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Trigonometric Functions-I
MODULE - IV
Functions
Notes
16
TRIGONOMETRIC FUNCTIONS - I
We have read about trigonometric ratios in our earlier classes. Recall that we defined the
ratios of the sides of a right triangle as follows :
sin θ =
c
a
c
, cos θ = , tan θ =
b
b
a
and cosec θ =
b
b
a
, sec θ = , cot θ =
c
a
c
We also developed relationships between these
trigonometric ratios as
Fig.16.1
sin 2 θ + cos2 θ =1 , sec 2 θ = 1 +tan 2 θ, cosec 2 θ = 1 +cot 2 θ
We shall try to describe this knowledge gained so far in terms of functions, and try to
develop this lesson using functional approach.
In this lesson, we shall develop the science of trigonometry using functional approach.
We shall develop the concept of trigonometric functions using a unit circle. We shall
discuss the radian measure of an angle and also define trigonometric functions of the type
y = sin x, y = cos x, y = tan x, y = cot x, y = sec x, y = cosec x, y = a sin x, y = b cos x,
etc., where x, y are real numbers.
We shall draw the graphs of functions of the type
y = sin x, y = cos x, y = tan x, y = cotx, y = secx, and y = cosecx y = a sin x,
y = a cos x.
OBJECTIVES
After studying this lesson, you will be able to :
define positive and negative angles;
•
define degree and radian as a measure of an angle;
•
convert measure of an angle from degrees to radians and vice-versa;
•
MATHEMATICS
61
Trigonometric Functions-I
MODULE - IV
Functions
•
state the formula l = r θ where r and θ have their usual meanings;
•
solve problems using the relation l = r θ ;
•
define trigonometric functions of a real number;
draw the graphs of trigonometric functions; and
interpret the graphs of trigonometric functions.
•
Notes
•
EXPECTED BACKGROUND KNOWLEDGE
•
Definition of an angle.
•
Concepts of a straight angle, right angle and complete angle.
•
Circle and its allied concepts.
•
Special products : (a ± b )2 = a 2 + b 2 ± 2ab , (a ± b )3 = a 3 ± b3 ± 3ab ( a ± b )
•
Knowledge of Pythagoras Theorem and Py thagorean numbers.
16.1 CIRCULAR MEASURE OF ANGLE
An angle is a union of two rays with the common end point. An angle is formed by the rotation
of a ray as well. Negative and positive angles are formed according as the rotation is clockwise or anticlock-wise.
16.1.1 A Unit Circle
It can be seen easily that when a line segment makes one complete rotation, its end point
describes a circle. In case the length of the rotating line be one unit then the circle described will
be a circle of unit radius. Such a circle is termed as unit circle.
16.1.2 A Radian
A radian is another unit of measurement of an angle other than degree.
A radian is the measure of an angle subtended at the centre of a circle by an arc equal in length
to the radius (r) of the circle. In a unit circle one radian will be the angle subtended at the centre
of the circle by an arc of unit length.
Fig. 16.2
Note : A radian is a constant angle; implying that the measure of the angle subtended by
an are of a circle, with length equal to the radius is always the same irrespective of the
radius of the circle.
62
MATHEMATICS
Trigonometric Functions-I
16.1.3 Relation between Degree and Radian
An arc of unit length subtends an angle of 1 radian. The circumference 2 π (Q r = 1) subtend
MODULE - IV
Functions
an angle of 2π radians.
Hence 2π radians = 360°
π radians = 180°
Notes
π
radians = 90°
2
π
radians = 45°
4
°
°
 360 
 180 
1 radian = 
=


 2π 
 π 
or 1° =
2π
π
radians =
radians
360
180
Example 16.1 Convert
(i) 90° into radians
(iii)
(ii) 15° into radians
π
radians into degrees.
6
(iv)
π
radians into degrees.
10
Solution :
(i)
⇒
(ii)
2π
radians
360
90° = 2π × 90 radiansor
360
1° =
15° =
2π
× 15 radians
360
90° =
π
radians
2
or
15° =
π
radians
12
°
 360 
(iii) 1 radian = 

 2π 
°
π
 360 π 
× 
radians = 
6
 2π 6 
π
radians = 30°
6
°
π
 360 π 
× 
(iv)
radians = 
10
 2π 10 
π
radians = 18°
10
MATHEMATICS
63
Trigonometric Functions-I
MODULE - IV
Functions
CHECK YOUR PROGRESS 16.1
1.
Convert the following angles (in degrees) into radians :
(i) 60° (ii) 15°
(iii) 75° (iv) 105°
(v) 270°
Convert the following angles into degrees:
2.
Notes
(i)
π
4
(ii)
π
(iii)
12
π
20
(iv)
π
60
(v)
2π
3
3.
The angles of a triangle are 45°, 65° and 70°. Express these angles in radians
4.
The three angles of a quadrilateral are
5.
Find the angle complementary to
π π 2π
, ,
. Find the fourth angle in radians.
6 3 3
π
.
6
16.1.4 Relation Between Length of an Arc and Radius of the Circle
An angle of 1 radian is subtended by an arc whose length is equal to the radius of the circle. An
angle of 2 radians will be substened if arc is double the radius.
An angle of 2½ radians willbe subtended if arc is 2½ times the radius.
All this can be read from the following table :
Length of the arc (l)
Angle subtended at the
centre of the circle θ (in radians)
r
1
2r
2
(2½)r
2½
4r
4
Therefore, θ =
l
r
or l = r θ
where r = radius of the circle,
θ = angle substended at the centre in radians
and
l = length of the arc.
The angle subtended by an arc of a circle at the centre of the circle is given by the ratio of the
length of the arc and the radius of the circle.
Note : In arriving at the above relation, we have used the radian measure of the angle
and not the degree measure. Thus the relation θ = l is valid only when the angle is
r
measured in radians.
64
MATHEMATICS
Trigonometric Functions-I
Example 16.2 Find the angle in radians subtended by an arc of length 10 cm at the centre of MODULE - IV
Functions
a circle of radius 35 cm.
l = 10cm and r = 35cm.
Solution :
θ =
θ=
or
l
radians
r
or
θ=
10
radians
35
Notes
2
radians
7
Example 16.3 If D and C represent the number of degrees and radians in an angle prove that
D C
=
180 π
°
°
 360   180 
1 radian = 
 or 

 2π   π 
Solution :
°
180 

∴ C radians =  C ×
π 

Since D is the degree measure of the same angle, therefore,
180
D=C ×
π
which implies
D C
=
180 π
Example 16.4 A railroad curve is to be laid out on a circle. What should be the radius of a
circular track if the railroad is to turn through an angle of 45° in a distance of 500m?
Solution : Angle θ is given in degrees. To apply the formula l = rθ , θ must be changed to
radians.
θ = 45 =
° 45×
π
radians
180
π
radians
4
l = 500 m
....(1)
=
....(2)
l = r θ gives r =
∴
r=
500
m
π
4
= 500 ×
MATHEMATICS
l
θ
[using (1) and (2)]
4
m
π
65
Trigonometric Functions-I
MODULE - IV
Functions
= 2000 ×0.32 m
1

 = 0.32 
π

= 640 m
Example 16.5 A train is travelling at the rate of 60 km per hour on a circular track. Through
5
what
angle
will
it
turn
in
15
seconds
if
the
radius
of
the
track
is
km.
Notes
6
Solution : The speed of the train is 60 km per hour. In 15 seconds, it will cover
60 × 15
km
60 × 60
1
km
4
=
∴
We have,
l =
1
km
4
and
r =
5
km
6
1
l 4
θ= = radians
r 5
6
∴
=
3
radians
10
CHECK YOUR PROGRESS 16.2
1.
2.
Express the following angles in radians :
(a) 30°
(b) 60°
(c) 150°
Express the following angles in degrees :
(a)
3.
4.
π
5
(b)
π
6
(c)
π
9
Find the angle in radians and in degrees subtended by an arc of length 2.5 cm at the
centre of a circle of radius 15 cm.
A train is travelling at the rate of 20 km per hour on a circular track. Through what angle
will it turn in 3 seconds if the radius of the track is
66
1
of a km?.
12
5.
A railroad curve is to be laid out on a circle. What should be the radius of the circular
track if the railroad is to turn through an angle of 60° in a distance of 100 m?
6.
Complete the following table for l, r, θ having their usual meanings.
MATHEMATICS
Trigonometric Functions-I
l
r
θ
(a)
1.25m
.......
135°
(b)
30 cm
.......
π
4
(c)
(d)
0.5 cm
.........
2.5 m
6m
........
120°
(e)
.........
150 cm
π
15
(f)
150 cm
40 m
........
(g)
........
12 m
π
6
(h)
(i)
1.5 m
25 m
0.75 m
........
........
75°
MODULE - IV
Functions
Notes
16.2 TRIGONOMETRIC FUNCTIONS
While considering, a unit circle you must have noticed that for every real number between 0 and
2 π , there exists a ordered pair of numbers x and y. This ordered pair (x, y) represents the
coordinates of the point P.
(i)
(iii)
(ii)
(iv)
Fig. 16.3
MATHEMATICS
67
Trigonometric Functions-I
MODULE - IV If we consider θ= 0 on the unit circle, we will have a point whose coordinates are (1,0).
Functions
π
If θ =
, then the corresponding point on the unit circle will have its coordinates (0,1).
2
In the above figures you can easily observe that no matter what the position of the point,
corresponding to every real number θ we have a unique set of coordinates (x, y). The values of
x and y will be negative or positive depending on the quadrant in which we are considering the
Notes
point.
Considering a point P (on the unit circle) and the corresponding coordinates (x, y), we define
trigonometric functions as :
sin θ = y , cos θ = x
tan θ =
x
y
(forx ≠0) , cot θ = (for y ≠0)
y
x
sec θ =
1
1
(for x ≠0) , cosec θ = (for y ≠0)
y
x
Now let the point P move from its original position in anti-clockwise direction. For various
positions of this point in the four quadrants, various real numbers θ will be generated. Wee
summarise, the above discussion as follows. For values of θ in the :
I
II
III
IV
or
quadrant, both x and y are positve.
quadrant, x will be negative and y will be positive.
quadrant, x as well as y will be negative.
quadrant, x will be positive and y will be negative.
I quadrant
II quadrant
III quadrant
All positive
sin positive
tan positive
cosec positive cot positive
Where what is positive can be rememebred by :
All
sin
tan
Quardrant
I
II
III
If (x, y) are the coordinates of a point P on a
unit circle and θ , the real number generated
by the position of the point, then sin θ = y and
cos θ = x. This means the coordinates of the
point P can also be written as (cos θ , sin θ )
From Fig. 16.5, you can easily see that the
values of x will be between −1 and + 1 as P
moves on the unit circle. Same will be true for
y also.
Thus, for all P on the unit circle
68
IV quadrant
cos positive
sec positive
cos
IV
P (cos q, sinq)
Fig. 16.4
MATHEMATICS
Trigonometric Functions-I
−1 ≤ x ≤ 1
and −1 ≤ y ≤ 1
Thereby, we conclude that for all real numbers θ
−1 ≤ cos θ ≤ 1 and
MODULE - IV
Functions
−1 ≤ sin θ ≤ 1
In other words, sin θ and cos θ can not be numerically greater than 1
Example 16.6 What will be sign of the following ?
(i) sin
7π
18
(ii) cos
4π
9
(iii) tan
Notes
5π
9
Solution :
(i) Since
7π
7π
lies in the first quadrant, the sign of sin
will be posilive.
18
18
(ii) Since
4π
4π
lies in the first quadrant, the sign of cos
will be positive.
9
9
(iii) Since
5π
5π
lies in the second quadrant, the sign of tan will be negative.
9
9
Example 16.7 Write the values of
(i) sin
π
π
(ii) cos 0 (iii) tan
2
2
Solution : (i) From Fig.16.5, we can see that the coordinates of the point A are (0,1)
π
∴ sin =1 , as sin θ = y
2
Fig.16.5
(ii) Coordinates of the point B are (1, 0)
∴ cos 0 = 1 , as cos θ = x
π
π sin 2 1
(iii) tan =
= which is not defined
2 cos π 0
2
π
Thus tan is not defined.
2
MATHEMATICS
69
Trigonometric Functions-I
MODULE - IV Example 16.8 Write the minimum and maximum values of cos θ .
Functions Solution : We know that −1 ≤ cos θ ≤ 1
∴ The maximum value of cos θ is 1 and the minimum value of cos θ is −1.
CHECK YOUR PROGRESS 16.3
Notes
1.
2.
What will be the sign of the following ?
(i)
cos
2π
3
(ii)
tan
5π
6
(iii)
sec
2π
3
(iv)
sec
35π
18
(v)
tan
25π
18
(vi)
cot
3π
4
(vii)
cosec
(viii)
cot
7π
8
8π
3
Write the value of each of the following :
(i)
cos
π
2
(ii)
sin0
(iii)
cos
2π
3
(iv)
tan
3π
4
(v)
sec 0
(vi)
tan
π
2
(vii)
tan
3π
2
(viii)
c o s 2π
16.2.1 Relation Between Trigonometric Functions
By definition
x = cos θ
y = sin θ
As tan θ =
y
, (x ≠ 0)
x
=
sin θ
nπ
θ≠
,
cos θ
2
and cot θ =
x
, ( y ≠ 0)
y
i.e., cot θ =
cos θ
1
=
sin θ tan θ
Similarly, sec θ =
70
1
cos θ
( θ ≠ nπ )
Fig. 16.6
nπ

θ ≠ 

2 
MATHEMATICS
Trigonometric Functions-I
and cosec θ =
1
sin θ
MODULE - IV
Functions
(θ ≠ n π )
Using Pythagoras theorem we have, x 2 + y 2 =1
i.e., (cos θ )2 + (sin θ)2 =1
Notes
or, cos θ + sin θ =1
2
2
Note : ( cos θ ) 2 is written as cos 2 θ and ( sin θ ) 2 as sin2 θ
Again x 2 + y 2 =1
2
2
 y   1
, for x ≠ 0
or 1 +   = 
x
x
or, 1 + ( tan θ) =(sec θ)
2
2
i.e. sec 2 θ = 1 +tan 2 θ
Similarly, cosec2 θ = 1 +cot 2 θ
Example 16.9 Prove that sin 4 θ + cos 4 θ =1 −2sin 2 θcos2 θ
Solution : L.H.S. = sin 4 θ +cos4 θ
= sin 4 θ + cos 4 θ + 2sin 2 θ cos 2 θ − 2sin 2 θ cos 2 θ
= ( sin 2 θ + cos 2 θ ) − 2sin 2 θ cos 2 θ
2
= 1 − 2sin 2 θ cos2 θ ( Q sin 2 θ + cos 2 θ = 1)
= R.H.S.
Example 16.10 Prove that
Solution :
1 − sin θ
= sec θ −tan θ
1 + sin θ
1 − sin θ
L.H.S. = 1 + sin θ
=
=
MATHEMATICS
(1 − sin θ) (1 − sin θ)
(1 + sin θ) (1 − sin θ)
(1 − sin θ )2
1 − sin 2 θ
71
Trigonometric Functions-I
MODULE - IV
Functions
(1 − sin θ )2
=
cos 2 θ
=
1 − sin θ
cos θ
=
1
sin θ
−
cos θ cos θ
Notes
= sec θ − tan θ =R.H.S.
Example 16.11 If sin θ =
21
1
, prove that sec θ + tan θ = 2 , given that θ lies in the first
29
2
quadrant.
sin θ =
Solution :
21
29
Also, sin 2 θ + cos2 θ =1
441 400
2
2
=
⇒ cos θ = 1 −sin θ =1 −
841 841
⇒ cos θ =
∴ tan θ =
2
 20
= 
 29 
20
( cos θ is positive as θ lies in the first quardrant)
29
21
20
∴ sec θ + tan θ =
29 21 29 + 21
+
=
20 20
20
=
5
1
=2
=R.H.S.
2
2
CHECK YOUR PROGRESS 16.4
72
1.
Prove that sin 4 θ − cos 4 θ = sin 2 θ −cos 2 θ
2.
If tan θ =
3.
If cosec θ=
b
, find the other five trigonometric functions, if θ lies in the first quardrant.
a
4.
Prove that
1 + cos θ
= cosec θ +cot θ
1 − cos θ
1
, find the other five trigonometric functions.
2
MATHEMATICS
Trigonometric Functions-I
MODULE - IV
Functions
5
13
5.
If cot θ + cosec θ =1.5 , show that cos θ =
6.
If tan θ + sec θ = m , find the value of cos θ
7.
Prove that ( tan A + 2) ( 2 t a n A +1) =5 t a n A +
2sec2 A
8.
Prove that sin 6 θ + cos 6 θ =1 −3sin 2 θcos2 θ
9.
Prove that
cos θ
sin θ
+
=cos θ +sin θ
1 − tan θ 1 − cot θ
10.
Prove that
tan θ
sin θ
+
= cot θ +cosec θsec
⋅ θ
1 + cos θ 1 −cos θ
Notes
16.3 TRIGONOMETRIC FUNCTIONS OF SOME
SPECIFIC REAL NUMBERS
The values of the trigonometric functions of 0,
π π π
π
, , and are summarised below in the
6 4 3
2
form of a table :
0
π
6
π
4
π
3
π
2
sin
0
1
2
1
2
3
2
1
cos
1
3
2
1
2
1
2
0
tan
0
1
3
1
Function
Real
Numbers
→
↓
3
Not defined
As an aid to memory, we may think of the following pattern for above mentioned values of sin
function :
0
,
4
1
,
4
2
,
4
3
,
4
4
4
On simplification, we get the values as given in the table. The values for cosines occur in the
reverse order.
Example 16.12 Find the value of the following :
(a)
sin
π
π
π
π
sin − cos cos
4
3
4
3
MATHEMATICS
(b)
4tan 2
π
π
π
− cosec 2 − cos 2
4
6
3
73
Trigonometric Functions-I
MODULE - IV
Functions
Solution :
(a)
sin
π
π
π
π
sin − cos cos
4
3
4
3
1  1
 1   3  
=     −   
 2  2   2   2
Notes
=
2
(b) 4tan
3 −1
2 2
π
π
π
− cosec 2 − cos 2
4
6
3
= 4 (1)
2
2  1
−( 2 ) − 
 2
2
1
1
= 4 −4 − = −
4
4
π
π
and B = , verify that
3
6
cos (A + B) =cosAcosB −s i n A s i n B
Example 16.13 If A =
Solution : L.H.S. = cos ( A +B)
π
 π π
= cos  +  = cos = 0
3 6
2
R.H.S. = cos
=
=
∴
74
π
π
π
π
cos −sin sin
3
6
3
6
1 3
3 1
⋅ −
⋅
2 2
2 2
3
3
−
=0
4
4
L.H.S. = 0 = R.H.S.
cos (A + B) = cos A cos B − sin A sin B
MATHEMATICS
Trigonometric Functions-I
MODULE - IV
Functions
CHECK YOUR PROGRESS 16.5
1.
Find the value of
(i)
sin 2
π
π
π
+ tan 2 +tan 2
6
4
3
sin 2
(ii)
2π
π
2π
π
cos − sin
sin
(iii) cos
(iv)
3
3
3
3
π

(v)  sin + sin

6
2.
π
π
π
π
+ cosec 2 +sec 2 −cos 2
3
6
4
3
Notes
π
π
π
π
4cot 2 + cosec 2 +sec 2 tan 2
3
4
3
4
π
π
1
π
 cos −cos  +
4
3
4 4
Show that
π
π 
π
π
π

π
=sec2 sec 2
 1 + tan tan  + tan −tan

6
3 
6
3
6
3
3.
Taking A =
(i)
4.
tan ( A + B ) =
If θ =
(i)
π
π
, B = , verify that
3
6
tan A + t a n B
1 − tan A tan B
(ii) cos ( A + B) = cosAcosB −s i n A s i n B
π
, verify the following :
4
sin2θ = 2sin θcos θ
(ii)
cos2 θ = cos 2 θ −sin2 θ
= 2cos2 θ −1
= 1 −2sin 2 θ
5.
If A =
(i)
π
, verify that
6
cos2A = 2cos 2 A −1
(ii)
tan2A =
2tanA
1 − tan 2 A
(iii) sin2A = 2 s i n A c o s A
16.4 GRAPHS OF TRIGONOMETRIC FUNCTIONS
Given any function, a pictorial or a graphical representation makes a lasting impression on the
minds of learners and viewers. The importance of the graph of functions stems from the fact that
this is a convenient way of presenting many properties of the functions. By observing the graph
we can examine several characteristic properties of the functions such as (i) periodicity, (ii)
intervals in which the function is increasing or decreasing (iii) symmetry about axes, (iv) maximum
and minimum points of the graph in the given interval. It also helps to compute the areas enclosed
by the curves of the graph.
MATHEMATICS
75
Trigonometric Functions-I
MODULE - IV 16.4.1 Variations of sinθ as θ Varies Continuously From 0 to 2π .
Functions Let X'OX and Y'OY be the axes of
coordinates.With centre O and radius
O P = unity, draw a circle. Let OP
starting from OX and moving in
anticlockwise direction make an angle
Notes θ with the x-axis, i.e. ∠ XOP = θ. Draw
PM ⊥ X'OX, then sinθ = MP as OP=1.
∴ The variations of sin θ are the same as those of
MP.
I Quadrant :
As θ increases continuously from 0 to
Fig. 16.7
π
2
PM is positive and increases from 0 to 1.
∴ sin θ is positive.
π 
II Quadrant  , π 
2 
In this interval, θ lies in the second quadrant.
Therefore, point P is in the second quadrant. Here PM
= y is positive, but decreases from 1 to 0 as θ varies
from
Fig. 16.8
π
to π . Thus sin θ is positive.
2
 3π 
III Quadrant  π,

 2
In this interval, θ lies in the third quandrant. Therefore,
point P can move in the third quadrant only. Hence
PM = y is negative and decreases from 0 to −1 as θ
Fig. 16.9
3π
varies from π to
. In this intervalsinθ decreases from 0 to −1. In this interval sin θ is
2
negative.
IV Quadrant
 3π

 2 , 2 π


In this interval, θ lies in the fourth quadrant. Therefore,
point P can move in the fourth quadrant only. Here again
PM = y is negative but increases from -1 to 0 as
3π
to 2π . Thus sin θ is negative in this
θ varies from
2
interval.
76
Fig. 16.10
MATHEMATICS
Trigonometric Functions-I
MODULE - IV
Let X'OX and Y'OY be the two coordinate axes of reference. The values of θ are to be measured Functions
16.4.2 Graph of sin θ as θ varies from 0 to 2π .
along x-axis and the values of sine θ are to be measured along y-axis.
2 = 1.41,
(Approximate value of
θ
0
π
6
sin θ
0
.5
1
3
=.707,
=.87 )
2
2
π
3
π
2
2π
3
5π
6
π
7π
6
4π 3π
3
2
.87
1
.87
.5
0
−.5 −.87
Notes
5π
3
11π
6
2π
−1 −.87
−.5
0
Fig. 16.11
Some Observations
(i) Maximum value of sin θ is 1.
(ii) Minimum value of sin θ is −1.
(iii) It is continuous everywhere.
π
3π
π
3π
and from
to 2π . It is decreasing from to
.
2
2
2
2
With the help of the graph drawn in Fig. 16.12 we can always draw another graph.
y = sin θ in the interval of [ 2 π, 4 π] ( see Fig. 16.11)
1)
(iv) It is increasing from 0 to
What do you observe ?
The graph of y = sin θ in the interval [ 2 π, 4 π] is the same as that in 0 to 2π . Therefore, this
graph can be drawn by using the property sin (2π + θ) =sin θ. Thus, sin θ repeats itself when
θ is increased by 2π . This is known as the periodicity of sin θ .
Fig. 16.12
MATHEMATICS
77
Trigonometric Functions-I
MODULE - IV We shall discuss in details the periodicity later in this lesson.
Functions Example 16.14 Draw the graph of y = sin 2 θ .
Solution :
θ :
0
π
6
π
4
π
3
π
2
2π
3
3π
4
5π
6
π
2θ :
0
π
3
π
2
2π
3
π
4π
3
3π
2
5π
3
2π
sin2θ : 0
.87
1
.87
0
−.87
−1 −.87
0
Notes
Fig. 16.13
The graph is similar to that of y = sin θ
Some Observations
1.
2.
The other graphs of sin θ , like a sin θ , 3 sin 2 θ can be drawn applying the same
method.
Graph of sin θ , in other intervals namely [ 4 π , 6 π] , [ −2 π , 0 ] , [ −4 π , −2 π] ,
can also be drawn easily. This can be done with the help of properties of allied
angles: sin (θ + 2 π) = sin θ
, sin (θ − 2 π ) = sin θ. i.e., θ repeats itself when
increased or decreased by 2 π .
CHECK YOUR PROGRESS 16.6
78
1.
What are the maximum and minimum values of sin θ in [0, 2π] ?
2.
Explain the symmetry in the graph of sin θ in [0, 2π]
3.
Sketch the graph of y = 2 sin θ , in the interval
[0, π]
MATHEMATICS
Trigonometric Functions-I
For what values of θ in
4.
(a)
5.
−1
2
(b)
MODULE - IV
Functions
[ π, 2π] , sin θ becomes
− 3
2
Sketch the graph of y = sin x in the interval of
[ −π , π]
Notes
16.4.3 Graph of cosθ as θ Varies From 0 to 2π
As in the case of sin θ , we shall also discuss the changes in the values of cos θ when θ assumes
values in the intervals 0 , π  ,  π , π ,
 2  2


 

 3π  and  3π , 2π .
2

π , 2 




I Quadrant : In the interval 0 , π  , point P lies in
 2


the first quadrant, therefore, OM = x is positive but
π
decreases from 1 to 0 as θ increases from 0 to .
2
Thus in this interval cos θ decreases from 1 to 0.
∴
cos θ is positive in this quadrant.
π
II Quadrant : In the interval  , π , point P lies in
2

the second quadrant and therefore point M lies on the
negative side of x-axis. So in this case OM = x is
negative and decreases from 0 t o − 1 as θ increases
Fig.16.14
π
to π . Hence in this inverval cos θ decreases
2
from 0 to − 1 .
from
∴
cos θ is negative.
3π 

III Quadrant : In the interval  π ,
, point P lies
2 

in the third quadrant and therefore, OM = x remains
negative as it is on the negative side of x-axis. Therefore
OM = x is negative but increases from−1 to 0 as θ
Fig. 16.15
3π
. Hence in this interval cos θ
2
increases from -1 to 0.
increases from π to
∴
cos θ is negative.
IV Quadrant : In the interval  3π , 2π , point P lies
2



MATHEMATICS
Fig. 16.16
79
Trigonometric Functions-I
MODULE - IV in the fourth quadrant and M moves on the positive
Functions side of x-axis.Therefore OM = x is positive. Also it
increases from 0 to 1 as θ increases from
3π
to 2π .
2
Thus in this interval cos θ increases from 0 to 1.
Notes
cos θ is positive.
∴
Let us tabulate the values of cosines of some suitable
values of θ .
Fig. 16.17
θ
cos θ
π
6
π
3
π
2
2π
3
5π
6
π
7π
6
4π
3
3π
2
5π
3
11π
2π
6
1 .87
.5
0
0.5 −.87
−1
−.87
−.5
0
0.5
.87 1
0
Fig. 16.18
Let X'OX and Y'OY be the axes. Values of θ are measured along x-axis and those of cosθ
along y-axis.
Some observations
(i) Maximum value of cos θ = 1.
(ii) Minimum value of cos θ = −1.
(iii) It is continuous everywhere.
(iv) cos (θ + 2 π) = cos .θ Also (θ − 2π ) = cos θ. Cos θ repeats itself when θ is
increased or decreased by 2π . It is called periodicity of cos θ . We shall discuss in
details about this in the later part of this lesson.
(v) Graph of cos θ in the intervals [2π , 4 π] [4 π, 6 π] [ −2 π , 0] , will be the same as
in [0, 2π] .
Example 16.15 Draw the graph of cos θ as θ varies from − π to π . From the graph read
the values of θ when cos θ = ±0.5 .
80
MATHEMATICS
Trigonometric Functions-I
Solution :
−π
θ :
−5π
6
cos θ : −1.0 −0.87
−2π
3
−0.5
−π
2
−π
3
0
.50
−π
6
0
π
6
π
3
−.87 1.0
0.87
0.5
π
2
2π
3
5π
6
MODULE - IV
Functions
π
0 −0.5 −0.87 −1
Notes
cos θ = 0.5
when θ =
π −π
,
3 3
cos θ = − 0.5
when θ = 2π , −2 π
3
3
Fig. 16.19
Example 16.16 Draw the graph of cos 2θ in the interval 0 to π .
Solution :
θ
0
π
6
π
4
π
3
π
2
2π
3
3π
4
5π
6
π
2θ
0
π
3
π
2
2π
3
π
4π
3
3π
2
5π
3
2π
c o s 2θ
1
0.5
0
−0.5
−1
−0.5
0
0.5
1
Fig. 16.20
MATHEMATICS
81
Trigonometric Functions-I
MODULE - IV
Functions
CHECK YOUR PROGRESS 16.7
−π
π
to .
4
4
Draw the graph of y = 3 cos θ as θ varies from 0 to 2π .
Sketch the graph of y = cos θ as θ varies from
1. (a)
(b)
Notes
Draw the graph of y = cos 3θ from − π to π and read the values of θ when
cos θ = 0.87 and cos θ = −0.87.
(c)
(d)
 π 3π 
Does the graph of y = cos θ in  ,
 lie above x-axis or below x-axis?
2 2 
(e)
Draw the graph of y = cos θ in [2π , 4 π]
16.4.4 Graph of tan θ as θ Varies from 0 to 2π
sin θ
In I Quadrant :
tan θ can be written as cos θ
Behaviour of tan θ depends upon the behaviour of sin θ and
1
cos θ
In I quadrant, sin θ increases from 0 to 1, cos θ decreases from 1 to 0
1
But
increases from 1 indefintely (and write it as increasses from 1 to ∞ )
cos θ
∴ tan θ increases from 0 to ∞ . (See the table and graph of tan θ ).
sin θ
tan θ =
In II Quadrant :
cos θ
sin θ decreases from 1 to 0.
cos θ decreases from 0 to − 1 .
tan θ is negative and increases from −∞ to 0
sin θ
tan θ =
In III Quadrant :
cos θ
sinθ decreases from 0 to − 1
cos θ increases from −1 t o 0
tan θ is positive and increases from 0 to ∞
∴
tan θ > 0
sin θ
cos θ
sin θ increases from −1 t o 0
cos θ increases from 0 to 1
tan θ is negative and increases form −∞ to0
Graph of tan θ
In IV Quadrant :
82
θ
0
tan θ
0
π
6
π
3
.58 1.73
tan θ =
2π
π
π
− 0° + 0°
3
2
2
+∞
-1.73
-.58
5π
6
π
7π
6
4π
3
0
.58
1.73 +∞
5π 11π
3π
3π
− 0°
+ 0°
2π
3
6
2
2
−∞
-1.73
-.58
0 0
MATHEMATICS
Trigonometric Functions-I
MODULE - IV
Functions
Notes
Fig. 16.21
Observations
(i)
tan(180+° θ )= tanθ . Therefore, the complete graph of tan θ consists of
infinitely many repetitions of the same to the left as well as to the right.
(ii) Since tan(−θ ) = − tan θ , therefore, if (θ ,tan θ ) is any point on the graph then
( −θ, − tan θ) will also be a point on the graph.
(iii) By above results, it can be said that the graph of y = tan θ is symmetrical in
opposite quadrants.
(iv) tan θ may have any numerical value, positve or negative.
π 3π
(v) The graph of tan θ is discontinuous (has a break) at the points θ = ,
.
2 2
(vi) As θ passes through these values, tan θ suddenly changes from + ∞ to − ∞.
16.4.5 Graph of cot θ as θ Varies From 0 to 2π
1
1
The behaviour of cot θ depends upon the behaviour of cos θ and sinθ as cot θ= cos θ
sin θ
We discuss it in each quadrant.
I Quadrant :
1
cot θ = cos θ×
sin θ
cos θ decreases from 1 to 0
sin θ increases from 0 to 1
∴ cot θ
also decreases from + ∞to 0 but cot θ > 0.
II Quadrant :
1
cot θ = cos θ×
sin θ
cos θ decreases from 0 t o − 1
sin θ decreases from 1 to 0
MATHEMATICS
83
Trigonometric Functions-I
MODULE - IV
Functions
⇒ cot θ < 0 or cot θ decreases from 0 to − ∞
III Quadrant :
1
cot θ = cos θ×
sin θ
cos θ increases from −1 t o 0
sin θ decreases from 0 to − 1
Notes
∴ cot θ
decreases from + ∞ to 0.
IV Quadrant :
1
cot θ = cos θ×
sin θ
cos θ increases from 0 to 1
sin θ increases from −1 t o 0
∴ cot θ < 0
cot θ decreases from 0 to − ∞
Graph of cot θ
θ
0
π
6
π π 2π
3 2 3
5π
6
π −0 π +0
cot θ ∞ 1.73 .58 0 -.58 -1.73 −∞
7 π 4π 3π
6 3 2
5π
3
11π
6
2π
+ ∞ 1.73 .58 0 -.58 -1.73 −∞
Fig 16.22
Observations
(i) Since cot ( π + θ) =cot θ, the complete graph of cot θ consists of the portion from
θ = 0 to θ= π or θ =
84
π
3π
to θ= .
2
2
MATHEMATICS
Trigonometric Functions-I
(ii) cot θ can have any numerical value - positive or negative.
(iii) The graph of cot θ is discontinuous, i.e. it breaks at 0 , π, 2 π, .
MODULE - IV
Functions
(iv) As θ takes values 0, π, 2 π, cot θ suddently changes from − ∞ to + ∞
Notes
CHECK YOUR PROGRESS 16.8
1.
(a) What is the maximum value of tan θ ?
(b) What changes do you observe in tan θ at
π 3π
,
?
2 2
(c) Draw the graph of y = tan θ from −π to π . Find from the graph the value of θ for
which tan θ = 1.7.
2.
(a) What is the maximum value of cot θ ?
(b) Find the value of θ when cot θ = −1 , from the graph.
16.4.6 To Find the Variations And Draw The Graph of sec θ As θ Varies From 0 to 2π .
Let X'OX and Y'OY be the axes of coordinates. With
centre O, draw a circle of unit radius.
Let P be any point on the circle. Join OP and draw
PM ⊥ X'OX.
Fig. 16.23
OP
1
sec θ =
=
OM OM
∴ Variations will depend upon OM.
I Quadrant : sec θ is positive as OM is positive.
Also sec 0 = 1 and sec
π
π
= ∞ when we approach
2
2
from the right.
∴ As θ varies from 0 to
π
, sec θ increases from 1 to
2
Fig. 16.24
∞.
II Quadrant : sec θ is negative as OM is negative.
π
π
=−∞ when we approach
from the left. Also sec
2
2
π = − 1.
sec
∴ As θ varies from
MATHEMATICS
π
to π , sec θ changes from
2
Fig.16.25
85
Trigonometric Functions-I
MODULE - IV
Functions
− ∞ to − 1.
It is observed that as θ passes through
π
, sec θ changes
2
from + ∞ to − ∞ .
Notes
III Quadrant : sec θ is negative as OM is negative.
sec π = − 1 and sec
3π
= − ∞ when the angle approaches
2
3π
in the counter clockwise direction. As θ varies from
2
π to
3π
, sec θ decreases from −1 t o − ∞ .
2
Fig.16.26
IV Quadrant : sec θ is positive as OM is positive. when θ
3π
, sec θ is positive and very large.
2
Also sec 2 π = 1. Hence sec θ decreases from ∞ to 1 as
is slightly greater than
3π
θ varies from
to 2 π .
2
It may be observed that as
θ passes through
3π
; sec θ changes from − ∞ to + ∞.
2
Fig.16.27
Graph of sec θ as θ varies from 0 to 2π
θ
0
π
6
cot θ
1
1.15
π π
2π
π
−0 +0
3 2
3
2
2
+ ∞ −∞
5π
6
π
7π
6
4π 3π
5π 11π
3π
−0
+0
3 2
3 6
2
-2 -1.15 -1 -1.15 -2
−∞
+∞
2
2π
1.15
1
Fig. 16.28
86
MATHEMATICS
Trigonometric Functions-I
MODULE - IV
Functions
Observations
(a)
sec θ cannot be numerically less than 1.
(b)
Graph of sec θ is discontinuous, discontinuties (breaks) occuring at
(c)
π
3π
and
.
2
2
Notes
π
3π
As θ passes through and
, see θ changes abruptly from + ∞ to − ∞ and then
2
2
from − ∞ to + ∞ respectively..
16.4.7 Graph of cosec θ as θ Varies From 0 to 2π
Let X'OX and Y'OY be the axes of coordinates. With
centre O draw a circle of unit radius. Let P be any
point on the circle. Join OP and draw PM
perpendicular to X'OX.
cosec θ =
OP
1
=
MP MP
∴ The variation ofcosec θ will depend upon MP.
Fig. 16.29
I Quadrant : cosec θ is positive as MP is positive.
π
= 1 when θ is very small, MP is also small and
2
therefore, the value of cosec θ is very large.
cosec
∴ As θ varies from 0 to
π
, cosec θ decreases from
2
∞ to 1.
Fig. 16.30
II Quadrant : PM is positive. Therefore, cosec θ is
π
= 1 and cosec π = ∞ when the
2
revolving line approaches π in the counter clockwise
direction.
positive. cosec
∴ As θ varies from
π
to π , cosec θ increases from
2
1 to ∞ .
Fig. 16.31
III Quadrant :PM is negative
∴ cosec θ is negative. When θ is slightly greater than π ,
MATHEMATICS
87
Trigonometric Functions-I
MODULE - IV cosec θ is very large and negative.
Functions
Also cosec
3π
= − 1.
2
3π
π to
As
θ
varies
from
, cosec θ changes from
∴
Notes
2
− ∞ to − 1 .
It may be observed that as θ passes through π , cosec θ
changes from + ∞ to − ∞ .
Fig. 16.32
IV Quadrant :
PM is negative.
Therefore, cosec θ = − ∞ as θ approaches 2π .
∴ as θ varies from
3π
to 2π , cosecθ varies from −
2
1 to −∞ .
Graph of cosec θ
Fig. 16.33
θ
0
π
6
π
3
π
2
2π
3
cosec θ
∞
2 1.15 1 1.15
5π
π −0 π +0
6
2
+∞
−∞
7π
6
4π
3
3π
2
5π 11π
2π
3
6
−2 −1.15 −1 −1.15 −2 −∞
Fig. 16.34
Observations
88
(a)
cosec θ cannot be numerically less than 1.
(b)
Graph of cosec θ is discountinous and it has breaks at θ = 0, π, 2 π.
(c)
As θ passes through π , coses θ changes from + ∞ to −∞ . The values at 0 and
2π are + ∞ and −∞ respectively..
MATHEMATICS
Trigonometric Functions-I
Example 16.17 Trace the changes in the values of sec θ as θ lies in −π to π.
Soluton :
MODULE - IV
Functions
Notes
Fig. 16.35
CHECK YOUR PROGRESS 16.9
1.
(a) Trace the changes in the values of sec θ when θ lies between −2 π and 2π and
draw the graph between these limits.
(b) Trace the graph of cosec θ ,when θ lies between −2 π and 2π .
16.5 PERIODICITY OF THE TRIGONOMETRIC FUNCTIONS
From your daily experience you must have observed things repeating themselves after regular
intervals of time. For example, days of a week are repeated regularly after 7 days and months
of a year are repeated regularly after 12 months. Position of a particle on a moving wheel is
another example of the type. The property of repeated occurence of things over regular intervals
is known as periodicity.
Definition : A function f (x) is said to be periodic if its value is unchanged when the value of the
variable in increased by a constant, that is if f (x + p) = f (x) for all x.
If p is smallest positive constant of this type, then p is called the period of the function f (x).
1
If f (x) is a periodic function with period p, then f ( x ) is also a periodic function with period p.
16.5.1 Periods of Trigonometric Functions
(i)
(ii)
sinx = sin ( x +2n π) ; n = 0, ± 1, ± 2, .....
cosx = cos ( x +2n π) ; n = 0, ± 1, ± 2 ,.....
Also there is no p, lying in 0 to 2π , for which
sinx = sin ( x +p )
cosx = cos ( x +p ) , for all x
MATHEMATICS
89
Trigonometric Functions-I
MODULE - IV ∴
Functions
2π is the smallest positive value for which
⇒
sin ( x + 2 π) =sinx and
cos ( x + 2 π) =cosx
sin x and cos x each have the period 2π .
(iii)
The period of cosec x is also 2π because cosec x =
(iv)
The period of sec x is also 2π as sec x =
(v)
Also tan ( x + π) =t a n x .
Notes
1
.
sin x
1
.
cosx
Suppose p (0 < p < π
) is the period of tan x, then
tan ( x + p ) = tanx, for all x.
Put x = 0, then tan p = 0, i.e., p = 0 or π .
⇒ the period of tan x is π .
∴ p can not values between 0 and π for which tanx = tan ( x +p )
∴ The period of tanx is π
(vi)
Since cot x =
1
, therefore, the period of cot x is also π .
tanx
Example 16.18 Find the period of each the following functions :
(a) y = 3 sin 2x
(b) y = cos
x
2
(c) y = tan
x
4
Solution :
(a)
(b)
(c)
Period is
2π
, i.e., π .
2
1
2π
y = cos x , therefore period =
=4 π
1
2
2
Period of y = tan
x π
=
=4 π
4 1
4
CHECK YOUR PROGRESS 16.10
1.
90
Find the period of each of the following functions :
(a) y = 2 sin 3x
(b) y = 3 cos 2x
(c) y = tan 3x
(d) y = sin 2 2x
MATHEMATICS
Trigonometric Functions-I
MODULE - IV
Functions
WH
LET US SUM UP
l
l
l
An angle is generated by the rotation of a ray.
The angle can be negative or positive according as rotation of the ray is clockwise or Notes
anticlockwise.
A degree is one of the measures of an angle and one complete rotation generates an
angle of 360°.
l
An angle can be measured in radians, 360° being equivalent to 2π radians.
l
If an arc of length l subtends an angle of θ radians at the centre of the circle with radius
r, we have l = r θ .
l
If the coordinates of a point P of a unit circle are (x, y) then the six trigonometric functions
x
are defined as sin θ = y , cos θ = x , tan θ = y , cot θ = , sec θ = 1 and
y
x
x
cosec θ =
1
.
y
The coordinates (x, y) of a point P can also be written as ( cos θ, sin θ ) .
Here θ is the angle which the line joining centre to the point P makes with the positive
direction of x-axis.
l
The values of the trigonometric functions sin θ and cos θ when θ takes values 0,
π π π π
, , , are given by
6 4 3 2
l
0
π
6
π
4
π
3
π
2
sin
0
1
2
1
2
3
2
1
cos
1
3
2
1
2
1
2
0
Graphs of sin θ , cos θ are continous every where
—
Maximum value of both sin θ and cos θ is 1.
—
Minimum value of both sin θ and cos θ is -1.
—
Period of these functions is 2π .
MATHEMATICS
91
Trigonometric Functions-I
MODULE - IV
Functions
l
tan θ and cot θ can have any value between −∞ and + ∞.
—
The function tan θ has discontinuities (breaks) at
3π
π
and
in ( 0, 2π ) .
2
2
Its period is π .
The graph of cot θ has discontinuities (breaks) at 0, π , 2π . Its period is π .
sec θ cannot have any value numerically less than 1.
—
—
Notes
l
(i)
(ii)
π
3π
and
. It repeats itself after 2π .
2
2
cosec θ cannot have any value between −1 and +1.
It has breaks at
It has discontinuities (breaks) at 0, π , 2π .
It repeats itself after 2π .
SUPPORTIVE WEB SITES
l
l
http://www.wikipedia.org
http://mathworld.wolfram.com
TERMINAL EXERCISE ONS
1.
A train is moving at the rate of 75 km/hour along a circular path of radius 2500 m.
Through how many radians does it turn in one minute ?
2.
Find the number of degrees subtended at the centre of the circle by an arc whose length
is 0.357 times the radius.
3.
The minute hand of a clock is 30 cm long. Find the distance covered by the tip of the hand
in 15 minutes.
4.
Prove that
(a)
(c)
1 − sin θ
= sec θ −tan θ
1 + sin θ
tan θ
cot θ
−
= 2sin θcos θ
1 + tan 2 θ 1 + cot 2 θ
(b)
1
= sec θ − tan θ
sec θ + tan θ
(d)
1 + sin θ
2
= ( tan θ + sec θ)
1 − sin θ
(e) sin8 θ − cos8 θ = ( sin 2 θ − cos 2 θ)( 1 − 2sin 2 θcos 2 θ)
(f)
5.
92
sec 2 θ + cosec2 θ = tan θ +cot θ
If θ =
π
, verify that
4
sin3θ = 3sin θ − 4sin 3 θ
MATHEMATICS
Trigonometric Functions-I
6.
Evaluate :
(a)
sin
25π
6
(d)
sin
17
π
4
(b)
sin
(e)
21π
4
(c)
cos
 3π
tan  
 4 
19
π
3
Notes
3π
π
to x =
.
2
2
7.
Draw the graph of cos x from x = −
8.
Define a periodic function of x and show graphically that the period of tan x is π , i.e. the
position of the graph from x = π to 2π is repetition of the portion from x = 0 to π .
MATHEMATICS
MODULE - IV
Functions
93
Trigonometric Functions-I
MODULE - IV
Functions
ANSWERS
CHECK YOUR PROGRESS 16.1
Notes
π
3
1.
(i)
2.
(i) 45°
3.
(ii)
π
12
(iii)
(ii) 15°
π 13π 14π
,
,
4 36
36
5π
12
(iv)
(iii) 9°
(v)
(iv) 3°
5π
6
4.
7π
12
3π
2
(v) 120°
π
3
5.
CHECK YOUR PROGRESS 16.2
1.
2.
3.
6.
(a)
π
6
(a) 36°
(b)
π
3
(b) 30°
5π
6
(c) 20°
1
radian; 9.55°
6
4.
(a) 0.53 m
(d) 12.56 m
(g) 6.28 m
(b) 38.22 cm (c) 0.002 radian
(e) 31.4 cm (f) 3.75 radian
(h) 2 radian
(i) 19.11 m.
(c)
1
radian
5
5.
95.54 m
CHECK YOUR PROGRESS 16.3
1.
(i) − ive
(v) + ive
(ii) − ive
(vi) − ive
(iii) − ive
(vii) + ive
2.
(i) zero
(ii) zero
(iii) −
(v) 1
(vi) Not defined
(vii) Not defined
(iv) + ive
(viii) − ive
1
2
(iv) − 1
(viii) 1
CHECK YOUR PROGRESS 16.4
2.
sin θ =
3.
sin θ =
tan θ =
94
1
2
, cos θ =
, cot θ = 2 , cosec θ =
5
5
a
, cos θ =
b
a
b2
−
a2
,
b2 − a 2
, sec θ =
b
cot θ =
b2 − a2
a
5 , sec θ =
b
b2 − a2
5
2
,
6.
2m
1 + m2
MATHEMATICS
Trigonometric Functions-I
MODULE - IV
Functions
CHECK YOUR PROGRESS 16.5
1.
(i) 4
1
4
(ii) 6
1
2
(iii) −1
(iv)
22
3
(v) Zero
CHECK YOUR PROGRESS 16.6
1.
1, −1
3.
Notes
Graph of y = 2 sin θ ,
[ 0, π ]
Fig. 16.36
4.
(a)
7 π 11π
,
6
6
(b)
4π 5 π
,
3
3
5.
y = sinxfrom − π to π
Fig. 16.37
CHECK YOUR PROGRESS 16.7
1.
(a) y = cos θ, −
π
π
to
4
4
Fig. 16.38
MATHEMATICS
95
Trigonometric Functions-I
(b) y = 3cos θ; 0 to 2 π
MODULE - IV
Functions
Notes
Fig. 16.39
(c) y = cos3 θ, −π to π
cos θ = 0.87
θ=
π
π
, −
6
6
cos θ = −0.87
θ=
5π
5π
, −
6
6
Fig. 16.40
 π 3 π
(d) Graph of y = cos θ in  ,  lies below the x-axis.
2 2 
(e) y = cos θ
θ lies in 2π to 4π
Fig. 16.41
CHECK YOUR PROGRESS 16.8
1.
96
(a) Infinite
(b) At
π 3π
,
there are breaks in graphs.
2 2
MATHEMATICS
Trigonometric Functions-I
MODULE - IV
Functions
y = t a n 2 θ, − π to π
(c)
π
At θ = ,tan θ = 1.7
3
2.
(a) Infinite
(b) cot θ = −1 at θ =
3π
4
Notes
CHECK YOUR PROGRESS 16.9
1.
y = sec θ
(a)
Fig. 16.42
Points of discontinuity of sec2θ are at
π 3π
,
in the interval [ 0, 2π ] .
4 4
(b) In tracing the graph from 0 to −2 π , use cosec ( −θ ) = −cosec θ.
CHECK YOUR PROGRESS 16.10
1.
(a) Period is
2π
3
(d) y = sin 2 2x =
(b) Period is
1 − cos4x 1
=
2
2
2π
= π
2
(c) Period of y is
π
3
1
2π π
− cos4x ; Period of y is
i.e
2
4
2
 x + 1
π
(e) y = 3cot 
 , Period of y is = 3 π
3
1


3
TERMINAL EXERCISE
1.
1
radian
2
6.
(a)
1
2
MATHEMATICS
2.
(b) −
20.45°
1
2
15π cm
3.
(c) −1 (d)
1
2
(e)
1
2
97
Trigonometric Functions-I
MODULE - IV
Functions 7.
Notes
Fig. 16.43
8.
y = sec θ
Fig. 16.44
98
MATHEMATICS