Ch 6 Trigonometry

TERMINOLOGY

6 Trigonometry

Angle of depression: The angle between the horizontal and the line of sight when looking down to an object below

Angle of elevation: The angle between the horizontal and the line of sight when looking up to an object above

Angles of any magnitude: Angles can be measured around a circle at the centre to fi nd the trigonometric ratios of angles of any size from 0c to 360c and beyond

Bearing: The direction relative to north. Bearings may be written as true bearings (clockwise from North) or as compass bearings (using N, S, E and W)

Complementary angles: Two or more angles that add up to 90c

Cosecant: The reciprocal ratio of sine (sin). It is the hypotenuse over the opposite side in a right triangle

Cotangent: The reciprocal ratio of tangent (tan). It is the adjacent over the opposite side in a right triangle

Secant: The reciprocal ratio of cosine (cos). It is the hypotenuse over the adjacent side in a right triangle

Trigonometric identities: A statement that is true for all trigonometric values in the domain. Relationships between trigonometric ratios

ch6.indd 290 7/10/09 5:44:55 PM

291Chapter 6 Trigonometry

INTRODUCTION

TRIGONOMETRY IS USED IN many fi elds, such as building, surveying and navigating. Wave theory also uses trigonometry.

This chapter revises basic right-angled triangle problems and applies them to real-life situations. Some properties of trigonometric ratios, angles greater than 90c and trigonometric equations are introduced. You will also study trigonometry in non-right-angled triangles.

Ptolemy (Claudius Ptolemaeus), in the second century, wrote He mathe matike syntaxis (or Almagest as it is now known) on astronomy. This is considered to be the fi rst treatise on trigonometry, but was based on circles and spheres rather than on triangles. The notation ‘chord of an angle’ was used rather than sin, cos or tan.

Ptolemy constructed a table of sines from 0c to 90c in steps of a quarter of a degree. He also calculated a value of r to 5 decimal places, and established the relationship for sin ( )X Y! and cos ( ) .X Y!

DID YOU KNOW?

Trigonometric Ratios

In similar triangles, pairs of corresponding angles are equal and sides are in proportion. For example:

the • hypotenuse is the longest side, and is always opposite the right angle the • opposite side is opposite the angle marked in the triangle the • adjacent side is next to the angle marked

In any triangle containing an angle of ,30c the ratio of : : .AB AC 1 2= Similarly, the ratios of other corresponding sides will be equal. These ratios of sides form the basis of the trigonometric ratios.

In order to refer to these ratios, we name the sides in relation to the angle being studied:

You studied similar triangles in Geometry in Chapter 4.

ch6.indd 291ch6.indd 291 8/11/09 10:58:08 AM8/11/09 10:58:08 AM

292 Maths In Focus Mathematics Extension 1 Preliminary Course

The opposite and adjacent sides vary according to where the angle is marked. For example:

The trigonometric ratios are

You can learn these by their initials SOH , CAH , TOA .

What about S ome O ld H ags C an’t A lways H ide T heir O ld A ge?

DID YOU KNOW? Trigonometry, or triangle measurement , progressed from the study of geometry in ancient Greece. Trigonometry was seen as applied mathematics. It gave a tool for the measurement of planets and their motion. It was also used extensively in navigation, surveying and mapping, and it is still used in these fi elds today.

Trigonometry was crucial in the setting up of an accurate calendar, since this involved measuring the distances between the Earth, sun and moon.

sin

cos

tan

hypotenuse

opposite

hypotenuse

adjacent

adjacent

opposite

Sine

Cosine

Tangent

i

i

i

=

=

=

As well as these ratios, there are three inverse ratios,

cosecsin

seccos

cottan

1

1

1

oppositehypotenuse

adjacent

hypotenuse

oppositeadjacent

Cosecant

Secant

Cotangent

ii

ii

ii

= =

= =

= =

fff

pp

p

ch6.indd 292 6/25/09 10:02:33 PM


EXAMPLES

1. Find ,sina tana and .seca

Solution

sin

tan

sec cos

AB

BC

AC

5

3

4

53

43

1

45

hypotenuse

opposite side

adjacent side

hypotenuse

opposite

adjacent

opposite

adjacent

hypotenuse

a

a

a a

= =

= =

= =

=

=

=

=

=

=

=

2. If 72,sini = fi nd the exact ratios of ,cosi tani and .coti

Solution

By Pythagoras’ theorem:

7 2

49 4

45

c a b

a

a

a

a 45

2 2 2

2 2 2

2

2

`

= +

= +

= +

=

=

CONTINUED

To fi nd the other ratios you need to fi nd the

adjacent side.

ch6.indd 293 7/12/09 2:08:42 AM


cos

tan

cottan

745

452

1

245

hypotenuse

adjacent

adjacent

opposite

i

i

ii

=

=

=

=

=

=

Complementary angles

, ,ABC B A 90In if then c+ +i iD = = -

sin

cos

tan

sec

cosec

cot

cb

ca

ab

ac

bc

ba

i

i

i

i

i

i

=

=

=

=

=

=

(angle sum of a Δ)

( )

( )

( )

( )

( )

( )

sin

cos

tan

sec

cosec

cot

ca

cb

ba

bc

ac

ab

90

90

90

90

90

90

c

c

c

c

c

c

i

i

i

i

i

i

- =

- =

- =

- =

- =

- =

From these ratios come the results.

( )

( )

( )

( )

( )

( )

°

°

°

°

°

°

sin cos

cos sin

sec cosec

cosec sec

tan cot

cot tan

90

90

90

90

90

90

i i

i i

i i

i i

i i

i i

= -

= -

= -

= -

= -

= -

ch6.indd 294 6/26/09 2:55:18 AM


1. Write down the ratios of ,cos sini i and .tani

2. Find ,sin cotb b and .secb

3. Find the exact ratios of ,sin tanb b and .cosb

4. Find exact values for ,cos tanx x and .cosecx

EXAMPLES

1. Simplify 50 40 .tan cotc c-

Solution

tan cotcot

tan cot tan tan

50 90 5040

50 40 50 50

0

`

c c c

c

c c c c

= -

=

- = -

=

] g

2. Find the value of m if .sec cosec m55 2 15c c= -] g

Solution

90 5535

sec coseccosec

m

m

m

55

2 15 35

2 50

25

`

c c c

c

= -

=

- =

=

=

] g

Check this by substituting m into the equation.

6.1 Exercises

Check this answer on your calculator.

ch6.indd 295 7/10/09 4:06:34 AM


5. If ,tan34

i = fi nd cos i and .sin i

6. If 32,cosi = fi nd exact values for

,tani seci and .sini

7. If 61,sini = fi nd the exact ratios

of cosi and .tani

8. If 0.7,cosi = fi nd exact values for tani and .sini

9. ABCD is a right-angled isosceles triangle with ABC 90c+ = and 1.AB BC= =

Find the exact length of (a) AC . Find (b) .BAC+ From the triangle, write down (c)

the exact ratios of 45 , 45sin cosc c and .ctan45

10.

Using Pythagoras’ theorem, (a) fi nd the exact length of AC .

Write down the exact ratios of (b) 30 , 30sin cosc c and 30 .tan c

Write down the exact ratios of (c) 60 , 60sin cosc c and 60 .tan c

11. Show .sin cos67 23c c=

12. Show .sec cosec82 8c c=

13. Show .tan cot48 42c c=

14. Simplify (a) cos sin61 29c c+

(b) 90sec cosec ci i- -] g (c) 70 20 2 70tan cot tanc c c+ -

(d) 3555

cossin

c

c

(e) 25

25 65cot

cot tanc

c c+

15. Find the value of x if .sin cos x80 90c c= -] g

16. Find the value of y if .tan cot y22 90c c= -^ h

17. Find the value of p if .cos sin p49 10c c= +^ h

18. Find the value of b if .sin cos b35 30c c= +] g

19. Find the value of t if .cot tant t2 5 3 15c c+ = -] ]g g

20. Find the value of k if .tan cotk k15 2 60c c- = +] ]g g

Hint: Change 0.7 to a fraction.

Trigonometric ratios and the calculator

Angles are usually given in degrees and minutes. In this section you will practise rounding off angles and fi nding trigonometric ratios on the calculator.

Angles are usually given in degrees and minutes in this course. The calculator uses degrees, minutes and seconds, so you need to round off.

utes ree

onds ute( )

60 1 (60 1)min degsec min

60 1 60 1c= =

= =

lm l

In normal rounding off, you round up to the next number if the number to the right is 5 or more. Angles are rounded off to the nearest degree by rounding up if there are 30 minutes or more. Similarly, angles are rounded off to the nearest minute by rounding up if there are 30 seconds or more.

ch6.indd 296 6/25/09 10:03:15 PM


EXAMPLES

Round off to the nearest minute. 1. 23 12 22c l m

Solution

23 12 22 23 12c c=l m l

2. 59 34 41c l m

Solution

59 34 41 59 35c c=l m l

3. 16 54 30c l m

Solution

16 54 30 16 55c c=l m l

, ,,% KEY

This key changes decimal angles into degrees, minutes and seconds

and vice versa.

Some calculators have

deg or dms keys.

EXAMPLES

1. Change 58 19c l into a decimal.

Solution

, ,, , ,, , ,,% % %58 19Press = So .58 19 58 31666667c =l

2. Change 45.236c into degrees and minutes.

Solution

, ,,%.45 236Press SHIFT= So .45 236 45 14c c= l

If your calculator does not give these answers, check

the instructions for its use.

Because 30 seconds is half a minute, we round

up to the next minute.

ch6.indd 297 7/10/09 5:45:27 PM


In order to use trigonometry in right-angled triangle problems, you need to fi nd the ratios of angles on your calculator.

EXAMPLES

1. Find ,cos 58 19c l correct to 3 decimal places.

Solution

, ,, , ,,% %58 19Press COS = So .cos 58 19 0 525c =l

2. Find ,sin 38 14c l correct to 3 decimal places.

Solution

, ,, , ,,% %38 14Press SIN = So .sin 38 14 0 619c =l

3. If 0.348,tani = fi nd i in degrees and minutes.

Solution

This is the reverse of fi nding trigonometric ratios. To fi nd the angle, given the ratio, use the inverse key .tan 1-^ h , ,,%TAN .0 348Press SHIFT SHIFT1 =-

.

( . )

tan

tan

0 348

0 34819 11

1

c

i

i

=

=

=

-

l

4. Find i in degrees and minutes if . .cos 0 675i =

Solution

, ,,%.0 675Press SHIFT COS SHIFT1 =-

.

( . )cos

cos0 675

0 67547 33

1

c

ii

===

-

l

6.2 Exercises

1. Round off to the nearest degree. (a) °47 13 12l m (b) °81 45 43l m (c) °19 25 34l m (d) °76 37 19l m (e) °52 29 54l m

2. Round off to the nearest minute. (a) °47 13 12l m (b) °81 45 43l m (c) °19 25 34l m (d) °76 37 19l m (e) °52 29 54l m

If your calculator doesn't give this answer, check that it is in degree mode.

ch6.indd 298 7/10/09 4:06:37 AM


3. Change to a decimal. (a) 77 45c l (b) 65 30c l (c) 24 51c l (d) 68 21c l (e) 82 31c l

4. Change into degrees and minutes. (a) 59.53c (b) 72.231c (c) 85.887c (d) 46.9c (e) 73.213c

5. Find correct to 3 decimal places. (a) 39 25sin c l (b) cos 45 51c l (c) 18 43tan c l (d) 68 06sin c l (e) 54 20tan c l

6. Find i in degrees and minutes if (a) .sin 0 298i = (b) .tan 0 683i = (c) .cos 0 827i = (d) .tan 1 056i = (e) .cos 0 188i =

Right-angled Triangle Problems

Trigonometry is used to fi nd an unknown side or angle of a triangle.

Finding a side

We can use trigonometry to fi nd a side of a right-angled triangle.

EXAMPLES

1. Find the value of x , correct to 1 decimal place.

Solution

°.

°.

. °

.

. .

cos

cos

cos

cos

x

x

x

x

23 4911 8

23 4911 8

11 8 23 49

10 8 1

11 8 11 8

hypotenuse

adjacent

cm to decimal point`

# #

i =

=

=

=

=

l

l

l

^ h

CONTINUED

ch6.indd 299 7/12/09 2:11:33 AM


2. Find the value of y , correct to 3 signifi cant fi gures.

Solution

c15c

15

15

15

c

c

c

c

15

15

c

.

.

.

.

.

.

sin

sin

sin

sin

sin

sin

sin sin

y

yy

y

y

y y

41 15 9 7

41 9 7

41 9 7

41 9 7

419 7

14 7 3

41 41

hypotenuse

opposite

m to significant figures

# #

i =

=

=

=

=

=

=

l

l

l

l

l

l l

^ h

6.3 Exercises

1. Find the values of all pronumerals, correct to 1 decimal place.

(a)

(b)

(c)

(d)

ch6.indd 300 7/12/09 2:13:27 AM


(e)

(f)

(g)

(h)

(i)

(j)

(k)

x

31c12l

5.4 cm

(l)

x4.7 cm

37c22l

(m) x

6.3 cm

72c18l

(n)

23 mm

63c14l

x

(o)

3.7 m

39c47l

y

(p)

14.3 cm46c5l

k

(q)

4.8 m74c29l

h

ch6.indd 301 7/10/09 4:07:29 AM


(r) 0.45 m

68c41ld

(s) 5.75 cm

19c17l

x

(t) 17.3 m

6c3l

b

2. A roof is pitched at 60c. A room built inside the roof space is to have a 2.7 m high ceiling. How far in from the side of the roof will the wall for the room go?

60c

2.7 m

x

3. A diagonal in a rectangle with breadth 6.2 cm makes an angle of 73c with the vertex as shown. Find the length of the rectangle correct to 1 decimal place.

73c

6.2 cm

4. Hamish is standing at an angle of 67c from a goalpost and 12.8 m away as shown. How far does he need to kick a football for it to reach the goal?

x

67c

12.8 m

5. Square ABCD with side 6 cm has line CD produced to E as shown so that EAD 64 12c+ = l . Evaluate the length, correct to 1 decimal place, of

(a) CE (b) AE

E

6 cm

64c12l

B

A

C

D

6. A right-angled triangle with hypotenuse 14.5 cm long has one interior angle of 43 36c l. Find the lengths of the other two sides of the triangle.

ch6.indd 302 6/25/09 10:04:09 PM


7. A right-angled triangle ABC with the right angle at A has B 56 44c+ = l and 26AB = mm. Find the length of the hypotenuse.

8. A triangular fence is made for a garden inside a park. Three holes A , B and C for fence posts are made at the corners so that A and B are 10.2 m apart, AB and CB are perpendicular, and angle CAB is 59 54c l. How far apart are A and C ?

9. Triangle ABC has 46BAC c+ =

and .ABC 54c+ = An altitude is drawn from C to meet AB at point D . If the altitude is 5.3 cm long, fi nd, correct to 1 decimal place, the length of sides

(a) AC (b) BC (c) AB

10. A rhombus has one diagonal 12 cm long and the diagonal makes an angle of 28 23c l with the side of the rhombus.

Find the length of the side of (a) the rhombus.

Find the length of the other (b) diagonal.

11. Kite ABCD has diagonal 15.8BD = cm as shown. If ABD+ = 57 29 andc l

72 51DBC c+ = l, fi nd the length of the other diagonal AC.

B

A

C

D72c51l

57c29l

15.8 cm

Finding an angle

Trigonometry can also be used to fi nd one of the angles in a right-angled triangle.

EXAMPLES

1. Find the value of ,i in degrees and minutes.

CONTINUED

ch6.indd 303 7/12/09 2:11:46 AM


Solution

..

7.35.8

cos

7 35 8hypotenuse

adjacent

1

i =

-cos

37 23

`

c

i =

=

=

l

c m

2. Find the value of ,a in degrees and minutes.

Solution

..

.

.

tan

tan

2 14 9

2 14 9

66 48

adjacent

opposite

1`

c

a

a

=

=

=

=

-

l

c m

6.4 Exercises

1. Find the value of each pronumeral, in degrees and minutes.

(a)

(b)

ch6.indd 304 7/10/09 4:07:33 AM


(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

2.4 cm

3.8 cm

a

(l)

8.3 cm

5.7 cm

i

(m)

6.9 mm

11.3 mm

i

(n)

3 m

7 m

i

ch6.indd 305 6/25/09 10:04:35 PM


(o)

5.1 cm

11.6 cm

b

(p)

15 m

13 ma

(q)

7.6 cm

4.4 cmi

(r)

14.3 cm8.4 cm

a

(s)

3 m

5 m

i

(t)

10.3 cm

18.9 cmc

2. A kite is fl ying at an angle of i above the ground as shown. If the kite is 12.3 m above the ground and has 20 m of string, fi nd angle i .

12.3 m20 m

i

3. A fi eld is 13.7 m wide and Andre is on one side. There is a gate on the opposite side and 5.6 m along from where Andre is. At what angle will he walk to get to the gate?

Gate

Andre

5.6 m

13.7 m

i

4. A 60 m long bridge has an opening in the middle and both sides open up to let boats pass underneath. The two parts of the bridge fl oor rise up to a height of 18 m. Through what angle do they move?

18 m

i60 m

5. An equilateral triangle ABC with side 7 cm has an altitude AD that is 4.5 cm long. Evaluate the angle the altitude makes with vertex A DAB+] g .

ch6.indd 306 8/1/09 10:36:54 PM


6. Rectangle ABCD has dimensions 18 cm # 7 cm. A line AE is drawn so that E bisects DC .

How long is line (a) AE ? (Answer to 1 decimal place).

Evaluate (b) DEA+ .

7. A 52 m tall tower has wire stays on either side to minimise wind movement. One stay is 61.3 m long and the other is 74.5 m long as shown. Find the angles that the tower makes with each stay.

52 m

61.3 m 74.5 m

ba

8. (a) The angle from the ground up to the top of a pole is 41c when standing 15 m on one side of it. Find the height h of the pole, to the nearest metre.

If Seb stands 6 m away on the (b) other side, fi nd angle i .

41c

h

6 m 15 mi

9. Rectangle ABCD has a line BE drawn so that AEB 90c+ = and 1DE = cm. The width of the rectangle is 5 cm.

5 cm

BA

CE

D1 cm

Find (a) BEC+ . Find the length of the (b)

rectangle.

10. A diagonal of a rhombus with side 9 cm makes an angle of 16cwith the side as shown. Find the lengths of the diagonals.

16c

9 cm

11. (a) Kate is standing at the side of a road at point A , 15.9 m away from an intersection. She is at an angle of 39c from point B on the other side of the road. What is the width w of the road?

(b) Kate walks 7.4 m to point C . At what angle is she from point B ?

w

B

CA7.4 m

15.9 m

39c i

ch6.indd 307 6/25/09 10:04:54 PM


Applications

DID YOU KNOW?

The Leaning Tower of Pisa was built as a belfry for the cathedral nearby. Work started on the tower in 1174, but when it was only half completed the soil underneath one side of it subsided. This made the tower lean to one side. Work stopped, and it wasn’t until 100 years later that architects found a way of completing the tower. The third and fi fth storeys were built close to the vertical to compensate for the lean. Later a vertical top storey was added.

The tower is about 55 m tall and 16 m in diameter. It is tilted about 5 m from the vertical, and tilts by an extra 0.6 cm each year.

Class Investigation

Discuss some of the problems with the Leaning Tower of Pisa.

Find the angle at which it is tilted from the vertical. • Work out how far it will be tilted in 10 years. • Use research to fi nd out if the tower will fall over, and if so, when. •

Angle of elevation

The angle of elevation is used to measure the height of tall objects that cannot be measured directly, for example a tree, cliff, tower or building.

ch6.indd 308 6/25/09 10:05:02 PM


Class Exercise

Stand outside the school building and look up to the top of the building. Think about which angle your eyes pass through to look up to the top of the building.

The angle of elevation, ,i is the angle measured when looking from the ground up to the top of the object. We assume that the ground is horizontal.

EXAMPLE

The angle of elevation of a tree from a point 50 m out from its base is .38 14c l Find the height of the tree, to the nearest metre.

Solution

We assume that the tree is vertical!

tan

tan

tan

h

h

h

h

38 1450

38 1450

50 38 1439

50 50# #

c

c

c

Z

=

=

=

l

l

l

So the tree is 39 m tall, to the nearest metre.

A clinometer is used to measure the angle of

elevation or depression.

ch6.indd 309 7/12/09 2:13:43 AM


Angle of depression

The angle of depression is the angle formed when looking down from a high place to an object below.

Class Exercise

If your classroom is high enough, stand at the window and look down to something below the window. If the classroom is not high enough, fi nd a hill or other high place. Through which angle do your eyes pass as you look down?

The angle of depression, ,i is the angle measured when looking down from the horizontal to an object below.

EXAMPLES

1. The angle of depression from the top of a 20 m building to a boy below is .c 961 3 l How far is the boy from the building, to 1 decimal place?

Solution

ch6.indd 310 7/10/09 5:45:49 PM


39

39c

39

39

39

3939

c

c

c

c

c

c c

( , )

61 39.

61

tan

tan

tan

tan

tan

tan tan

DAC ACB

AD BC

x

xx

x

x

x x

61

61 20

61 20

61 20

61 20

20

10 8

61

alternate angles

# #

+ +

<

Z

=

=

=

=

=

=

=

l

l

l

l

l

l

l l

So the boy is 10.8 m from the building.

2. A bird sitting on top of an 8 m tall tree looks down at a possum 3.5 m out from the base of the tree. Find the angle of elevation to the nearest minute.

Solution

3.5 m

8 m

AB

C D

i

The angle of depression is i

AB DCBDC

Since horizontal linesalternate angles+

<i=

]^ gh

.

.

tan

tan

3 58

3 58

66

1`

c

i

i

=

=

=

-

22l

c m

ch6.indd 311 7/10/09 5:46:08 PM


Bearings

Bearings can be described in different ways: For example, N70 Wc :

Start at north and measure 70o around towards the west.

True bearings measure angles clockwise from north

EXAMPLES

1. Sketch the diagram when M is on a bearing of 315c from P .

Solution

2. X is on a bearing of 030c from Y . Sketch this diagram.

Solution

3. A house is on a bearing of 305c from a school. What is the bearing of the school from the house?

Measure clockwise, starting at north.

All bearings have 3 digits so 30° becomes 030° for a bearing.

We could write 315o T for true bearings.

ch6.indd 312 7/10/09 5:46:30 PM


Solution

The diagram below shows the bearing of the house from the school.

North

School

House

305c

To fi nd the bearing of the school from the house, draw in North from the house and use geometry to fi nd the bearing as follows:

S

H

N1

N2NN

305c

The bearing of the school from the house is N HS2+ .

360 305

180 55 ( )

N SH

N HS N H N S

55

125

angle of revolution

cointerior angles,

1

2 2 1

c c

c

c c

c

+

+ <

= -

=

= -

=

^ h

So the bearing of the school from the house is 125c .

4. A plane leaves Sydney and fl ies 100 km due east, then 125 km due north. Find the bearing of the plane from Sydney, to the nearest degree.

CONTINUED

ch6.indd 313 7/12/09 2:12:52 AM


Solution

c

.

( . )51 ( )

tan

tan

x

x

x

100125

1 25

1 25

90

90 51

39

to the nearest degree

1

c c

c c

c

i

=

=

=

=

= -

= -

=

-

So the bearing of the plane from Sydney is .°039

5. A ship sails on a bearing of °140 from Sydney for 250 km. How far east of Sydney is the ship now, to the nearest km?

Solution

cos

cos

cos

x

x

x

x

140 90

50

50250

50250

250 50

161

250 250# #

c c

c

c

c

c

Z

i = -

=

=

=

=

So the ship is 161 km east of Sydney, to the nearest kilometre.

A navigator on a ship uses a sextant to measure angles.

Could you use a different triangle for this question?

ch6.indd 314 6/26/09 2:58:04 AM


6.5 Exercises

1. Draw a diagram to show the bearing in each question .

A boat is on a bearing of 100(a) c from a beach house.

Jamie is on a bearing of 320(b) c from a campsite.

A seagull is on a bearing of (c) 200c from a jetty.

Alistair is on a bearing of (d) 050c from the bus stop.

A plane is on a bearing of (e) 285c from Broken Hill .

A farmhouse is on a bearing (f) of 012c from a dam.

Mohammed is on a bearing of (g) 160c from his house.

A mine shaft is on a bearing (h) of 080c from a town.

Yvonne is on a bearing of (i) 349c from her school.

A boat ramp is on a bearing of (j) 280c from an island.

2. Find the bearing of X from Y in each question in 3 fi gure (true) bearings .

X

Y

North

112c

(a)

X

35c

Y

North

South

EastWest

(b)

X

10cY

North

South

EastWest

(c)

23c

X

Y

North

South

EastWest

(d)

X

Y

North

South

EastWest

(e)

ch6.indd 315 6/26/09 2:58:34 AM


3. Jack is on a bearing of 260c from Jill. What is Jill’s bearing from Jack?

4. A tower is on a bearing of 030c from a house. What is the bearing of the house from the tower?

5. Tamworth is on a bearing of 340c from Newcastle. What is the bearing of Newcastle from Tamworth?

6. The angle of elevation from a point 11.5 m away from the base of a tree up to the top of the tree is 42 12c l. Find the height of the tree to one decimal point.

7. Geoff stands 25.8 m away from the base of a tower and measures the angle of elevation as .39 20c l Find the height of the tower to the nearest metre.

8. A wire is suspended from the top of a 100 m tall bridge tower down to the bridge at an angle of elevation of 52c. How long is the wire, to 1 decimal place?

9. A cat crouches at the top of a 4.2 m high cliff and looks down at a mouse 1.3 m out from the foot (base) of the cliff. What is the angle of depression, to the nearest minute?

10. A plane leaves Melbourne and fl ies on a bearing of 065c for 2500 km.

How far north of Melbourne (a) is the plane?

How far east of Melbourne (b) is it?

What is the bearing of (c) Melbourne from the plane?

11. The angle of elevation of a tower is 39 44c l when measured at a point 100 m from its base. Find the height of the tower, to 1 decimal place.

12. Kim leaves his house and walks for 2 km on a bearing of .155c How far south is Kim from his house now, to 1 decimal place?

13. The angle of depression from the top of an 8 m tree down to a rabbit is .43 52c l If an eagle is perched in the top of the tree, how far does it need to fl y to reach the rabbit, to the nearest metre?

14. A girl rides a motorbike through her property, starting at her house. If she rides south for 1.3 km, then rides west for 2.4 km, what is her bearing from the house, to the nearest degree?

15. A plane fl ies north from Sydney for 560 km, then turns and fl ies east for 390 km. What is its bearing from Sydney, to the nearest degree?

16. Find the height of a pole, correct to 1 decimal place, if a 10 m rope tied to it at the top and stretched out straight to reach the ground makes an angle of elevation of .67 13c l

ch6.indd 316 7/12/09 2:17:05 AM


17. The angle of depression from the top of a cliff down to a boat 100 m out from the foot of the cliff is .59 42c l How high is the cliff, to the nearest metre?

18. A group of students are bushwalking. They walk north from their camp for 7.5 km, then walk west until their bearing from camp is .320c How far are they from camp, to 1 decimal place?

19. A 20 m tall tower casts a shadow 15.8 m long at a certain time of day. What is the angle of elevation from the edge of the shadow up to the top of the tower at this time?

15.8 m

20 m

20. A fl at verandah roof 1.8 m deep is 2.6 m up from the ground. At a certain time of day, the sun makes an angle of elevation of .72 25c l How much shade is provided on the ground by the verandah roof at that time, to 1 decimal place?

21. Find the angle of elevation of a .15 9 m cliff from a point 100 m

out from its base.

22. A plane leaves Sydney and fl ies for 2000 km on a bearing of 195 .c How far due south of Sydney is it?

23. The angle of depression from the top of a 15 m tree down to a pond is .25 41c l If a bird is perched in the top of the tree, how far does it need to fl y to reach the pond, to the nearest metre?

24. A girl starting at her house, walks south for 2.7 km then walks east for 1.6 km. What is her bearing from the house, to the nearest degree?

25. The angle of depression from the top of a tower down to a car 250 m out from the foot of the tower is .38 19c l How high is the tower, to the nearest metre?

26. A hot air balloon fl ies south for 3.6 km then turns and fl ies east until it is on a bearing of 127c from where it started. How far east does it fl y?

27. A 24 m wire is attached to the top of a pole and runs down to the ground where the angle of elevation is .22 32c l Find the height of the pole.

Untitled-1 317 7/25/09 2:58:33 AM


28. A train depot has train tracks running north for 7.8 km where they meet another set of tracks going east for 5.8 km into a station. What is the bearing of the depot from the station, to the nearest degree?

29. Jessica leaves home and walks for 4.7 km on a bearing of .075c She then turns and walks for 2.9 km on a bearing of 115c and she is then due east of her home.

How far north does Jessica (a) walk?

How far is she from home? (b)

30. Builder Jo stands 4.5 m out from the foot of a building and looks up at to the top of the building where the angle of elevation is 71c. Builder Ben stands at the top of the building looking down at his wheelbarrow that is 10.8 m out from the foot of the building on the opposite side from where Jo is standing.

Find the height of the (a) building.

Find the angle of depression (b) from Ben down to his wheelbarrow.

Exact Ratios

A right-angled triangle with one angle of °45 is isosceles. The exact length of its hypotenuse can be found.

c a b

AC

AC

1 12

2

2 2 2

2 2 2

= +

= +

=

=

This means that the trigonometric ratios of 45c can be written as exact ratios.

Pythagoras’ theorem is used to fi nd the length of the hypotenuse.

sin

cos

tan

452

1

452

1

45 1

c

c

c

=

=

=

ch6.indd 318 7/10/09 4:09:19 AM


This angle is commonly used; for example, °45 is often used for the pitch of a roof. The triangle with angles of °60 and °30 can also be written with exact sides.

2 13

AD

AD 3

2 2 2= -

=

=

Halve the equilateral triangle to get .ABDT

60

60

60

°

°

°

sin

cos

tan

23

21

3

=

=

=

30sin

cos

tan

21

3023

303

1

c

c

c

=

=

=

It may be easier to remember the triangle

rather than all these ratios.

DID YOU KNOW?

The ratios of all multiples of these angles follow a pattern:

A 0c 30c 45c 60c 90c 120c 135c 150c

sin A 20

21

22

23

24

23

22

21

cos A 24

23

22

21

20

2

1-

22-

2

3-

The rules of the pattern are:

for sin • A , when you reach 4, reverse the numbers

for cos • A , when you reach 0, change signs and reverse

ch6.indd 319 6/25/09 10:09:36 PM


EXAMPLES

1. Find the exact value of °.sec 45

Solution

°°

seccos

4545

1

211

2

=

=

=

2. A boat ramp is to be made with an angle of 30c and base length 5 m. What is the exact length of the surface of the ramp?

Solution

cos

cos

cos

xx

x

30 5

30 5

305

23

5

53

2

310

310 3

#

c

c

c

=

=

=

=

=

=

=

So the exact length of the ramp is .3

10 3m

ch6.indd 320 6/25/09 10:09:45 PM


6.6 Exercises

Find the exact value in all questions, with rational denominator where relevant.

1. Evaluate (a) sin cos60 60c c+

(b) cos sin45 452 2c c+ (c) cosec 45c (d) sec2 60c (e) cot cot30 60c c+

(f) tan tan60 30c c-

(g) sin sin60 452 2c c+

(h) sin cos cos sin45 30 45 30c c c c+

(i) tan3 30c

(j) tan tan

tan tan1 45 60

45 60c c

c c

-

+

(k) cos cos sin sin30 60 30 60c c c c-

(l) cos sin30 302 2c c+ sec cosec2 45 30c c-(m)

(n) sinsin

452 60

c

c

(o) tan1 302 c+

(p) coscos

1 451 45

c

c

+

-

(q) seccot

6030c

c

(r) sin 45 12 c -

(s) cosec5 602 c

(t) sec

tan45

2 602 c

c-

2. Find the exact value of all pronumerals

(a)

(b)

(c)

3. A 2.4 m ladder reaches 1.2 m up a wall. At what angle is it resting against the wall?

4. A 2-person tent is pitched at an angle of .45c Each side of the tent is 2 m long. A pole of what height is needed for the centre of the tent?

5. If the tent in the previous question was pitched at an angle of ,60c how high would the pole need to be?

6. The angle of elevation from a point 10 m out from the base of a tower to the top of the tower is .30c Find the exact height of the tower, with rational denominator.

cos 45 ( )cos 452 2c c=

ch6.indd 321 6/25/09 10:09:56 PM


7. The pitch of a roof is 45c and spans a length of 12 .m

What is the length (a) l of the

roof? If a wall is placed inside the (b)

roof one third of the way along from the corner, what height will the wall be?

8. A 1.8 m ladder is placed so that it makes a 60c angle where it meets

the fl oor. How far out from the wall is it?

9. Find the exact length of AC .

10. The angle of depression from the top of a 100 m cliff down to a boat at the foot of the cliff is 30 .c How far out from the cliff is theboat?

Angles of Any Magnitude

The angles in a right-angled triangle are always acute. However, angles greater than 90c are used in many situations, such as in bearings. Negative angles are also used in areas such as engineering and science.

We can use a circle to fi nd trigonometric ratios of angles of any magnitude (size) up to and beyond 360 .c

Investigation

(a) Copy and complete the table for these acute angles 1. (between 0c and 90c).

x 0c 10c 20c 30c 40c 50c 60c 70c 80c 90c

sin x

cos x

tan x

(b) Copy and complete the table for these obtuse angles (between 90c and 180c).

x 100c 110c 120c 130c 140c 150c 160c 170c 180c

sin x

cos x

tan x

ch6.indd 322 6/25/09 10:10:05 PM


(c) Copy and complete the table for these refl ex angles (between 180c and 270c).

x 190c 200c 210c 220c 230c 240c 250c 260c 270c

sin x

cos x

tan x

(d) Copy and complete the table for these refl ex angles (between 270c and 360c).

x 280c 290c 300c 310c 320c 330c 340c 350c 360c

sin x

cos x

tan x

What do you notice about their signs? Can you see any patterns? 2. Could you write down any rules for the sign of sin, cos and tan for different angle sizes? Draw the graphs of 3. siny x= , cosy x= and tany x= for .x0 360c c# # For tany x= , you may need to fi nd the ratios of angle close to and either side of 90c and 270c.

Drawing the graphs of the trigonometric ratios can help us to see the change in signs as angles increase.

We divide the domain 0c to 360c into 4 quadrants:

1 st quadrant: 0c to 90c 2 nd quadrant: 90c to 180c 3 rd quadrant: 180c to 270c 4 th quadrant: 270c to 360c

EXAMPLES

1. Describe the sign of sin x in each section (quadrant) of the graph .siny x=

Solution

We can sketch the graph using the table below or using the values from the tables in the investigation above for more accuracy.

x 0c 90c 180c 270c 360c

y 0 1 0 -1 0

CONTINUED

ch6.indd 323 7/12/09 2:09:48 AM


y

90c 180c 270c 360c

1

-1

y = sin x

x

The graph is above the x -axis for the fi rst 2 quadrants, then below for the 3 rd and 4 th quadrants. This means that sin x is positive in the 1 st and 2 nd quadrants and negative in the 3 rd and 4 th quadrants.

2. Describe the sign of cos x in each section (quadrant) of the graph of .cosy x=

Solution


x 0c 90c 180c 270c 360c

y 1 0 -1 0 1

y

90c 180c 270c 360c

1

-1

y = cos x

x

The graph is above the x -axis in the 1 st quadrant, then below for the 2 nd and 3 rd quadrants and above again for the 4 th quadrant.

ch6.indd 324 6/25/09 10:10:21 PM


This means that cos x is positive in the 1 st and 4 th quadrants and negative in the 2 nd and 3 rd quadrants.

3. Describe the sign of tan x in each section (quadrant) of the graph tany x= .

Solution


x 0c 90c 180c 270c 360c

y 0 No result 0 No result 0

Neither tan 90c nor tan 270c exists (we say that they are undefi ned). Find the tan of angles close to these angles, for example tan 89c 59l and tan 90c 01l, tan 279c 59l and tan 270 .01c l

There are asymptotes at 90c and 270 .c On the left of 90c and 270c, tan x is positive and on the right, the ratio is negative.

y

x90c 180c 270c 360c

y = tanx

The graph is above the x -axis in the 1 st quadrant, below for the 2 nd , above for the 3 rd and below for the 4 th quadrant.

This means that tan x is positive in the 1 st and 3 rd quadrants and negative in the 2 nd and 4 th quadrants.

You will see why these ratios are undefi ned later

on in this chapter.

To show why these ratios have different signs in different quadrants, we look at angles around a unit circle (a circle with radius 1 unit).

We use congruent triangles when fi nding angles of any magnitude. Page 326 shows an example of congruent triangles all with angles of 20c inside a circle with radius 1 unit.

ch6.indd 325 7/12/09 2:17:45 AM


y

x

1 unit 1 unit

20c20c 20c

20c

1 unit 1 unit

If we divide the circle into 4 quadrants, we notice that the x - and y -values have different signs in different quadrants. This is crucial to notice when looking at angles of any magnitude and explains the different signs you get when fi nding sin, cos and tan for angles greater than 90c.

Quadrant 1

Looking at the fi rst quadrant (see diagram below), notice that x and y are both positive and that angle i is turning anticlockwise from the x -axis.

(x, y)

1 unit

First quadrant

y

x

y

xi

Point ( x , y ) forms a triangle with sides 1, x and y , so we can fi nd the trigonometric ratios for angle i .

The angle at the x -axis is 0 and the angle at the y -axis is 90c, with all other angles in this quadrant between these two angles .

ch6.indd 326 6/25/09 10:10:39 PM


Investigation

Since cos xi = and sin yi = , we can write the point ( x , y ) as (cos i , sin i ) .

The polar coordinates (cos i , sin i ) give a circle.

The polar coordinates ,sin sinA a c B bi i+] ]g g6 @ form a shape called a Lissajous fi gure. These are sometimes called a Bowditch curve and they are often used as logos, for example the ABC logo.

Use the Internet to research these and other similar shapes.

Use a graphics calculator or a computer program such as Autograph to draw other graphs with polar coordinates using variations of sin i and cos i .

These are called polar coordinates.

Quadrant 2

In the second quadrant, angles are between 90c and 180 .c If we take the 1 st quadrant coordinates ( x , y ), where x 02 and 0y2 and

put them in the 2 nd quadrant, we notice that all x values are negative in the second quadrant and y values are positive.

So the point in the 2 nd quadrant will be (- x , y )

x

y

0c

90c

180c

(-x, y)

1 unit

Second quadrant

y

x

180c-i

i

siny

y1

i =

=

cos x

x1

i =

=

tan xy

i =

ch6.indd 327 7/10/09 5:47:20 PM


Since cos xi = , cos i will negative in the 2 nd quadrant. Since sin yi = , sin i will be positive in the 2 nd quadrant.

tan xy

i = so it will be negative (a positive number divided by a negative number).

To have an angle of i in the triangle, the angle around the circle is 180c - i .

Quadrant 3

In the third quadrant, angles are between 180c and 270 .c

90c

270c

x

y

0c180c

(-x, -y)

1 unit

Third quadrant

i

y

x180c + i

Notice that x and y are both negative in the third quadrant, so cos i and sin i will be both negative.

tan xy

i = so will be positive (a negative divided by a negative number).

To have an angle of i in the triangle, the angle around the circle is 180c + i .

Quadrant 4

In the fourth quadrant, angles are between 270c and 360 .c

90c

270c

x

y

0c180c

(x, -y)

1 unit

Fourth quadrant

y

x

360c - i360ci

ch6.indd 328 6/25/09 10:10:53 PM


While y remains negative in the fourth quadrant, x is positive again, so sin i is negative and cos i is positive.

tan xy

i = so will be negative (a negative divided by a positive number)

For an angle i in the triangle, the angle around the circle is 360c - i .

ASTC rule

Putting all of these results together gives a rule for all four quadrants that we usually call the ASTC rule.

4th quadrant

1st quadrant

3rd quadrant

2nd quadrant

S A

T C180c + i 360c - i

360c

180c - i i

90c

270c

0c180c

y

x

You could remember this rule as A ll S tations

T o C entral or A S illy T rigonometry C oncept, or

you could make up your own!

This rule also works for the reciprocal trigonometric ratios. For example, where cos is positive, sec is also positive, where sin is positive, so is cosec and where tan is positive, so is cot.

We can summarise the ASTC rules for all 4 quadrants:

A: ALL ratios are positive in the 1 st quadrant S: Sin is positive in the 2 nd quadrant (cos and tan are negative) T: Tan is positive in the 3 rd quadrant (sin and cos are negative) C: Cos is positive in the 4 th quadrant (sin and tan are negative)

First quadrant: Angle i : sin i is positive cos i is positive tan i is positive

ch6.indd 329 6/25/09 10:11:00 PM


Second quadrant: Angle 180c i- : sin sin180c i i- =] g cos cos180c i i- = -] g tan tan180c i i- = -] g

Third quadrant: Angle 180c i+ : sin sin180c i i+ = -] g cos cos180c i i+ = -] g tan tan180c i i+ =] g

Fourth quadrant: Angle 360c i- : sin sin360c i i- = -] g cos cos360c i i- =] g tan tan360c i i- = -] g

EXAMPLES

1. Find all quadrants where (a) sin 02i (b) cos 01i (c) tan cos0 0and1 2i i

Solution

(a) sin 02i means sin i is positive. Using the ASTC rule, sin i is positive in the 1 st and 2 nd quadrants.

cos (b) i is positive in the 1 st and 4 th quadrants, so cos i is negative in the 2 nd and 3 rd quadrants.

tan (c) i is positive in the 1 st and 3 rd quadrants so tan i is negative in the 2 nd and 4 th quadrants. Also cos i is positive in the 1 st and 4 th quadrants. So tan i 1 0 and cos i 2 0 in the 4 th quadrant.

ch6.indd 330 6/25/09 10:11:09 PM


2. Find the exact ratio of tan 330c .

Solution

First we fi nd the quadrant that 330c is in. It is in the 4 th quadrant.

The angle inside the triangle in the 4 th quadrant is 30c and tan is negative in the 4 th quadrant.

tan tan330 30

31

c c= -

= -

3. Find the exact value of sin 225c .

Solution

The angle in the triangle in the 3 rd quadrant is 45c and sin is negative in the 3 rd quadrant.

CONTINUED

Notice that 30 3 0 .360 3c c c=-

Notice that 1 0 .45 2258 c c c=+

330c 30c

y

x

60c

30c

2

1

:3

ch6.indd 331 7/13/09 1:32:06 PM


225c45c

y

x

sin sin225 45

21

c c= -

= -

4. Find the exact value of cos 510c .

Solution

To fi nd cos 510c, we move around the circle more than once.

510c

150c30c

y

x

510 360 150

510 360 150

So

c c c

c c c

- =

= +

45c

45c

1

1:2

ch6.indd 332 6/25/09 10:11:27 PM


The angle is in the 2 nd quadrant where cos is negative. The triangle has 30c in it.

cos cos510 30

23

c c= -

= -

5. Simplify cos (180c + x ) .

Solution

180c + x is an angle in the 3 rd quadrant where cos is negative. So cos cosx x180c + = -] g

6. If sin 53x = - and cos x 2 0, fi nd the value of tan x and sec x .

Solution

sin x 1 0 in the 3 rd and 4 th quadrants and cos x 2 0 in the 1 st and 4 th quadrants. So sin x 1 0 and cos x 2 0 in the 4 th quadrant. This means that tan x 1 0 and sec x 2 0.

sin xhypotenuse

opposite=

So the opposite side is 3 and the hypotenuse is 5.

35

y

xx

By Pythagoras’ theorem, the adjacent side is 4.

sec x is the reciprocal of cos x so is positive in the

4 th quadrant .

This is a 3-4-5 triangle .

Notice that 180 .30 150c c c=-

CONTINUED

60c

30c

2

1

:3

ch6.indd 333 7/13/09 2:24:05 PM


So tan x43

= -

sec cosx x

1

45

=

=

The ASTC rule also works for negative angles. These are measured in the opposite way (clockwise) from positive angles as shown.

4th quadrant

1st quadrant

3rd quadrant

2nd quadrant

S A

T C-i

0

-(180c+ i )

-(180c- i )

-(360c- i )

-180c

y

x

-90c

-270c

-360c

The only difference with this rule is that the angles are labelled differently.

EXAMPLE

Find the exact value of tan (-120c) .

Solution

Moving around the circle the opposite way, the angle is in the 3 rd quadrant, with 60c in the triangle.

y

x120c

60c

Notice that 180 0 1 0 .( 6 ) 2c c c=- - -

ch6.indd 334 7/10/09 5:47:39 PM


Tan is positive in the 3 rd quadrant.

tan tan120 60

3

c c- =

=

] g

6.7 Exercises

1. Find all quadrants where (a) cos 02i

(b) tan 02i

(c) sin 02i

(d) tan 01i

(e) sin 01i

(f) cos 01i

(g) sin 01i and tan 02i

(h) cos 01i and tan 02i

(i) sin 02i and tan 01i

(j) sin 01i and tan 01i

2. (a) Which quadrant is the angle 240c in?

Find the exact value of cos (b) 240c .


Find the exact value of sin (b) 315c .


Find the exact value of (b) tan 120c .

5. (a) Which quadrant is the angle -225c in?

Find the exact value of (b) sin (-225c) .

6. (a) Which quadrant is the angle -330c in?

Find the exact value of (b) cos (-330c) .

7. Find the exact value of each ratio. tan 225(a) c cos 315(b) c tan 300(c) c sin 150(d) c cos 120(e) c sin 210(f) c cos 330(g) c tan 150(h) c sin 300(i) c cos 135(j) c

8. Find the exact value of each ratio. cos ((a) -225c) cos ((b) -210c) tan ((c) -300c) cos ((d) -150c) sin ((e) -60c) tan ((f) -240c) cos ((g) -300c) tan ((h) -30c) cos ((i) -45c) sin ((j) -135c)

60c

30c

2

1

:3

ch6.indd 335 6/25/09 10:52:20 PM


Trigonometric Equations

Whenever you fi nd an unknown angle in a triangle, you solve a trigonometric equation e.g. .cos x 0 34= . You can fi nd this on your calculator.

Now that we know how to fi nd the trigonometric ratios of angles of any magnitude, there can be more than one solution to a trigonometric equation if we look at a larger domain.

9. Find the exact value of cos 570(a) c tan 420(b) c sin 480(c) c cos 660(d) c sin 690(e) c tan 600(f) c sin 495(g) c cos 405(h) c tan 675(i) c sin 390(j) c

10. If tan43

i = and cos 01i , fi nd

sin i and cos i as fractions.

11. Given sin74

i = and tan 01i ,

fi nd the exact value of cos i and tan i .

12. If sin x 1 0 and tan x85

= - , fi nd

the exact value of cos x and cosec x.

13. Given cos 52x = and ,tan x 01

fi nd the exact value of cosec x , cot x and tan x .

14. If cos x 1 0 and sin x 1 0, fi nd cos x and sin x in surd form with

rational denominator if tan x75

= .

15. If sin94

i = - and

270 360c c1 1i , fi nd the exact

value of tan i and sec i .

16. If cos83

i = - and

° °180 2701 1i , fi nd the exact value of tan x , sec x and cosec x .

17. Given sin 0.3x = and tan x 1 0, express sin (a) x as a fraction fi nd the exact value of cos (b) x

and tan x .

18. If tan 1.2a = - and ° °270 3601 1i , fi nd the exact values of cot a , sec a and cosec a .

19. Given that 0.7cos i = - and 90 180c c1 1i , fi nd the exact value of sin i and cot i .

20. Simplify (a) sin 180c i-] g (b) cos x360c -] g (c) tan 180c b+^ h (d) sin 180c a+] g (e) tan 360c i-] g (f) sin i-] g (g) cos a-] g (h) tan x-] g

Use Pythagoras’ theorem to fi nd the third side .

This is called the principle solution.

ch6.indd 336 6/25/09 10:12:05 PM


EXAMPLES

1. Solve cos x23

= in the domain ° °x0 360# # .

Solution

23

is a positive ratio and cos is positive in the 1 st and 4 th quadrants .

So there are two possible answers. In the 1 st quadrant, angles are in the form of i and in the 4 th quadrant angles are in the form of 360c - i .

cos 3023

c =

But there is also a solution in the 4 th quadrant where the angle is 360c - i .

cos x23

For =

,,

x 30 360 3030 330c c cc c

= -=

2. Solve sin x2 1 02 =- for .x0 360c c# #

Solution

sin

sin

sin

sin

x

x

x

x

2 1 0

2 1

21

2

1

21

2

2

2

!

!

- =

=

=

=

=

Since the ratio could be positive or negative, there are solutions in all 4 quadrants. 1 st quadrant: angle i 2 nd quadrant: angle 180c - i 3 rd quadrant: angle 180c + i 4 th quadrant: angle 360c - i

60c

30c

2

1

:3

This is called the principle solution.

CONTINUED

ch6.indd 337 7/12/09 2:13:12 AM


, , ,

, , ,

sin

x

452

1

45 180 45 180 45 360 45

45 135 225 315

c

c c c c c c c

c c c c

=

= - + -

=

3. Solve tan x 3= for x180 180c c# #- .

Solution

3 is a positive ratio and tan is positive in the 1 st and 3 rd quadrants . So there are two possible answers. In the domain x180 180c c# #- , we use positive angles for x0 180c c# # and negative angles for .x180 0c c# #-

In the 1 st quadrant, angles are in the form of i and in the 3 rd quadrant angles are in the form of 180c i- -^ h . tan 60 3c = But there is also a solution in the 3 rd quadrant where the angle is

180c i- -^ h .

120c

,,

tan xx

360 180 6030

Forc c c

c

=

= - -

= -

] g

45c

45c

1

1:2

60c

30c

2

1

:3

4th quadrant

1st quadrant

3rd quadrant

2nd quadrant

S A

T C-(180c - i) -i

180c - i

90c

-90c

0c0c

180c-180c

y

x

i

ch6.indd 338 6/25/09 10:12:23 PM


4. Solve sin x2 2 1 0- = for 0 360xc c# # .

Solution

Notice that the angle is 2 x but the domain is for x . If 0 360xc c# # then we multiply each part by 2 to get the domain for 2 x .

0 2 720xc c# #

This means that we can fi nd the solutions by going around the circle twice!

sin

sin

sin

sin

x

x

x

2 2 1 0

2 2 1

221

3021

c

- =

=

=

=

Sin is positive in the 1 st and 2 nd quadrants. First time around the circle, 1 st quadrant is i and the 2 nd quadrant is 180c i- . Second time around the circle, we add 360c to the angles. So 1 st quadrant answer is 360c i+ and the 2 nd quadrant answer is 360 180c c i+ -] g or 540c i- .

,, ,

, , ,

, , ,

x

x

2 30 180 30 360 30 540 30

30 150 390 510

15 75 195 255

So c c c c c c c

c c c c

c c c c

= - + -

=

=`

The trigonometric graphs can also help solve some trigonometric equations.

EXAMPLE

Solve cos x 0= for 0 360xc c# # . cos 90 0c = However, looking at the graph of cosy x= shows that there is another solution in the domain 0 360xc c# # .

0

90 , 270

cos x

x

For

c c

=

=

Notice that these solutions lie inside the original domain of

.0 x 360c c# #

60c

30c

2

1

:3

y

90c 180c 270c 360c

1

-1

x

ch6.indd 339 6/25/09 10:55:57 PM


Investigation

Here are the 3 trigonometric graphs that you explored earlier in the chapter.

siny x=

cosy x=

tany x=

Use the values in the sin, cos and tan graphs to fi nd values for the inverse trigonometric functions in the tables below and then sketch the inverse trigonometric functions.

For example sin °270 1= -

cosec 270

11

1

So c =-

= -

Some values will be undefi ned, so you will need to fi nd values near them in order to see where the graph goes.

cosecy x=

x 0c 90c 180c 270c 360c

sin x

cosec x

ch6.indd 340 6/25/09 10:12:45 PM


Here are the graphs of the inverse trigonometric functions.

cosecy x=

secy x=

coty x=

secy x=

x 0c 90c 180c 270c 360c

cos x

sec x

coty x=

x 0c 90c 180c 270c 360c

tan x

cot x

y

y = cotx

x90c 180c 270c 360c

360c-1

1

0

ch6.indd 341 7/12/09 2:09:33 AM


1. Solve for .0 360c c# #i (a) .sin 0 35i =

(b) cos21

i = -

(c) tan 1i = -

(d) sin23

i =

(e) tan3

1i = -

(f) cos2 3i =

(g) tan 2 3i =

(h) sin2 3 1i = -

(i) cos2 2 1 0i - =

(j) tan 3 12 i =

2. Solve for .180 180c c# #i- (a) .cos 0 187i =

(b) sin21

i =

(c) tan 1i =

(d) sin23

i = -

(e) tan3

1i = -

(f) tan3 12 i =

(g) tan 2 1i =

(h) sin2 3 12 i =

(i) 1 0tani + =

(j) tan 2 32 i =

3. Sketch cosy x= for 0 360 .xc c##

4. Evaluate .sin 270c

5. Sketch tany x= for 0 360 .xc c# #

6. Solve 0tanx = for 0 360 .xc c# #

7. Evaluate .cos180c

8. Find the value of .sin90c

9. Solve cosx 1= for .x0 360c c# #

10. Sketch siny x= for .x180 180c c# #-

11. Evaluate .cos 270c

12. Solve sin x 1 0+ = for .x0 360c c# #

13. Solve cos x 12 = for .x0 360c c# #

14. Solve sin x 0= for .x0 360c c# #

15. Solve sin x 1= for .x360 360c c# #-

16. Sketch secy x= for .x0 360c c# #

17. Sketch coty x= for

.x0 360c c# #

6.8 Exercises

Trigonometric Identities

Trigonometric identities are statements about the relationships of trigonometric ratios. You have already met some of these—the reciprocal ratios, complementary angles and the rules for the angle of any magnitude.

ch6.indd 342 6/25/09 10:13:03 PM


cosecsin

seccos

cottan

1

1

1

ii

ii

ii

=

=

=

Reciprocal ratios

Complementary angles

sin cos

cosec sec

tan cot

90

90

90

c

c

c

i i

i i

i i

= -

= -

= -

]]]

ggg

Angles of any magnitude

sin sin

cos cos

tan tan

180

180

180

c

c

c

i i

i i

i i

- =

- = -

- = -

]]]

ggg

( )

( )

( )

sin sin

cos cos

tan tan

180

180

180

c

c

c

i i

i i

i i

+ = -

+ = -

+ =

( )

( )

( )

sin sin

cos cos

tan tan

360

360

360

c

c

c

i i

i i

i i

- = -

- =

- = -

)

)

cos

tan

i i

i i

=

= -

) sini i= -(

(

(

sin

cos

tan

-

-

-

In this section you will learn some other identities, based on the unit circle. In the work on angles of any magnitude, we defi ned sin i as the y -coordinate of P and cos i as the x -coordinate of P .

ch6.indd 343 6/25/09 10:13:11 PM


tan

cossinxy

i

ii

=

=

tancossin

iii

=

cot

tan

sincos

1i

i

ii

=

=

cotsincos

iii

=

Pythagorean identities

The circle has equation .x y 12 2+ = Substituting cosx i= and siny i= into 1x y2 2+ = gives

cos sin 12 2i i+ =

This is an equation so can be rearranged to give

sin coscos sin

11

2 2

2 2i ii i

==

--

There are two other identities that can be derived from this identity.

tan sec1 2 2i i+ =

Remeber that cos2 i means (cos ) 2i .

ch6.indd 344 6/25/09 10:13:17 PM


Proof

cos sin

coscos

cossin

costan sec

1

1

1

2 2

2

2

2

2

2

2 2

i i

i

i

i

i

i

i i

+ =

+ =

+ =

This identity can be rearranged to give

tan sec

sec tan

1

1

2 2

2 2

i i

i i

=

=

-

-

cot cosec12 2i i+ =

Proof

cos sin

sincos

sinsin

sincot cosec

1

1

1

2 2

2

2

2

2

2

2 2

i i

i

i

i

i

i

i i

+ =

+ =

+ =

This identity can be rearranged to give

cot cosec

cosec cot

1

1

2 2

2 2

i i

i i

= -

= -

These are called Pythagorean identities since the equation

of the circle comes from Pythagoras’ rule (see Chapter 5).

EXAMPLES

1. Simplify .sin coti i

Solution

sin cot sin

sincos

cos

#i i iii

i

=

=

2. Simplify sin sec90c b b-^ h where b is an acute angle .

Solution

1

sin sec coscos

90 1#c b b b

b- =

=

^ h

CONTINUED

ch6.indd 345 7/12/09 2:12:38 AM


3. Simplify .sin sin cos4 2 2i i i+

Solution

sin sin cos sin sin cos

sin

sinsin

1

4 2 2 2 2 2

2

2

i i i i i i

i

ii

+ = +

=

=

=

^

]

h

g

4. Prove .cot tan cosec secx x x x+ =

Solution

cot tan

sincos

cossin

sin coscos sin

sin cos

sin coscosec sec

x x

xx

xx

x xx x

x x

x xx x

1

1 1

LHS

RHS

2 2

#

= +

= +

=+

=

=

=

=

cot tan cosec secx x x x+ =`

5. Prove that .sin

coscosx

xx

11

12

-=

+

Solution

sincos

coscos

cos coscos

cos

xx

xx

x xx

x

1

11

1 11

11

LHS

RHS

2

2

=-

=-

-

=+ -

-

=+

=

] ]g g

11sin

coscosx

xx

12

`-

+=

ch6.indd 346 7/25/09 5:21:14 PM


6.9 Exercises

1. Simplify (a) sin 90c i-] g (b) tan 360c i-] g (c) cos i-] g (d) cot 90c i-] g (e) sec 180c a+] g

2. Simplify (a) tan cosi i (b) tan coseci i (c) sec cotx x (d) sin x1 2-

(e) cos1 2 a-

(f) cot x 12 +

(g) tan x1 2+

(h) 1sec2 i -

(i) cot5 52 i +

(j) cosec x

12

(k) sin cosec2 2a a

(l) cot cot cos2i i i-

3. Prove that (a) cos sinx x12 2- = -

(b) sec tancos

sin1i i

ii

+ =+

(c) tansin

3 31

322

aa

+ =-

(d) sec tanx x2 2-

cosec cotx x2 2= -

(e) sin cosx x 3-] g +2sin cosx x

sin cos sin cosx x x x2 2

2= - -

(f) cot sec2i i+

sin cossin sin1 22

i ii i

=- +

(g) cos cot902 c i i-] g

= sin i cos i

(h) ( )( )cosec cot cosec cotx x x x 1+ - =

( )cos

sin cos

tan cos

1i2

2 2

2 2

i

i i

i i

-

= +

( )cosec

cotcos

tan cot

sec

1j

b

bb

b b

b

+-

=+

4. If cosx 2 i= and siny 2 i= , show that 4x y2 2+ = .

5. Show that 81x y2 2+ = if

cosx 9 i= and y = 9 sin i.

Non-right-angled Triangle Results

A non-right-angled triangle is named so that its angles and opposite sides have the same pronumeral. There are two rules in trigonometry that refer to non-right-angled triangles. These are the sine rule and the cosine rule .

ch6.indd 347 7/10/09 5:49:02 PM


Proof

In ,ABCT draw perpendicular AD and call it h .

From ,ABDT

sinsin

B ch

h c B`

=

=

(1)

From ,ACDT

sin

sin

Cbh

h b C`

=

=

(2)

From (1) and (2),

sin sin

sin sin

c B b C

bB

cC

=

=

Similarly, drawing a perpendicular from C it can be proven that

.sin sina

Ab

B=

or

sin sin sina

Ab

Bc

C= = Use this rule for fi nding an angle.

Use this rule for fi nding a side.

sin sin sinAa

Bb

Cc

= =

Sine rule

ch6.indd 348 6/25/09 10:13:48 PM


EXAMPLES

1. Find the value of x , correct to 1 decimal place.

Solution

Name the sides a and b, and angles A and B.

.

.

.

.

sin sin

sin sin

sin sin

sinsin

sin sin

Aa

Bb

x

x

x

43 21 79 1210 7

43 21 79 1210 7

79 1210 7 43 21

7 5

43 21 43 21

cm

# #

c c

c c

c

c

c c

Z

=

=

=

=

l l

l l

l

l

l l

2. Find the value of y , to the nearest whole number.

Solution

( )

sin sin

sin sin

sin sin

sinsin

sin sin

Y

Aa

Bb

y

y

y

180 53 24

103

103 538

103 538

538 103

10

103 103# #

c c c

c

c c

c c

c

c

c c

+

Z

= - +

=

=

=

=

=

You can rename the triangle ABC or just make

sure you put sides with their opposite angles

together.

The sine rule uses 2 sides and 2 angles, with 1

unknown.

You need to fi nd Y+ fi rst, as it is opposite y .

CONTINUED

ch6.indd 349 7/12/09 2:10:06 AM


3. Find the value of ,i in degrees and minutes.

Solution

1-

. .

. .86 11

..

..

6.7 6.7

sin sin

sin sin

sin sin

sin sin

sin sin

aA

bB

6 7 8 386 11

6 7 8 3

8 36 7 86 11

8 36 7 86 11

53 39

# #

c

c

c

c

cZ

i

i

i

i

=

=

=

=

=

l

l

l

l

l

c m

Since sin x is positive in the fi rst 2 quadrants, both acute angles (between 0c and 90c) and obtuse angles (between 90c and 180c) give positive sin ratios. e.g. .sin50 0 766c= and .sin130 0 766c = This affects the sine rule, since there is no way of distinguishing between an acute angle and an obtuse angle. When doing a question involving an obtuse angle, we need to use the 2 nd quadrant angle of 180c - i rather than relying on the calculator to give the correct answer.

EXAMPLE

Angle i is obtuse. Find the value of ,i in degrees and minutes.

ch6.indd 350 7/12/09 2:14:09 AM


6.10 Exercises

1. Evaluate all pronumerals, correct to 1 decimal place.

(a)

(b)

(c)

(d)

(e)

Solution

. .

. .

..

..

. .

sin sin

sin sin

sin sin

sin sin

sin sin

aA

bB

11 9 5 415 49

11 9 5 415 49

5 411 9 15 49

5 411 9 15 49

36 55

180 36 55

143 05

11 9 11 9

acute angleBut is obtuse

1

`

# #

c

c

c

c

c

c c

c

i

i

i

i

i

i

=

=

=

=

=

=

= -

=

-

l

l

l

l

l

l

l

c^

mh

ch6.indd 351 7/12/09 2:14:22 AM


2. Find the value of all pronumerals, in degrees and minutes.

(a)

(b)

(c)

(d)

((e) i is obtuse)

3. Triangle ABC has an obtuse angle at A . Evaluate this angle to the nearest minute if AB = 3.2 cm,

BC = 4.6 cm and .ACB 33 47c+ = l

4. Triangle EFG has FEG 48c+ = , EGF 32c+ = and FG = 18.9 mm. Find the length of

the shortest side (a) the longest side. (b) .

5. Triangle XYZ has ,XYZ 51c+ = YXZ 86c+ = and XZ = 2.1 m. Find the length of

the shortest side (a) the longest side .(b)

6. Triangle XYZ has XY = 5.4 cm, ZXY 48c+ = and .XZY 63c+ = Find the length of XZ .

7. Triangle ABC has BC = 12.7 m, ABC 47c+ = and ACB 53c+ = as shown. Find the lengths of

(a) AB (b) AC.

53c47c

12.7 mCB

A

8. Triangle PQR has sides PQ = 15 mm, QR = 14.7 mm and PRQ 62 29c+ = l . Find to the nearest minute

(a) QPR+ (b) .PQR+

9. Triangle ABC is isosceles with AB = AC . BC is produced to D as shown. If AB = 8.3 cm, BAC+ = 52c and ADC 32c+ = fi nd the length of

4.9

3.7

21c31l i

The shortest side is opposite the smallest angle and the longest side is opposite the largest angle .

ch6.indd 352 7/12/09 2:14:34 AM


(a) AD (b) BD.

32c

52c8.3 cm

DBC

A

10. Triangle ABC is equilateral with side 63 mm. A line is drawn from A to BC where it meets BC at D and .DAB 26 15c+ = l Find the length of

(a) AD (b) DC.

Cosine rule

cosc a b ab C22 2 2= + -

Similarly

cosa b c bc A22 2 2= + -

cosb a c ac B22 2 2= + -

Proof

BCD

cbp

x a - x

A

In triangle ABC , draw perpendicular CD with length p and let CD = x . Since BC = a , BD = a - x From triangle ACD

b x p2 2 2= + (1)

cos

cos

Cbx

b C x`

=

=

(2)

From triangle DAB

c p a xp a ax xp x a ax

22

2 2 2

2 2 2

2 2 2

= + -

= + - +

= + + -

] g

(3)

ch6.indd 353 6/25/09 10:15:14 PM


Substitute (1) into (3):

c b a ax22 2 2= + - (4)

Substituting (2) into (4):

cos

cos

c b a a b C

b a ab C

2

2

2 2 2

2 2

= + -

= + -

] g

DID YOU KNOW?

Pythagoras’ theorem is a special case of the cosine rule when the triangle is right angled.

cosc a b 2ab C2 2 2= -+

When C = 90c

2 cos 90

2 0

c a b ab

a b ab

a b

2 2 2

2 2

2 2

=

=

- c+

+ -

= +

] g

EXAMPLE

Find the value of x , correct to the nearest whole number.

Solution

.99 79Z

10Z

5.6 6.4 2(5.6)(6.4) 112 32

.

cos

cos

c a b ab C

x

x

2

99 79

2 2 2

2 2 2 c

= + -

= + -

=

l

The cosine rule uses 3 sides and 1 angle, with 1 unknown.

, ,, , ,,% %

Press 5.6 6.4 2 5.6 6.4

cos 112 32

x x2 2# #

#

+ -

= =

ch6.indd 354 7/12/09 2:14:50 AM


When fi nding an unknown angle, it is easier to change the subject of this formula to cos C.

cos

cos

cos

cos

cos

cos

cos cos

c a b ab C

c a b ab C

c ab C a b

c ab C a b

ab C a b c

ab C a b c

ab C ab C

c c

ab ab

2

2

2

2

2

2

2 2

2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2

= + -

= + -

+ = +

+ = +

= + -

=+ -

+ +

- -

Subtract the square of the side opposite the

unknown angle.

EXAMPLES

1. Find ,i in degrees and minutes.

Solution

cos

cos

cos

Cab

a b c2

2 5 65 6 3

6052

6052

29 56

2 2 2

2 2 2

1

cZ

i

i

=+ -

=+ -

=

= -

l

] ]

c

g g

m

2. Evaluate BAC+ in degrees and minutes .

C

4.5 cm

A

B8.4 cm

6.1 cm

CONTINUED

cosCab

a b c2

2 2 2

=+ -

Similarly

cos Abc

b c a2

2 2 2

=+ -

cos Bac

a c b2

2 2 2

=+ -

ch6.indd 355 7/12/09 2:10:24 AM


Solution

1-

. .. . .

.

.

cos

cos

cos

Cab

a b c

BAC

BAC

2

2 4 5 6 14 5 6 1 8 4

0 2386

0 2386103 48

2 2 2

2 2 2

c

+

+

=+ -

=+ -

= -

= -

= l

] ]]g g

g

Notice that the negative sign tells us that the angle will be obtuse.

6.11 Exercises

1. Find the value of all pronumerals, correct to 1 decimal place.

(a)

(b)

(c)

(d)

(e)

2. Evaluate all pronumerals correct to the nearest minute

(a)

(b)

(c)

ch6.indd 356 7/12/09 2:15:11 AM


(d)

(e)

3. Kite ABCD has AB = 12.9 mm, CD = 23.8 mm and ABC 125c+ = as shown. Find the length of diagonal AC.

12.9 mm 125c

23.8 mm

A

B

C

D

4. Parallelogram ABCD has sides 11 cm and 5 cm, and one interior angle .79 25c l Find the length of the diagonals .

5. Quadrilateral ABCD has sides AB = 12 cm, BC = 10.4 cm, CD = 8.4 cm and AD = 9.7 cm with .ABC 63 57c+ = l

Find the length of diagonal (a) AC Find (b) DAC+ Find (c) .ADC+

6. Triangle XYZ is isosceles with XY = XZ = 7.3 cm and

YZ = 5.9 cm. Find the value of all angles, to the nearest minute .

7. Isosceles trapezium MNOP has MP = NO = 12 mm, MN = 8.9 mm, OP = 15.6 mm and 11 15 .NMP 9c l+ =

Find the length of diagonal (a) NP. Find (b) .NOP+

8. Given the fi gure below, fi nd the length of

(a) AC (b) AD.

8.4 cm

42c8l

A

B

C

D

101c38l

3.7 cm

9.9 cm

9. In a regular pentagon ABCDE with sides 8 cm, fi nd the length of diagonal AD .

10. A regular hexagon ABCDEF has sides 5.5 cm.

Find the length of (a) AD. Find (b) .ADF+

ch6.indd 357 7/10/09 4:12:05 AM


Application s

The sine and cosine rules can be used in solving problems.

Use the sine rule to fi nd:

a 1. side, given one side and two angles an 2. angle, given two sides and one angle

Use the cosine rule to fi nd: a 1. side, given two sides and one angle an 2. angle, given three sides

EXAMPLES

1. The angle of elevation of a tower from point A is .72c From point B , 50 m further away from the tower than A , the angle of elevation is .47c

Find the exact length of (a) AT . Hence, or otherwise, fi nd the height (b) h of the tower to 1 decimal place.

Solution

)

(a)

(

sin sin

sin sin

sinsin

BAT

BTA

Aa

Bb

AT

AT

180 72108

180 47 10825

47 2550

2550 47

straight angle

angle sum of

`

c c

c

c c c

c

c c

c

c

+

+ T

= -

=

= - +

=

=

=

=

^]

hg

Use BTAT to fi nd AT.

ch6.indd 358 6/25/09 10:15:53 PM


( )

50 47

.

sin

sin

sinsin sin

ATh

h AT

72

72

2572

82 3

b

m

`

#

c

c

c

cc

Z

=

=

=

2. A ship sails from Sydney for 200 km on a bearing of ,040c then sails on a bearing of 157c for 345 km.

How far from Sydney is the ship, to the nearest km? (a) What is the bearing of the ship from Sydney, to the nearest degree? (b)

Solution

)

( )

(

SAN

SAB

180 40140

360 140 15763

a cointerior angles

angle of revolution`

c c

c

c c c

c

+

+

= -

=

= - +

=

^^

hh

( ) ( )

.

.

cos

cos

c a b ab C

x

x

2

200 345 2 200 345 6396374 3

96374 3310

2 2 2

2 2 2 c

Z

Z

= + -

= + -

=

So the ship is 310 km from Sydney.

( )

.

sin sin

sin sin

sin sin

b aA

bB

345 31063

310345 63

0 99

82

`

c

c

c

Z

Z

i

i

i

=

=

=

40 82

122

The bearing from Sydney c c

c

= +

=

Use right-angled ATOT to fi nd h . Do not use the

sine rule.

To fi nd the bearing, measure TSB.+

ch6.indd 359 6/25/09 10:16:00 PM


6.12 Exercises

1. Find the lengths of the diagonals of a parallelogram with adjacent sides 5 cm and 8 cm and one of its angles .32 42c l

2. A car is broken down to the north of 2 towns. The car is 39 km from town A and 52 km from town B .If A is due west of B and the 2 towns are 68 km apart, what is the bearing of the car from (a) town A (b) town B , to the nearest degree?

3. The angle of elevation to the top of a tower is 54 37c l from a point 12.8 m out from its base. The tower is leaning at an angle of 85 58c l as shown. Find the height of the tower.

54c37l 85c58l

12.8 m

4. A triangular park has sides 145.6 m, 210.3 m and 122.5 m. Find the size of the largest interior angle of the park.

5. A 1.5 m high fence leans outwards from a house at an angle of 102c. A boy sits on top of the fence and the angle of depression from him down to the house is .32 44c l How far from the fence is the house?

6. Football posts are 3.5 m apart. If a footballer is standing 8 m

from one post and 11 m from the other, fi nd the angle within which the ball must be kicked to score a goal, to the nearest degree.

7. A boat is sinking 1.3 km out to sea from a marina. Its bearing is 041c from the marina and 324c from a rescue boat. The rescue boat is due east of the marina.

How far, correct to 2 decimal (a) places, is the rescue boat from the sinking boat?

How long will it take the (b) rescue boat, to the nearest minute, to reach the other boat if it travels at 80 km/h?

8. The angle of elevation of the top of a fl agpole from a point a certain distance away from its base is .20c After walking 80 m towards the fl agpole, the angle of elevation is .75c Find the height of the fl agpole, to the nearest metre.

9. A triangular fi eld ABC has sides 85AB m= and 50 .AC m= If B is on a bearing of 065c from A and C is on a bearing of 166c from A , fi nd the length of BC , correct to the nearest metre.

10. (a) Find the exact value of AC in the diagram.

Hence, or otherwise, fi nd the (b) angle ,i correct to the nearest minute.

ch6.indd 360 7/12/09 2:15:27 AM


11. Find the value of h , correct to 1 decimal place.

12. A motorbike and a car leave a service station at the same time. The motorbike travels on a bearing of 080c and the car travels for 15.7 km on a bearing of 108c until the bearing of the motorbike from the car is .310c How far, correct to 1 decimal place, has the motorbike travelled?

13. A submarine is being followed by two ships, A and B , 3.8 km apart, with A due east of B . If A is on a bearing of 165c from the submarine and B is on a bearing of 205c from the submarine, fi nd the distance from the submarine to both ships.

14. A plane fl ies from Dubbo on a bearing of 1 93 c for 852 km, then turns and fl ies on a bearing of 2 58 cuntil it is due west of Dubbo. How far from Dubbo is the plane, to the nearest km?

15. A triangular roof is 16.8 m up to its peak, then 23.4 m on the other side with a 125c angle at the peak as shown. Find the length of the roof.

125c23.4 m16.8 m

16. Rhombus ABCD with side 8 cm has diagonal BD 11.3 cm long. Find .DAB+

17. Zeke leaves school and runs for 8.7 km on a bearing of 338c, then turns and runs on a bearing of 061c until he is due north of school. How far north of school is he?

18. A car drives due east for 83.7 km then turns and travels for 105.6 km on a bearing of 029c. How far is the car from its starting point?

19. The fi gure below shows the diagram that a surveyor makes to measure a triangular piece of land. Find its perimeter.

58c1l132c31l

14.3 m

11.4 m

13.9 m

20. A light plane leaves Sydney and fl ies for 1280 km on a bearing of

.050c It then turns and fl ies for 3215 km on a bearing of .149c How far is the plane from Sydney, to the nearest km?

21. Trapezium ABCD has AD BC; , with AB = 4.6 cm, BC = 11.3 cm, CD = 6.4 cm, DAC 2 303c+ = l and ABC 78c+ = .

Find the length of (a) AC. Find (b) ADC+ to the nearest

minute .

22. A plane leaves Adelaide and fl ies for 875 km on a bearing of

.056c It then turns and fl ies on a bearing of i for 630 km until it is due east of Adelaide. Evaluate i to the nearest degree.

ch6.indd 361 7/10/09 4:12:51 AM


Similarly,

sin

sin

A ac B

A bc A

21

21

=

=

Proof

From ,BCDD

sin

sin

sin

C ah

h a C

A bh

ba C

21

21

`

=

=

=

=

23. Quadrilateral ABCD has AB = AD = 7.2 cm, BC = 8.9 cmand CD = 10.4 cm, with DAB 107c+ =

Find the length of diagonal (a) BD. Find (b) BCD+ .

24. Stig leaves home and travels on a bearing of 248c for 109.8 km. He then turns and travels for 271.8 km on a bearing of 143c. Stig then turns and travels home on a bearing of a .

How far does he travel on the (a) fi nal part of his journey?

Evaluate (b) a .

25. A wall leans inwards and makes an angle of 88c with the fl oor.

A 4 m long ladder leans against (a) the wall with its base 2.3 m out from the wall. Find the angle that the top of the ladder makes with the wall.

A longer ladder is placed the (b) same distance out from the wall and its top makes an angle of 31c with the wall.

How long is this (i) ladder?

How much further (ii) does it reach up the wall than the fi rst ladder?

Area

To fi nd the area of a triangle, you need to know its perpendicular height. Trigonometry allows us to fi nd this height in terms of one of the angles in the triangle.

sinA ab C21

=

ch6.indd 362 7/10/09 4:12:53 AM


EXAMPLE

Find the area of ABCD correct to 2 decimal places.

Solution

( . ) ( . )

.

sin

sin

A ab C21

21 4 3 5 8 112 34

11 52 units2

c

Z

=

= l

To fi nd the area, use 2 sides and their included angle.

6.13 Exercises

1. Find the area of each triangle correct to 1 decimal place. (a)

(b)

(c)

(d)

(e)

ch6.indd 363 7/12/09 2:15:42 AM


2. Calculate the exact area of .ABCD

3. Find the area of OABD correct to 1 decimal place ( O is the centre of the circle).

4. Find the area of a parallelogram with sides 3.5 cm and 4.8 cm, and one of its internal angles ,67 13c l correct to 1 decimal place.

5. Find the area of kite ABCD , correct to 3 signifi cant fi gures.

6. Find the area of the sail, correct to 1 decimal place.

7. Find the area of a regular hexagon with sides 4 cm, to the nearest .cm2

8. Calculate the area of a regular pentagon with sides 12 mm.

9. The fi gure below is made from a rectangle and isosceles triangle with AE = AB as shown.

10.5 cm

84c

A

B

CD

14.3 cm

E

Find the length of (a) AE. Find the area of the fi gure .(b)

10. Given the following fi gure,

9.4 cm

44c

A

B C D6.7 cm

58c

36c

Find the length of (a) AC Find the area of triangle (b) ACD Find the area of triangle (c) ABC .

ch6.indd 364 7/12/09 2:15:58 AM


Trigonometry in Three Dimensions

EXAMPLES

1. From point X , 25 m due south of the base of a tower, the angle of elevation is .47c Point Y is 15 m due east of the tower. Find:

the height, (a) h , of the tower, correct to 1 decimal place the angle of elevation, (b) ,i of the tower from point Y .

Solution

From (a) XTOD

.

tan

tan

h

h

h

4725

25 47

26 8

c

c

=

=

=

So the tower is 26.8 m high.

From (b) YTOD

.

.

tan

tan

1526 8

1526 8

60 46

1`

c

i

i

=

=

=

-

l

c m

So the angle of elevation from Y is 60 46 .c l

2. A cone has a base diameter of 18 cm and a slant height of 15 cm. Find the vertical angle at the top of the cone.

Solution

The radius of the base is 9 cm.

36 52

sin

sin

159

1591

`

c

i

i

=

=

=

-

l

c m

2

73 44

Vertical angle

c

i=

= l

Use the full value of 26.80921775 for a more

accurate answer to (b).

ch6.indd 365 7/12/09 2:16:45 AM


6.14 Exercises

1. A gymnastics bar is supported by wires as shown below.

If one wire is inclined at an (a) angle of 55c to the horizontal and is 1.4 m out from the base of the bar, fi nd the height of the bar, to the nearest metre.

The second wire is inclined at (b) an angle of 68c to the horizontal. How long is the wire (to 1 decimal place)?

The third wire is 2.2 m long. (c) What is its angle of elevation?

2. A pole has two supporting ropes, 2.5 m and 3.1 m long.

If the 3.1 m rope makes an (a) angle of elevation of ,38c fi nd the length of the pole, correct to 1 decimal place.

What angle of elevation does (b) the other rope make?

3. A 25 11 8cm cm cm# # cardboard box contains an insert (the shaded area) made of foam.

Find the area of foam in the (a) insert, to the nearest .cm2

Find the angle, (b) ,i the insert makes at the corner of the box.

4. A cone has radius 7 cm and a slant height of 13 cm. Find the vertical angle at the top of the cone, in degrees and minutes.

5. From a point 15 m due north of a tower, the angle of elevation of the tower is 32c

Find the height of the tower, (a) to the nearest metre.

Find the angle of elevation (b) of the tower at a point 20 m due east of the tower.

6. A pole is seen from two points A and B . The angle of elevation from A is .58c If CAB 52c+ = and ,ABC 34c+ = and A and B are 100 m apart, fi nd:

how far (a) A is from the foot of the pole, to the nearest metre.

the height of the pole, to (b) 1 decimal place.

ch6.indd 366 6/25/09 10:16:56 PM


7. Two straight paths up to the top of a cliff are inclined at angles of 25c and 22c to the horizontal.

If path 1 is 114 m long, fi nd (a) the height of the cliff, to the nearest metre.

Find the length of path 2, to (b) 1 decimal place.

If the paths meet at (c) 47c at the base of the cliff, fi nd their distance apart at the top of the cliff, correct to 1 decimal place.

8. David walks along a straight road. At one point he notices a tower on

a bearing of 053c with an angle of elevation of .21c After walking 230 m, the tower is on a bearing of ,342c with an angle of elevation of .26c Find the height of the tower correct to the nearest metre.

9. A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fi t inside a 12 cm high box. At what angle must it be tilted?

10. A hot air balloon fl ying at 950 m/h at a constant altitude

of 3000 m is observed to have an angle of elevation of .78c After 20 minutes, the angle of elevation is .73c Calculate the angle through which the observer has turned during those 20 minutes.

Sums and Differences of Angles

Sums and differences

Angles can be expressed as sums or differences of other angles. This enables us to simplify or evaluate some angles that normally would be too hard to simplify.

cos cos cos sin sinx y x y x y- = +^ h

ch6.indd 367 7/10/09 6:28:00 PM


Proof

By the distance formula:

( ) ( )( )

cos cos sin sin

cos cos cos cos sin sin sin sin

cos sin cos sin cos cos sin sincos cos sin sin

d x x y y

AB x y x y

x x y y x x y y

x x y y x y x yx y x y

2 2

2 22 2

22 1

22 1

2

2 2 2

2 2 2 2

2 2 2 2

= - + -

= - + -

= - + + - +

= + + + - -

= - +

_ _^ ^

i ih h

(1)

By the cosine rule:

( ) ( ) ( )( )

cos

coscos

a b ab C

AB x yx y

c 2

1 1 2 1 12 2

2 2 2

2 2 2

= + -

= + - -

= - -

(2)

From (1) and (2):

cos cos cos sin sin

cos cos cos sin sin

x y x y x y

x y x y x y

2 2 2 2

`

- - = - +

- = +

^ ^^

h hh

Remember: cos x coordinate= -i and sin y coordinate.= -i

cos cos cos sin sinx y x y x y+ = -^ h

Proof

Substitute y- for .y

y

sin y

y y

( )

( ( )) ( ) ( )

( ) ( )

( )

cos cos cos sin sin

cos cos cos sin sin

cos cos cos sin

cos cos cos sin sin

x y x y x y

x x x

x y x y x

x y x y x y

- = +

- - = - + -

+ = + -

+ = -

sin sin cos cos sinx y x y x y+ = +^ h

ch6.indd 368 6/25/09 10:17:13 PM


Proof

Substitute 90 .x xforc -

( )

( ) ( ) ( )

( ( ))

( )

cos cos cos sin sin

cos cos cos sin sin

cos sin cos cos sin

sin sin cos cos sin

x y x y x y

x y x y x y

x y x y x y

x y x y x y

90 90 90

90

c c c

c

- = +

- - = - + -

- + = +

+ = +

sin sin cos cos sinx y x y x y- = -^ h

Proof

Substitute .y yfor-

( )

( ( )) ( ) ( )

( ) ( )

( )

sin sin cos cos sin

sin sin cos cos sin

sin sin cos cos sin

sin sin cos cos sin

x y x y x y

x y x y x y

x y x y x y

x y x y x y

+ = +

+ - = - + -

- = + -

- = -

tantan tan

tan tanx y

x y

x y

1+ =

-

+^ h

Proof

( )

( )

tan

cos cos sin sinsin cos cos sin

cos coscos cos sin sin

cos cossin cos cos sin

tantan tan

tan tan

cos

sinx y

x y

x y

x y x yx y x y

x yx y x y

x yx y x y

x yx y

x y

1

+ =+

+

=-

+

=-

+

+ =-

+

^^

hh

Divide top and bottom by

.cos x cos y

tantan tan

tan tanx y

x y

x y

1- =

+

-^ h

ch6.indd 369 6/25/09 10:17:21 PM


Proof

Substitute y- for y .

))

)

)

y

y

y

( )

( (

(

(

tantan tan

tan tan

tantan tan

tan tan

tantan tan

tan tan

tantan tan

tan tan

x yx y

x y

xx y

x y

xx y

x y

xx y

x y

1

1

1

1

+ =-

+

+ - =- -

+ -

- =- -

-

- =+

-

^^

^hh

h

EXAMPLES

1. Simplify .sin cos cos sin2 2i i i i-

Solution

( )sin cos cos sin sin

sin2 2 2i i i i i i

i- = -

=

2. Find the exact value of .cos75c

Solution

)45c+(cos cos

cos cos sin sin

75 30

30 45 30 45

23

21

21

21

2 2

3 1

2 2

3 1

2

2

46 2

# #

#

c c

c c c c

=

= -

= -

=-

=-

=-

3. Simplify .cos sin60 60c ci i+ + +] ]g g

Solution

( ) ( )cos sin

cos cos sin sin sin cos cos sin

cos sin sin cos

cos sin

cos sin

60 60

60 60 60 60

21

23

21

23

21

23

23

21

21 3

21 3

# # # #

c c

c c c c

i i

i i i i

i i i i

i i

i i

+ + +

= - + +

= - + +

= + + - +

=+

+-

d dd d

n nn n

ch6.indd 370 7/10/09 4:13:02 AM


Ratios of double angles

By using the sum of angles, we can fi nd the trigonometric ratios for double angles.

sin sin cosx x x2 2=

Proof

( )sin sin

sin cos cos sin

sin cos

x x x

x x x x

x x

2

2

= +

= +

=

cos cos sin

sin

cos

x x x

x

x

2

1 2

2 1

2 2

2

2

= -

= -

= -

Proof

( )

( )

( )

cos cos

cos cos sin sin

cos sin

sin sin

sin

cos

cos

x x x

x x x x

x x

x x

x

x

x

2

1

1 2

1 2 1

2 1

2 2

2 2

2

2

2

= +

= -

= -

= - -

= -

= - -

= -

Remember:

.sin x cos x 12 2+ =

tantan

tanxx

x21

22

=-

Proof

( )tan tan

tan tantan tan

tantan

tan

x x x

x xx x

xx

x

2

1

21

22

= +

=-

+

=-

ch6.indd 371 6/25/09 10:17:36 PM


EXAMPLES

1. Simplify .cos sin2 22 2i i-

Solution

2 2 2(2 )

4cos sin cos

cos

2 2i i i

i

- =

=

2. If 74,sinx = fi nd the exact value of .sin x2

Solution

sin sin cos

AC

ACx x x

7 433

332 2

274

733

498 33

2 2 2

`

# #

= -

=

=

=

=

=

PROBLEM

Ulug Beg (1393–1449) used the relation sin sin sin41 3 33 i i i= -] g to

draw up a table of sine ratios. Can you prove this relation?

6.15 Exercises

1. Expand (a) sin a b-] g (b) cos p q+^ h (c) tan a b+^ h (d) ( )sin x 20c+ (e) tan x48c +] g (f) cos 2i a-] g (g) ( )cos x 75c+ (h) tan x y5 7-^ h (i) sin 4a b-^ h (j) tan 3a b-^ h

2. Simplify (a) sin cos cos sina b a b+

(b) tan tan

tan tan1 36 29

36 29c c

c c

-

+

(c) cos cos sin sin28 27 28 27c c c c- (d) sin cos cos sinx y x y2 3 2 3+

(e) tan tan

tan tan1 3

3i i

i i+

-

(f) sin cos cos sin74 42 74 42c c c c- (g) sin sina b a b+ -+] ]g g (h) sin sinx y x y+ --^ ^h h (i) cos cosx y x y- +-^ ^h h (j) cos cosm n m n-+ +] ]g g

ch6.indd 372 6/25/09 10:17:46 PM


3. Find the exact value of (a) sin 75c (b) cos 15c (c) tan 75c (d) tan 105c (e) cos 105c (f) sin 15c (g) sin 105c (h) tan 285c (i) ( ) ( )sin cosx x30 30c c+ + + (j) cos cosy y45 45c c- + +^ ^h h

4. Simplify

.tan tan

tan tan

x y x y

x y x y

1 - + -

+ + -

^ ^^ ^

h hh h

5. If sin x32

= and ,cos y43

= fi nd

the exact value of (a) sin x y+^ h (b) cos x y-^ h (c) tan x y+^ h

6. By taking 2 ,i i i= + fi nd an expression for

(a) sin 2i (b) cos 2i (c) tan 2i

7. By writing 3i as 2 ,i i+ fi nd an expression in terms of i for

(a) sin 3i (b) cos 3i (c) tan 3i

8. (a) Simplify .tan tan

tan tan1 7 3

7 3i i

i i+

-

Find an expression for sin (b) 4i in terms of 7i and 3 .i

9. Find an expression for cos x9 in terms of 2x and 7 .x

10. Find the exact value of (a) cos cos sin sin23 22 23 22c c c c-

(b) tan tan

tan tan1 85 25

85 25c c

c c

+

-

(c) sin cos180 60c c cos sin180 60c c+

(d) cos cos290 80c c sin sin290 80c c+

(e) tan tan

tan tan1 11 19

11 19c c

c c

-

+

11. If sin x53

= and ,cos y135

= fi nd

the value of (a) cos x (b) sin y (c) sin x y-^ h (d) tan y (e) tan x y+^ h

12. (a) Write an expression for .cos cosx y x y+ + -^ ^h h

Hence write an expression for (b) .cos cos50 65c c

13. Find an expression for (a) sin sinx y x y+ + -^ ^h h (b) cos cosx y x y+ - -^ ^h h (c) sin sinx y x y+ - -^ ^h h (d) cos sinx y x y+ + -^ ^h h (e) tan tanx y x y+ + -^ ^h h (f) tan tanx y x y+ - -^ ^h h

14. Expand (a) sin b2 (b) tan 2i (c) cos 2i (d) ( )sin x y2+ (e) ( )cos 2a b+ (f) ( )tan x y2+ (g) ( )sin 2i d- (h) ( )cos 2i c- (i) ( )tan x z2- (j) ( )sin x y2 2-

15. Simplify (a) cos sinx x2 3 3 (b) cos siny y7 72 2-

(c) tan

tan1 5

2 52 i

i

-

(d) sin y1 2 2- (e) sin cos6 6i i (f) sin cosx x 2+] g (g) cos2 3 12 a -

(h) sin1 2 402 c-

(i) tan

tan

1

22 b

b

-

(j) sin cosx x3 3 2-] g

ch6.indd 373 6/25/09 10:17:53 PM


16. Find the exact value of (a) . .cos sin22 5 22 5c c (b) cos sin30 302 2c c-

(c) tan

tan1 15

2 152 c

c

-

(d) sin cos2 75 75c c

(e) tan

tan1 120

2 1202 c

c

-

(f) sin1 2 1652 c-

(g) .cos2 22 5 12 c-

(h) tan

tan1

22 i

i

- where .112 5ci =

(i) . .sin cos67 5 67 5c c (j) cos sin2 105 105c c

17. If 85,cosx = fi nd the exact value

of cos x2 and .sin x2

18. If sin53

a = and ,tan512

b = fi nd

the exact values of (a) sin a b+^ h (b) cos 2a (c) sin 2b (d) tan a b-^ h

19. Express sin 4i in terms of .i

20. (a) Simplify .cos

sinx

x1 2

2+

Hence, fi nd the exact value of (b) .tan 15c

21. Find the exact value of tan 2221c

by using the expression for .tan x2

22. Prove

(a) sin sin tan21 22 i i i=

(b) tansin

cos2

1ii

i=

-

23. Show that .sin sin sin sin7 4 11 32 2i i i i- =

24. Prove that .cos cos cos3 4 33i i i= -

25. Find an expression for sin x3 in terms of .sin x

Further Trigonometric Equations

Some trigonometric equations are diffi cult to solve. However, there are some expressions that can be used to solve them.

Ratios in terms of tan i2

If ,tan t2i

= then tant

t1

22

i =-

Proof

tantan

tan

tantan

tan

AA

A

A

tt

21

2

12

22 2

12

where

2

2

2

` ii

i

i

=-

=

-

=

=-

ch6.indd 374 7/10/09 4:13:24 AM


If ,tan t2i

= then sint

t1

22

i =+

Proof

tan t t2 1i

= =

2 2

sin sin cos

sin sin cos

A A A

A

t

t

t

tt

2 2

2 2

21 1

1

12

where

2 2

2

` ii i

i

=

= =

=+ +

=+

e eo o

The hypotenuse is t1 2+ by Pythagoras’

theorem.

These ratios for sin2

i

and cos2

i come from the

triangle above.

If ,tan t2i

= then costt

11

2

2

i =+

-

The ratios for cos2

i and

sin2

i come from the

previous triangle.

Proof

2

cos cos sin

cos cos sin

A A A

A

t t

t

t tt

tt

2

2 2

1

1

1

11

1

11

where

2 2

2 2

2

2

2

2

2 2

2

2

2

` ii i

i

= -

= - =

=+

-+

=+

-+

=+

-

e eo o

ch6.indd 375 7/10/09 4:13:26 AM


EXAMPLES

1. Find the exact value of .tan

tan1 15

2 152 c

c

+

Solution

sintt

12

2i =

+ where tant

2i

=

tantan sin

1 152 15 30

21

So2 c

cc

+=

=

2. Prove that .cot cot tan2

22

ii

i- =

Solution

LHS 2

RHS

cot cot

tan tan

tan

tan

ttt

t

t tt

t tt

tt

tt

t

2

2

1 2

1

122

2

12

2 1

1 1

1 1

2

where

2

2

2

2

2

ii

i i

i

i

= -

= -

= -

-

=

= --

= --

=- +

=

=

=

=

^ h

cot cot tan2

22

`i

ii

- =

There is also another expression that will help solve some further trigonometric equations.

sin cos sin

tan

a b r

r a b ab

where

and2 2

i i i a

a

+ = +

= + =

] g

Untitled-1 376 7/25/09 3:42:57 AM


Proof

( )

( )

LHS

sin

sin cos cos sin

sin cos

sin cos

r

a b

a ba b

a

a b

b

a b

RHS2 2

2 2

2 2 2 2# #

i a

i a i a

i i

i i

= +

= + +

= ++

++

= +

=

e o

sin cos sin

tan

a b r

r a b abwhere and2 2

` i i i a

a

+ = +

= + =

] g

If tanab

,=a then the

hypotenuse is a b2 2+ by Pythagoras’ theorem.

EXAMPLES

1. Write sin cosx x3 + in the form sinr x a+] g .

Solution

sin cos sina b ri i i a+ = +] g where r a b2 2= + and tan ab

a =

:

,

sin cosx x

a b

r a b

3

3 1

3 13 1

42

For

2 2

2 2

+

= =

= +

= +

= +

=

=

tan

tan

ab

31

31

30

1

c

a

a

=

=

=

=

-e o

So sin cosx x3 + = 2 sin( x + 30c)

CONTINUED

Untitled-1 377 7/25/09 3:44:46 AM


2. Write sin cos3 2i i+ in the form sinr i a+] g . Solution

( )sin cos sina b ri i i a+ = + where r a b2 2= + and tan ab

a =

For sin cos3 2i i+ :

3, 2a b= =

r a b

3 29 4

13

2 2

2 2

= +

= +

= +

=

°

tan

tan

ab

32

32

33 41

1

l

a

a

=

=

=

=

- c m

So ( )sin cos sin3 2 13 33 41ci i i+ = + l

Class Investigation

Can you fi nd similar results for these?

• sin cosa bi i-

• cos cosa bi i+

• cos sina bi i-

6.16 Exercises

1. Simplify

(a) 1

2tt

2-

(b) 11

tt

2

2

+

-

(c) tan

tan1 10

2 102 c

c

-

(d) tantan

1 251 25

2

2

c

c

+

-

(e) tan

tan1

22 i

i

+

(f) tan

tan

12

12

2

2

i

i

+

-

2. Find the exact value of

(a) tan

tan1 30

2 302 c

c

+

ch6.indd 378 6/25/09 10:18:36 PM


(b) ..

tantan

1 22 51 22 5

2

2

c

c

+

-

(c) tantan

1 301 30

2

2

c

c

+

-

(d) tan

tan1 90

2 902 c

c

-

3. Write each expression in terms of

t where .tant2i

=

(a) cosec i (b) sec i (c) cot i (d) sin cosi i+ (e) 1 tani+

(f) 12

tan tanii

+

(g) cos sin3 4i i+

(h) sin cossin cos

11

i ii i

+ -

+ +

(i) tan seci i+

(j) sin 2i

4. Prove .sin cossin cos t

11

i ii i

+ +

+ -=

5. Find an expression for sin cos2 2i i- in terms of t .

6. Write each expression in the form sinr i a+] g .

(a) sin cos2 i i+ (b) sin cos3i i+ (c) sin cosi i+ (d) sin cos5 2i i+ (e) sin cos4 i i+ (f) sin cos3 i i+ (g) sin cos2 3i i+ (h) sin cos4 7i i+ (i) sin cos5 4i i+ (j) sin cos3 5i i+

7. Write each expression in the form sinr i a-] g.

(a) sin cosi i- (b) sin cos2i i- (c) sin cos3i i- (d) sin cos3 i i- (e) sin cos5 2i i-

8. Write the expression cos sin3 i i+ in the form cosr i a-] g .

9. Write the expression cos sin3i i- in the form cosr i a+] g .

10. Write the expression sin cos9 2i i+ in the form .

(a) sinr i a+] g (b) cosr i a-] g

EXAMPLES

1. Solve sin cos2 i i= for .0 360c c# #i

Solution

sin cos2 i i=

Dividing both sides by cos i : (check cos 0i = does not give a solution)

.

cossin

coscos

tan

tan

2

2 1

0 5

ii

ii

i

i

=

=

=

CONTINUED

We can use these results to help solve some trigonometric equations.

ch6.indd 379 7/12/09 2:11:03 AM


Since tan i is positive in the fi rst and third quadrants:

,,

26 34 180 26 3426 34 206 34c c cc c

i = +=

l ll l

2. Solve cos cos2i i= for .0 360c c# #i

Solution

cos cos

cos cos

cos coscos cos

2

2 1

2 1 02 1 1 0

2

2

i i

i i

i i

i i

=

- =

- - =

+ - =] ]g g

,

,

cos cos

cos cos

cos

2 1 0 1 0

2 1 1

21 0 360

120 240

or`

c c

c c

i i

i i

i i

i

+ = - =

= - =

= - =

=

, ,,0 120 240 360solutions are` c c c ci =

3. Solve sin cosx x3 1+ = for .x0 360c c# #

Solution (Method 1)

Use the result for .sin cosa x b x+ For ,sin cosx x a3 3+ = and 1b =

2

r a b

3 1

2 2

2 2

= +

= +

=

^ h

`

tan ab

31

30c

a

a

=

=

=

sin cos sinx x x3 2 30` c+ = +] g Solving:

sin cossin

sin

x x xx x

x

3 1 0 3602 30 1 30 30 390

3021

forfor

c c

c c c c

c

# #

# #

+ =

+ = +

+ =

]]

gg

` , ,

, ,

, ,

x

x

30 30 180 30 360 30

30 150 390

0 120 360

c c c c c c

c c c

c c c

+ = - +

=

=

Sine is positive in the fi rst and second quadrants.

ch6.indd 380 6/25/09 10:18:54 PM


Solution (Method 2)

Use the results for 2

.tant i=

sin cosx x

tt

tt

tt t

t t t

t t

t t

3 1

31

211 1

12 3 1

1

2 3 1 1

0 2 2 3

2 3

2 2

2

2

2

2 2

2

+ =

++

+

-=

+

+ -=

+ - = +

= -

= -

d

^

n

h

`

`

,

0 ,120 , 360

tan tan

t t

t tx x x

x x

x

2 0 3 0

0 3

20

23 0

2180

20 180

260

or

for c c

c c c

c c c

# #

= - =

= =

= =

= =

=

Test x 180c= separately:

)1(sin cos3 180 180 0

11

c c

!

+ = + -= -

x 180` c= is not a solution

Solutions are , , .x 0 120 360c c c=

General solutions of trigonometric equations

Often the solutions of trigonometric equations are restricted, for example, to 0 360 .c c## i If the solutions are not restricted, then they can be described by a general formula.

EXAMPLE

Find all solutions for .sin23

i =

Solution

CONTINUED

ch6.indd 381 7/12/09 2:11:19 AM


,540 60c c- g,360 60c c+ -

, , , ,

360 360 60

, , , , ,

,

( ), ( ), [ ( )],

,

sin sin 60

60 180 60 360 60 360 180 60

60 180 60 360 60 540 60 720 60

180 60 360 60 360 180 60

180 60

If can also be negative

c

c c c c c c c c

c c c

c c c c c c c c c

c c c c c c c

c c

i

i

i

i

=

= - + + -

+ +

= - + - +

= - + - - - - -

= - - -

g

g

g

1

60sin sin

n n180 60

So the general solution for is

where is an integer.n#

c

c c

i

i

=

= + -] g

In general, the solution for sin sini a= is given by )1 a(n180 ni = + - where n is an integer.

EXAMPLE

Find all solutions for .cos2

1i =

Solution

, , , ,

360 360 45 ,

, , , , ,

cos cos45

45 360 45 360 45 360 360 45

45 360 45 360 45 720 45 720 45

c

c c c c c c c c

c c c f

c c c c c c c c c f

i

i

=

= - + + -

+ +

= - + - +

, ( ), ( ), [ ( )],

, , , ,

45 360 45 360 45 360 360 45

45 360 45 360 45 720 45

If can also be negative,

c c c c c c c c f

c c c c c c c f

i

i = - - - - + - + -

= - - + - - - +

cos cos

n n

45

360 45

So the general solution for is

where is an integer.# !

c

c c

i

i

=

=

In general, the solution for cos cosi a= is given by n360 !i a= where n is an integer.

Sin is positive in the 1st and 2nd quadrants.

Cos is positive in the 1st and 4th quadrants.

ch6.indd 382 7/25/09 3:40:44 AM


6.17 Exercises

In general, the solution for tan tani a= is given by 180ni a= + where n is an integer.

EXAMPLE

Find all solutions for .tan 1i =

Solution

,540 45c c f+,360 45c c+ -,180 45c c+ -

, , ,

360 180 45 , 360 360 45 ,

, , , , ,

( ), ( ), [ ( )],

45 180 45 360 45

45 180 45 360 45 540 45 720 45

180 45 360 45 360 180 45

tan tan 45

If can also be negative, then

c

c c c c c

c c c c c c f

c c c c c c c c c f

c c c c c c c f

i

i

i

i

=

= + +

+ + + +

= + + + +

= - - - - - + -

= -

.n n180 45

The general solution for tan tan 45 is

where is an integer#

c

c c

i

i

=

= +

1. Solve for 0 360 .xc c##

(a) sin cosx x=

(b) cos sinx x3=

(c) sin sinx x2 =

(d) tan tanx x 02 - =

(e) sin sinx x2 1 02 - - =

(f) sin cosx x2 3 3 02 + - =

(g) sin cot sinx x x 0- =

(h) cos x 1 02 - = (i)

2 1 0

sin tan tan

sin

x x x

x

2 -

+ - =

(j) cos cosx x3 7 4 02 - + =

2. Solve for .0 360c c# #i

(a) sin cos3 4 0i i+ =

(b) 3cos sin5 12i i- = -

(c) sin cos3 0i i- =

(d) 1sin cosi i+ = -

(e) sin cos4 3 0i i- + =

(f) sin cos 1i i- =

(g) cos sin2 1i i+ =

(h) sin cos225

i i- =

(i) cos sin3 5 2 0i i- + =

(j) cos sin2 1 0i i+ + =

Tan is positive in the 1st and 3rd quadrants.

ch6.indd 383 6/25/09 10:19:14 PM


3. Find the general solution for

(a) sin21

i =

(b) tan 3a =

(c) cos23

i =

(d) 1sin x2 = -

(e) tan 1 0i + =

(f) cos2 12 b =

(g) sin4 32 c =

(h) tan3

1i =

(i) .cos 0 245i =

(j) .sin 0 399a =

4. Solve sin x2 4523

c- =] g for

.x180 180c c# #-

5. Find the general solutions of 2 .sin cosx x=

6. Solve sin2 sinx = x for .x180 180c c# #-

7. Find the general solutions of (a) sin 0i = (b) 1cosx = (c) 0tanx = (d) 1sini = - (e) 0cosa =

8. For each question solve for (i) x0 360c c# # and fi nd the general solutions (ii)

(a) sin x2 1 0- = (b) cos x4 3 0- = (c) sin cosx x3= (d) sin cosx x3 0+ = (e) sin cosx x 2+ =

9. Find the general solutions of – .sin sinx x2 1 02 + =

10. (a) Solve cos 2 x = cos x for x0 360c c# # .

(b) Find the general solutions of cos 2 x = cos x .

ch6.indd 384 7/10/09 4:14:03 AM


1. Find the exact value of cosi and sini if

.tan53

i =

2. Simplify

(a) sin cotx x

(b) cos

cos sin40

40 50c

c c+

(c) cot A1 2+

(d) 11

tt

2

2

+

- where tant2i

=

(e) sin1 2 102 i-

3. Evaluate to 2 decimal places. (a) sin 39 54c l (b) tan 61 30c l (c) cos 19 2c l

4. Find i to the nearest minute if (a) .sin 0 72i = (b) .cos 0 286i =

(c) tan75

i =

5. Prove that .sin

cos sin12 2 2

2

ii

i-

= +

6. Find the value of b if .sin cosb b2 30 c= -] g

7. Find the exact value of (a) cos 315c (b) sin 60c-] g (c) tan 120c (d) sin cos2 105 105c c

(e) sin sinx y x178when and- =^ h

cos y135

=

8. Solve cos x2 1= - for .x0 360c c# #

9. Sketch the graph of ,cosy x= and hence solve cos x 0= for 0 360 .xc c##

10. A ship sails on a bearing of 215c from port until it is 100 km due south of port. How far does it sail, to the nearest km?

11. Find the length of AB as a surd.

12. Evaluate x , correct to 2 signifi cant fi gures. (a)

(b)

13. Evaluate i to the nearest minute.

(a)

(b)

(c)

Test Yourself 6

ch6.indd 385 7/12/09 2:16:27 AM


14. Find the area of triangle MNO .

15. Solve for x180 180c c# # .-

(a) sin x432 =

(b) tan x23

1=

(c) tan tanx x3 2 =

16. If sec45

i = - and ,tan 02i fi nd sin i and .cot i

17. Jacquie walks south from home for 3.2 km, then turns and walks west for 1.8 km. What is the bearing, to the nearest degree, of

Jacquie from her home? (a) her home from where Jacquie is now? (b)

18. Find the general solution of .sin cos6 8 5i i- =

19. The angle of elevation from point B to the top of a pole is 39 ,c and the angle of elevation from D , on the other side of the pole, is . B42c and D are 20 m apart.

(a) Find an expression for the length of AD . Find the height of the pole, to (b)

1 decimal place.

20. A plane fl ies from Orange for 1800 km on a bearing of .300c It then turns and fl ies for 2500 km on a bearing of .205c How far is the plane from Orange, to the nearest km?

21. Find the exact value of (a) sin 75c (b) cos105c (c) .sin cos22 30 22 30c cl l

22. Find the general solutions of 2 cos (a) x – 1 = 0 tan (b) x = 1

sin (c) x = 23

.

23. Solve 3 1sin cosi i+ = for 0 360c c# #i .

24. Evaluate a in the fi gure below .

12 mm

10 mm

4 mm

a

25. (a) Simplify cos x cos y – sin x sin y . (b) Show that cos 2 x = 1 – sin 2 x .

ch6.indd 386 7/10/09 4:14:09 AM


Challenge Exercise 6

1. Two cars leave an intersection at the same time, one travelling at 70 km/h along one road and the other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection?

2. A ship sails from port on a bearing of ,055c then turns and sails on a bearing of 153c for 29.1 km, when it is due east of port. How far, to 1 decimal place, is the ship from its starting point?

3. Evaluate x correct to 3 signifi cant fi gures.

4. (a) Find an exact expression for the length of AC .

(b) Hence, or otherwise, fi nd the value of h correct to 1 decimal place.

5. A man walks 3.8 km on a bearing of 134c from a house. He then walks 2.9 km on a bearing of .029c How far is he from the house, to 1 decimal place?

6. Simplify .sin tanx x360 90$c c- -] ]g g

7. Find the exact area of .ABCD

8. Find the exact value of ( ) .cos 315c-

9. Solve 2 1 0tan x - = for .x0 360c c# #

10. Find i to the nearest minute.

11. The angle of depression from the top of a 4.5 m mast of a boat down to a fi sh is 56 28 .c l How far down, to 1 decimal place, does a pelican sitting at the top of the mast need to fl y to catch the fi sh?

12. Solve 2 ( 10 ) 1cos ci + = - for .0 360c c# #i

13. Two roads meet at an angle of 74 .c Find the distance, correct to 3 signifi cant fi gures, between two cars, one 6.3 km from the intersection along one road and the other 3.9 km along the other road.

ch6.indd 387 7/12/09 2:16:13 AM


14. Find the exact value of ,cosi given

sin95

i = and .cos 01i

15. From the top of a vertical pole the angle of depression to a man standing at the foot of the pole is 43 .c On the other side of the pole is another man, and the angle of depression from the top of the pole to this man is 52 .c The men are standing 58 m apart. Find the height of the pole, to the nearest metre.

16. Show that

.sin sin

cos sin costan

1 11

i i

i i ii

+ -

+= +] ]

]g g

g

17. If 3x = sin i and ,cosy 3 2i= - eliminate i to fi nd an equation relating x and y .

18. From point A , 93 m due south of the base of a tower, the angle of elevation is 35 .c Point B is 124 m due east of the tower. Find

the height of the tower, to the (a) nearest metre

the angle of elevation of the tower (b) from point B .

19. ABCD is a triangular pyramid with 7 , 10 , 8 ,BC CD BD AB ACm m m= = = = and 67 .ACB c+ = Calculate

(a) BCD+ length (b) AB , to the nearest metre .

20. A cone has a base diameter of 14 cm and a perpendicular height of 26 cm. Find the vertical angle at the top of the cone.

21. Show that .cos cos sin sin cos6 4 6 4 2 5 12i i i i i- = -

22. A cable car 100 m above the ground is seen to have an angle of elevation of 65c when it is on a bearing of 345 .c After a minute, it has an angle of elevation of 69c and is on a bearing of 025 .c Find how far it travels in that minute, and its speed in .ms 1-

23. Solve cos sin2 0i i- = for .0 360c c# #i

24. Find the general solutions of .1sini = -

25. Simplify cosec cos 1i i -] g by expressing

it in terms of .tant2ic m

ch6.indd 388 7/10/09 4:44:21 AM

ch6.indd 389 6/25/09 10:20:20 PM

Ch 6 Trigonometry

Documents

Transcript of Ch 6 Trigonometry