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Trigonomet
angle of elevation angle of depression
P
Trigonometry I must go down to the seas again, to the lonely sea and the sky, And all I ask is a tall ship and a star to steer her by. John Masefield
Trigonometry background
Angles of elevation and depression The angle of elevation is the angle between the horizontal and a direction above the horizontal (see figure 7.1). The angle of depression is the angle between the horizontal and a direction below the horizontal (see figure 7.2). Figure 7.1 Figure 7. Bearing The bearing (or compass bearing) is the direction measured as an angle from north, clockwise (see figure 7.3). N 150° W E 216 Figure 7.
this
directi
on is
S
a
beari
ng of
150°
AD
2
2
2
AD 3.
217
45°
From triangle ABD,
sin 60
; cos 60
tan 60 3;
sin 30
cos 30
tan 30
Without using a calculator, find the value of cos 60°sin 30° cos
2
(Note that cos
2
30° means (cos 30°)
2
SOLUTION
2
cos 60°sin 30° cos
2
(ii) The angle 45°
In figure 7.6, triangle PQR is a right-angled isosceles triangle with equal sides
of length 1 unit.
Q
1
P R
Figure 7.6 1
Using Pythagoras’ theorem, PQ 2.
This gives
sin 45
; cos 45
; tan 45 1.
(iii) The angles 0° and 90°
Although you cannot have an angle of 0° in a triangle (because one side would
be lying on top of another), you can still imagine what it might look like. In
figure 7.7, the hypotenuse has length 1 unit and the angle at X is very small.
218
X
Figure
7.
adjacent
Z
opposite
Y
EXAMPLE 7.
Trigonomet ry
P
an angle
x
sin 60°
AB 2l
3 l
Positive and negative angles
Unless given in the form of bearings, angles are measured from the x axis (see
figure 7.8). Anticlockwise is taken to be positive and clockwise to be negative.
of +135°
x
an angle of –30°
Figure 7.
In the diagram, angles ADB and CBD are right angles, angle BAD = 60°, AB = 2 l
and BC = 3 l.
Find the angle θ.
B 3 l
C
θ
2 l
Figure 7.
60°
A
D
SOLUTION
First, find an expression for BD.
In triangle ABD,
BD
AB
BD 2 l sin 60°
2 l
220
EXAMPLE 7.
Trigonomet ry
P
θ
3
3 l
10 cm
30°
45°
C
30°
In triangle BCD, tan
BD
BC
3 l 7
θ = tan
1 In the triangle PQR, PQ = 17 cm, QR = 15 cm and PR = 8 cm.
(i) Show that the triangle is right-angled.
(ii) Write down the values of sin Q , cos Q and tan Q , leaving your answers
as fractions.
(iii) Use your answers to part (ii) to show that
(a) sin
2
Q cos
2
Q 1
(b) tan Q
sin Q
cos Q
2 Without using a calculator, show that:
(i) sin 60°cos 30° + cos 60°sin 30° = 1
(ii) sin
2
30° sin
2
45° sin
2
(iii) 3sin
2
30° cos
2
3 In the diagram, AB 10 cm, angle BAC 30°, angle BCD 45° and
angle BDC 90°.
B
(i) Find the length of BD.
(ii) Show that AC 5
cm.
A D
C
4 In the diagram, OA 1 cm, angle AOB angle BOC angle COD 30° and
angle OAB angle OBC angle OCD 90°.
D
(i) Find the length of OD giving your
answer in the form a 3.
(ii) Show that the perimeter of OABCD
is
5
1 3
cm.
B
O A
221
EXERCISE 7A
7A Exercise
P
x
cos
1
θ
y
P(x, y)
1
y
θ
O x x
First look at the right-angled triangle in figure 7.10 which has hypotenuse
of unit length.
P
y
O
x
Figure 7.
Figure 7.
This gives rise to the definitions:
sin
y
y ; cos
x
x ; tan
y
1 1 x
Now think of the angle θ being situated at the origin, as in figure 7.11, and
allow θ to take any value. The vertex marked P has co-ordinates ( x , y ) and can
now be anywhere on the unit circle.
You can now see that the definitions above can be applied to any angle θ, whether
it is positive or negative, and whether it is less than or greater than 90°
sin y , cos x , tan
y
x
For some angles, x or y (or both) will take a negative value, so the sign of sin θ,
cos θ and tan θ will vary accordingly.
ACTIVITY 7.1 Draw x and y axes. For each of the four quadrants formed, work out the sign
of sin θ, cos θ and tan θ, from the definitions above.
Identities involving sin θ , cos θ and tan θ
Since tan θ
y
and y sin θ and x cos θ it follows that
tan θ
sin
cos
It would be more accurate here to use the identity sign, , since the relationship
is true for all values of θ
tan θ
sin
An identity is different from an equation since an equation is only true for
certain values of the variable, called the solution of the equation. For example, tan
θ 1 is
223
Trigonometrical
functions
for
angles
of
any
size
P
7
P
an equation: it is true when θ 45° or 225°, but not when it takes any other value in the range 0° θ 360°. By contrast, an identity is true for all values of the variable, for example tan 30 sin 30 , tan 72 sin 72 , tan(– 339 ) sin(– 399
cos 30 and so on for all values of the angle. cos
cos(–399) In this book, as in mathematics generally, we often use an equals sign where it would be more correct to use an identity sign. The identity sign is kept for situations where we really want to emphasise that the relationship is an identity and not an equation. Another useful identity can be found by applying Pythagoras’ theorem to any point P( x , y ) on the unit circle y 2
OP
2 (sin θ) 2
This is written as sin 2 θ cos 2 θ 1. You can use the identities tanθ sin θ and sin 2 θ cos 2 θ 1 to prove other identities are true. cosθ There are two methods you can use to prove an identity; you can use either method or a mixture of both. Method 1 When both sides of the identity look equally complicated you can work with both the left-hand side (LHS) and the right-hand side (RHS) and show that LHS – RHS = 0. Prove the identity cos 2 θ – sin 2 θ 2 cos 2 θ – 1. 224 SOLUTION Both sides look equally complicated, so show LHS – RHS = 0. So you need to show cos 2 θ – sin 2 θ – 2 cos 2 θ + 1 0. Simplifying: cos 2 θ – sin 2 θ – 2 cos 2 θ + 1 – cos 2 θ – sin 2 θ + 1 –(cos 2 θ + sin 2 θ) + 1
Using sin 2 θ + cos 2 θ = 1. 0 as required EXAMPLE 7. Trigonomet ry
7
P
y sin θ P 3 P 4 P 2
5 P 1 P 0 P 12 x O θ 7 P 11 P 10 –1 P 9 P 8 P 11 P 8 P 10 9 P P 7 P 0 90 ° 180 ° 270 ° 360 ° P 1 P 6 P 12 P 5 P 2 P 4 3 P sin θ
O 180° 360° 540° 720°θ
10 Prove the identity
sin x
cos x
cos x 1 sin x cos x [Cambridge AS & A Level Mathematics 9709, Paper 1 Q2 November 2008] sin x sin x 2 11 Prove the identity
sin x
sin x 2 tan x. [Cambridge AS & A Level Mathematics 9709, Paper 1 Q1 June 2009]
The sine and cosine graphs
In figure 7.12, angles have been drawn at intervals of 30° in the unit circle, and the resulting y co-ordinates plotted relative to the axes on the right. They have been joined with a continuous curve to give the graph of sin θ for 0° θ 360°. P P 6 P Figure 7. The angle 390° gives the same point P 1 on the circle as the angle 30°, the angle 420° gives point P 2 and so on. You can see that for angles from 360° to 720° the sine wave will simply repeat itself, as shown in figure 7.13. This is true also for angles from 720° to 1080° and so on. Since the curve repeats itself every 360° the sine function is described as periodic , with period 360°. Figure 7. 226 In a similar way you can transfer the x co-ordinates on to a set of axes to obtain the graph of cos θ. This is most easily illustrated if you first rotate the circle through 90° anticlockwise. Trigonometry
cos θ
P 0
P 1
P 12
P 11
P 2
P
P
P 10
O
3
9
90°
P
180° 270° 360°θ
4
P 5
P 8
P 6
P 7
y
y = c
1
θ
y= sin θ
O 20° 90 ° 120° 180 ° 270 ° 360 °
10° 210 °
os θ
Figure 7.14 shows the circle in this new orientation, together with the resulting
graph.
Figure 7.
For angles in the interval 360° θ 720°, the cosine curve will repeat itself. You
can see that the cosine function is also periodic with a period of 360°.
Notice that the graphs of sin θ and cos θ have exactly the same shape. The cosine
graph can be obtained by translating the sine graph 90° to the left, as shown in
figure 7.15.
Figure 7.
From the graphs it can be seen that, for example
cos 20° sin 110°, cos 90° sin 180°, cos 120° sin 210°, etc.
In general
cos θ sin (θ 90°).
●
? 1 What do the graphs of sin θ and cos θ look like for negative angles?
2 Draw the curve of sin θ for 0° θ 90°.
Using only reflections, rotations and translations of this curve, how can
you generate the curves of sin θ and cos θ for 0° θ 360°?
227
The
sine
and
cosine
graphs
y
P
3
P
4
P
2
P
5
P
1
P
0
P
12
x
P
11
P
6
P
7
P
8
P
10
P
9
P
y
1
300°420°270°660°780°θ
Solving equations using graphs of trigonometrical functions
P
Suppose that you want to solve the equation cos θ 0.5.
You press the calculator keys for cos
1
0.5 (or arccos 0.5 or invcos 0.5), and the
answer comes up as 60°.
However, by looking at the graph of y cos θ (your own or figure 7.18) you can
see that there are in fact infinitely many roots to this equation.
Figure 7.
You can see from the graph of y cos θ that the roots for cos θ 0.5 are:
θ ..., 420°, 300°, 60°, 60°, 300°, 420°, 660°, 780°, ....
The functions cosine, sine and tangent are all many-to-one mappings, so their
inverse mappings are one-to-many. Thus the problem ‘find cos 60°’ has only
one solution, 0.5, whilst ‘find θ such that cos θ = 0.5’ has infinitely many
solutions.
Remember, that a function has to be either one-to-one or many-to-one; so in
order to define inverse functions for cosine, sine and tangent, a restriction has
to be placed on the domain of each so that it becomes a one-to-one mapping.
This means your calculator only gives one of the infinitely many solutions to
the equation cos θ = 0.5. In fact, your calculator will always give the value of the
solution between:
0° θ 180° (cos)
90° θ 90° (sin)
90° θ 90° (tan).
The solution that your calculator gives you is called principal value.
Figure 7.19 shows the graphs of cosine, sine and tangent together with their
principal values. You can see from the graph that the principal values cover
the whole of the range ( y values) for each function.
Solving
equations
using
graphs
of
trigonometrical
functions
229
3
3
sin θ
1
O
30° 150° θ
y
y = 3tan θ
3
° 270
° 180
90
° O
180° – 90
Find values of θ in the interval 360° θ 360° for which sin θ 0.5.
P
SOLUTION
sin θ 0.5 sin
0.5 = 30° θ 30°. Figure 7.20 shows the graph of sin θ.
Figure 7.
The values of θ for which sin θ 0.5 are 330°, 210°, 30°, 150°.
Solve the equation 3tan θ 1 for 180° θ 180°.
SOLUTION
3tan θ 1
tan θ
1
θ tan
(
1
)
θ 18.4° to 1 d.p. (calculator).
Figure 7.
From figure 7.21, the other answer in the range
is
θ 18.4° 180°
The values of θ are 18.4° or 161.6° to 1 d.p.
231
EXAMPLE 7.
EXAMPLE 7.
Solving
equations
using
graphs
of
trigonometrical
functions
This is a quadratic equation like x2 – x – 2 = 0.
tan θ
2
O
° 180 ° 270 ° 360°θ
90
P1 ●
? How can you find further roots of the equation 3tan θ 1, outside the range
180° θ 180°?
Find values of θ in the interval 0
θ 360
for which tan
2
θ tan θ 2.
SOLUTION
First rearrange the equation.
tan
2
θ tan θ 2
tan
2
θ tan θ 2 0
(tan θ 2)(tan θ 1) 0
tan θ 2 or tan θ 1.
tan θ 2 θ 63.
(calculator)
or θ 63.
(see figure 7.22)
tan θ 1 θ 45
(calculator).
This is not in the range 0° θ 360° so figure 7.22 is used to give
θ 45° 180° 135°
or θ 45° 360° 315°.
Figure 7.
The values of θ are 63.4°, 135°, 243.4°, 315°.
232
EXAMPLE 7.
Trigonomet ry
