INTRODUCTION TO TRIGONOMETRY 113

INTRODUCTION TO
TRIGONOMETRY 8
There is perhaps nothing which so occupies the
middle position of mathematics as trigonometry.
– J.F. Herbart (1890)
8.1 Introduction
You have already studied about triangles, and in particular, right triangles, in your
earlier classes. Let us take some examples from our surroundings where right triangles
can be imagined to be formed. For instance :
1. Suppose the students of a school are
visiting Qutub Minar. Now, if a student
is looking at the top of the Minar, a right
triangle can be imagined to be made,
as shown in Fig 8.1. Can the student
find out the height of the Minar, without
actually measuring it?
2. Suppose a girl is sitting on the balcony
of her house located on the bank of a Fig. 8.1
river. She is looking down at a flower
pot placed on a stair of a temple situated
nearby on the other bank of the river.
A right triangle is imagined to be made
in this situation as shown in Fig.8.2. If
you know the height at which the
person is sitting, can you find the width
of the river?
Fig. 8.2

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114 MATHEMATICS

3. Suppose a hot air balloon is flying in

the air. A girl happens to spot the
balloon in the sky and runs to her
mother to tell her about it. Her mother
rushes out of the house to look at the
balloon.Now when the girl had spotted
the balloon intially it was at point A.
When both the mother and daughter
came out to see it, it had already
travelled to another point B. Can you
find the altitude of B from the ground? Fig. 8.3
In all the situations given above, the distances or heights can be found by using
some mathematical techniques, which come under a branch of mathematics called
‘trigonometry’. The word ‘trigonometry’ is derived from the Greek words ‘tri’
(meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact,
trigonometry is the study of relationships between the sides and angles of a triangle.
The earliest known work on trigonometry was recorded in Egypt and Babylon. Early
astronomers used it to find out the distances of the stars and planets from the Earth.
Even today, most of the technologically advanced methods used in Engineering and
Physical Sciences are based on trigonometrical concepts.
In this chapter, we will study some ratios of the sides of a right triangle with
respect to its acute angles, called trigonometric ratios of the angle. We will restrict
our discussion to acute angles only. However, these ratios can be extended to other
angles also. We will also define the trigonometric ratios for angles of measure 0° and
90°. We will calculate trigonometric ratios for some specific angles and establish
some identities involving these ratios, called trigonometric identities.

8.2 Trigonometric Ratios

In Section 8.1, you have seen some right triangles
imagined to be formed in different situations.
Let us take a right triangle ABC as shown
in Fig. 8.4.
Here,  CAB (or, in brief, angle A) is an
acute angle. Note the position of the side BC
with respect to angle A. It faces  A. We call it
the side opposite to angle A. AC is the
hypotenuse of the right triangle and the side AB
is a part of  A. So, we call it the side
Fig. 8.4
adjacent to angle A.

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INTRODUCTION TO TRIGONOMETRY 115

Note that the position of sides change

when you consider angle C in place of A
(see Fig. 8.5).
You have studied the concept of ‘ratio’ in
your earlier classes. We now define certain ratios
involving the sides of a right triangle, and call
them trigonometric ratios.
The trigonometric ratios of the angle A
in right triangle ABC (see Fig. 8.4) are defined
as follows :
Fig. 8.5
side opposite to angle A BC
sine of  A = 
hypotenuse AC

side adjacent to angle A AB

cosine of  A = 
hypotenuse AC
side opposite to angle A BC
tangent of  A = 
side adjacent to angle A AB
1 hypotenuse AC
cosecant of  A =  
sine of  A side opposite to angle A BC
1 hypotenuse AC
secant of  A =  
cosine of  A side adjacent to angle A AB
1 side adjacent to angle A AB
cotangent of  A =  
tangent of  A side opposite to angle A BC
The ratios defined above are abbreviated as sin A, cos A, tan A, cosec A, sec A
and cot A respectively. Note that the ratios cosec A, sec A and cot A are respectively,
the reciprocals of the ratios sin A, cos A and tan A.
BC
BC AC sin A cos A .
Also, observe that tan A =   and cot A =
AB AB cos A sin A
AC
So, the trigonometric ratios of an acute angle in a right triangle express the
relationship between the angle and the length of its sides.
Why don’t you try to define the trigonometric ratios for angle C in the right
triangle? (See Fig. 8.5)

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116 MATHEMATICS

The first use of the idea of ‘sine’ in the way we use

it today was in the work Aryabhatiyam by Aryabhata,
in A.D. 500. Aryabhata used the word ardha-jya
for the half-chord, which was shortened to jya or
jiva in due course. When the Aryabhatiyam was
translated into Arabic, the word jiva was retained as
it is. The word jiva was translated into sinus, which
means curve, when the Arabic version was translated
into Latin. Soon the word sinus, also used as sine,
became common in mathematical texts throughout
Europe. An English Professor of astronomy Edmund
Gunter (1581–1626), first used the abbreviated Aryabhata
notation ‘sin’. C.E. 476 – 550
The origin of the terms ‘cosine’ and ‘tangent’ was much later. The cosine function
arose from the need to compute the sine of the complementary angle. Aryabhatta
called it kotijya. The name cosinus originated with Edmund Gunter. In 1674, the
English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.

Remark : Note that the symbol sin A is used as an

abbreviation for ‘the sine of the angle A’. sin A is not
the product of ‘sin’ and A. ‘sin’ separated from A
has no meaning. Similarly, cos A is not the product of
‘cos’ and A. Similar interpretations follow for other
trigonometric ratios also.
Now, if we take a point P on the hypotenuse
AC or a point Q on AC extended, of the right triangle
ABC and draw PM perpendicular to AB and QN
perpendicular to AB extended (see Fig. 8.6), how
will the trigonometric ratios of  A in  PAM differ
from those of  A in  CAB or from those of  A in Fig. 8.6
 QAN?
To answer this, first look at these triangles. Is  PAM similar to  CAB? From
Chapter 6, recall the AA similarity criterion. Using the criterion, you will see that the
triangles PAM and CAB are similar. Therefore, by the property of similar triangles,
the corresponding sides of the triangles are proportional.
AM AP MP
So, we have =  
AB AC BC

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INTRODUCTION TO TRIGONOMETRY 117

MP BC
From this, we find =  sin A .
AP AC
AM AB MP BC
Similarly,  = cos A,   tan A and so on.
AP AC AM AB
This shows that the trigonometric ratios of angle A in  PAM not differ from
those of angle A in  CAB.
In the same way, you should check that the value of sin A (and also of other
trigonometric ratios) remains the same in  QAN also.
From our observations, it is now clear that the values of the trigonometric
ratios of an angle do not vary with the lengths of the sides of the triangle, if
the angle remains the same.
Note : For the sake of convenience, we may write sin2A, cos2A, etc., in place of
(sin A)2, (cos A)2, etc., respectively. But cosec A = (sin A)–1  sin–1 A (it is called sine
inverse A). sin–1 A has a different meaning, which will be discussed in higher classes.
Similar conventions hold for the other trigonometric ratios as well. Sometimes, the
Greek letter  (theta) is also used to denote an angle.
We have defined six trigonometric ratios of an acute angle. If we know any one
of the ratios, can we obtain the other ratios? Let us see.
1
If in a right triangle ABC, sin A = ,
3
BC 1
then this means that  , i.e., the
AC 3
lengths of the sides BC and AC of the triangle
ABC are in the ratio 1 : 3 (see Fig. 8.7). So if
BC is equal to k, then AC will be 3k, where
Fig. 8.7
k is any positive number. To determine other
trigonometric ratios for the angle A, we need to find the length of the third side
AB. Do you remember the Pythagoras theorem? Let us use it to determine the
required length AB.
AB2 = AC2 – BC2 = (3k)2 – (k)2 = 8k2 = (2 2 k)2
Therefore, AB =  2 2 k
So, we get AB = 2 2 k (Why is AB not – 2 2 k ?)
AB 2 2 k 2 2
Now, cos A =  
AC 3k 3
Similarly, you can obtain the other trigonometric ratios of the angle A.

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118 MATHEMATICS

Remark : Since the hypotenuse is the longest side in a right triangle, the value of
sin A or cos A is always less than 1 (or, in particular, equal to 1).
Let us consider some examples.

4
Example 1 : Given tan A = , find the other
3
trigonometric ratios of the angle A.
Solution : Let us first draw a right  ABC
(see Fig 8.8).
BC 4
Now, we know that tan A =  .
AB 3
Therefore, if BC = 4k, then AB = 3k, where k is a
positive number.
Fig. 8.8
Now, by using the Pythagoras Theorem, we have
AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2
So, AC = 5k
Now, we can write all the trigonometric ratios using their definitions.
BC 4k 4
sin A =  
AC 5k 5
AB 3k 3
cos A =  
AC 5k 5
1 3 1 5 1 5
Therefore, cot A =  , cosec A =  and sec A =  
tan A 4 sin A 4 cos A 3

Example 2 : If  B and  Q are

acute angles such that sin B = sin Q,
then prove that  B =  Q.
Solution : Let us consider two right
triangles ABC and PQR where
sin B = sin Q (see Fig. 8.9).
AC Fig. 8.9
We have sin B =
AB
PR
and sin Q =
PQ

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INTRODUCTION TO TRIGONOMETRY 119

AC PR
Then =
AB PQ

AC AB
Therefore, =  k , say (1)
PR PQ
Now, using Pythagoras theorem,
BC = AB2  AC2

and QR = PQ2 – PR 2

BC AB2  AC 2 k 2 PQ 2  k 2 PR 2 k PQ 2  PR 2
So, =   k (2)
QR PQ 2  PR 2 PQ 2  PR 2 PQ2  PR 2
From (1) and (2), we have
AC AB BC
= 
PR PQ QR
Then, by using Theorem 6.4,  ACB ~  PRQ and therefore,  B =  Q.

Example 3 : Consider  ACB, right-angled at C, in

which AB = 29 units, BC = 21 units and  ABC = 
(see Fig. 8.10). Determine the values of
(i) cos2  + sin2 ,
(ii) cos2  – sin2 
Solution : In  ACB, we have

AC = AB2  BC 2 = (29) 2  (21) 2 Fig. 8.10

= (29  21) (29  21)  (8) (50)  400  20 units

AC 20 , BC 21
So, sin  =  cos  =  
AB 29 AB 29
2 2
 20   21  202  212 400  441
Now, (i) cos  + sin  =      
2 2   1,
 29   29  292 841
2 2
 21   20  (21  20) (21  20) 41
and (ii) cos  – sin  =      
2 2  .
 29   29 
2
29 841

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Example 4 : In a right triangle ABC, right-angled at B,

if tan A = 1, then verify that
2 sin A cos A = 1.

BC
Solution : In  ABC, tan A = =1 (see Fig 8.11)
AB

i.e., BC = AB
Fig. 8.11
Let AB = BC = k, where k is a positive number.

Now, AC = AB2  BC 2

= ( k ) 2  (k ) 2  k 2

BC 1 AB 1
Therefore, sin A =  and cos A = 
AC 2 AC 2

 1  1 
So, 2 sin A cos A = 2     1, which is the required value.
 2  2 
Example 5 : In  OPQ, right-angled at P,
OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12).
Determine the values of sin Q and cos Q.

Solution : In  OPQ, we have

OQ2 = OP2 + PQ2

i.e., (1 + PQ)2 = OP2 + PQ2 (Why?)

i.e., 1 + PQ2 + 2PQ = OP2 + PQ2

i.e., 1 + 2PQ = 72 (Why?)

i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm
Fig. 8.12
7 24
So, sin Q = and cos Q = 
25 25

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INTRODUCTION TO TRIGONOMETRY 121

EXERCISE 8.1
1. In  ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C
2. In Fig. 8.13, find tan P – cot R.
3,
3. If sin A = calculate cos A and tan A.
4
4. Given 15 cot A = 8, find sin A and sec A.
13 ,
5. Given sec  = calculate all other trigonometric ratios. Fig. 8.13
12
6. If  A and  B are acute angles such that cos A = cos B, then show that  A =  B.

7, (1  sin ) (1  sin ) ,
7. If cot  = evaluate : (i) (ii) cot2 
8 (1  cos ) (1  cos )

1  tan 2 A
8. If 3 cot A = 4, check whether = cos2 A – sin2A or not.
1 + tan 2 A
1 ,
9. In triangle ABC, right-angled at B, if tan A = find the value of:
3
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
10. In  PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of
sin P, cos P and tan P.
11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
12
(ii) sec A = for some value of angle A.
5
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
4
(v) sin  = for some angle .
3

8.3 Trigonometric Ratios of Some Specific Angles

From geometry, you are already familiar with the construction of angles of 30°, 45°,
60° and 90°. In this section, we will find the values of the trigonometric ratios for these
angles and, of course, for 0°.

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122 MATHEMATICS

Trigonometric Ratios of 45°

In  ABC, right-angled at B, if one angle is 45°, then
the other angle is also 45°, i.e.,  A =  C = 45°
(see Fig. 8.14).
So, BC = AB (Why?)
Now, Suppose BC = AB = a.
Fig. 8.14
Then by Pythagoras Theorem, AC2 = AB2 + BC2 = a2 + a2 = 2a2,

and, therefore, AC = a 2 
Using the definitions of the trigonometric ratios, we have :

side opposite to angle 45° BC a 1

sin 45° =   
hypotenuse AC a 2 2

side adjacent to angle 45° AB a 1

cos 45° =   
hypotenuse AC a 2 2

side opposite to angle 45° BC a

tan 45° =   1
side adjacent to angle 45° AB a

1 1 1
Also, cosec 45° =  2 , sec 45° =  2 , cot 45° =  1.
sin 45 cos 45 tan 45

Trigonometric Ratios of 30° and 60°

Let us now calculate the trigonometric ratios of 30°
and 60°. Consider an equilateral triangle ABC. Since
each angle in an equilateral triangle is 60°, therefore,
 A =  B =  C = 60°.
Draw the perpendicular AD from A to the side BC
(see Fig. 8.15).
Now  ABD   ACD (Why?) Fig. 8.15
Therefore, BD = DC
and  BAD =  CAD (CPCT)
Now observe that:
 ABD is a right triangle, right- angled at D with  BAD = 30° and  ABD = 60°
(see Fig. 8.15).

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INTRODUCTION TO TRIGONOMETRY 123

As you know, for finding the trigonometric ratios, we need to know the lengths of the
sides of the triangle. So, let us suppose that AB = 2a.

1
Then, BD = BC = a
2
and AD2 = AB2 – BD2 = (2a)2 – (a)2 = 3a2,

Therefore, AD = a 3
Now, we have :
BD a 1 AD a 3 3
sin 30° =   , cos 30° =  
AB 2a 2 AB 2a 2
BD a 1
tan 30° =   .
AD a 3 3
1 1 2
Also, cosec 30° =  2, sec 30° = 
sin 30 cos 30 3
1
cot 30° =  3.
tan 30
Similarly,
AD a 3 3 1
sin 60° =   , cos 60° = , tan 60° = 3,
AB 2a 2 2
2 , 1
cosec 60° = sec 60° = 2 and cot 60° = 
3 3

Trigonometric Ratios of 0° and 90°

Let us see what happens to the trigonometric ratios of angle
A, if it is made smaller and smaller in the right triangle ABC
(see Fig. 8.16), till it becomes zero. As  A gets smaller and
smaller, the length of the side BC decreases.The point C gets
closer to point B, and finally when  A becomes very close
to 0°, AC becomes almost the same as AB (see Fig. 8.17). Fig. 8.16

Fig. 8.17

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When  A is very close to 0°, BC gets very close to 0 and so the value of
BC
sin A = is very close to 0. Also, when  A is very close to 0°, AC is nearly the
AC
AB
same as AB and so the value of cos A = is very close to 1.
AC
This helps us to see how we can define the values of sin A and cos A when
A = 0°. We define : sin 0° = 0 and cos 0° = 1.
Using these, we have :

sin 0° 1 ,
tan 0° = = 0, cot 0° = which is not defined. (Why?)
cos 0° tan 0°
1 1 ,
sec 0° = = 1 and cosec 0° = which is again not defined.(Why?)
cos 0 sin 0
Now, let us see what happens to the trigonometric ratios of  A, when it is made
larger and larger in  ABC till it becomes 90°. As  A gets larger and larger,  C gets
smaller and smaller. Therefore, as in the case above, the length of the side AB goes on
decreasing. The point A gets closer to point B. Finally when  A is very close to 90°,
 C becomes very close to 0° and the side AC almost coincides with side BC
(see Fig. 8.18).

Fig. 8.18

When  C is very close to 0°,  A is very close to 90°, side AC is nearly the
same as side BC, and so sin A is very close to 1. Also when  A is very close to 90°,
 C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0.
So, we define : sin 90° = 1 and cos 90° = 0.
Now, why don’t you find the other trigonometric ratios of 90°?
We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60°
and 90° in Table 8.1, for ready reference.

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INTRODUCTION TO TRIGONOMETRY 125

Table 8.1

A 0° 30° 45° 60° 90°

1 1 3
sin A 0 1
2 2 2

3 1 1
cos A 1 0
2 2 2

1
tan A 0 1 3 Not defined
3

2
cosec A Not defined 2 2 1
3

2
sec A 1 2 2 Not defined
3
1
cot A Not defined 3 1 0
3
Remark : From the table above you can observe that as  A increases from 0° to
90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0.
Let us illustrate the use of the values in the table above through some examples.

Example 6 : In  ABC, right-angled at B,

AB = 5 cm and  ACB = 30° (see Fig. 8.19).
Determine the lengths of the sides BC and AC.
Solution : To find the length of the side BC, we will
choose the trigonometric ratio involving BC and the
given side AB. Since BC is the side adjacent to angle
C and AB is the side opposite to angle C, therefore Fig. 8.19
AB
= tan C
BC
5 1
i.e., = tan 30° =
BC 3
which gives BC = 5 3 cm

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To find the length of the side AC, we consider

AB
sin 30° = (Why?)
AC

1 5
i.e., =
2 AC
i.e., AC = 10 cm
Note that alternatively we could have used Pythagoras theorem to determine the third
side in the example above,

i.e., AC = AB2  BC 2  52  (5 3) 2 cm = 10cm.

Example 7 : In  PQR, right - angled at

Q (see Fig. 8.20), PQ = 3 cm and PR = 6 cm.
Determine  QPR and  PRQ.
Solution : Given PQ = 3 cm and PR = 6 cm.

PQ
Therefore, = sin R
PR
Fig. 8.20
3 1
or sin R = 
6 2
So,  PRQ = 30°
and therefore,  QPR = 60°. (Why?)
You may note that if one of the sides and any other part (either an acute angle or any
side) of a right triangle is known, the remaining sides and angles of the triangle can be
determined.
1 1
Example 8 : If sin (A – B) = , cos (A + B) = , 0° < A + B  90°, A > B, find A
2 2
and B.
1
Solution : Since, sin (A – B) = , therefore, A – B = 30° (Why?) (1)
2
1
Also, since cos (A + B) = , therefore, A + B = 60° (Why?) (2)
2
Solving (1) and (2), we get : A = 45° and B = 15°.

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INTRODUCTION TO TRIGONOMETRY 127

EXERCISE 8.2
1. Evaluate the following :

(i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan2 45° + cos2 30° – sin2 60°

cos 45° sin 30° + tan 45° – cosec 60°

(iii) sec 30° + cosec 30° (iv)
sec 30° + cos 60° + cot 45°

5 cos 2 60  4 sec 2 30  tan 2 45

(v)
sin 2 30  cos 2 30

2. Choose the correct option and justify your choice :

2 tan 30
(i) 
1  tan 2 30
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°

1  tan 2 45
(ii) 
1  tan 2 45
(A) tan 90° (B) 1 (C) sin 45° (D) 0

(iii) sin 2A = 2 sin A is true when A =

(A) 0° (B) 30° (C) 45° (D) 60°

2 tan 30
(iv) 
1  tan 2 30
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

1
3. If tan (A + B) = 3 and tan (A – B) = ; 0° < A + B  90°; A > B, find A and B.
3

4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin  increases as  increases.
(iii) The value of cos  increases as  increases.
(iv) sin  = cos  for all values of .
(v) cot A is not defined for A = 0°.

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8.4 Trigonometric Identities

You may recall that an equation is called an identity
when it is true for all values of the variables involved.
Similarly, an equation involving trigonometric ratios
of an angle is called a trigonometric identity, if it is
true for all values of the angle(s) involved.
In this section, we will prove one trigonometric
identity, and use it further to prove other useful
trigonometric identities. Fig. 8.21
In  ABC, right-angled at B (see Fig. 8.21), we have:
AB2 + BC2 = AC 2 (1)
2
Dividing each term of (1) by AC , we get

AB2 BC 2 AC2
 =
AC 2 AC 2 AC2
2 2 2
 AB   BC   AC 
i.e.,     =  
 AC   AC   AC 
i.e., (cos A)2 + (sin A)2 = 1
i.e., cos2 A + sin2 A = 1 (2)

This is true for all A such that 0°  A  90°. So, this is a trigonometric identity.
Let us now divide (1) by AB2. We get

AB2 BC 2 AC2
 =
AB2 AB2 AB2
2 2 2
 AB   BC   AC 
or,     =  
 AB   AB   AB 
i.e., 1 + tan2 A = sec 2 A (3)
Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and
sec A are not defined for A = 90°. So, (3) is true for all A such that 0°  A  90°.
Let us see what we get on dividing (1) by BC2. We get

AB2 BC2 AC2

 =
BC2 BC2 BC2

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INTRODUCTION TO TRIGONOMETRY 129

2 2 2
 AB   BC   AC 
i.e.,     =  
 BC   BC   BC 

i.e., cot2 A + 1 = cosec 2 A (4)

Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for
all A such that 0° < A  90°.
Using these identities, we can express each trigonometric ratio in terms of other
trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the
values of other trigonometric ratios.
Let us see how we can do this using these identities. Suppose we know that
1
tan A =  Then, cot A = 3.
3

1 4, 2 3
Since, sec2 A = 1 + tan2 A = 1   sec A = , and cos A = 
3 3 3 2

3 1
Again, sin A = 1  cos2 A  1   . Therefore, cosec A = 2.
4 2

Example 9 : Express the ratios cos A, tan A and sec A in terms of sin A.

Solution : Since cos2 A + sin2 A = 1, therefore,

cos2 A = 1 – sin2 A, i.e., cos A =  1  sin A

2

This gives cos A = 1  sin 2 A (Why?)

sin A sin A 1 1
Hence, tan A = = and sec A = 
cos A 1 – sin 2 A cos A 1  sin 2 A

Example 10 : Prove that sec A (1 – sin A)(sec A + tan A) = 1.

Solution :

 1   1 sin A 
LHS = sec A (1 – sin A)(sec A + tan A) =   (1  sin A)   
 cos A   cos A cos A 

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130 MATHEMATICS

(1  sin A) (1 + sin A) 1  sin 2 A

= 
cos2 A cos2 A
cos2 A
=  1 = RHS
cos2 A

cot A – cos A cosec A – 1

Example 11 : Prove that 
cot A + cos A cosec A + 1

cos A
 cos A
cot A – cos A sin A
Solution : LHS = 
cot A + cos A cos A
 cos A
sin A
 1   1 
cos A  1    1
 sin A    sin A   cosec A – 1
= = RHS
 1   1  cosec A + 1
cos A   1   1
 sin A   sin A 

sin   cos   1 1
Example 12 : Prove that  , using the identity
sin   cos   1 sec   tan 
sec2  = 1 + tan2 .
Solution : Since we will apply the identity involving sec  and tan , let us first
convert the LHS (of the identity we need to prove) in terms of sec  and tan  by
dividing numerator and denominator by cos 

sin  – cos  + 1 tan   1  sec 

LHS = 
sin  + cos  – 1 tan   1  sec 

(tan   sec )  1 {(tan   sec )  1} (tan   sec )

= 
(tan   sec )  1 {(tan   sec )  1} (tan   sec )

(tan 2   sec 2 )  (tan   sec )

=
{tan   sec   1} (tan   sec )

– 1  tan   sec 
=
(tan   sec   1) (tan   sec )

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INTRODUCTION TO TRIGONOMETRY 131

–1 1
=  ,
tan   sec  sec   tan 
which is the RHS of the identity, we are required to prove.

EXERCISE 8.3
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
2. Write all the other trigonometric ratios of  A in terms of sec A.
3. Choose the correct option. Justify your choice.
(i) 9 sec2 A – 9 tan2 A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan  + sec ) (1 + cot  – cosec ) =
(A) 0 (B) 1 (C) 2 (D) –1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A (B) sin A (C) cosec A (D) cos A

1  tan 2 A
(iv) 
1 + cot 2 A
(A) sec2 A (B) –1 (C) cot2 A (D) tan2 A
4. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.

1  cos  cos A 1  sin A

(i) (cosec  – cot )2 = 1  cos  (ii)   2 sec A
1 + sin A cos A
tan  cot 
(iii)   1  sec  cosec 
1  cot  1  tan 
[Hint : Write the expression in terms of sin  and cos ]
1  sec A sin 2 A
(iv)  [Hint : Simplify LHS and RHS separately]
sec A 1 – cos A
cos A – sin A + 1
(v)  cosec A + cot A, using the identity cosec2 A = 1 + cot2 A.
cos A + sin A – 1
1  sin A sin   2 sin 3 
(vi)  sec A + tan A (vii)  tan 
1 – sin A 2 cos3   cos 
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

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132 MATHEMATICS

1
(ix) (cosec A – sin A) (sec A – cos A)  tan A + cot A

[Hint : Simplify LHS and RHS separately]

2
 1  tan 2 A   1  tan A 
(x)  2    = tan2 A
 1 + cot A   1 – cot A 

8.5 Summary
In this chapter, you have studied the following points :
1. In a right triangle ABC, right-angled at B,
side opposite to angle A , side adjacent to angle A
sin A = cos A =
hypotenuse hypotenuse
side opposite to angle A
tan A = .
side adjacent to angle A
1 1 1 , sin A
2. cosec A = ; sec A = ; tan A = tan A = .
sin A cos A cot A cos A
3. If one of the trigonometric ratios of an acute angle is known, the remaining trigonometric
ratios of the angle can be easily determined.
4. The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°.
5. The value of sin A or cos A never exceeds 1, whereas the value of sec A (0° £ A < 90°) or
cosec A (0° < A £ 90º) is always greater than or equal to 1.
6. sin2 A + cos2 A = 1,
sec2 A – tan2 A = 1 for 0° £ A < 90°,
cosec2 A = 1 + cot2 A for 0° < A £ 90º.

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SOME APPLICATIONS OF TRIGONOMETRY 133

SOME APPLICATIONS OF
TRIGONOMETRY 9
9.1 Heights and Distances
In the previous chapter, you have studied about trigonometric ratios. In this chapter,
you will be studying about some ways in which trigonometry is used in the life around
you.
Let us consider Fig. 8.1 of prvious chapter, which is redrawn below in Fig. 9.1.

Fig. 9.1
In this figure, the line AC drawn from the eye of the student to the top of the
minar is called the line of sight. The student is looking at the top of the minar. The
angle BAC, so formed by the line of sight with the horizontal, is called the angle of
elevation of the top of the minar from the eye of the student.
Thus, the line of sight is the line drawn from the eye of an observer to the point
in the object viewed by the observer. The angle of elevation of the point viewed is

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134 MATHEMATICS

the angle formed by the line of sight with the horizontal when the point being viewed is
above the horizontal level, i.e., the case when we raise our head to look at the object
(see Fig. 9.2).

Fig. 9.2
Now, consider the situation given in Fig. 8.2. The girl sitting on the balcony is
looking down at a flower pot placed on a stair of the temple. In this case, the line of
sight is below the horizontal level. The angle so formed by the line of sight with the
horizontal is called the angle of depression.
Thus, the angle of depression of a point on the object being viewed is the angle
formed by the line of sight with the horizontal when the point is below the horizontal
level, i.e., the case when we lower our head to look at the point being viewed
(see Fig. 9.3).

Fig. 9.3
Now, you may identify the lines of sight, and the angles so formed in Fig. 8.3.
Are they angles of elevation or angles of depression?
Let us refer to Fig. 9.1 again. If you want to find the height CD of the minar
without actually measuring it, what information do you need? You would need to know
the following:
(i) the distance DE at which the student is standing from the foot of the minar

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SOME APPLICATIONS OF TRIGONOMETRY 135

(ii) the angle of elevation,  BAC, of the top of the minar

(iii) the height AE of the student.
Assuming that the above three conditions are known, how can we determine the
height of the minar?
In the figure, CD = CB + BD. Here, BD = AE, which is the height of the student.
To find BC, we will use trigonometric ratios of  BAC or  A.
In  ABC, the side BC is the opposite side in relation to the known  A. Now,
which of the trigonometric ratios can we use? Which one of them has the two values
that we have and the one we need to determine? Our search narrows down to using
either tan A or cot A, as these ratios involve AB and BC.

BC AB ,
Therefore, tan A = or cot A = which on solving would give us BC.
AB BC
By adding AE to BC, you will get the height of the minar.

Now let us explain the process, we have just discussed, by solving some problems.

Example 1 : A tower stands vertically on the ground. From a point on the ground,
which is 15 m away from the foot of the tower, the angle of elevation of the top of the
tower is found to be 60°. Find the height of the tower.
Solution : First let us draw a simple diagram to
represent the problem (see Fig. 9.4). Here AB
represents the tower, CB is the distance of the point
from the tower and  ACB is the angle of elevation.
We need to determine the height of the tower, i.e.,
AB. Also, ACB is a triangle, right-angled at B.
To solve the problem, we choose the trigonometric
ratio tan 60° (or cot 60°), as the ratio involves AB
and BC.
AB
Now, tan 60° =
BC

AB
i.e., 3 =
15 Fig. 9.4
i.e., AB = 15 3

Hence, the height of the tower is 15 3 m.

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136 MATHEMATICS

Example 2 : An electrician has to repair an

electric fault on a pole of height 5 m. She needs
to reach a point 1.3m below the top of the pole
to undertake the repair work (see Fig. 9.5). What
should be the length of the ladder that she should
use which, when inclined at an angle of 60° to
the horizontal, would enable her to reach the
required position? Also, how far from the foot
of the pole should she place the foot of the
ladder? (You may take 3 = 1.73)
Solution : In Fig. 9.5, the electrician is required to
reach the point B on the pole AD.
So, BD = AD – AB = (5 – 1.3)m = 3.7 m. Fig. 9.5
Here, BC represents the ladder. We need to find
its length, i.e., the hypotenuse of the right triangle BDC.
Now, can you think which trigonometic ratio should we consider?
It should be sin 60°.

BD 3.7 3
So, = sin 60° or =
BC BC 2

3.7  2
Therefore, BC = = 4.28 m (approx.)
3
i.e., the length of the ladder should be 4.28 m.

DC 1
Now, = cot 60° =
BD 3

3.7
i.e., DC = = 2.14 m (approx.)
3
Therefore, she should place the foot of the ladder at a distance of 2.14 m from
the pole.

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SOME APPLICATIONS OF TRIGONOMETRY 137

Example 3 : An observer 1.5 m tall is 28.5 m away

from a chimney. The angle of elevation of the top
of the chimney from her eyes is 45°. What is the
height of the chimney?
Solution : Here, AB is the chimney, CD the
observer and  ADE the angle of elevation (see
Fig. 9.6). In this case, ADE is a triangle, right-angled
at E and we are required to find the height of the
chimney.
We have AB = AE + BE = AE + 1.5
Fig. 9.6
and DE = CB = 28.5 m
To determine AE, we choose a trigonometric ratio, which involves both AE and
DE. Let us choose the tangent of the angle of elevation.

AE
Now, tan 45° =
DE

AE
i.e., 1=
28.5
Therefore, AE = 28.5
So the height of the chimney (AB) = (28.5 + 1.5) m = 30 m.

Example 4 : From a point P on the ground the angle of elevation of the top of a 10 m
tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation
of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the
distance of the building from the point P. (You may take 3 = 1.732)
Solution : In Fig. 9.7, AB denotes the height of the building, BD the flagstaff and P
the given point. Note that there are two right triangles PAB and PAD. We are required
to find the length of the flagstaff, i.e., DB and the distance of the building from the
point P, i.e., PA.

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138 MATHEMATICS

Since, we know the height of the building AB, we

will first consider the right  PAB.

AB
We have tan 30° =
AP

1 10
i.e., =
3 AP

Therefore, AP = 10 3 Fig. 9.7

i.e., the distance of the building from P is 10 3 m = 17.32 m.

Next, let us suppose DB = x m. Then AD = (10 + x) m.

AD 10  x
Now, in right  PAD, tan 45° = 
AP 10 3

10  x
Therefore, 1=
10 3

i.e., x = 10  
3  1 = 7.32

So, the length of the flagstaff is 7.32 m.

Example 5 : The shadow of a tower standing

on a level ground is found to be 40 m longer
when the Sun’s altitude is 30° than when it is
60°. Find the height of the tower.
Solution : In Fig. 9.8, AB is the tower and
BC is the length of the shadow when the
Sun’s altitude is 60°, i.e., the angle of
elevation of the top of the tower from the tip
Fig. 9.8
of the shadow is 60° and DB is the length of
the shadow, when the angle of elevation
is 30°.

Now, let AB be h m and BC be x m. According to the question, DB is 40 m longer

than BC.

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SOME APPLICATIONS OF TRIGONOMETRY 139

So, DB = (40 + x) m
Now, we have two right triangles ABC and ABD.
AB
In  ABC, tan 60° =
BC
h
or, 3 = x (1)

AB
In  ABD, tan 30° =
BD
1 h
i.e., = (2)
3 x  40

From (1), we have h= x 3


Putting this value in (2), we get x 3  3 = x + 40, i.e., 3x = x + 40

i.e., x = 20

So, h = 20 3 [From (1)]

Therefore, the height of the tower is 20 3 m.

Example 6 : The angles of depression of the top and the bottom of an 8 m tall building
from the top of a multi-storeyed building are 30°
and 45°, respectively. Find the height of the multi-
storeyed building and the distance between the
two buildings.
Solution : In Fig. 9.9, PC denotes the multi-
storyed building and AB denotes the 8 m tall
building. We are interested to determine the height
of the multi-storeyed building, i.e., PC and the
distance between the two buildings, i.e., AC.
Look at the figure carefully. Observe that PB is
a transversal to the parallel lines PQ and BD.
Therefore, QPB and PBD are alternate
angles, and so are equal. Fig. 9.9
So PBD = 30°. Similarly,  PAC = 45°.
In right  PBD, we have

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140 MATHEMATICS

PD 1
= tan 30° = or BD = PD 3
BD 3
In right  PAC, we have

PC
= tan 45° = 1
AC
i.e., PC = AC
Also, PC = PD + DC, therefore, PD + DC = AC.

Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD = PD 3 (Why?)

8

8  
3 1
 4  3  1 m.
 
This gives PD =
3 1 3 1 3 1

So, the height of the multi-storeyed building is 4 

3  1  8 m = 4  3 + 3  m
and the distance between the two buildings is also 4  3  3  m.

Example 7 : From a point on a bridge across a river, the angles of depression of

the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge
is at a height of 3 m from the banks, find the width of the river.
Solution : In Fig 9.10, A and B
represent points on the bank on
opposite sides of the river, so that
AB is the width of the river. P is a
point on the bridge at a height of 3
m, i.e., DP = 3 m. We are
interested to determine the width Fig. 9.10
of the river, which is the length of
the side AB of the D APB.
Now, AB = AD + DB
In right  APD,  A = 30°.

PD
So, tan 30° =
AD

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SOME APPLICATIONS OF TRIGONOMETRY 141

1 3
i.e., = or AD = 3 3 m
3 AD
Also, in right  PBD,  B = 45°. So, BD = PD = 3 m.

Now, AB = BD + AD = 3 + 3 3 = 3 (1 + 3 ) m.

Therefore, the width of the river is 3  

3 1 m.

EXERCISE 9.1
1. A circus artist is climbing a 20 m long rope, which is
tightly stretched and tied from the top of a vertical
pole to the ground. Find the height of the pole, if
the angle made by the rope with the ground level is
30° (see Fig. 9.11).
2. A tree breaks due to storm and the broken part
bends so that the top of the tree touches the ground
making an angle 30° with it. The distance between Fig. 9.11
the foot of the tree to the point where the top
touches the ground is 8 m. Find the height of
the tree.
3. A contractor plans to install two slides for the children to play in a park. For the children
below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and
is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have
a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What
should be the length of the slide in each case?
4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m
away from the foot of the tower, is 30°. Find the height of the tower.
5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is
temporarily tied to a point on the ground. The inclination of the string with the ground
is 60°. Find the length of the string, assuming that there is no slack in the string.
6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of
elevation from his eyes to the top of the building increases from 30° to 60° as he walks
towards the building. Find the distance he walked towards the building.
7. From a point on the ground, the angles of elevation of the bottom and the top of a
transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.
Find the height of the tower.

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142 MATHEMATICS

8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the
angle of elevation of the top of the statue is 60° and from the same point the angle of
elevation of the top of the pedestal is 45°. Find the height of the pedestal.
9. The angle of elevation of the top of a building from the foot of the tower is 30° and the
angle of elevation of the top of the tower from the foot of the building is 60°. If the tower
is 50 m high, find the height of the building.
10. Two poles of equal heights are standing opposite each other on either side of the road,
which is 80 m wide. From a point between them on the road, the angles of elevation of
the top of the poles are 60° and 30°, respectively. Find the height of the poles and the
distances of the point from the poles.
11. A TV tower stands vertically on a bank
of a canal. From a point on the other
bank directly opposite the tower, the
angle of elevation of the top of the
tower is 60°. From another point 20 m
away from this point on the line joing
this point to the foot of the tower, the
angle of elevation of the top of the
tower is 30° (see Fig. 9.12). Find the
height of the tower and the width of Fig. 9.12
the canal.
12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is
60° and the angle of depression of its foot is 45°. Determine the height of the tower.
13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of
depression of two ships are 30° and 45°. If one ship is exactly behind the other on the
same side of the lighthouse, find the distance between the two ships.
14. A 1.2 m tall girl spots a balloon moving
with the wind in a horizontal line at a
height of 88.2 m from the ground. The
angle of elevation of the balloon from
the eyes of the girl at any instant is
60°. After some time, the angle of
elevation reduces to 30° (see Fig. 9.13).
Find the distance travelled by the
balloon during the interval. Fig. 9.13
15. A straight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30°, which is approaching the foot of the

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SOME APPLICATIONS OF TRIGONOMETRY 143

tower with a uniform speed. Six seconds later, the angle of depression of the car is found
to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

9.2 Summary
In this chapter, you have studied the following points :
1. (i) The line of sight is the line drawn from the eye of an observer to the point in the
object viewed by the observer.
(ii) The angle of elevation of an object viewed, is the angle formed by the line of sight
with the horizontal when it is above the horizontal level, i.e., the case when we raise
our head to look at the object.
(iii) The angle of depression of an object viewed, is the angle formed by the line of sight
with the horizontal when it is below the horizontal level, i.e., the case when we lower
our head to look at the object.
2. The height or length of an object or the distance between two distant objects can be
determined with the help of trigonometric ratios.

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 Chapter 3
TRIGONOMETRIC FUNCTIONS
vA mathematician knows how to solve a problem,
he can not solve it. – MILNE v

3.1 Introduction
The word ‘trigonometry’ is derived from the Greek words
‘trigon’ and ‘metron’ and it means ‘measuring the sides of
a triangle’. The subject was originally developed to solve
geometric problems involving triangles. It was studied by
sea captains for navigation, surveyor to map out the new
lands, by engineers and others. Currently, trigonometry is
used in many areas such as the science of seismology,
designing electric circuits, describing the state of an atom,
predicting the heights of tides in the ocean, analysing a
musical tone and in many other areas.
In earlier classes, we have studied the trigonometric Arya Bhatt
ratios of acute angles as the ratio of the sides of a right (476-550)
angled triangle. We have also studied the trigonometric identities and application of
trigonometric ratios in solving the problems related to heights and distances. In this
Chapter, we will generalise the concept of trigonometric ratios to trigonometric functions
and study their properties.
3.2 Angles
Angle is a measure of rotation of a given ray about its initial point. The original ray is
Vertex

Fig 3.1

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44 MATHEMATICS

called the initial side and the final position of the ray after rotation is called the
terminal side of the angle. The point of rotation is called the vertex. If the direction of
rotation is anticlockwise, the angle is said to be positive and if the direction of rotation
is clockwise, then the angle is negative (Fig 3.1).
The measure of an angle is the amount of
rotation performed to get the terminal side from
the initial side. There are several units for
measuring angles. The definition of an angle Fig 3.2
suggests a unit, viz. one complete revolution from the position of the initial side as
indicated in Fig 3.2.
This is often convenient for large angles. For example, we can say that a rapidly
spinning wheel is making an angle of say 15 revolution per second. We shall describe
two other units of measurement of an angle which are most commonly used, viz.
degree measure and radian measure.
th
 1 
3.2.1 Degree measure If a rotation from the initial side to terminal side is   of
 360 
a revolution, the angle is said to have a measure of one degree, written as 1°. A degree is
divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is
called a minute, written as 1′, and one sixtieth of a minute is called a second, written as 1″.
Thus, 1° = 60′, 1′ = 60″
Some of the angles whose measures are 360°,180°, 270°, 420°, – 30°, – 420° are
shown in Fig 3.3.

Fig 3.3

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 TRIGONOMETRIC FUNCTIONS 45

3.2.2 Radian measure There is another unit for measurement of an angle, called
the radian measure. Angle subtended at the centre by an arc of length 1 unit in a
unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig
3.4(i) to (iv), OA is the initial side and OB is the terminal side. The figures show the
1 1
angles whose measures are 1 radian, –1 radian, 1 radian and –1 radian.
2 2

(i) (ii)
(iii)

(iv)
Fig 3.4 (i) to (iv)
We know that the circumference of a circle of radius 1 unit is 2π. Thus, one
complete revolution of the initial side subtends an angle of 2π radian.
More generally, in a circle of radius r, an arc of length r will subtend an angle of
1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre.
Since in a circle of radius r, an arc of length r subtends an angle whose measure is 1
l
radian, an arc of length l will subtend an angle whose measure is radian. Thus, if in
r
a circle of radius r, an arc of length l subtends an angle θ radian at the centre, we have
l
θ = or l = r θ.
r

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46 MATHEMATICS

P
3.2.3 Relation between radian and real numbers
2
Consider the unit circle with centre O. Let A be any point
on the circle. Consider OA as initial side of an angle.
Then the length of an arc of the circle will give the radian 1
measure of the angle which the arc will subtend at the
centre of the circle. Consider the line PAQ which is
1 A 0
tangent to the circle at A. Let the point A represent the O
real number zero, AP represents positive real number and
AQ represents negative real numbers (Fig 3.5). If we
−1
rope the line AP in the anticlockwise direction along the
circle, and AQ in the clockwise direction, then every real
number will correspond to a radian measure and −2
conversely. Thus, radian measures and real numbers can Fig 3.5 Q
be considered as one and the same.

3.2.4 Relation between degree and radian Since a circle subtends at the centre
an angle whose radian measure is 2π and its degree measure is 360°, it follows that
2π radian = 360° or π radian = 180°
The above relation enables us to express a radian measure in terms of degree
measure and a degree measure in terms of radian measure. Using approximate value
22
of π as , we have
7
180°
1 radian = = 57° 16′ approximately.
π

π
Also 1° = radian = 0.01746 radian approximately.
180

The relation between degree measures and radian measure of some common angles
are given in the following table:

Degree 30° 45° 60° 90° 180° 270° 360°

π π π π 3π
Radian π 2π
6 4 3 2 2

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 TRIGONOMETRIC FUNCTIONS 47

Notational Convention
Since angles are measured either in degrees or in radians, we adopt the convention
that whenever we write angle θ°, we mean the angle whose degree measure is θ and
whenever we write angle β, we mean the angle whose radian measure is β.
Note that when an angle is expressed in radians, the word ‘radian’ is frequently
π π
omitted. Thus, π = 180° and = 45° are written with the understanding that π and
4 4
are radian measures. Thus, we can say that
π
180 ×
Radian measure = Degree measure

180
Degree measure = × Radian measure
π
Example 1 Convert 40° 20′ into radian measure.
Solution We know that 180° = π radian.
1 π 121 121π
Hence 40° 20′ = 40 degree = × radian = radian.
3 180 3 540
121π
Therefore 40° 20′ = radian.
540
Example 2 Convert 6 radians into degree measure.
Solution We know that π radian = 180°.
180 1080 × 7
Hence 6 radians = × 6 degree = degree
π 22
7 7 × 60
= 343 degree = 343° + minute [as 1° = 60′]
11 11
2
= 343° + 38′ + minute [as 1′ = 60″]
11
= 343° + 38′ + 10.9″ = 343°38′ 11″ approximately.
Hence 6 radians = 343° 38′ 11″ approximately.
Example 3 Find the radius of the circle in which a central angle of 60° intercepts an
22
arc of length 37.4 cm (use π = ).
7

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48 MATHEMATICS

60π π
Solution Here l = 37.4 cm and θ = 60° = radian =
180 3
l
Hence, by r = , we have
θ
37.4×3 37.4×3×7
r= = = 35.7 cm
π 22
Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in
40 minutes? (Use π = 3.14).
Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore,
2 2
in 40 minutes, the minute hand turns through of a revolution. Therefore, θ = × 360°
3 3
4π
or radian. Hence, the required distance travelled is given by
3
4π
l = r θ = 1.5 × cm = 2π cm = 2 × 3.14 cm = 6.28 cm.
3
Example 5 If the arcs of the same lengths in two circles subtend angles 65°and 110°
at the centre, find the ratio of their radii.
Solution Let r1 and r2 be the radii of the two circles. Given that
π 13π
θ1 = 65° = × 65 = radian
180 36
π 22π
and θ2 = 110° = × 110 = radian
180 36
Let l be the length of each of the arc. Then l = r1θ1 = r2θ2, which gives
13π 22π r1 22
× r1 = × r2 , i.e., r =
36 36 2 13
Hence r1 : r2 = 22 : 13.

EXERCISE 3.1
1. Find the radian measures corresponding to the following degree measures:
(i) 25° (ii) – 47°30′ (iii) 240° (iv) 520°

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 TRIGONOMETRIC FUNCTIONS 49

2. Find the degree measures corresponding to the following radian measures

22
(Use π = ).
7
11 5π 7π
(i) (ii) – 4 (iii) (iv)
16 3 6
3. A wheel makes 360 revolutions in one minute. Through how many radians does
it turn in one second?
4. Find the degree measure of the angle subtended at the centre of a circle of
22
radius 100 cm by an arc of length 22 cm (Use π = ).
7
5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of
minor arc of the chord.
6. If in two circles, arcs of the same length subtend angles 60° and 75° at the
centre, find the ratio of their radii.
7. Find the angle in radian through which a pendulum swings if its length is 75 cm
and th e tip describes an arc of length
(i) 10 cm (ii) 15 cm (iii) 21 cm
3.3 Trigonometric Functions
In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of
sides of a right angled triangle. We will now extend the definition of trigonometric
ratios to any angle in terms of radian measure and study them as trigonometric functions.
Consider a unit circle with centre
at origin of the coordinate axes. Let
P (a, b) be any point on the circle with
angle AOP = x radian, i.e., length of arc
AP = x (Fig 3.6).
We define cos x = a and sin x = b
Since ∆OMP is a right triangle, we have
OM2 + MP2 = OP2 or a2 + b2 = 1
Thus, for every point on the unit circle,
we have
a2 + b2 = 1 or cos2 x + sin2 x = 1
Since one complete revolution
subtends an angle of 2π radian at the
π
centre of the circle, ∠AOB = , Fig 3.6
2

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50 MATHEMATICS

3π π
∠AOC = π and ∠AOD = . All angles which are integral multiples of are called
2 2
quadrantal angles. The coordinates of the points A, B, C and D are, respectively,
(1, 0), (0, 1), (–1, 0) and (0, –1). Therefore, for quadrantal angles, we have
cos 0° = 1 sin 0° = 0,
π π
cos =0 sin =1
2 2
cosπ = − 1 sinπ = 0
3π 3π
cos =0 sin = –1
2 2
cos 2π = 1 sin 2π = 0
Now, if we take one complete revolution from the point P, we again come back to
same point P. Thus, we also observe that if x increases (or decreases) by any integral
multiple of 2π, the values of sine and cosine functions do not change. Thus,
sin (2nπ + x) = sin x , n ∈ Z , cos (2nπ + x) = cos x , n ∈ Z
Further, sin x = 0, if x = 0, ± π, ± 2π , ± 3π, ..., i.e., when x is an integral multiple of π
π 3π 5π
and cos x = 0, if x = ± ,± ,± , ... i.e., cos x vanishes when x is an odd
2 2 2
π
multiple of . Thus
2
π, where n is any integer
sin x = 0 implies x = nπ,
π
cos x = 0 implies x = (2n + 1) , where n is any integer.
2
We now define other trigonometric functions in terms of sine and cosine functions:
1
cosec x = , x ≠ nπ, where n is any integer.
sin x
1 π
sec x = , x ≠ (2n + 1) , where n is any integer.
cos x 2
sin x π
tan x = , x ≠ (2n +1) , where n is any integer.
cos x 2
cos x
cot x = , x ≠ n π, where n is any integer.
sin x

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 TRIGONOMETRIC FUNCTIONS 51

We have shown that for all real x, sin2 x + cos2 x = 1

It follows that

1 + tan2 x = sec2 x (why?)

1 + cot2 x = cosec2 x (why?)

In earlier classes, we have discussed the values of trigonometric ratios for 0°,
30°, 45°, 60° and 90°. The values of trigonometric functions for these angles are same
as that of trigonometric ratios studied in earlier classes. Thus, we have the following
table:
π π π π 3π
0° π 2π
6 4 3 2 2
1 1 3
sin 0 1 0 –1 0
2 2 2
3 1 1
cos 1 0 –1 0 1
2 2 2
1 not not
tan 0 1 3 0 0
3 defined defined

The values of cosec x, sec x and cot x

are the reciprocal of the values of sin x,
cos x and tan x, respectively.
3.3.1 Sign of trigonometric functions
Let P (a, b) be a point on the unit circle
with centre at the origin such that
∠AOP = x. If ∠AOQ = – x, then the
coordinates of the point Q will be (a, –b)
(Fig 3.7). Therefore
cos (– x) = cos x
and sin (– x) = – sin x
Since for every point P (a, b) on
the unit circle, – 1 ≤ a ≤ 1 and Fig 3.7

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52 MATHEMATICS

– 1 ≤ b ≤ 1, we have – 1 ≤ cos x ≤ 1 and –1 ≤ sin x ≤ 1 for all x. We have learnt in

π
previous classes that in the first quadrant (0 < x < ) a and b are both positive, in the
2
π
second quadrant ( < x <π) a is negative and b is positive, in the third quadrant
2
3π 3π
(π < x < ) a and b are both negative and in the fourth quadrant ( < x < 2π) a is
2 2
positive and b is negative. Therefore, sin x is positive for 0 < x < π, and negative for
π π 3π
π < x < 2π. Similarly, cos x is positive for 0 < x < , negative for < x < and also
2 2 2
3π
positive for < x < 2π. Likewise, we can find the signs of other trigonometric
2

functions in different quadrants. In fact, we have the following table.

I II III IV

sin x + + – –

cos x + – – +

tan x + – + –

cosec x + + – –

sec x + – – +

cot x + – + –

3.3.2 Domain and range of trigonometric functions From the definition of sine
and cosine functions, we observe that they are defined for all real numbers. Further,
we observe that for each real number x,
– 1 ≤ sin x ≤ 1 and – 1 ≤ cos x ≤ 1
Thus, domain of y = sin x and y = cos x is the set of all real numbers and range
is the interval [–1, 1], i.e., – 1 ≤ y ≤ 1.

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 TRIGONOMETRIC FUNCTIONS 53

1
Since cosec x =
sin x , the domain of y = cosec x is the set { x : x ∈ R and
x ≠ n π, n ∈ Z} and range is the set {y : y ∈ R, y ≥ 1 or y ≤ – 1}. Similarly, the domain
π
of y = sec x is the set {x : x ∈ R and x ≠ (2n + 1)
, n ∈ Z} and range is the set
2
{y : y ∈ R, y ≤ – 1or y ≥ 1}. The domain of y = tan x is the set {x : x ∈ R and
π
x ≠ (2n + 1) , n ∈ Z} and range is the set of all real numbers. The domain of
2
y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real
numbers.
π
We further observe that in the first quadrant, as x increases from 0 to , sin x
2
π
increases from 0 to 1, as x increases from to π, sin x decreases from 1 to 0. In the
2
3π
third quadrant, as x increases from π to , sin x decreases from 0 to –1and finally, in
2
3π
the fourth quadrant, sin x increases from –1 to 0 as x increases from to 2π.
2
Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we
have the following table:

I quadrant II quadrant III quadrant IV quadrant

sin increases from 0 to 1 decreases from 1 to 0 decreases from 0 to –1 increases from –1 to 0

cos decreases from 1 to 0 decreases from 0 to – 1 increases from –1 to 0 increases from 0 to 1

tan increases from 0 to ∞ increases from –∞to 0 increases from 0 to ∞ increases from –∞to 0

cot decreases from ∞ to 0 decreases from 0 to–∞ decreases from ∞ to 0 decreases from 0to –∞

sec increases from 1 to ∞ increases from –∞to–1 decreases from –1to–∞ decreases from ∞ to 1

cosec decreases from ∞ to 1 increases from 1 to ∞ increases from –∞to–1 decreases from–1to–∞

Remark In the above table, the statement tan x increases from 0 to ∞ (infinity) for
π π
0<x< simply means that tan x increases as x increases for 0 < x < and
2 2

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54 MATHEMATICS

π
assumes arbitraily large positive values as x approaches to . Similarly, to say that
2
cosec x decreases from –1 to – ∞ (minus infinity) in the fourth quadrant means that
3π
cosec x decreases for x ∈ ( , 2π) and assumes arbitrarily large negative values as
2
x approaches to 2π. The symbols ∞ and – ∞ simply specify certain types of behaviour
of functions and variables.
We have already seen that values of sin x and cos x repeats after an interval of
2π. Hence, values of cosec x and sec x will also repeat after an interval of 2π. We

Fig 3.8

Fig 3.9

Fig 3.10 Fig 3.11

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 TRIGONOMETRIC FUNCTIONS 55

Fig 3.12 Fig 3.13

shall see in the next section that tan (π + x) = tan x. Hence, values of tan x will repeat
after an interval of π. Since cot x is reciprocal of tan x, its values will also repeat after
an interval of π. Using this knowledge and behaviour of trigonometic functions, we can
sketch the graph of these functions. The graph of these functions are given above:

Example 6 If cos x = – 3 , x lies in the third quadrant, find the values of other five
5
trigonometric functions.
3 5
Solution Since cos x = − , we have sec x = −
5 3
2 2 2 2
Now sin x + cos x = 1, i.e., sin x = 1 – cos x
9 16
or sin2 x = 1 – =
25 25
4
Hence sin x = ±
5
Since x lies in third quadrant, sin x is negative. Therefore
4
sin x = –
5
which also gives
5
cosec x = –
4

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Further, we have
sin x 4 cos x 3
tan x = = and cot x = = .
cos x 3 sin x 4
5
Example 7 If cot x = – , x lies in second quadrant, find the values of other five
12
trigonometric functions.

5 12
Solution Since cot x = – , we have tan x = –
12 5
144 169
Now sec2 x = 1 + tan2 x = 1 + =
25 25
13
Hence sec x = ±
5
Since x lies in second quadrant, sec x will be negative. Therefore
13
sec x = – ,
5
which also gives
5
cos x = −
13
Further, we have
12 5 12
sin x = tan x cos x = (– ) × (– )=
5 13 13
1 13
and cosec x = = .
sin x 12

31π
Example 8 Find the value of sin .
3
Solution We know that values of sin x repeats after an interval of 2π. Therefore

31π π π 3
sin = sin (10π + ) = sin = .
3 3 3 2

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 TRIGONOMETRIC FUNCTIONS 57

Example 9 Find the value of cos (–1710°).

Solution We know that values of cos x repeats after an interval of 2π or 360°.
Therefore, cos (–1710°) = cos (–1710° + 5 × 360°)
= cos (–1710° + 1800°) = cos 90° = 0.

EXERCISE 3.2
Find the values of other five trigonometric functions in Exercises 1 to 5.

1
1. cos x = – , x lies in third quadrant.
2
3
2. sin x = , x lies in second quadrant.
5
3
3. cot x = , x lies in third quadrant.
4
13
4. sec x = , x lies in fourth quadrant.
5
5
5. tan x = – , x lies in second quadrant.
12
Find the values of the trigonometric functions in Exercises 6 to 10.
6. sin 765° 7. cosec (– 1410°)
19π 11π
8. tan 9. sin (– )
3 3
15π
10. cot (– )
4
3.4 Trigonometric Functions of Sum and Difference of Two Angles
In this Section, we shall derive expressions for trigonometric functions of the sum and
difference of two numbers (angles) and related expressions. The basic results in this
connection are called trigonometric identities. We have seen that

1. sin (– x) = – sin x
2. cos (– x) = cos x
We shall now prove some more results:

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58 MATHEMATICS

3. cos (x + y) = cos x cos y – sin x sin y

Consider the unit circle with centre at the origin. Let x be the angle P4OP1and y be
the angle P1OP2. Then (x + y) is the angle P4OP2. Also let (– y) be the angle P4OP3.
Therefore, P 1 , P 2 , P 3 and P 4 will have the coordinates P 1 (cos x, sin x),
P2 [cos (x + y), sin (x + y)], P3 [cos (– y), sin (– y)] and P4 (1, 0) (Fig 3.14).

Fig 3.14
Consider the triangles P1OP3 and P2OP4. They are congruent (Why?). Therefore,
P1P3 and P2P4 are equal. By using distance formula, we get
P 1P 32 = [cos x – cos (– y)]2 + [sin x – sin(–y]2
= (cos x – cos y)2 + (sin x + sin y)2
= cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y + 2sin x sin y
= 2 – 2 (cos x cos y – sin x sin y) (Why?)
Also, P 2P 42 = [1 – cos (x + y)] 2 + [0 – sin (x + y)]2
= 1 – 2cos (x + y) + cos2 (x + y) + sin2 (x + y)
= 2 – 2 cos (x + y)

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Since P 1 P 3 = P2P4, we have P1P32 = P2P42.

Therefore, 2 –2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y).
Hence cos (x + y) = cos x cos y – sin x sin y
4 . cos (x – y) = cos x cos y + sin x sin y
Replacing y by – y in identity 3, we get
cos (x + (– y)) = cos x cos (– y) – sin x sin (– y)
or cos (x – y) = cos x cos y + sin x sin y
π
5. cos ( – x ) = sin x
2
π
If we replace x by and y by x in Identity (4), we get
2
π π π
cos ( − x ) = cos cos x + sin sin x = sin x.
2 2 2
π
6. sin ( – x ) = cos x
2
Using the Identity 5, we have
π π  π 
sin (− x ) = cos  −  − x   = cos x.
2 2  2 
7. sin (x + y) = sin x cos y + cos x sin y
We know that
π   π 
sin (x + y) = cos  − (x + y )  = cos  ( − x) − y 
2   2 
π π
= cos ( − x ) cos y + sin ( − x) sin y
2 2
= sin x cos y + cos x sin y
8. sin (x – y) = sin x cos y – cos x sin y
If we replace y by –y, in the Identity 7, we get the result.
9. By taking suitable values of x and y in the identities 3, 4, 7 and 8, we get the
following results:
π π
cos ( + x ) = – sin x sin ( + x ) = cos x
2 2
π – x) = – cos x
cos (π π – x) = sin x
sin (π

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60 MATHEMATICS

π + x) = – cos x
cos (π π + x) = – sin x
sin (π
cos (2ππ – x) = cos x sin (2ππ – x) = – sin x
Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin
x and cos x.
π
10. If none of the angles x, y and (x + y) is an odd multiple of , then
2
tan x + tan y
tan (x + y) =
1 – tan x tan y

π
Since none of the x, y and (x + y) is an odd multiple of , it follows that cos x,
2
cos y and cos (x + y) are non-zero. Now
sin( x + y ) sin x cos y + cos x sin y
tan (x + y) = = .
cos( x + y ) cos x cos y − sin x sin y
Dividing numerator and denominator by cos x cos y, we have

sin x cos y cos x sin y

+
cos x cos y cos x cos y
tan (x + y) =
cos x cos y sin x sin y
−
cos x cos y cos x cos y

tan x + tan y
= 1 – tan x tan y

tan x – tan y
11. tan ( x – y) =
1 + tan x tan y
If we replace y by – y in Identity 10, we get
tan (x – y) = tan [x + (– y)]
tan x + tan (− y ) tan x − tan y
= =
1 − tan x tan ( − y ) 1+ tan x tan y
12. If none of the angles x, y and (x + y) is a multiple of π, then
cot x cot y – 1
cot ( x + y) =
cot y + cot x

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 TRIGONOMETRIC FUNCTIONS 61

Since, none of the x, y and (x + y) is multiple of π, we find that sin x sin y and
sin (x + y) are non-zero. Now,
cos ( x + y ) cos x cos y – sin x sin y
cot ( x + y) = =
sin ( x + y ) sin x cos y + cos x sin y
Dividing numerator and denominator by sin x sin y, we have
cot x cot y – 1
cot (x + y) =
cot y + cot x

cot x cot y + 1
13. cot (x – y) = if none of angles x, y and x–y is a multiple of π
cot y – cot x
If we replace y by –y in identity 12, we get the result

2 2 2
1 – tan 2 x
2
14. cos 2x = cos x – sin x = 2 cos x – 1 = 1 – 2 sin x =
1 + tan 2 x
We know that
cos (x + y) = cos x cos y – sin x sin y
Replacing y by x, we get
cos 2x = cos2x – sin2 x
= cos2 x – (1 – cos2 x) = 2 cos2x – 1
Again, cos 2x = cos2 x – sin2 x
= 1 – sin2 x – sin2 x = 1 – 2 sin2 x.

2
cos2 x − sin 2 x
2
We have cos 2x = cos x – sin x =
cos2 x + sin 2 x
Dividing numerator and denominator by cos2 x, we get
1 – tan 2 x π
cos 2x = 2 ,
x ≠ n π + , where n is an integer
1 + tan x 2

2tan x π
15. sin 2x = 2 sinx cos x = 2 x ≠ n π + , where n is an integer
1 + tan x 2
We have
sin (x + y) = sin x cos y + cos x sin y
Replacing y by x, we get sin 2x = 2 sin x cos x.
2sin x cos x
Again sin 2x =
cos2 x + sin 2 x

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Dividing each term by cos2 x, we get

2tan x
sin 2x =
1 +tan 2 x

2tan x π
16. tan 2x = 2 if 2 x ≠ n π + , where n is an integer
1 – tan x 2
We know that
tan x + tan y
tan (x + y) = 1 – tan x tan y

2 tan x
Replacing y by x , we get tan 2 x =
1− tan 2 x
17. sin 3x = 3 sin x – 4 sin3 x
We have,
sin 3x = sin (2x + x)
= sin 2x cos x + cos 2x sin x
= 2 sin x cos x cos x + (1 – 2sin2 x) sin x
= 2 sin x (1 – sin2 x) + sin x – 2 sin3 x
= 2 sin x – 2 sin3 x + sin x – 2 sin3 x
= 3 sin x – 4 sin3 x
18. cos 3x = 4 cos3 x – 3 cos x
We have,
cos 3x = cos (2x +x)
= cos 2x cos x – sin 2x sin x
= (2cos2 x – 1) cos x – 2sin x cos x sin x
= (2cos2 x – 1) cos x – 2cos x (1 – cos2 x)
= 2cos3 x – cos x – 2cos x + 2 cos3 x
= 4cos3 x – 3cos x.
3 tan x – tan 3 x π
19. tan 3 x = if 3 x ≠ n π + , where n is an integer
1 – 3tan 2 x 2
We have tan 3x =tan (2x + x)
2tan x
+ tan x
tan 2 x + tan x 1 – tan 2 x
= =
1 – tan 2 x tan x 1 – 2tan x . tan x
1 – tan 2 x

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 TRIGONOMETRIC FUNCTIONS 63

2tan x + tan x – tan 3 x 3 tan x – tan 3 x

= =
1 – tan 2 x – 2tan 2 x 1 – 3tan 2 x

x+ y x– y
20. (i) cos x + cos y = 2cos cos
2 2
x+ y x– y
(ii) cos x – cos y = – 2sin sin
2 2
x+ y x– y
(iii) sin x + sin y = 2sin cos
2 2
x+ y x– y
(iv) sin x – sin y = 2cos sin
2 2
We know that
cos (x + y) = cos x cos y – sin x sin y ... (1)
and cos (x – y) = cos x cos y + sin x sin y ... (2)
Adding and subtracting (1) and (2), we get
cos (x + y) + cos(x – y) = 2 cos x cos y ... (3)
and cos (x + y) – cos (x – y) = – 2 sin x sin y ... (4)
Further sin (x + y) = sin x cos y + cos x sin y ... (5)
and sin (x – y) = sin x cos y – cos x sin y ... (6)
Adding and subtracting (5) and (6), we get
sin (x + y) + sin (x – y) = 2 sin x cos y ... (7)
sin (x + y) – sin (x – y) = 2cos x sin y ... (8)
Let x + y = θ and x – y = φ. Therefore
 θ+ φ   θ−φ 
x =  and y =  
 2   2 
Substituting the values of x and y in (3), (4), (7) and (8), we get
 θ+φ   θ −φ 
cos θ + cos φ = 2 cos   cos  
 2   2 

θ+φ θ – φ
cos θ – cos φ = – 2 sin   sin  
 2   2 

 θ+φ   θ−φ 
sin θ + sin φ = 2 sin   cos  
 2   2 

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64 MATHEMATICS

 θ+φ   θ−φ 
sin θ – sin φ = 2 cos   sin  
 2   2 
Since θ and φ can take any real values, we can replace θ by x and φ by y.
Thus, we get
x+ y x− y x+ y x− y
cos x + cos y = 2 cos cos ; cos x – cos y = – 2 sin sin ,
2 2 2 2
x+ y x− y x+ y x− y
sin x + sin y = 2 sin cos ; sin x – sin y = 2 cos sin .
2 2 2 2
Remark As a part of identities given in 20, we can prove the following results:
21. (i) 2 cos x cos y = cos (x + y) + cos (x – y)
(ii) –2 sin x sin y = cos (x + y) – cos (x – y)
(iii) 2 sin x cos y = sin (x + y) + sin (x – y)
(iv) 2 cos x sin y = sin (x + y) – sin (x – y).
Example 10 Prove that
π π 5π π
3sin sec − 4sin cot =1
6 3 6 4
Solution We have
π π 5π π
L.H.S. = 3sin sec − 4sin cot
6 3 6 4
1  π π
=3× × 2 – 4 sin  π −  × 1 = 3 – 4 sin
2  6 6
1
=3–4× = 1 = R.H.S.
2
Example 11 Find the value of sin 15°.
Solution We have
sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
1 3 1 1 3 –1
= × − × = .
2 2 2 2 2 2
13 π
Example 12 Find the value of tan .
12

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 TRIGONOMETRIC FUNCTIONS 65

Solution We have
13 π  π π π π
tan = tan  π +  = tan = tan  − 
12  12  12 4 6

π π 1−
1
tan − tan
4 6 3 = 3 −1 = 2 − 3
= =
π π 1 3 +1
1 + tan tan 1+
4 6 3

Example 13 Prove that

sin ( x + y ) tan x + tan y
=
sin ( x − y ) tan x − tan y .

Solution We have
sin (x + y ) sin x cos y + cos x sin y
L.H.S. = =
sin (x − y ) sin x cos y − cos x sin y
Dividing the numerator and denominator by cos x cos y, we get
sin ( x + y ) tan x + tan y
=
sin ( x − y ) tan x − tan y .

Example 14 Show that

tan 3 x tan 2 x tan x = tan 3x – tan 2 x – tan x
Solution We know that 3x = 2x + x
Therefore, tan 3x = tan (2x + x)
tan 2 x + tan x
or tan 3x =
1– tan 2 x tan x
or tan 3x – tan 3x tan 2x tan x = tan 2x + tan x
or tan 3x – tan 2x – tan x = tan 3x tan 2x tan x
or tan 3x tan 2x tan x = tan 3x – tan 2x – tan x.
Example 15 Prove that
π  π 
cos  + x  + cos  − x  = 2 cos x
4  4 
Solution Using the Identity 20(i), we have

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66 MATHEMATICS

π  π 
L.H.S. = cos  + x  + cos  − x 
4  4 

π π  π π 
 4 +x+ 4 −x  4 + x – ( 4 − x) 
= 2cos   cos  
 2   2 
   

π 1
= 2 cos cos x = 2 × cos x = 2 cos x = R.H.S.
4 2

cos 7 x + cos 5 x
Example 16 Prove that = cot x
sin 7 x – sin 5 x

Solution Using the Identities 20 (i) and 20 (iv), we get

7 x + 5x 7 x − 5x
2cos cos cos x
2 2 = cot x = R.H.S.
L.H.S. = =
7 x + 5x 7 x − 5x sin x
2cos sin
2 2

sin 5 x − 2sin 3x + sin x

Example 17 Prove that = = tan x
cos5 x − cos x

Solution We have

sin 5 x − 2sin 3 x + sin x sin 5 x + sin x − 2sin 3x

L.H.S. = =
cos5 x − cos x cos5 x − cos x

2sin 3 x cos 2 x − 2sin 3 x sin 3 x (cos 2 x − 1)

= =–
– 2sin 3x sin 2x sin 3x sin 2x

1− cos 2 x 2sin 2 x
= = = tan x = R.H.S.
sin 2 x 2sin x cos x

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 TRIGONOMETRIC FUNCTIONS 67

EXERCISE 3.3
Prove that:
π π π 1 π 7π π 3
1. sin2 + cos2 – tan2 = – 2. 2sin2 + cosec2 cos 2 =
6 3 4 2 6 6 3 2
π 5π π 2 3π π π
3. cot
2
+ cosec + 3tan 2 = 6 4. 2sin + 2cos 2 + 2sec 2 = 10
6 6 6 4 4 3
5. Find the value of:
(i) sin 75° (ii) tan 15°

Prove the following:

π  π  π  π 
6. cos  − x  cos  − y  − sin  − x  sin  − y  = sin ( x + y )
 4   4   4   4 

π 
tan  + x  2
 4  =  1 + tan x cos (π + x) cos ( − x )
7.   8. = cot 2 x
π   1 − tan x π 
tan  − x  sin (π − x) cos  + x 
4  2 

 3π    3π  
9. cos  + x  cos (2 π + x)  cot  − x  + cot (2π + x)  = 1
 2    2  
10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
 3π   3π 
11. cos  + x  − cos  − x  = − 2 sin x
 4   4 
12. sin2 6x – sin2 4x = sin 2x sin 10x 13. cos2 2x – cos2 6x = sin 4x sin 8x
14. sin2 x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
cos 9 x − cos 5 x sin 2 x sin 5x + sin 3 x
16. =− 17. = tan 4 x
sin 17 x − sin 3 x cos 10 x cos 5x + cos 3 x
sin x − sin y x−y sin x + sin 3 x
18. = tan 19. = tan 2 x
cos x + cos y 2 cos x + cos 3 x
sin x − sin 3x cos 4 x + cos 3x + cos 2 x
20. = 2 sin x 21. = cot 3x
2
sin x − cos x
2
sin 4 x + sin 3x + sin 2 x

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22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

4tan x (1 − tan 2 x)
23. tan 4 x = 24. cos 4x = 1 – 8sin2 x cos2 x
1 − 6 tan 2 x + tan 4 x

25. cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1

Miscellaneous Examples
3 12
Example 18 If sin x = , cos y = − , where x and y both lie in second quadrant,
5 13
find the value of sin (x + y).
Solution We know that
sin (x + y) = sin x cos y + cos x sin y ... (1)
9 16
Now cos2 x = 1 – sin2 x = 1 – =
25 25
4
Therefore cos x = ± .
5
Since x lies in second quadrant, cos x is negative.
4
Hence cos x = −
5
144 25
Now sin2y = 1 – cos2y = 1 – =
169 169
5
i.e. sin y = ± .
13
5
Since y lies in second quadrant, hence sin y is positive. Therefore, sin y = . Substituting
13
the values of sin x, sin y, cos x and cos y in (1), we get
3  12   4  5 36 20 56
sin( x + y ) = ×  −  +  − × = − − =− .
5  13   5  13 65 65 65
Example 19 Prove that
x 9x 5x
cos 2 x cos − cos 3 x cos = sin 5 x sin .
2 2 2
Solution We have

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 TRIGONOMETRIC FUNCTIONS 69

1  x 9x 
L.H.S. =  2cos 2 x cos − 2cos cos 3x 
2  2 2 

1  x  x  9x   9x 
=  cos  2 x +  + cos  2 x −  − cos  + 3x  − cos  − 3x  
2  2  2  2   2 
1 5x 3x 15x 3x  1  5x 15x 
 cos + cos − cos − cos  =  cos − cos
2 
=
2 2 2 2 2  2 2

  5 x 15 x   5 x 15 x 
1  2 + 2   2 − 2 
 −2sin   sin  
= 2  2   2 
    

 5x  5x
= − sin 5x sin  −  = sin 5x sin = R.H.S.
 2 2
π
Example 20 Find the value of tan .
8
π π
Solution Let x = . Then 2 x = .
8 4
2 tan x
Now tan 2 x = 2
1 − tan x
π
2tan
π 8
tan =
or 4 1 − tan 2 π
8

π 2y
Let y = tan . Then 1 =
8 1− y2
or y2 + 2y – 1 = 0

−2 ± 2 2
Therefore y= = − 1± 2
2

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π π
Since lies in the first quadrant, y = tan is positve. Hence
8 8
π
tan = 2 −1 .
8

3 3π x x x
Example 21 If tan x = , π < x < , find the value of sin , cos and tan .
4 2 2 2 2
3π
Solution Since π < x < , cos x is negative.
2
π x 3π
Also < < .
2 2 4
x x
Therefore, sin is positive and cos is negative.
2 2
9 25
Now sec2 x = 1 + tan2 x = 1 + =
16 16
16 4
Therefore cos2 x = or cos x = – (Why?)
25 5
x 4 9
Now 2 sin 2 = 1 – cos x = 1 + = .
2 5 5
x 9
Therefore sin2 =
2 10
x 3
or sin = (Why?)
2 10
x 4 1
Again 2cos2 = 1+ cos x = 1 − =
2 5 5
x 1
Therefore cos2 =
2 10
x 1
or cos =− (Why?)
2 10

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 TRIGONOMETRIC FUNCTIONS 71
x
sin  − 10 
x 2 3
Hence tan = = ×  = – 3.
2 x 10  1 
cos
2
Example 22
 π  π 3
Prove that cos2 x + cos2  x +  + cos 2  x −  = .
 3  3 2

Solution We have

 2π   2π 
1 + cos  2 x +  1 + cos  2 x − 
L.H.S. = 1 + cos 2 x +  3 
+  3  .
2 2 2

1  2π   2π  
 3 + cos 2 x + cos  2 x +  + cos  2 x − 
3  
=
2  3  

1 2π 
=  3 + cos 2 x + 2cos 2 x cos 
2 3

1  π 
=  3 + cos 2 x + 2cos 2 x cos  π −  
2  3 

1 π
=  3 + cos 2 x − 2cos 2 x cos 
2 3
1 3
= [3 + cos 2x − cos 2 x ] = = R.H.S.
2 2

Miscellaneous Exercise on Chapter 3

Prove that:
π 9π 3π 5π
1. 2 cos cos + cos + cos =0
13 13 13 13
2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
x+ y
3. (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2
2

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72 MATHEMATICS

x −y
4. (cos x – cos y)2 + (sin x – sin y)2 = 4 sin2
2
5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
(sin 7x + sin 5x ) + (sin 9x + sin 3x )
6. = tan 6x
(cos 7x + cos 5x ) + (cos 9x + cos 3x )
x 3x
7. sin 3x + sin 2x – sin x = 4sin x cos cos
2 2
x x x
Find sin , cos and tan in each of the following :
2 2 2
4 1
8. tan x = − , x in quadrant II 9. cos x = − , x in quadrant III
3 3
1
10. sin x = , x in quadrant II
4

Summary
® If in a circle of radius r, an arc of length l subtends an angle of θ radians, then
l=rθ
π
® Radian measure = × Degree measure
180
180
® Degree measure = π × Radian measure
® cos2 x + sin2 x = 1
® 1 + tan2 x = sec2 x
® 1 + cot2 x = cosec2 x
® cos (2nπ + x) = cos x
® sin (2nπ + x) = sin x
® sin (– x) = – sin x
® cos (– x) = cos x
® cos (x + y) = cos x cos y – sin x sin y
® cos (x – y) = cos x cos y + sin x sin y
π
® cos ( 2 − x ) = sin x

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 TRIGONOMETRIC FUNCTIONS 73

π
® sin ( 2 − x ) = cos x
® sin (x + y) = sin x cos y + cos x sin y
® sin (x – y) = sin x cos y – cos x sin y
π  π 
® cos  2 + x  = – sin x sin  + x  = cos x
2 
cos (π – x) = – cos x sin (π – x) = sin x
cos (π + x) = – cos x sin (π + x) = – sin x
cos (2π – x) = cos x sin (2π – x) = – sin x

π
® If none of the angles x, y and (x ± y) is an odd multiple of 2
, then

tan x + tan y
tan (x + y) =
1 − tan x tan y
tan x − tan y
® tan (x – y) = 1 + tan x tan y
® If none of the angles x, y and (x ± y) is a multiple of π, then
cot x cot y − 1
cot (x + y) = cot y + cot x

cot x cot y + 1
® cot (x – y) = cot y − cot x

1 – tan 2 x
® cos 2x = cos2 x – sin2 x = 2cos2 x – 1 = 1 – 2 sin2 x = 1 + tan 2 x

2 tan x
® sin 2x = 2 sin x cos x =
1 + tan 2 x

2tanx
® tan 2x = 1 − tan 2 x
® sin 3x = 3sin x – 4sin3 x
® cos 3x = 4cos3 x – 3cos x

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3tan x − tan 3 x
® tan 3x = 1− 3tan 2 x

x+ y x− y
® (i) cos x + cos y = 2cos cos
2 2
x+ y x− y
(ii) cos x – cos y = – 2sin sin
2 2
x+ y x− y
(iii) sin x + sin y = 2 sin cos
2 2
x+ y x− y
(iv) sin x – sin y = 2cos sin
2 2
® (i) 2cos x cos y = cos ( x + y) + cos ( x – y)
(ii) – 2sin x sin y = cos (x + y) – cos (x – y)
(iii) 2sin x cos y = sin (x + y) + sin (x – y)
(iv) 2 cos x sin y = sin (x + y) – sin (x – y).

Historical Note
The study of trigonometry was first started in India. The ancient Indian
Mathematicians, Aryabhatta (476), Brahmagupta (598), Bhaskara I (600) and
Bhaskara II (1114) got important results. All this knowledge first went from
India to middle-east and from there to Europe. The Greeks had also started the
study of trigonometry but their approach was so clumsy that when the Indian
approach became known, it was immediately adopted throughout the world.
In India, the predecessor of the modern trigonometric functions, known as
the sine of an angle, and the introduction of the sine function represents the main
contribution of the siddhantas (Sanskrit astronomical works) to the history of
mathematics.
Bhaskara I (about 600) gave formulae to find the values of sine functions
for angles more than 90°. A sixteenth century Malayalam work Yuktibhasa
(period) contains a proof for the expansion of sin (A + B). Exact expression for
sines or cosines of 18°, 36°, 54°, 72°, etc., are given by
Bhaskara II.

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The symbols sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, etc., were
suggested by the astronomer Sir John F.W. Hersehel (1813) The names of Thales
(about 600 B.C.) is invariably associated with height and distance problems. He
is credited with the determination of the height of a great pyramid in Egypt by
measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known
height, and comparing the ratios:
H h
= = tan (sun’s altitude)
S s
Thales is also said to have calculated the distance of a ship at sea through
the proportionality of sides of similar triangles. Problems on height and distance
using the similarity property are also found in ancient Indian works.

—v —

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18 MATHEMATICS

Chapter 2
INVERSE TRIGONOMETRIC
FUNCTIONS
v Mathematics, in general, is fundamentally the science of
self-evident things. — FELIX KLEIN v
2.1 Introduction
In Chapter 1, we have studied that the inverse of a function
f, denoted by f –1, exists if f is one-one and onto. There are
many functions which are not one-one, onto or both and
hence we can not talk of their inverses. In Class XI, we
studied that trigonometric functions are not one-one and
onto over their natural domains and ranges and hence their
inverses do not exist. In this chapter, we shall study about
the restrictions on domains and ranges of trigonometric
functions which ensure the existence of their inverses and
observe their behaviour through graphical representations.
Besides, some elementary properties will also be discussed.
The inverse trigonometric functions play an important Aryabhata
(476-550 A. D.)
role in calculus for they serve to define many integrals.
The concepts of inverse trigonometric functions is also used in science and engineering.
2.2 Basic Concepts
In Class XI, we have studied trigonometric functions, which are defined as follows:
sine function, i.e., sine : R → [– 1, 1]
cosine function, i.e., cos : R → [– 1, 1]
π
tangent function, i.e., tan : R – { x : x = (2n + 1) , n ∈ Z} → R
2
cotangent function, i.e., cot : R – { x : x = nπ, n ∈ Z} → R
π
secant function, i.e., sec : R – { x : x = (2n + 1) , n ∈ Z} → R – (– 1, 1)
2
cosecant function, i.e., cosec : R – { x : x = nπ, n ∈ Z} → R – (– 1, 1)

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 INVERSE TRIGONOMETRIC FUNCTIONS 19

We have also learnt in Chapter 1 that if f : X→Y such that f (x) = y is one-one and
onto, then we can define a unique function g : Y→X such that g (y) = x, where x ∈ X
and y = f (x), y ∈ Y. Here, the domain of g = range of f and the range of g = domain
of f. The function g is called the inverse of f and is denoted by f –1. Further, g is also
one-one and onto and inverse of g is f. Thus, g –1 = (f –1)–1 = f. We also have
(f –1 o f ) (x) = f –1 (f (x)) = f –1(y) = x
and (f o f –1) (y) = f (f –1(y)) = f (x) = y
Since the domain of sine function is the set of all real numbers and range is the
−π π
closed interval [–1, 1]. If we restrict its domain to  ,  , then it becomes one-one
 2 2
and onto with range [– 1, 1]. Actually, sine function restricted to any of the intervals
 −3π π  ,  −π π  ,  π 3π  etc., is one-one and its range is [–1, 1]. We can,
 2 , 2   2 , 2   2 , 2 
 
therefore, define the inverse of sine function in each of these intervals. We denote the
inverse of sine function by sin–1 (arc sine function). Thus, sin–1 is a function whose
 −3π −π   −π π 
domain is [– 1, 1] and range could be any of the intervals  , , , or
 2 2   2 2 
 π 3π 
 2 , 2  , and so on. Corresponding to each such interval, we get a branch of the
 
 −π π 
function sin–1. The branch with range  ,  is called the principal value branch,
 2 2
whereas other intervals as range give different branches of sin–1. When we refer
to the function sin–1, we take it as the function whose domain is [–1, 1] and range is
 −π π   −π π 
 2 , 2  . We write sin : [–1, 1] →  2 , 2 
–1
   
From the definition of the inverse functions, it follows that sin (sin–1 x) = x
π π
if – 1 ≤ x ≤ 1 and sin–1 (sin x) = x if − ≤ x ≤ . In other words, if y = sin–1 x, then
2 2
sin y = x.
Remarks
(i) We know from Chapter 1, that if y = f (x) is an invertible function, then x = f –1 (y).
Thus, the graph of sin–1 function can be obtained from the graph of original
function by interchanging x and y axes, i.e., if (a, b) is a point on the graph of
sine function, then (b, a) becomes the corresponding point on the graph of inverse

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20 MATHEMATICS

of sine function. Thus, the graph of the function y = sin–1 x can be obtained from
the graph of y = sin x by interchanging x and y axes. The graphs of y = sin x and
y = sin–1 x are as given in Fig 2.1 (i), (ii), (iii). The dark portion of the graph of
y = sin–1 x represent the principal value branch.
(ii) It can be shown that the graph of an inverse function can be obtained from the
corresponding graph of original function as a mirror image (i.e., reflection) along
the line y = x. This can be visualised by looking the graphs of y = sin x and
y = sin–1 x as given in the same axes (Fig 2.1 (iii)).

Fig 2.1 (i)

Fig 2.1 (ii) Fig 2.1 (iii)

Like sine function, the cosine function is a function whose domain is the set of all
real numbers and range is the set [–1, 1]. If we restrict the domain of cosine function
to [0, π], then it becomes one-one and onto with range [–1, 1]. Actually, cosine function

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 INVERSE TRIGONOMETRIC FUNCTIONS 21

restricted to any of the intervals [– π, 0], [0,π], [π, 2π] etc., is bijective with range as
[–1, 1]. We can, therefore, define the inverse of cosine function in each of these
intervals. We denote the inverse of the cosine function by cos–1 (arc cosine function).
Thus, cos–1 is a function whose domain is [–1, 1] and range
could be any of the intervals [–π, 0], [0, π], [π, 2π] etc.
Corresponding to each such interval, we get a branch of the
function cos–1. The branch with range [0, π] is called the principal
value branch of the function cos–1. We write
cos–1 : [–1, 1] → [0, π].
The graph of the function given by y = cos–1 x can be drawn
in the same way as discussed about the graph of y = sin–1 x. The
graphs of y = cos x and y = cos–1 x are given in Fig 2.2 (i) and (ii).

Fig 2.2 (i) Fig 2.2 (ii)

Let us now discuss cosec–1x and sec–1x as follows:

1
Since, cosec x = , the domain of the cosec function is the set {x : x ∈ R and
sin x
x ≠ nπ, n ∈ Z} and the range is the set {y : y ∈ R, y ≥ 1 or y ≤ –1} i.e., the set
R – (–1, 1). It means that y = cosec x assumes all real values except –1 < y < 1 and is
not defined for integral multiple of π. If we restrict the domain of cosec function to
 π π
 − 2 , 2  – {0}, then it is one to one and onto with its range as the set R – (– 1, 1). Actually,
 
 −3π −π   −π π 
cosec function restricted to any of the intervals  ,  − {−π} ,  ,  – {0},
 2 2   2 2
 π 3π 
 2 , 2  − {π} etc., is bijective and its range is the set of all real numbers R – (–1, 1).

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Thus cosec–1 can be defined as a function whose domain is R – (–1, 1) and range could
 −3π − π   −π π 
be any of the intervals  ,  − {− π} ,  ,  − {0} ,  π , 3π  − {π} etc. The
 2 2   2 2 2 2 
 
 −π π 
function corresponding to the range  ,  − {0} is called the principal value branch
 2 2
of cosec–1. We thus have principal branch as
 −π π 
cosec–1 : R – (–1, 1) →  ,  − {0}
 2 2
–1
The graphs of y = cosec x and y = cosec x are given in Fig 2.3 (i), (ii).

Fig 2.3 (i) Fig 2.3 (ii)

1 π
Also, since sec x = , the domain of y = sec x is the set R – {x : x = (2n + 1) ,
cos x 2
n ∈ Z} and range is the set R – (–1, 1). It means that sec (secant function) assumes
π
all real values except –1 < y < 1 and is not defined for odd multiples of . If we
2
π
restrict the domain of secant function to [0, π] – { }, then it is one-one and onto with
2

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 INVERSE TRIGONOMETRIC FUNCTIONS 23

its range as the set R – (–1, 1). Actually, secant function restricted to any of the
−π π 3π
intervals [–π, 0] – { }, [0, π] –   , [π, 2π] – { } etc., is bijective and its range
2
–1
2 2
is R – {–1, 1}. Thus sec can be defined as a function whose domain is R– (–1, 1) and
−π π 3π
range could be any of the intervals [– π, 0] – { }, [0, π] – { }, [π, 2π] – { } etc.
2 2 2
Corresponding to each of these intervals, we get different branches of the function sec–1.
π
The branch with range [0, π] – { } is called the principal value branch of the
2
function sec–1. We thus have
π
sec–1 : R – (–1,1) → [0, π] – { }
2
The graphs of the functions y = sec x and y = sec-1 x are given in Fig 2.4 (i), (ii).

Fig 2.4 (i) Fig 2.4 (ii)

Finally, we now discuss tan–1 and cot–1

We know that the domain of the tan function (tangent function) is the set
π
{x : x ∈ R and x ≠ (2n +1) , n ∈ Z} and the range is R. It means that tan function
2
π
is not defined for odd multiples of . If we restrict the domain of tangent function to
2

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24 MATHEMATICS

 −π π 
 ,  , then it is one-one and onto with its range as R. Actually, tangent function
 2 2 
 −3π −π   −π π   π 3 π 
restricted to any of the intervals  ,  ,  , ,  ,  etc., is bijective
 2 2   2 2 2 2 
and its range is R. Thus tan–1 can be defined as a function whose domain is R and
 −3π −π   −π π   π 3π 
range could be any of the intervals  ,  ,  ,  ,  ,  and so on. These
 2 2   2 2 2 2 
 −π π 
intervals give different branches of the function tan–1. The branch with range  , 
 2 2
is called the principal value branch of the function tan–1.
We thus have
 −π π 
tan–1 : R →  , 
 2 2
The graphs of the function y = tan x and y = tan–1x are given in Fig 2.5 (i), (ii).

Fig 2.5 (i) Fig 2.5 (ii)

We know that domain of the cot function (cotangent function) is the set
{x : x ∈ R and x ≠ nπ, n ∈ Z} and range is R. It means that cotangent function is not
defined for integral multiples of π. If we restrict the domain of cotangent function to
(0, π), then it is bijective with and its range as R. In fact, cotangent function restricted
to any of the intervals (–π, 0), (0, π), (π, 2π) etc., is bijective and its range is R. Thus
cot –1 can be defined as a function whose domain is the R and range as any of the

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 INVERSE TRIGONOMETRIC FUNCTIONS 25

intervals (–π, 0), (0, π), (π, 2π) etc. These intervals give different branches of the
function cot –1. The function with range (0, π) is called the principal value branch of
the function cot –1. We thus have
cot–1 : R → (0, π)
The graphs of y = cot x and y = cot–1x are given in Fig 2.6 (i), (ii).

Fig 2.6 (i) Fig 2.6 (ii)

The following table gives the inverse trigonometric function (principal value
branches) along with their domains and ranges.
 π π
sin–1 : [–1, 1] → − 2 , 2 
 
cos –1 : [–1, 1] → [0, π]

 π π
cosec –1 : R – (–1,1) →  − 2 , 2  – {0}
 
π
sec –1 : R – (–1, 1) → [0, π] – { }
2
 −π π 
tan –1 : R →  , 
 2 2
cot –1 : R → (0, π)

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26 MATHEMATICS

A Note 1
1. sin–1x should not be confused with (sin x)–1. In fact (sin x)–1 = and
sin x
similarly for other trigonometric functions.
2. Whenever no branch of an inverse trigonometric functions is mentioned, we
mean the principal value branch of that function.
3. The value of an inverse trigonometric functions which lies in the range of
principal branch is called the principal value of that inverse trigonometric
functions.
We now consider some examples:
 1 
Example 1 Find the principal value of sin–1  .
 2
 1  1
Solution Let sin–1   = y. Then, sin y = .
 2 2
 −π π 
We know that the range of the principal value branch of sin–1 is  ,  and
2 2
π 1  1  π
sin   = . Therefore, principal value of sin–1   is
4 2  2 4
 − 1 
Example 2 Find the principal value of cot–1  
 3
 −1 
Solution Let cot–1   = y. Then,
 3
−1 π  π  2π 
cot y = = − cot   = cot  π −  = cot  
3 3  3  3 
We know that the range of principal value branch of cot–1 is (0, π) and
 2π  −1  −1  2π
cot   = . Hence, principal value of cot–1   is
 3  3  3 3

EXERCISE 2.1
Find the principal values of the following:

 1  3
1. sin–1  −  2. cos–1  2  3. cosec–1 (2)
 2  

 1
4. tan–1 (− 3) 5. cos–1  −  6. tan–1 (–1)
 2

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 INVERSE TRIGONOMETRIC FUNCTIONS 27

 2   1 
7. sec–1   8. cot–1 ( 3) 9. cos–1  − 
 3  2
10. cosec–1 ( − 2 )
Find the values of the following:
1 1 1 1
11. tan–1(1) + cos–1 − + sin–1 − 12. cos–1 + 2 sin–1
2 2 2 2
13. If sin–1 x = y, then
π π
(A) 0 ≤ y ≤ π (B) − ≤ y≤
2 2
π π
(C) 0 < y < π (D) − < y<
2 2
14. tan–1 3 − sec −1 ( − 2 ) is equal to

π π 2π
(A) π (B) − (C) (D)
3 3 3
2.3 Properties of Inverse Trigonometric Functions
In this section, we shall prove some important properties of inverse trigonometric
functions. It may be mentioned here that these results are valid within the principal
value branches of the corresponding inverse trigonometric functions and wherever
they are defined. Some results may not be valid for all values of the domains of inverse
trigonometric functions. In fact, they will be valid only for some values of x for which
inverse trigonometric functions are defined. We will not go into the details of these
values of x in the domain as this discussion goes beyond the scope of this textbook.
Let us recall that if y = sin–1x, then x = sin y and if x = sin y, then y = sin–1x. This
is equivalent to
−π π
sin (sin–1 x) = x, x ∈ [– 1, 1] and sin–1 (sin x) = x, x ∈  , 
 2 2
 
For suitable values of domain similar results follow for remaining trigonometric
functions.

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We now consider some examples.

Example 3 Show that

1 1
(i) sin–1 ( 2 x 1 − x 2 ) = 2 sin–1 x, − ≤x≤
2 2
1
(ii) sin–1 ( 2 x 1 − x 2 ) = 2 cos–1 x, ≤ x ≤1
2
Solution
(i) Let x = sin θ. Then sin–1 x = θ. We have

(
sin–1 ( 2 x 1 − x 2 ) = sin–1 2sin θ 1 − sin 2 θ )
= sin–1 (2sinθ cosθ) = sin–1 (sin2θ) = 2θ
= 2 sin–1 x
(ii) Take x = cos θ, then proceeding as above, we get, sin–1 ( 2 x 1 − x 2 ) = 2 cos–1 x

cos x − 3π π
Example 4 Express tan −1 , < x < in the simplest form.
1 − sin x 2 2
Solution We write

 x x 
 cos 2 − sin 2 
 cos x  2 2
tan −1   = tan 
–1

 1 − sin x  2 x 2 x
 cos + sin − 2sin cos
x x

 2 2 2 2 

 x x  x x 
  cos 2 + sin 2  cos 2 − sin 2  
–1   
= tan 
  x x2 
  cos − sin  
 2 2

 x x  x
cos + sin  1 + tan 
–1 2 2 –1  2
= tan  = tan 
x x x
 cos − sin   1 − tan 
 2 2  2

–1   π x  π x
= tan  tan  +   = +
  4 2  4 2

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 INVERSE TRIGONOMETRIC FUNCTIONS 29

–1  1 
Example 5 Write cot  2  , x > 1 in the simplest form.
 x −1 

Solution Let x = sec θ, then x2 − 1 = sec2 θ − 1 = tan θ

–1 1
Therefore, cot = cot–1 (cot θ) = θ = sec–1 x, which is the simplest form.
x −1
2

EXERCISE 2.2
Prove the following:
 1 1
1. 3sin–1 x = sin–1 (3x – 4x3), x ∈  – , 
 2 2
1 
2. 3cos–1 x = cos–1 (4x3 – 3x), x ∈  , 1
2 
Write the following functions in the simplest form:

1 + x2 − 1  1 − cos x 
3. tan −1 ,x≠0 4. tan −1   , 0 < x < π
x  1 + cos x 
 cos x − sin x  −π 3π
5. tan −1  , <x<
 cos x + sin x  4 4
−1 x
6. tan , |x| < a
a − x2
2

 3a 2 x − x 3  −a a
tan −1  3 <x<
7. 2  , a > 0;
 a − 3ax  3 3
Find the values of each of the following:
  1 
8. tan –1  2 cos  2sin –1  
  2 

1 2x –1 1 − y 
2

9. tan sin –1 + cos  , | x | < 1, y > 0 and xy < 1

2 1 + x2 1 + y2 

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30 MATHEMATICS

Find the values of each of the expressions in Exercises 16 to 18.

–1  2π   3π 
10. sin  sin  11. tan –1  tan 
 3   4 
 3 3
12. tan  sin –1 + cot –1 
 5 2
−1  7π 
13. cos  cos  is equal to
 6 
7π 5π π π
(A) (B) (C) (D)
6 6 3 6
π 1 
14. sin  − sin −1 ( − )  is equal to
3 2 
1 1 1
(A) (B) (C) (D) 1
2 3 4
15. tan −1 3 − cot −1 (− 3) is equal to
π
(A) π (B) − (C) 0 (D) 2 3
2

Miscellaneous Examples

−1 3π
Example 6 Find the value of sin (sin )
5

−1 3π 3π
Solution We know that sin −1 (sin x) = x . Therefore, sin (sin )=
5 5
3π  π π 
But ∉ − , , which is the principal branch of sin–1 x
5  2 2 
3π 3π 2π 2π  π π 
However sin ( ) = sin(π − ) = sin and ∈ − ,
5 5 5 5  2 2 

3π 2π 2 π
Therefore sin −1 (sin ) = sin −1 (sin ) =
5 5 5

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 INVERSE TRIGONOMETRIC FUNCTIONS 31

Miscellaneous Exercise on Chapter 2

Find the value of the following:

–1  13π   7π 
1. cos  cos  2. tan –1  tan 
 6   6 
Prove that
3 24 8 3 77
3. 2sin
–1
= tan –1 4. sin
–1
+ sin –1 = tan –1
5 7 17 5 36

–1 4 12 33 12 3 56
5. cos + cos –1 = cos –1 6. cos
–1
+ sin –1 = sin –1
5 13 65 13 5 65
63 5 3
7. tan –1 = sin –1 + cos –1
16 13 5
Prove that
1 1− x
8. tan –1 x = cos –1 , x ∈ [0, 1]
2 1+ x

 1 + sin x + 1 − sin x  x  π
9. cot –1   = , x ∈  0, 
 1 + sin x − 1 − sin x  2  4

 1+ x − 1− x  π 1 1
tan –1   = − cos x , − ≤ x ≤ 1 [Hint: Put x = cos 2θ]
–1
10.
 1+ x + 1− x  4 2 2
Solve the following equations:
1 − x 1 –1
11. 2tan–1 (cos x) = tan–1 (2 cosec x) 12. tan –1 = tan x,( x > 0)
1+ x 2
13. sin (tan–1 x), | x | < 1 is equal to
x 1 1 x
(A) (B) (C) (D)
1 − x2 1 − x2 1 + x2 1 + x2

π
14. sin–1 (1 – x) – 2 sin–1 x = , then x is equal to
2

1 1 1
(A) 0, (B) 1, (C) 0 (D)
2 2 2

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Summary
® The domains and ranges (principal value branches) of inverse trigonometric
functions are given in the following table:
Functions Domain Range
(Principal Value Branches)
 −π π 
y = sin–1 x [–1, 1]  2 , 2 
y = cos–1 x [–1, 1] [0, π]
 −π π 
y = cosec–1 x R – (–1,1)  2 , 2  – {0}
 
π
y = sec–1 x R – (–1, 1) [0, π] – { }
2
 π π
y = tan–1 x R − , 
 2 2
y = cot–1 x R (0, π)
1
® sin–1x should not be confused with (sin x)–1. In fact (sin x)–1 =
sin x
and
similarly for other trigonometric functions.
® The value of an inverse trigonometric functions which lies in its principal
value branch is called the principal value of that inverse trigonometric
functions.
For suitable values of domain, we have
® y = sin–1 x ⇒ x = sin y ® x = sin y ⇒ y = sin–1 x
® sin (sin–1 x) = x ® sin–1 (sin x) = x

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Historical Note
The study of trigonometry was first started in India. The ancient Indian
Mathematicians, Aryabhata (476A.D.), Brahmagupta (598 A.D.), Bhaskara I
(600 A.D.) and Bhaskara II (1114 A.D.) got important results of trigonometry. All
this knowledge went from India to Arabia and then from there to Europe. The
Greeks had also started the study of trigonometry but their approach was so
clumsy that when the Indian approach became known, it was immediately adopted
throughout the world.
In India, the predecessor of the modern trigonometric functions, known as
the sine of an angle, and the introduction of the sine function represents one of
the main contribution of the siddhantas (Sanskrit astronomical works) to
mathematics.
Bhaskara I (about 600 A.D.) gave formulae to find the values of sine functions
for angles more than 90°. A sixteenth century Malayalam work Yuktibhasa
contains a proof for the expansion of sin (A + B). Exact expression for sines or
cosines of 18°, 36°, 54°, 72°, etc., were given by Bhaskara II.
The symbols sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, etc., were suggested
by the astronomer Sir John F.W. Hersehel (1813) The name of Thales
(about 600 B.C.) is invariably associated with height and distance problems. He
is credited with the determination of the height of a great pyramid in Egypt by
measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known
height, and comparing the ratios:
H h
= = tan (sun’s altitude)
S s
Thales is also said to have calculated the distance of a ship at sea through
the proportionality of sides of similar triangles. Problems on height and distance
using the similarity property are also found in ancient Indian works.

—v —

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