Transcript Document
Trigonometry Review
(I)
Introduction
By convention, angles are measured from the initial line
or the x-axis with respect to the origin.
P
If OP is rotated counter-clockwise
positive angle
from the x-axis, the angle so formed
x
O
is positive.
But if OP is rotated clockwise
from the x-axis, the angle so
formed is negative.
O
x
negative angle
P
1
(II)
Degrees & Radians
Angles are measured in degrees or radians.
Given a circle with radius r, the
angle subtended by an arc of length r
measures 1 radian.
r
1
c
r
rad 180
Care with calculator! Make sure your
calculator is set to radians when you are making
radian calculations.
2
r
(III) Definition of trigonometric ratios
y
P(x, y)
r
y
sin
sin
cos
tan
hyp
adj
hyp
opp
adj
1
sin
x
x
opp
1
Note:
y
cosec
r
x
r
y
x
sec
sin
cos
1
sin
1 Do not write
1
1
cos , tan .
cos
1
cos
cot
tan sin 3
Graph of y=sin x
y sin x
1
0
90
180
270
360
1
4
Graph of y=cos x
y cos x
1
0
90
180
270
360
1
5
Graph of y=tan x
y tan x
0
90
180
270
360
6
From the above definitions, the signs of sin , cos
& tan in different quadrants can be obtained.
These are represented in the following diagram:
sin +ve
2nd
3rd
tan +ve
All +ve
1st
4th
cos +ve
7
(IV) Trigonometrical ratios of special angles
What are special angles?
90
,
0,
180 , 270 , 360
30 , 45 , 60
Trigonometrical ratios of these angles are
worth exploring
8
y sin x
1
0
2
1
sin 0 0
3
2
sin 2 0
sin 0
sin 0° 0 sin 1
2
sin 180° 0
sin 90° 1
2
sin 360° 0
3
sin 1
2
sin 270° 1
9
1
y cos x
0
2
1
cos 0° 1
cos 2 1
cos 360° 1
cos 1
cos 0 1
2
3
2
cos 180° 1
cos 0
2
cos 90° 0
3
cos 0
2
cos 270° 0
10
y tan x
0
tan 0 0
tan 0° 0
2
3
2
tan 0
tan 180° 0
2
tan 2 0
tan 360° 0
tan is undefined.
2
3
tan is undefined.
2
tan 90° is undefined
tan 270° is undefined
11
Using the equilateral triangle
(of side length 2 units) shown
on the right, the following exact
values can be found.
1
sin 30 sin
6 2
3
sin 60 sin
3 2
1
cos 60 cos
3 2
3
cos 30 cos
6 2
1
tan 30 tan
6
3
tan 60 tan 3
3
12
1
2
sin 45 sin
4
2
2
1
2
cos 45 cos
4
2
2
tan 45 tan 1
4
Complete the table. What do you observe?
13
14
Important properties:
2nd quadrant
sin( ) sin
1st quadrant
cos( ) cos
sin(2 ) sin
cos( 2 ) cos
tan( ) tan
tan(2 ) tan
3rd quadrant
sin( ) sin
cos( ) cos
tan( ) tan
15
Important properties:
4th quadrant
sin(2 ) sin
cos( 2 ) cos
tan( 2 ) tan
sin() sin
cos( ) cos
tan( ) tan
In the diagram, is acute.
However, these
relationships are true for
all sizes of .
16
Complementary angles
Two angles that sum up to 90° or radians are called
2
complementary angles.
E.g.: 30° & 60° are complementary angles.
and are complementary angles.
2
Recall:
1
sin 30 cos 60
2
1
tan 30 cot 60
3
3
sin cos
3
6 2
tan 60 cot 30 3
17
We say that sine & cosine are complementary
functions.
Also, tangent & cotangent are complementary
functions.
sin
40
cos
50
E.g.:
3
cos
sin
8
8
3
tan cot
8
8
cot 35 tan 55
18
E.g. 1: Simplify
(i) sin 210
5
(ii) cos
3
2
(iii) tan(– )
3
3
(iv) sin( )
2
Solution:
(a) sin 210 sin(180°+30) - sin 30 =
1
2
3rd quadrant
210° = 180°+30°
19
(b) cos
5
4th
1
cos( 2 ) cos
3
3 2
3
5
3
quadrant
2 3
2
2
tan( ) tan( )
(c) tan
3
3
3
( tan
3
)
3
20
E.g. 2:
If sin x = 0.6, cos x = 0.8, find
(a) sin (3 x)
(b) cos (4 x).
Soln :
sin (3 - x)
cos (4 + x)
sin (2 - x)
cos (2 + x)
sin ( - x)
sin x
0.6
cos x
0.8
21
(V) Basic Angle
The basic angle is defined to be the positive, acute angle
between the line OP & its projection on the x-axis. For
any general angle, there is a basic angle associated with it.
Let denotes the basic angle.
So 0 90 or 0 .
2
P
O
P
O
180°
or
22
(0 90 or 0 )
2
O
P
– 180°
or –
O
P
360° or
2
23
E.g.:
55
basic 55
(1st quadrant)
(1st quadrant)
P
O
4
basic
4
24
E.g.:
130 (2nd quadrant)
basic 180 130 50
2
3
basic
3
O
180°
or
(2nd quadrant)
2
P
3
25
E.g.:
200 (3rd quadrant)
basic 200 180 20
5
4
P
(3rd quadrant)
basic
5
4
O
– 180°
or –
4
26
E.g.:
300 (4th quadrant)
basic 360 300 60
O
11
6
basic 2
6
P
360°
or 2
(4th quadrant)
11
6
27
Principal Angle & Principal Range
Example: sinθ = 0.5
2
2
Principal range
Restricting y= sinθ inside the principal range makes it a
one-one function, i.e. so that a unique θ= sin-1y exists
28
E.g. 3(a):
Since
sin (
3
1
)
2
2
3
sin ( )
2
is positive, it is in the 1st or 2nd quadrant
Basic angle, α =
4
3
Therefore
2
4
5
(inadmissib le)
4
Hence,
. Solve for θ if 0
or
3
2
4
or
3
4
3
4
29
E.g. 3(b): cos (2 250 ) 0.8 . Solve for θ if 0 180
0
Since cos(2 25 ) is negative, it is in the 2nd or 3rd quadrant
Basic angle, α = 36.870o
Therefore 2 25 180 36.870
or
2 25 180 36.870
59.1
or
95.9
Hence, 59.1 or 95.9
30
(VI) 3 Important Identities
P(x, y)
By Pythagoras’ Theorem,
x2 y 2 r 2
2
2
x y
1
r r
x
y
Since sin A
and cos A ,
r
r
sin A2 cos A2 1
sin2 A cos2 A 1
O
r
y
A
x
Note:
sin 2 A (sin A)2
cos 2 A (cos A)2
31
(VI) 3 Important Identities
(1)
sin2 A + cos2 A 1
Dividing (1) throughout by cos2 A,
tan 2 x = (tan x)2
(2)
tan2 A +1 sec2 A
1
Dividing (1) throughout by sin2 A,
(3)
1+
cot2 A
csc2 A
cos2 A
1
cos A
(sec A)
2
2
2
sec A
32
(VII) Important Formulae
(1)
Compound Angle Formulae
sin( A B) sin A cos B cos A sin B
sin( A B) sin A cos B cos A sin B
cos( A B) cos A cos B sin A sin B
cos( A B) cos A cos B sin A sin B
tan A tan B
tan( A B)
1 tan A tan B
tan A tan B
tan( A B)
1 tan A tan B
33
E.g. 4: It is given that tan A = 3. Find, without using calculator,
(i) the exact value of tan , given that tan ( + A) = 5;
(ii) the exact value of tan , given that sin ( + A) = 2 cos ( – A)
Solution:
(i)
Given tan ( + A) 5 and tan A 3,
tan tan A
tan( A)
1 tan tan A
tan 3
5
1 3 tan
5 15 tan tan 3
1
tan
8
34
Solution:
(ii)
Given sin ( + A) = 2 cos ( – A) & tan A 3,
sin cos A + cos sin A = 2[ cos cos A + sin sin A ]
(Divide by cos A on both sides)
sin + cos tan A = 2(cos + sin tan A)
sin + 3cos = 2(cos + 3sin )
5sin = cos
1
tan =
5
35
(2)
Double Angle Formulae
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = cos2 A – sin2 A
= 2 cos2 A – 1
= 1 – 2 sin2 A
(iii) tan 2 A
2 tan A
2
1 tan A
Proof:
sin 2 A
sin( A A)
sin A cos A cos Asin A
2 sin Acos A
cos 2 A cos( A A)
2
2
cos A sin A
2
2
cos A (1 cos A)
2 cos2 A 1
36
(3)
Triple Angle Formulae:
(i) cos 3A = 4 cos3 A – 3 cos A
Proof:
cos 3A = cos (2A + A)
= cos 2A cos A – sin 2A sin A
= ( 2cos2A 1)cos A – (2sin A cos A)sin A
= 2cos3A cos A – 2cos A sin2A
= 2cos3A cos A – 2cos A(1 cos2A)
= 4cos3A 3cos A
37
(ii) sin 3A = 3 sin A – 4 sin3 A
Proof:
sin 3A = sin (2A + A)
= sin 2A cos A + cos 2A sin A
= (2sin A cos A )cos A + (1 – 2sin2A)sin A
= 2sin A(1 – sin2A) + sin A – 2sin3A
= 3sin A – 4sin3A
38
E.g. 5: Given sin2 A 16 & A is obtuse, find,
25
without using calculators, the values of
(i) cos 4A
(ii) sin ½A
Solution:
16
2
Since sin A
25
4
sin A
5
4
But A is obtuse, sin A =
5
4
5
A
3
3
cos A
5
39
2
(i) cos 4 A 1 2 sin 2 A
1 2(2sin A cos A)
2
24
1 2
25
527
625
4
2
5
A
3
3
cos A
5
40
A
(ii) cos A = 1 – 2sin2 (
)
2
3
A
2
= 1 – 2sin (
)
5
2
2 A 4
sin
2 5
A
Since 90 A 180, 45 90.
2
A
A
st
i.e.
lies in the 1 quadrant. So sin 0
2
2
A
2
sin (
)=
2
5
41
E.g. 6: Prove the following identities:
4
2
(i)
cos 4 A 8 cos A 8 cos A 1
Recall:
cos 2A = cos2 A – sin2 A
= 2 cos2 A – 1
Solution:
(i) LHS = cos 4 A
2 cos 2 2 A 1
= 1 – 2 sin2 A
2(2 cos 2 A 1) 2 1
2( 4 cos 4 A 4 cos 2 A 1) 1
4
2
8 cos A 8 cos A 1
= RHS
42
1 cos 2 A
E.g. 6: Prove the following identities: (ii)
tan A
sin 2 A
Solution:
1 cos 2 A
(ii) LHS =
sin 2 A
1 (1 2 sin 2 A)
2 sin A cos A
2
2 sin A
2 sin A cos A
sin A
cos A
tan A = RHS
43
E.g. 6: Prove the following identities:
(iii) 1 cos cosec cot , where 0
1 cos
2
Solution:
1 cos
LHS
1 cos
(1 cos )(1 cos )
(1 cos )(1 cos )
(1 cos )2
1 cos
2
(1 cos )2
sin
2
44
(1 cos )2
sin 2
1 cos
sin
1 cos
sin
1
cos
sin sin
( Given 0
,
2
0 sin 1 and 0 cos 1.)
cos ec cot
RHS
45
E.g. 6: Prove the following identities:
3
3
cos
cos
3
sin
sin 3
(iv)
3
cos
sin
Solution:
cos3 cos 3 sin 3 sin 3
LHS =
cos
sin
cos 3
sin 3
2
2
cos
sin
cos
sin
sin 3 cos sin cos 3
1
sin cos
1
sin(3 )
1 sin 2
2
1 2 3 RHS
46
(5)
The Factor Formulae (Sum or difference of
similar trigo. functions)
Recall compound angles formulae:
sin( A B) sin A cos B cos A sin B ….
sin( A B) sin A cos B cos A sin B ….
cos( A B) cos A cos B sin A sin B ….
cos( A B) cos A cos B sin A sin B ….
+ : sin( A B) sin( A B) 2 sin A cos B
: sin( A B) sin( A B) 2 cos A sin B
+ : cos( A B) cos( A B) 2 cos A cos B
: cos( A B) cos( A B) 2 sin A sin B47
By letting X = A + B and
obtain the factor formulae:
Y = A – B, we
X Y X Y
(1) sin X sin Y 2 sin
cos
2 2
X Y X Y
(2) sin X sin Y 2 cos
sin
2 2
X Y X Y
(3) cos X cos Y 2 cos
cos
2 2
X Y X Y
(4) cos X cos Y 2 sin
sin
2 2 48
E.g. 8: Show that
2
(i)cos cos 3 cos 5 cos 3 ( 4 cos 1)
Solution:
(i) LHS
Using cos X cos Y
= cos + cos 3 + cos 5
= (cos 5 + cos ) + cos 3
= 2cos 3 cos 2 + cos 3
X Y X Y
2 cos
cos
2 2
= cos 3 [2cos2 + 1]
= cos 3 [ 2(2 cos2 – 1) + 1 ]
= cos 3 (4 cos2 – 1) = RHS
49
sin A sin B
A B
E.g. 8: Show that (ii)
cot
cos A cos B
2
Soln:
sin A sin B
(ii) LHS =
cos A cos B
A B A B
2 sin
cos
2 2
A B A B
2 sin
sin
2 2
A B
cos
A B
2
cot
= RHS
A B
2
sin
50
2
E.g. 8: Show that
(iii) sin + sin 3 + sin 5 + sin 7 = 16 sin cos2 cos2 2
Soln:
(iii) LHS = sin + sin 3 + sin 5 + sin 7
= (sin 3 + sin ) + (sin 7 + sin 5 )
4
2
12
2
2 sin cos 2 sin
cos
2
2
2
2
= 2sin 2 cos + 2sin 6 cos
= 2cos [ sin 6 + sin 2 ]
8
4
2 cos 2 sin cos
2
2
51
8
4
2 cos 2 sin cos
2
2
= 4 cos cos 2 sin 4
= 4 cos cos 2 [ 2 sin 2 cos 2 ]
= 8 cos cos2 2 sin 2
= 8 cos cos2 2 ( 2 sin cos )
= 16 sin cos2 cos2 2
= RHS
52