ π 2 --- π 4 --- π 4 --- π 2

--- 3π

4 --- x y 4 3 2 1 −1 −2 −3

y = −2 cos 3x

y = 1 − 2 cos (3x + ) π 12 ---− − 7π 36

--- 19π 36

---π

36

--- 13π 36

--- 25π 36

Your graphics calculator should show similar results for Examples 17 and 18, allowing for differences in scales.

The sine and cosine functions can be used to model a wide range of periodic phenomena to a high degree of accuracy.

A sinusoidal model has the shape of the sine graph. This can be sine or cosine, whichever is more convenient.

In a harbour, high tides occur 12 h 20 min apart. The depth of water at the entrance bar is 2.8 m at low tide and 8.4 m at high tide. A high tide occurs at 2 am on 3 October. a Sketch a rough version of the tidal heights.

b Find a sinusoidal model for the depth of the water, d, as a function of the time t hours since midnight on 3 October.

c Find the depth of water at 4 pm on 5 October.

d Use a graphics calculator to find the times on 3 October at which a ship with a draught of 5 m can enter the harbour.

Solution

a Low tides occur halfway between high tides, so the next low tide will be 6 h 10 min after 2 am, at 8:10 am.

The average depth will be halfway between the tides, so the first average depth will occur 3 h 5 min after 2 am, which is 5:05 am. The depth will be (8.4+2.8)2=5.6 m. We can continue adding 3 h 5 min to the last times to get the following times and depths.

Draw a sketch using the points found and knowledge of the shape of sine and cosine. Time 2:00 am 5:05 am 8:10 am 11:15 am 2:20 pm 5:25 pm 8:30 pm 11:35 pm

d (m) 8.4 5.6 2.8 5.6 8.4 5.6 2.8 5.6

b Since the graph starts near a maximum, it is more appropriate to use cosine than sine.

Write the cosine function. y = A cos B + D

Find the amplitude. Amplitude = (8.4 − 2.8) 2 m A = 2.8 m

Find the period. Period = 12 h

= h 2

4 6 8

0

d (m)

4 8 12 16 20 24 t (h)

x C

B ----+

⎝ ⎠

⎛ ⎞

1 3

---37 3

Turn Y1 off:

■ For the Casio, use the SEL key to select Y2.

■ For the Texas Instruments, place the cursor over Y₁ and press .

■ For the Sharp, do the same as for the Texas Instruments.

the function Y2.

Find the zeros of Y2:

■ For the Casio, use ROOT in G-Solv, and press the right cursor to get each value.

■ For the Texas Instruments, use 2:zero in CALC and appropriate guesses to get each value.

■ For the Sharp, use 5 X_Incpt in CALC repeatedly to get each value.

Use period = . =

Rearrange to find B. B =

Find the phase shift. Maximum is 2 hours after t = 0.

Phase shift =−2 h

Find the mean value. D = (8.4 + 2.8) 2 = 5.6 m

Choose variables. t = x, d = y

Use y = A cos B + D. d = 2.8 cos (t − 2) + 5.6

c At 4 pm on 5 October, t = 64 h. d = 2.8 cos (64 − 2) + 5.6

≈ 8.36 m

d Enter the function in Y₁. Y₁= 2.8 cos (6π 37 × (X − 2)) + 5.6 Use VARS to enter Y2. Y2= Y1− 5

The display on your graphics calculator should look something like this. There will be minor variations depending on the brand of calculator, but the major features of the graph will be the same.

Using CALC or G-Solv gives X = 5.5072446, X = 10.826089, X = 17.840578, X = 23.159422 Change to times and write answers. The ship could enter from midnight

to 5:30 am, from 10:50 am to 5:50 pm, and after 11:10 pm.

An alternative method is to draw graphs of Y1= 2.8 cos (6π 37 × (X − 2)) + 5.6 and Y₂= 5 using a graphics calculator, then find the intersections of the graphs in a similar way to the zeros.

2π B

--- 37

3 --- 2π

B

---6π 37

---C B

----x C

B ----+

⎝ ⎠

⎛ ⎞ 6π

37

---6π 37

---F1

ENTER

Exercise 8.4

Trigonometric functions and applications

1 For 0 x 2π, sketch a graph of each of the following and check with a graphics calculator. a y = 2 sin x b y = sin x c y = 3 cos x d y = 2 sin x + 1 2 For −π x π, sketch a graph of each of the following and check with a graphics

calculator.

a y = sin 2x b y = sin x c y = cos 3x d y = sin 2x + 3 3 Sketch a graph of each of the following and check with a graphics calculator, showing a

full cycle.

a y = 3 sin 4x b y = cos 3x c y = 5 sin 2x d y = sin 8x + 4 4 Sketch graphs of the following and check with a graphics calculator, showing a full cycle

of each.

a y = sin b y = sin c y = cos d y = cos

e y = 2 sin f y = 2 sin g y = 3 cos h y = 3 cos

5 Sketch graphs of the following and check with a graphics calculator, showing a full cycle of each.

a y = 4 sin 2 + 2 b y = 2 cos 3 − 3 c y = 5 sin 4 − 2

d y = 3 cos 2 + 4 e y = 4 − 2 cos 4 f y = 3 − 3 sin 2

To tune a guitar, the frets are used to play a note on one string that should be the same as the base note of the next string. When the strings are in tune, the notes will be the same. When the strings are quite out of tune, they sound quite different. When they are close, but not quite the same, the sound will seem to beat (fade and strengthen rhythmically). This can be shown quite easily on a graphics calculator.

1 Consider a note that should have a frequency

of 440 Hz. Store this in the memory in C. Store 442 in the memory in D.

Put in Y₁= sin (2π C X), Y₂= sin (2π D X) and Y₃= Y₁+ Y₂, and turn Y₁ and Y₂ off.

Set the (or V-Window) with X from 0 to 5 and Y from −3 to 3. the result. Change D to 441 and graph it again.

Change D to 440.2, change X to go from 0 to 20 and graph again.

Find what note corresponds to 440 Hz.

2 Try different values of C and D. 3 What do you find?

4 Write a short account of your results. WINDOW

GRAPH

Investigation

Musical beats

Additional Exercise 8.4 1 2 ---1 2 ---1 2 --- 1 3 ---x π 2 ---– ⎝ ⎠

⎛ ⎞ _x π

2 ---+

⎝ ⎠

⎛ ⎞ _x π

6 ---–

⎝ ⎠

⎛ ⎞ _x π

6 ---+ ⎝ ⎠ ⎛ ⎞ x π 3 ---– ⎝ ⎠

⎛ ⎞ _x π

3 ---+

⎝ ⎠

⎛ ⎞ _x π

4 ---–

⎝ ⎠

⎛ ⎞ _x π

4 ---+ ⎝ ⎠ ⎛ ⎞ x π 3 ---– ⎝ ⎠

⎛ ⎞ _x π

6 ---+

⎝ ⎠

⎛ ⎞ _x π

8 ---+ ⎝ ⎠ ⎛ ⎞ x π 4 ---– ⎝ ⎠

⎛ ⎞ _x π

3 ---+

⎝ ⎠

⎛ ⎞ _x π

4 ---–

⎝ ⎠

6 Sketch graphs and check with a graphics calculator, showing a full cycle of each. a y = 2 sin + 3 b y = 3 cos (4x +π) − 2 c y = 4 − 3 sin

d y = 5 cos − 3 e y = 3 − 5 cos f y = 4 − 3 sin

Modelling and problem solving

7 The depth of water, d(t) m, at a pier changes according to the rule d(t) = 5 + 2 cos t where t is time in hours from high tide. On 18 March, high tide was at 7:30 am.

a Draw a sketch showing the depth of water for the 24 hours after 7:30 am on 18 March. b When was the first low tide on that day?

c When was the water at a depth of 6 m? d How long is it between low tide and high tide?

8 The sales, S in 100s of units, of a seasonal product are modelled by S = 54.8 + 32.5 cos where t is the time in months (t = 1 is January and t = 12 is December).

a Draw a graph of the sales for a period of 12 months.

b Use the graph to determine the months for which sales exceed 6800 units.

9 In an unusual meteorological investigation, the temperature, T °C, in a town in central Queensland was found to fluctuate approximately according to the rule T = 25 + 6 sin 0.1πt where t is the number of hours after 10:00 am.

a Sketch a graph of the temperature fluctuations for a sufficient number of hours to be able to determine the maximum and minimum temperatures for that day and the next night. b Use the graph to determine the maximum and minimum temperatures.

c When did they occur?

d At what time was the temperature: i 27°C? ii 20°C? e Why was this unusual?

10 In Calcutta, the highest mean monthly temperature is 29.45°C in June and the lowest is 18.3°C in December. Find a model for the temperature throughout the year and graph it. 11 In Upernivik, Greenland, the average temperature varies between −23°C in January and 5°C

in July. Find and graph a model for the average temperature. 12 A large hoop is rolling along the ground. The vertical distance

above the ground of a point P on the rim of the hoop is given by y = 1.25 − 1.25 cos 2πt where y is in metres and t is in seconds. Find the first three times at which P is 2.00 m above the ground. 13 The time between successive high tides at a pier is

12 h 20 min. The average depth of water is 4 m, but at low tide it is 3.1 m. Write an equation to model this relationship and use it to find the lengths of time that a boat with a draught of 3.5 m and 4.5 m can use the pier. 14 An oscilloscope tracing signals from a microphone

shows a sinusoidal curve. The amplitude relates to the volume of the sound while the frequency gives the pitch. A violin string vibrates at 200 Hz (1 Hz is 1 vibration per second). The oscilloscope trace amplitude is 12 V. a Find the period of the sound wave.

b Write an equation for the sound wave. 2 x π

3 ---–

⎝ ⎠

⎛ ⎞ _{6 x} 3π

2 ---+

⎝ ⎠

⎛ ⎞

2 x 3π 4 ---–

⎝ ⎠

⎛ ⎞ _{3 x} π

3 ---–

⎝ ⎠

⎛ ⎞ _{4 x} π

4 ---+

⎝ ⎠

⎛ ⎞

π

6 ---⎝ ⎠ ⎛ ⎞

πt 6

■ The radian measure (c _{or rad) of a central angle} θ _{is the ratio of the} arc length (l) to the length of the radius (r) subtended the centre of the circle.

θ= or l = rθ

■ Radians can be converted to degrees using formulas:

Radians to degrees: θd= Degrees to radians θr =

■ Some radian to degree conversions should be committed to memory:

2π = 360° = 270° π = 180° = 90° = 60° = 45° = 30°

■ On a Cartesian coordinate plane, an angle is in standard position when the initial side of the angle is along the positive x-axis and the vertex of the angle is at the origin.

■ When drawn in standard position, angles are measured as positive in the anticlockwise direction and negative in the clockwise direction.

■ Coterminal angles differ by an integer multiple of 2π radians or 360°.

■ The trigonometric ratios may be defined in terms of the unit circle as:

sin θ= y = y-coordinate of P

cos θ= x = x-coordinate of P

tan θ= =

Standard triangles may be used to find

the trigonometric ratios for the angle 45° , 30° and 60° .

■ CAST can be used to determine the signs of trig ratios: 1st quadrant: All ratios are positive (A)

2nd quadrant: Sin only is positive (S) 3rd quadrant: Tan only is positive (T) 4th quadrant: Cos only is positive (C)

■ The reference angle (β) is the acute angle between the terminal side of an angle drawn in standard position (θ) and the x-axis.

θ

r l

l r

--180°

π

---θ_r π

180° ---θ_d

3π 2

--- π

2

--- π

3

--- π

4

--- π

6

---O y

x P(θ) = P(x, y)

1

θ

y

x B

A (1, 0)

y x

-- y-coordinate of P x-coordinate of P

---1 1

45°

2 2

1 30°

3

60°

π

4 ---⎝ ⎠ ⎛ ⎞

π

6 ---⎝ ⎠

⎛ ⎞ π

3 ---⎝ ⎠ ⎛ ⎞

S A

T C

y

x

■ The value of sin, cos or tan for any angle θ, may be found by: 1 Drawing θ in standard position.

2 Calculating the reference angle β:

3 Using CAST to determine the sign (+ or −).

4 Finding the value of the trig ratio for β including the sign. ■ POM is the reference triangle for reference angle β.

For the reference triangle POM, a2+ _b2= _r2

So r=

■ For the functions y = sin x and y = cos x:

– domain = real numbers – range is −1 y 1 – period = 2π – amplitude = 1

■ For the function y = tan x:

– domain = real numbers, except odd multiples of – range = real numbers

– period =π

– amplitude is infinite.

■ For the functions y = A sin (Bx + C) + D = A sin B + D

and y = A cos (Bx + C) + D = A cos B + D: – the amplitude is A

– the period is

– the value is called the phase shift and is the horizontal translation – the average (mean) value is D and is the vertical translation.

■ Sine and cosine graphs may be sketched using translations and changes of scale or by finding the five important points in a cycle.

y

x

P (a, b) O

θ β

a M

b r

a2_+b2

x y

1

−1

−π π 2π

y = cos x

3π 2

---π

2

---π

2

---x y

1

−1

−π π 2π

y = sin x

3π 2

---π

2

---π

2

---− −

x y

4

−4

−π π 2π

y = tan x

3π 2

---π

2

---π

2

---− π

2

---x C

B ----+

⎝ ⎠

⎛ ⎞

x C

B ----+

⎝ ⎠

⎛ ⎞

----Knowledge and procedures

1 Find the value of each of the following angles drawn in standard position.

2 Express the following in radians, leaving your answers in terms of π.

a 270° b 225° c 240°

d 160° e 330° f 288°

3 On a unit circle, an arc of length subtends an angle at the centre of:

A 120° B 135° C 60° D 150° E 210°

4 2.3c _{in degrees is given by:}

A 2.3 × B 2.3 × C π− 2.3 D 2.3 − E

5 Express in radians correct to 4 decimal places.

a 47° b 220.75°

6 Express in degrees, correct to 1 decimal place.

a b 0.43

7 Which angle is larger: or 30°?

8 Which of the following pairs are coterminal angles?

a 186° and −534° b −320° and 500° c and

9 Simplify the following and represent the resulting angles in standard location on the unit circle.

a π+ b π− c π+

10 If θ is drawn in standard position on the Cartesian plane, in which quadrants must the terminal side of θ lie for the following to be true?

a cos θ 0 b tan θ 0 c sin θ 0

11 Draw unit circles and use them to find sin 270° and cos −60°.

12 Draw standard triangles and use them to find tan and sin . Ex 8.1

θ

y

x

310° a

θ

y

x

b

θ

y

x

c

π

6

4π 3

θ

y

x

135° d

θ

y

x

e

θ

y

x

f

π

6

5π 6

Ex 8.1

Ex 8.1 5π

6

---Ex 8.1

π

180°

--- 180°

π

--- π

2

--- 2.3

2π

---Ex 8.1

Ex 8.1

3π 4

---Ex 8.1 π

3

---Ex 8.1

5π 6

---– 7π

6

---Ex 8.1

5π 6

--- π

3

--- 3π

4

---Ex 8.1

Ex 8.2

Ex 8.2 π

3

---13 Show the angle on a unit circle and give the third- and fourth-quadrant angles with cosines of the same magnitude.

14 The exact value of sin is:

A 240° B 60° C − D − E

15 cos (π+ x) =

A cos x B sin x C −cos x D −sin x E π+ cos x 16 Find the exact values of:

a sin 210° b cos −120° c tan 300°

d sin − e cos f tan −

17 In each of the following cases, verify that P lies on the unit circle. If θ is an angle drawn in standard position and its terminal side passes through P, find the exact values of sin θ, cos θ and tan θ without actually calculating the value of θ.

a P b P

18 In each of the following cases, θ is an angle drawn in standard position. Without actually calculating the value of θ, find the exact value of the other two trig ratios using the infor