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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
293Chapter 6 Trigonometry
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
294 Maths In Focus Mathematics Extension 1 Preliminary Course
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
295Chapter 6 Trigonometry
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
296 Maths In Focus Mathematics Extension 1 Preliminary Course
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
297Chapter 6 Trigonometry
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
298 Maths In Focus Mathematics Extension 1 Preliminary Course
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
299Chapter 6 Trigonometry
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
300 Maths In Focus Mathematics Extension 1 Preliminary Course
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
301Chapter 6 Trigonometry
(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
302 Maths In Focus Mathematics Extension 1 Preliminary Course
(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
303Chapter 6 Trigonometry
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
304 Maths In Focus Mathematics Extension 1 Preliminary Course
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
305Chapter 6 Trigonometry
(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
306 Maths In Focus Mathematics Extension 1 Preliminary Course
(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
307Chapter 6 Trigonometry
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
308 Maths In Focus Mathematics Extension 1 Preliminary Course
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
309Chapter 6 Trigonometry
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
310 Maths In Focus Mathematics Extension 1 Preliminary Course
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
311Chapter 6 Trigonometry
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
312 Maths In Focus Mathematics Extension 1 Preliminary Course
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
313Chapter 6 Trigonometry
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
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314 Maths In Focus Mathematics Extension 1 Preliminary Course
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?
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315Chapter 6 Trigonometry
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)
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316 Maths In Focus Mathematics Extension 1 Preliminary Course
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
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317Chapter 6 Trigonometry
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.
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318 Maths In Focus Mathematics Extension 1 Preliminary Course
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
=
=
=
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319Chapter 6 Trigonometry
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
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320 Maths In Focus Mathematics Extension 1 Preliminary Course
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
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321Chapter 6 Trigonometry
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=
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322 Maths In Focus Mathematics Extension 1 Preliminary Course
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
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323Chapter 6 Trigonometry
(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
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324 Maths In Focus Mathematics Extension 1 Preliminary Course
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
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 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.
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325Chapter 6 Trigonometry
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
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 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.
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326 Maths In Focus Mathematics Extension 1 Preliminary Course
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 .
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327Chapter 6 Trigonometry
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 =
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328 Maths In Focus Mathematics Extension 1 Preliminary Course
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
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329Chapter 6 Trigonometry
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
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330 Maths In Focus Mathematics Extension 1 Preliminary Course
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.
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331Chapter 6 Trigonometry
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
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332 Maths In Focus Mathematics Extension 1 Preliminary Course
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
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333Chapter 6 Trigonometry
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
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334 Maths In Focus Mathematics Extension 1 Preliminary Course
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=- - -
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335Chapter 6 Trigonometry
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 .
3. (a) Which quadrant is the angle 315c in?
Find the exact value of sin (b) 315c .
4. (a) Which quadrant is the angle 120c in?
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
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336 Maths In Focus Mathematics Extension 1 Preliminary Course
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
337Chapter 6 Trigonometry
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
338 Maths In Focus Mathematics Extension 1 Preliminary Course
, , ,
, , ,
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
339Chapter 6 Trigonometry
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
340 Maths In Focus Mathematics Extension 1 Preliminary Course
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
341Chapter 6 Trigonometry
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
342 Maths In Focus Mathematics Extension 1 Preliminary Course
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
343Chapter 6 Trigonometry
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
344 Maths In Focus Mathematics Extension 1 Preliminary Course
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
345Chapter 6 Trigonometry
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
346 Maths In Focus Mathematics Extension 1 Preliminary Course
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
347Chapter 6 Trigonometry
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
348 Maths In Focus Mathematics Extension 1 Preliminary Course
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
349Chapter 6 Trigonometry
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
350 Maths In Focus Mathematics Extension 1 Preliminary Course
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
351Chapter 6 Trigonometry
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
352 Maths In Focus Mathematics Extension 1 Preliminary Course
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
353Chapter 6 Trigonometry
(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
354 Maths In Focus Mathematics Extension 1 Preliminary Course
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
355Chapter 6 Trigonometry
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
356 Maths In Focus Mathematics Extension 1 Preliminary Course
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
357Chapter 6 Trigonometry
(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
358 Maths In Focus Mathematics Extension 1 Preliminary Course
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
359Chapter 6 Trigonometry
( )
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
360 Maths In Focus Mathematics Extension 1 Preliminary Course
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
361Chapter 6 Trigonometry
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
362 Maths In Focus Mathematics Extension 1 Preliminary Course
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
363Chapter 6 Trigonometry
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
364 Maths In Focus Mathematics Extension 1 Preliminary Course
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
365Chapter 6 Trigonometry
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
366 Maths In Focus Mathematics Extension 1 Preliminary Course
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
367Chapter 6 Trigonometry
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
368 Maths In Focus Mathematics Extension 1 Preliminary Course
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
369Chapter 6 Trigonometry
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
370 Maths In Focus Mathematics Extension 1 Preliminary Course
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
371Chapter 6 Trigonometry
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
372 Maths In Focus Mathematics Extension 1 Preliminary Course
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
373Chapter 6 Trigonometry
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
374 Maths In Focus Mathematics Extension 1 Preliminary Course
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
375Chapter 6 Trigonometry
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
376 Maths In Focus Mathematics Extension 1 Preliminary Course
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
377Chapter 6 Trigonometry
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
378 Maths In Focus Mathematics Extension 1 Preliminary Course
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
379Chapter 6 Trigonometry
(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
380 Maths In Focus Mathematics Extension 1 Preliminary Course
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
381Chapter 6 Trigonometry
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
382 Maths In Focus Mathematics Extension 1 Preliminary Course
,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
383Chapter 6 Trigonometry
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
384 Maths In Focus Mathematics Extension 1 Preliminary Course
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
385Chapter 6 Trigonometry
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
386 Maths In Focus Mathematics Extension 1 Preliminary Course
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
387Chapter 6 Trigonometry
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
388 Maths In Focus Mathematics Extension 1 Preliminary Course
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
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