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Transcript of 1 Trigonometry Review (I)Introduction By convention, angles are measured from the initial line or...
1
Trigonometry Review
(I) Introduction
By convention, angles are measured from the initial line or the x-axis with respect to the origin.
If OP is rotated counter-clockwisefrom the x-axis, the angle so formed is positive.
But if OP is rotated clockwisefrom the x-axis, the angle so formed is negative.
O
P
xnegative angle
P
O xpositive angle
2
(II) Degrees & Radians
Angles are measured in degrees or radians.
rr
r1c
Given a circle with radius r, the angle subtended by an arc of length r measures 1 radian.
Care with calculator! Make sure your calculator is set to radians when you are making radian calculations.
180rad
3
(III) Definition of trigonometric ratios
r
y
hyp
oppsin
r
x
hyp
adjcos
x
y
adj
opptan
cos
sin
sin
1 cosec
cos
1sec
sin
cos
tan
1cot
x
y P(x, y)
r y
x
Note:
1sin
sin
1
Do not write cos1tan1
7
From the above definitions, the signs of sin , cos & tan in different quadrants can be obtained. These are represented in the following diagram:
All +ve sin +ve
tan +ve
1st2nd
3rd 4th
cos +ve
8
What are special angles?
(IV) Trigonometrical ratios of special angles
,30 60,45
,90,0 ,180 360,270
Trigonometrical ratios of these angles are worth exploring
9
1
00sin
0sin
12
sin
02sin
12
3sin
sin 0° 0
sin 360° 0sin 180° 0
sin 90° 1 sin 270° 1
xy sin
0 2
23
1
10
10cos 1cos
02
cos
12cos
02
3cos
cos 0° 1
cos 360° 1
cos 180° 1
cos 90° 0cos 270°
1xy cos
0 2
23
1
11
00tan 0tan
undefined. is 2
tan
02tan
undefined. is 2
3tan
tan 180° 0
tan 0° 0
tan 90° is undefined tan 270° is undefined
tan 360° 0
xy tan
0 2
23
12
Using the equilateral triangle (of side length 2 units) shown on the right, the following exact values can be found.
2
3
3sin60sin
2
3
6cos30cos
2
1
6sin30sin
2
1
3cos60cos
33
tan60tan
3
1
6tan30tan
15
2nd quadrant sin)sin(
cos)cos(
tan)tan(
Important properties:Important properties:
3rd quadrant sin)sin(
cos)cos(
tan)tan(
1st quadrant sin)2sin(
cos)2cos(
tan)2tan(
or 2
16
Important properties:Important properties:
4th quadrant
sin)2sin( cos)2cos(
tan)2tan(or
or 2
sin)sin( cos)cos(
tan)tan(In the diagram, is acute. However, these relationships are true for all sizes of
17
Complementary angles
E.g.: 30° & 60° are complementary angles.
Two angles that sum up to 90° or radians are called complementary angles.
2
2
and are complementary angles.
Recall:
2
160cos30sin
2
3
6cos
3sin
3
160cot30tan 330cot60tan
18
We say that sine & cosine are complementary functions.
Also, tangent & cotangent are complementary functions.
E.g.: 50cos40sin
8
3cos
8
3tan
8cot
35cot 55tan
8sin
19
E.g. 1: Simplify
(i) sin 210 (ii) cos (iii) tan(– ) (iv) sin( )
sin(180°+30)
(a) sin 210
Solution:
2
1
210° = 180°+30°
3rd quadrant
3
53
2
2
3
- sin 30 =
21
sin (3 - x)
sin (2 - x)
sin ( - x)
sin x
0.6
cos (4 + x)
cos (2 + x)
0.8
cos x
Soln :
E.g. 2: If sin x = 0.6, cos x = 0.8, find
(a) sin (3 x) (b) cos (4 x).
22
(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.
.0or 900 So2
P
O
P
O
180° or
Let denotes the basic angle.
28
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
29
E.g. 3(a): sin . Solve for θ if 2
1)
2
3( 0
4
Basic angle, α =
Since sin is positive, it is in the 1st or 2nd quadrant )2
3(
42
3
42
3 orTherefore
4
3)(
4
5 orleinadmissib
Hence, 4
3
30
E.g. 3(b): cos . Solve for θ if
Since cos is negative, it is in the 2nd or 3rd quadrant )252( 0
Basic angle, α = 36.870o
870.36180252870.36180252 orTherefore
9.951.59 or
8.0)252( 0 1800
Hence, 9.951.59 or
31
ry
xA
O
P(x, y)By Pythagoras’ Theorem,
222 ryx
122
r
y
r
x
(VI) 3 Important Identities
sin2 A cos2 A 1
r
xA cos
r
yA sinSince and ,
1cossin 22 AA Note:
sin 2 A (sin A)2 cos 2 A (cos A)2
32
A2cos
1
(1) sin2 A + cos2 A 1
(2) tan2 A +1 sec2 A
(3) 1 + cot2 A csc2 A
tan 2 x = (tan x)2
(VI) 3 Important Identities
Dividing (1) throughout by cos2 A,
Dividing (1) throughout by sin2 A,
2)(sec A
2
cos
1
A
A2sec
33
(VII) Important Formulae
(1) Compound Angle Formulae
BABABA sincoscossin)sin( BABABA sincoscossin)sin(
BABABA sinsincoscos)cos( BABABA sinsincoscos)cos(
BA
BABA
tantan1
tantan)tan(
BA
BABA
tantan1
tantan)tan(
34
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,
tan31
3tan5
3tantan155
8
1tan
A
AA
tantan1
tantan)tan(
35
Solution:
sin + cos tan A = 2(cos + sin tan A)
sin + 3cos = 2(cos + 3sin )
(ii) Given sin ( + A) = 2 cos ( – A) & tan A 3,
5sin = cos
tan = 51
sin cos A + cos sin A = 2[ cos cos A + sin sin A ]
(Divide by cos A on both sides)
36
(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)
A
AA
2tan1
tan22tan
Proof:
)sin(
2sin
AA
A
AAAA sincoscossin
AAcossin2
)cos(2cos AAA
AA 22 sincos
)cos1(cos 22 AA
1cos2 2 A
37
(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
38
(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
39
E.g. 5: Given sin2 A & A is obtuse, find,
without using calculators, the values of
2516
(i) cos 4A (ii) sin ½A
Solution:
Since sin2 A 25
16
But A is obtuse, sin A =5
45
4Asin
5
3Acos
A 5
3
4
41
(ii) cos A = 1 – 2sin2 ( )2
A
5
3 = 1 – 2sin2 ( )
2
A
5
4
2sin2
A
2
Asin ( ) =
5
2
,18090 Since A .902
45 A
i.e. lies in the 1st quadrant. So 02
sin A
2
A
42
E.g. 6: Prove the following identities:
(i) 1cos8cos84cos 24 AAA
Solution:
1)1cos2(2 22 A
1)1cos4cos4(2 24 AA
1cos8cos8 24 AA
(i) A4cosLHS =
= RHS
12cos2 2 A
cos 2A = cos2 A – sin2 A
= 2 cos2 A – 1
= 1 – 2 sin2 A
Recall:
43
(ii) A
A
2sin
2cos1LHS =
AA
A
cossin2
)sin21(1 2
AA
A
cossin2
sin2 2
A
A
cos
sin
Atan = RHS
E.g. 6: Prove the following identities: (ii) AA
Atan
2sin
2cos1
Solution:
44
)cos1)(cos1(
)cos1)(cos1(
cos1
cos1
LHS
2
2
cos1
)cos1(
2
2
sin
)cos1(
20 where,cotcosec
cos1
cos1
E.g. 6: Prove the following identities:
(iii)
Solution:
46
LHS =
sin
3sinsin
cos
3coscos 33
RHS
3sin
3sinsin
cos
3coscos 33
E.g. 6: Prove the following identities:
(iv)
Solution:
sin
3sinsin
cos
3coscos 22
cossin
3cossincos3sin1
2sin
)3sin(1
21
321
47
(5) The Factor Formulae (Sum or difference of similar trigo. functions)
Recall compound angles formulae:
BABABA sincoscossin)sin( ….
BABABA sincoscossin)sin( ….
BABABA cossin2)sin()sin( + :
BABABA sinsincoscos)cos( ….
BABABA sinsincoscos)cos( ….
BABABA sincos2)sin()sin( :BABABA coscos2)cos()cos( + :BABABA sinsin2)cos()cos( :
48
By letting X = A + B and Y = A – B, we obtain the factor formulae:
2
cos2
sin2sinsin)1(YXYX
YX
2
sin2
cos2sinsin)2(YXYX
YX
2
cos2
cos2coscos)3(YXYX
YX
2
sin2
sin2coscos)4(YXYX
YX
49
Solution:(i) LHS
= cos + cos 3 + cos 5
= cos 3 (4 cos2 – 1) = RHS
= cos 3 [ 2(2 cos2 – 1) + 1 ]
= (cos 5 + cos ) + cos 3
= 2cos 3 cos 2 + cos 3
= cos 3 [2cos2 + 1]
2
cos2
cos2
coscos Using
YXYX
YX
E.g. 8: Show that)1cos4(3cos5cos3coscos 2 (i)
50
(ii)
BA
BA
coscos
sinsin
LHS =
2
cotBA
2sin
2sin2
2cos
2sin2
BABA
BABA
2sin
2cos
BA
BA
= RHS
2cot
coscos
sinsin BA
BA
BAE.g. 8: Show that (ii)
Soln:
51
(iii) LHS = sin + sin 3 + sin 5 + sin 7 = (sin 3 + sin ) + (sin 7 + sin 5 )
2
2cos
2
12sin2
2
2cos
2
4sin2
= 2sin 2 cos + 2sin 6 cos
= 2cos [ sin 6 + sin 2 ]
2
4cos
2
8sin2cos2
(iii) sin + sin 3 + sin 5 + sin 7 = 16 sin cos2 cos2 2E.g. 8: Show that
Soln: