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INTRODUCTION TO TRIGONOMETRY

113

INTRODUCTION TO
TRIGONOMETRY

8

There is perhaps nothing which so occupies the
middle position of mathematics as trigonometry
...
F
...
1 Introduction
You have already studied about triangles, and in particular, right triangles, in your
earlier classes
...
For instance :
1
...
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
...
Can the student
find out the height of the Minar, without
actually measuring it?
2
...
She is looking down at a flower
pot placed on a stair of a temple situated
nearby on the other bank of the river
...
8
...
If
you know the height at which the
person is sitting, can you find the width
of the river?

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Fig
...
1

Fig
...
2

114

MATHEMATICS

3
...
A girl happens to spot the
balloon in the sky and runs to her
mother to tell her about it
...
Now when the girl had spotted
the balloon intially it was at point A
...
Can you
find the altitude of B from the ground?

Fig
...
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’
...
In fact,
trigonometry is the study of relationships between the sides and angles of a triangle
...
Early
astronomers used it to find out the distances of the stars and planets from the Earth
...

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
...
However, these ratios can be extended to other
angles also
...
We will calculate trigonometric ratios for some specific angles and establish
some identities involving these ratios, called trigonometric identities
...
2 Trigonometric Ratios
In Section 8
...

Let us take a right triangle ABC as shown
in Fig
...
4
...
Note the position of the side BC
with respect to angle A
...
We call it
the side opposite to angle A
...
So, we call it the side
adjacent to angle A
...
8
...
8
...

You have studied the concept of ‘ratio’ in
your earlier classes
...

The trigonometric ratios of the angle A
in right triangle ABC (see Fig
...
4) are defined
as follows :
sine of  A =

side opposite to angle A BC

hypotenuse
AC

cosine of  A =
tangent of  A =

side adjacent to angle A AB

hypotenuse
AC
side opposite to angle A BC

side adjacent to angle A AB

cosecant of  A =
secant of  A =

Fig
...
5

1
hypotenuse
AC


sine of  A side opposite to angle A BC

1
hypotenuse
AC


cosine of  A side adjacent to angle A AB

cotangent of  A =

1
side adjacent to angle A AB


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
...


BC
BC AC sin A
cos A
...

Why don’t you try to define the trigonometric ratios for angle C in the right
triangle? (See Fig
...
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
...
500
...
When the Aryabhatiyam was
translated into Arabic, the word jiva was retained as
it is
...
Soon the word sinus, also used as sine,
became common in mathematical texts throughout
Europe
...


Aryabhata
C
...
476 – 550

The origin of the terms ‘cosine’ and ‘tangent’ was much later
...
Aryabhatta
called it kotijya
...
In 1674, the
English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’
...
sin A is not
the product of ‘sin’ and A
...
Similarly, cos A is not the product of
‘cos’ and A
...

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
...
6), how
will the trigonometric ratios of  A in  PAM differ
from those of  A in  CAB or from those of  A in
 QAN?

Fig
...
6

To answer this, first look at these triangles
...
Using the criterion, you will see that the
triangles PAM and CAB are similar
...

So, we have

AM
AP MP


=
AB
AC BC

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INTRODUCTION TO TRIGONOMETRY

117

MP
BC
 sin A
...

AP AC
AM AB

From this, we find
Similarly,

This shows that the trigonometric ratios of angle A in  PAM not differ from
those of angle A in  CAB
...

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
...
, in place of
(sin A)2, (cos A)2, etc
...
But cosec A = (sin A)–1  sin–1 A (it is called sine
inverse A)
...

Similar conventions hold for the other trigonometric ratios as well
...

We have defined six trigonometric ratios of an acute angle
...

1
If in a right triangle ABC, sin A = ,
3
BC 1
 , i
...
, the
then this means that
AC 3
lengths of the sides BC and AC of the triangle
ABC are in the ratio 1 : 3 (see Fig
...
7)
...
8
...
To determine other
trigonometric ratios for the angle A, we need to find the length of the third side
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


AC
3k
3
Similarly, you can obtain the other trigonometric ratios of the angle A
...

Let us consider some examples
...

Example 1 : Given tan A =

Solution : Let us first draw a right  ABC
(see Fig 8
...

Now, we know that tan A =

BC 4

...

Now, by using the Pythagoras Theorem, we have

Fig
...
8

AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 25k2
So,

AC = 5k

Now, we can write all the trigonometric ratios using their definitions
...

Solution : Let us consider two right
triangles ABC and PQR where
sin B = sin Q (see Fig
...
9)
...
8
...
4,  ACB ~  PRQ and therefore,  B =  Q
...
8
...
Determine the values of
(i) cos2  + sin2 ,
(ii) cos2  – sin2 
Solution : In  ACB, we have
AC =
=
So,

sin  =

AB2  BC 2 =

(29) 2  (21) 2

(29  21) (29  21)  (8) (50) 

Fig
...
10
400  20 units

AC 20 ,
BC 21

cos  =
 
AB 29
AB 29
2

2

2

2

20 2  212 400  441
 20   21 

 1,
Now, (i) cos  + sin  =      
841
292
 29   29 
2

2

(21  20) (21  20)
41
 21 
 20 

and (ii) cos  – sin  =      

...

Solution : In  ABC, tan A =

BC
=1
AB

i
...
,

(see Fig 8
...
8
...

Now,

AC =
=

Therefore,
So,

sin A =

AB2  BC 2

( k ) 2  (k ) 2  k 2
BC
1

AC
2

and

cos A =

AB
1

AC
2

 1  1 
2 sin A cos A = 2 

  1, which is the required value
...
8
...

Determine the values of sin Q and cos Q
...
e
...
e
...
e
...
e
...
8
...
1
1
...
Determine :
(i) sin A, cos A
(ii) sin C, cos C
2
...
8
...

3,
calculate cos A and tan A
...
Given 15 cot A = 8, find sin A and sec A
...
If sin A =

13 ,
calculate all other trigonometric ratios
...
8
...
If  A and  B are acute angles such that cos A = cos B, then show that  A =  B
...
Given sec  =

7
...

1 + tan 2 A
1 ,
9
...
If 3 cot A = 4, check whether

(ii) cos A cos C – sin A sin C
10
...
Determine the values of
sin P, cos P and tan P
...
State whether the following are true or false
...

(i) The value of tan A is always less than 1
...

(ii) sec A =
5
(iii) cos A is the abbreviation used for the cosecant of angle A
...

(v) sin  =

4
for some angle 
...
3 Trigonometric Ratios of Some Specific Angles
From geometry, you are already familiar with the construction of angles of 30°, 45°,
60° and 90°
...


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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
...
,  A =  C = 45°
(see Fig
...
14)
...

Then by Pythagoras Theorem, AC2 = AB2 + BC2 = a2 + a2 = 2a2,
and, therefore,

Fig
...
14

AC = a 2 

Using the definitions of the trigonometric ratios, we have :
sin 45° =

side opposite to angle 45° BC
a
1



hypotenuse
AC a 2
2

cos 45° =

side adjacent to angle 45° AB
a
1



hypotenuse
AC a 2
2

tan 45° =

side opposite to angle 45° BC a

 1
side adjacent to angle 45° AB a

Also, cosec 45° =

1
1
1
 2 , sec 45° =
 2 , cot 45° =
 1
...
Consider an equilateral triangle ABC
...

Draw the perpendicular AD from A to the side BC
(see Fig
...
15)
...
8
...
8
...

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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
...

Then,

BD =

and

1
BC = a
2

AD2 = AB2 – BD2 = (2a)2 – (a)2 = 3a2,
AD = a 3

Therefore,
Now, we have :

BD
a
1
AD a 3
3

 , cos 30° =


AB 2a 2
AB
2a
2
BD
a
1


tan 30° =

...

cot 30° =
tan 30

cosec 30° =

Similarly,
sin 60° =
cosec 60° =

AD a 3
3
1


, cos 60° = , tan 60° =
2
AB
2a
2

3,

2 ,
1

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
...
16), till it becomes zero
...
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
...
17)
...
8
...
8
...
Also, when  A is very close to 0°, AC is nearly the
sin A =
AC
AB
same as AB and so the value of cos A =
is very close to 1
...
We define : sin 0° = 0 and cos 0° = 1
...
(Why?)
cos 0°
tan 0°

1
1 ,
= 1 and cosec 0° =
which is again not defined
...
As  A gets larger and larger,  C gets
smaller and smaller
...
The point A gets closer to point B
...
8
...

sec 0° =

Fig
...
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
...

So, we define :

sin 90° = 1 and cos 90° = 0
...
1, for ready reference
...
1
A

0°

30°

45°

60°

90°

sin A

0

1
2

1
2

3
2

1

cos A

1

3
2

1
2

1
2

0

tan A

0

1
3

1

3

Not defined

Not defined

2

2

2
3

1

sec A

1

2
3

2

2

Not defined

cot A

Not defined

3

1

1
3

0

cosec A

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
...

Example 6 : In  ABC, right-angled at B,
AB = 5 cm and  ACB = 30° (see Fig
...
19)
...

Solution : To find the length of the side BC, we will
choose the trigonometric ratio involving BC and the
given side AB
...
e
...
8
...
e
...
e
...
e
...


Example 7 : In  PQR, right - angled at
Q (see Fig
...
20), PQ = 3 cm and PR = 6 cm
...

Solution : Given PQ = 3 cm and PR = 6 cm
...
8
...


(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
...

Solution : Since, sin (A – B) =

1
, therefore, A – B = 30° (Why?)
2

1
, therefore, A + B = 60°
2
Solving (1) and (2), we get : A = 45° and B = 15°
...
2
1
...
Choose the correct option and justify your choice :
(i)

2 tan 30

1  tan 2 30

(A)

sin 60°

(B) cos 60°

(C) tan 60°

(D) sin 30°

(B) 1

(C) sin 45°

(D) 0

(B) 30°

(C) 45°

(D) 60°

(B) sin 60°

(C) tan 60°

(D) sin 30°

1  tan 2 45

(ii)
1  tan 2 45

(A)

tan 90°

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

0°

2 tan 30

1  tan 2 30

(A)

cos 60°

3
...


4
...
Justify your answer
...

(ii) The value of sin  increases as  increases
...

(iv) sin  = cos  for all values of 
...


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128

MATHEMATICS

8
...

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
...


Fig
...
21

In  ABC, right-angled at B (see Fig
...
21), we have:
AB2 + BC2 = AC 2

(1)

Dividing each term of (1) by AC , we get
2

AB2 BC 2
AC2

=
AC 2 AC 2
AC2
2

i
...
,

2

 AB   BC 
 AC 


 
 = 
 AC 
 AC   AC 

i
...
,

(cos A)2 + (sin A)2 = 1

i
...
,

cos2 A + sin2 A = 1

2

(2)

This is true for all A such that 0°  A  90°
...

Let us now divide (1) by AB2
...
e
...
What about A = 90°? Well, tan A and
sec A are not defined for A = 90°
...

Let us see what we get on dividing (1) by BC2
...
e
...
e
...
Therefore (4) is true for
all A such that 0° < A  90°
...
e
...

Let us see how we can do this using these identities
...

2
3
1 4,

sec A =
, and cos A =

2
3
3 3

Since, sec2 A = 1 + tan2 A = 1 

Again, sin A = 1  cos2 A  1 

3 1

...

4 2

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

Solution :
LHS = sec A (1 – sin A)(sec A + tan A) =

 1 
 1
sin 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

Example 11 : Prove that

cot A – cos A cosec A – 1

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


Example 12 : Prove that
sec2  = 1 + tan2 
...


EXERCISE 8
...
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A
...
Write all the other trigonometric ratios of  A in terms of sec A
...
Choose the correct option
...

(i) 9 sec2 A – 9 tan2 A =
(A) 1

(B) 9

(C) 8

(D) 0

(C) 2

(D) –1

(B) sin A

(C) cosec A

(D) cos A

(B) –1

(C) cot2 A

(D) tan2 A

(ii) (1 + tan  + sec ) (1 + cot  – cosec ) =
(A) 0

(B) 1

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
1  tan 2 A

(iv)
1 + cot 2 A

(A) sec2 A

4
...

1  cos 
(i) (cosec  – cot )2 = 1  cos 

(iii)

(ii)

cos A
1  sin A

 2 sec A
1 + sin A
cos A

tan 
cot 

 1  sec  cosec 
1  cot  1  tan 

[Hint : Write the expression in terms of sin  and cos ]
1  sec A
sin 2 A

[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
...
5 Summary
In this chapter, you have studied the following points :
1
...

side adjacent to angle A
1
1
1 ,
sin A
cosec A =
; sec A =
; tan A =
tan A =

...

The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°
...

sin2 A + cos2 A = 1,
sec2 A – tan2 A = 1 for 0° £ A < 90°,
cosec2 A = 1 + cot2 A for 0° < A £ 90º
...

3
...

5
...


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