Term 3 : Unit 1Trigonometric Functions
Name : ____________ ( ) Class : _____ Date : _____
1.1 Trigonometric Ratios and General Angles
1.2 Trigonometric Ratios of Any Angles
Trigonometric Equations
Objectives
1.1 Trigonometric Ratios and General Angles
In this lesson, we will learn how to find the trigonometric ratios for
acute angles, particularly those for 30°, 45° and 60° (or
respectively in radians).
,3
and 4
,6
Trigonometric Ratios of Acute Angles
The three trigonometric ratios
are defined as
OPQ is a right angled triangle
Trigonometric Equations
adjacent
oppositehypotenuse
sin
cos
tan
oppositeoppositehypotenusehypotenuse
adjacentadjacent
PQ
OP
OQ
OP
PQ
OQ
Example 1
In the right-angled triangle ABC, tan θ = 2. Find sin θ and cos θ.
Solution
Trigonometric Equations
A B
C
θ
Since tan θ = ,1
2 2
1
BC = 2 units and AB = 1 unit.
By Pythagoras’ Theorem, AC = . units 5
5
AC
BCsin
5
2
AC
ABcos
5
1
Trigonometric Ratios of Special Angles
Draw a diagonal to the
square.
Draw a unit square.
Trigonometric Equations
The length of the diagonal is √2 and the angle is 45°.
1sin 45
2
1cos 45
2
tan 45 1
Trigonometric Ratios of Special Angles
Draw an equilateral triangle of side 2 cm.
Trigonometric Equations
The altitude bisects the base of the
triangle.
3sin 60
2
1cos 60
2
tan 60 3
Draw an altitude.
The length of the altitude is √3 and the angles are 60° and
30°.
3cos30
2
1sin 30
2
1tan 30
3
Trigonometric Ratios of Complementary Angles
In the right-angled triangle OPQ
but OPQ = 90°– θ
Trigonometric Equations
cosOQ
OP
sinPQ
OP
tanPQ
OQ
sin 90
cos 90
1
tan 90
OPQ = – θ
sin2
cos2
1
tan2
If θ is in radians
2
Trigonometric Equations
Example 2
Using the right-angled triangle in the diagram, show that sin(900 – θ) = cos θ. Hence, deduce the value of
SolutionP Q
R
θ
a
bc
900 – θ.70sin20cos
20cos
.)sin(90 and cos,In c
a
c
aPQR
Thus, sin(900 – θ) = cos θ.
sin 700 = sin (900 – 200)
= cos 200
2
1
20cos2
20cos
70sin20cos
20cos
Trigonometric Equations
O
y
x
P
O
y
x
P
180° –
P '
O
y
x
P
180° +
P '
O
y
x
P
360° –
P '
O
y
x
P
O
y
x
P
–
P '
O
y
x
P
– 180
P '
O
y
x
P
– – 180
P '
Consider angles in the Cartesian plane.
OP is rotated in an anticlockwise direction around the origin O. The basic (reference) angle that OP makes with the positive x–axis is α.
Now OP is rotated in the clockwise direction.
1st quadrant2nd quadrant
4th quadrant3rd quadrant
O
y
x
P
– 360
220
180
Example 3
Given that 00 < θ < 3600 and the basic angle for θ is 400, find the value of θ if it lies in the (a) 3rd quadrant, (b) 4th quadrant.
Solution(a) (b)
320
360
Trigonometric Equations
Using the complementary angle identity.
Substitute for sin θ.
Trigonometric Equations
cos 90 sin
1Given that sin , find the value of sin cos 90
2
sin cos 90 sin sin 1 1 1
2 2 4
1tan 90
tanA
A
Given that tan 2, find the value of 2 tan tan 90A A A
1 12 4
2 22
12 tan tan 90 2 tan
tanA A A
A
Using the complementar
y angle identity.
Substitute for tan A.
Example 4
Trigonometric Equations
1
2sin 45
cos30 sin 60 3 3
2 2
sin 45Without using a calculator, evaluate
cos30 sin 60
1 1
2 3 6
Example 5
Solution
Trigonometric Equations
70 , 180 70 , 180 70 , 360 70
Find all the angles between 0° and 360° which make a basic angle of 70°.
The angles are as follows:
70 , 110 , 250 , 290
Example 6
Solution
Trigonometric Equations
Objectives
In this lesson, we will learn how to
• extend the definitions of sine, cosine and tangent to any angle,
• determine the sign of a trigonometric ratio of an angle in a
quadrant,
• relate the trigonometric functions of any angle to that of its basic
(reference) angle and solve simple trigonometric equations.
1.2 Trigonometric Ratios of Any Angles
Trigonometric Ratios of Any AnglesThe three
trigonometric ratios are defined as
Trigonometric Equations
x
yr
sin
cos
tan
yyr
x
r
x
PQ
OP
OQ
OP
PQ
OQ
PQ = yOQ = x
22 yxrOP
Example 7
Find the values of cos θ, sin θ and tan θ when θ = 1350.
Solution
Trigonometric Equations
When θ = 1350, 1800 – θ = 450.(basic angle)
P has coordinates (1, -1)
and
2
1)1( 22
r
Trigonometric Equations
2
1135cos
2
1135sin
11
1135cos
x
y
O
P(a, b)
rb
Signs of Trigonometric Ratios in Quadrants
1st quadrant
Trigonometric Equations
x
y
O
P( – a, b)
rb
– a
θ = α
sinb
r
cosa
r
tanb
a
sin
cos
tan
P has coordinates ( a, b )
2nd quadrant
x
y
O
P( – a, – b)
r– b
– a
θ = ( 180° – α )
P has coordinates ( – a, b )
cosa
r
cos
tanb
a
tan
sinb
r
sin
tanb
a
x
y
O
P(a, – b)
r– b
a
3rd quadrant
θ = ( 180° + α )
P has coordinates ( – a, – b )
4th quadrant
θ = ( 360° – α )
P has coordinates ( a, – b ).
tanb
a
Trigonometric Equations
902
For positive ratios
Signs of Trigonometric Ratios in Quadrants
3270
2
180 0 0 360 2
In the four quadrantsS (sin θ) A ( all )
T (tan θ) C (cos θ)The signs are
summarised in this diagram.
x
y
O
120°
Trigonometric Equations
Example 8
Without using a calculator, evaluate cos 120°.
Solution
Basic angle,
120° is in the 2nd quadrant, so cosine is negative
180 120
60
cos120 cos60 1
2
AS
T C
Trigonometric Equations
Example 9
Find all the values of θ between 0° and 360° such that
sin θ = – 0.5.
Solution
For the basic angle,
Since sin θ < 0, θ is in the 3rd or 4th quadrant,
sin 0.5 30
180 or 360
180 30 or 360 30
x
y
O
AS
T C
Basic Trigonometric Equations
210 or 330
Trigonometric Equations
For the basic angle,
θ is in the 1st, 2nd, 3rd or 4th quadrant,
1sin
2 045
x
y
O
AS
T C
22sin 1 0 2 1sin
2
1sin
2
45 ,180 45 ,180 45 ,360 45
45 ,135 ,225 ,315
o o o o o o o
o o o o
3 5 7
, , or 4 4 4 4
Example 10
Find all the values of θ between 0o and 360o such that
2sin2 θ – 1 = 0.
Solution