Mathematics
Session 1
Trigonometric ratios and Identities
Topics
Measurement of Angles
Definition and Domain and Range
of Trigonometric Function
Compound Angles
Transformation of Angles
Measurement of Angles
J001
B
OA  InitialRay
OB  Ter minal Ray
O
A
Angle is considered as the figure obtained by
rotating initial ray about its end point.
Measure and Sign of an Angle
J001
Measure of an Angle :-
Amount of rotation from initial side
to terminal side.
Sign of an Angle :B

O

B’
A
Rotation anticlockwise –
Angle positive
Rotation clockwise –
Angle negative
Right Angle
J001
Y
O
X
Revolving ray describes one – quarter of a circle then
we say that measure of angle is right angle
Angle < Right angle  Acute Angle
Angle > Right angle  Obtuse Angle
Quadrants
J001
Y
II Quadrant
I Quadrant
( ,  )
( ,  )
O
X’
III Quadrant
X
IV Quadrant
( ,  )
( ,  )
X’OX – x - axis
Y’OY – y - axis
Y’
System of Measurement of Angle
J001
Measurement of Angle
Sexagesimal System
Centesimal System
or
or
British System
French System
Circular System
or
Radian Measure
System of Measurement of Angles
J001
Sexagesimal System (British System)
1 right angle = 90 degrees (=90o)
1 degree = 60 minutes (=60’)
1 minute = 60 seconds (=60”)
Centesimal System (French System)
1 right angle = 100 grades
(=100g)
1 grade = 100 minutes (=100’)
1 minute = 100 Seconds (=100”)
Is 1 minute of
sexagesimal
=
1 minute of
centesimal ?
NO
System of Measurement of Angle
Circular System
B
r
O
r
1c
r
A
If OA = OB = arc AB
Then  AOB  1radian( 1c )
J001
System of Measurement of Angle
Circular System
J001
C
1c
B
O
 AOC arc AC

 AOB arc ACB
A

1radian
r

2right angles r
 2right angles   radian
 180   radian
Relation Between Degree Grade
And Radian Measure of An Angle
D
900

G
100g

2C

OR
D
1800

G
200g

C

J002
Illustrative Problem
Find the grade and radian measures
of the angle 5o37’30”
Solution
o
'
 30   30   1 
30"  
   60  60    120 
60

 

 
 37 
and37 '  

 60 
o
o
o
37
1   45 

5o 37 '30 "   5 



60 120   8 

We know that
D
G
2C


90 100 
o
J002
Illustrative Problem
Find the grade and radian measures of the
angle 5o37’30”
Solution
g
 10

G
 D
 9

g
g
g
 10 45   225 




12.5
Ans





8   18 
 9
 

and R  
D
 180

c
c
45 

 



radian Ans

 180 8  32
J002
Relation Between Angle Subtended
by an Arc At The Center of Circle
C
B

O
J002
1c
A
Arc AC = r and Arc ACB = 
 AOC arc AC  1radian  r


 AOB arc ACB
 
r
Illustrative Problem
A horse is tied to a post by a rope. If
the horse moves along a circular path
always keeping the rope tight and
describes 88 meters when it has traced
out 72o at the center. Find the length of
rope. [ Take  = 22/7 approx.].
J002
Solution
Arc AB = 88 m and AP = ?
B
c
  2

  72o   72 

rad

180 
5

arc AB
 
r
AP
2 88
22

 AP  70 m
[ 
approx.]
5 AP
7
72o
P
A
Definition of Trigonometric Ratios
Y
P (x,y)
J003
r
y

O
y
sin  
r
x
cos  
r
y
tan  
x
x
M
x
cot  
y
r
sec  
x
r
cos ec  
y
X
r  x2  y2
Some Basic Identities
 sin   cos ec   1 ;   n,n I

 cos   s ec   1 ;   2n  1 ,n I
2

 tan   cot   1 ;   2n  1 ;   n,nI
2
 sin2  cos2   1
 sec2  tan2   1
 cos ec2  cot2   1
sin 

 tan  
;   2n  1 ,n  I
cos 
2
 cot  
cos 
;   n,n  I
sin 
Illustrative Problem
If tan2   1  e2 ,prove that
sec   tan3  .cos ec   2  e2



0




2 


J003
3
2
Solution
 sec   tan3  .cos ec 
cos ec 

 sec  1  tan3 
sec  




   1  tan  
 sec  1  tan3  cot 
2

2
 1  tan  1  tan

 2
3
e2 2

Proved
 sec  1  tan2 
2
3
2

Signs of Trigonometric Function In
All Quadrants
In First Quadrant
Y
P (x,y)
r

O
J004
x
y
sin    0
r
x
cos    0
r
y
tan    0
x
y
Here x >0, y>0, r  x2  y2 >0
X
M
x
cot    0
y
r
sec    0
x
r
co sec    0
y
Signs of Trigonometric Function In
All Quadrants
J004
In Second Quadrant
Y
P (x,y)
y
X’
r

x
y
0
r
x
cos    0
r
y
tan    0
x
sin  
Y’
X
Here x <0, y>0, r  x2  y2 >0
x
0
y
r
sec    0
x
r
co sec    0
y
cot  
Signs of Trigonometric Function In
All Quadrants
In Third Quadrant
Y
M
X’
O
P (x,y)
y
sin    0
r
x
cos    0
r
y
tan    0
x
Y’
J004

X
Here x <0, y<0, r  x2  y2 >0
x
cot    0
y
r
sec    0
x
r
co sec    0
y
Signs of Trigonometric Function In
All Quadrants
J004
In Fourth Quadrant

M
O
P (x,y)
Y’
y
sin    0
r
x
cos    0
r
y
tan    0
x
X
Here x >0, y<0, r  x2  y2 >0
x
cot    0
y
r
sec    0
x
r
co sec    0
y
Signs of Trigonometric Function In
All Quadrants
Y
I Quadrant
All Positive
II Quadrant
sin & cosec
are Positive
X’
O
X
III Quadrant
IV Quadrant
tan & cot are
Positive
cos & sec are
Positive
Y’
ASTC :- All Sin Tan Cos
J004
Illustrative Problem
12
,  lies in second
If cot  = 
5
quadrant, find the values of
other five trigonometric function
Solution Method : 1
J004
12
5
 tan   
5
12
169
sec2   1  tan2   sec2  
13
13 144
 sec   
 sec   
  liesinsec ondquadrant 
12
12
12
Whichgives cos   
13
5
12
5
Thensin   tan   cos   


12
13 13
13
 cos ec  
5
cot   
Illustrative Problem
J004
12  lies in second

,
5
If cot  =
quadrant, find the values of other five
trigonometric function
Solution Method : 2
P (-12,5)
5
X’
Y
sin  
r

-12
Y’
y
5

r 13
cos  
x 12

r
13
tan  
y
5

x 12
X
Here x = -12, y = 5 and r = 13
r
13

x 12
r 13
cos ec  
y
5
sec  
Domain and Range of Trigonometric
Function
Functions
Domain
Range
sin
R
[-1,1]
cos
R
[-1,1]
tan
cot
sec
cosec


R   :   (2n  1) 
2

R   :   n


R   :   (2n  1) 
2

R   :   n
R
R
R-(-1,1)
R-(-1,1)
J005
Illustrative problem
(x  y)2
sin  
Prove that
4xy
is possible for real values of x and
y only when x=y
2
sin2   1
Solution
(x  y)2
2

 1   x  y   4xy
4xy
2
2
  x  y   4xy  0   x  y   0
But for real values of x and y
2
is not less than zero
2
Pr oved
x  y
 x  y  0  x  y
J005
Trigonometric Function For Allied
Angles
If angle is multiple of 900 then
sin  cos;tan  cot; sec  cosec
If angle is multiple of 1800 then
sin  sin;cos  cos; tan  tan etc.
Trig. ratio
-
90o-
90o+
180o-
180o+ 360o- 360o+
sin
- sin
cos
cos
sin
- sin
- sin
sin
cos
cos
sin
- sin
- cos
- cos
cos
cos
tan
- tan
cot
- cot
-tan
tan
- tan
tan
Trigonometric Function For Allied
Angles
Trig. ratio
-
90o-
90o+ 180o-
cot
- cot
tan
sec
sec
cosec - cosec - sec
cosec
- cosec sec
-tan
sec
-cot
180o+
360o-
360o+
cot
- cot
cot
- sec
sec
sec
cosec -cosec - cosec cosec
Periodicity of Trigonometric
Function
If f(x+T) = f(x)  x,then T is called
period of f(x) if T is the smallest
possible positive number
J005
Periodicity : After certain value of
x the functional values repeats
itself
Period of basic trigonometric functions
sin (360o+) = sin  period of sin is 360o or 2
cos (360o+) = cos  period of cos is 360o or 2
tan (180o+) = tan  period of tan is 180o or 
Trigonometric Ratio of Compound
Angle
Angles of the form of A+B, A-B,
A+B+C, A-B+C etc. are called
compound angles
(I) The Addition Formula
 sin (A+B) = sinAcosB + cosAsinB
 cos (A+B) = cosAcosB - sinAsinB
tan A  tanB
1  tan A tanB
sin(A  B)
tan  A  B  
cos(A  B)
 tan  A  B  
Pr oof: 

sin A cosB  cos A sinB
cos A cosB  sin A sinB
J006
Trigonometric Ratio of Compound
Angle

sin A cosB  cos A sinB
cos A cosB  sin A sinB
Dividing Nr and Dr by cos A cosB
We get

tan A  tanB
1  tan A tanB
 cot  A  B  
Proved
cot B cot A  1
cot B  cot A
J006
Illustrative problem
Find the value of
(i) sin 75o
(ii) tan 105o
Solution
(i) Sin 75o = sin (45o + 30o)
= sin 45o cos 30o + cos 45o sin 30o


1
3
1 1
3 1


 
2
2
2 2
2 2
(ii) Ans:  2  3

Trigonometric Ratio of
Compound Angle
(I) The Difference Formula
 sin (A - B) = sinAcosB - cosAsinB
 cos (A - B) = cosAcosB + sinAsinB
tan A  tanB
 tan  A  B  
1  tan A tanB
cot B cot A  1
 cot  A  B   
cot A  cot B
Note :- by replacing B to -B in addition
formula we get difference formula
Illustrative problem
If tan (+) = a and tan ( - ) = b
Prove that tan2 
ab
1  ab
Solution
tan2  tan           


tan       tan     
1  tan      tan     
ab
1  ab
Some Important Deductions
 sin (A+B) sin (A-B) = sin2A - sin2B = cos2B - cos2A
 cos (A+B) cos (A-B) = cos2A - sin2B = cos2B - sin2A
tan A  tanB  tanC  tan A tanB tanC
 tan  A  B  C  
1  tan A tanB  tanB tanC  tanC tan A
To Express acos + bsin in the
form kcos or sin
acos +bsin


a
b
 a b 
cos  
sin  
2
2
2
2
a b
 a b

2
2
Let cos  
a
a2  b2
, then sin  
b
a2  b2
acos   b sin   a2  b2  cos  cos   sin  sin  
 a2  b2 cos     
 k cos ,where k  a2  b2 ,     
Similarly we get acos + bsin = sin
Illustrative problem
Find the maximum and minimum
values of 7cos + 24sin
Solution 7cos +24sin


7
24
 7  24 
cos  
sin  
2
2
2
2
7  24
 7  24

2
2
24
 7

 25 
cos  
sin  
25
 25

Let cos  
7
24
 sin  
25
25
 7 cos   24 sin   25  cos  cos   sin  sin  
Illustrative problem
Find the maximum and minimum value of
7cos + 24sin
Solution
 25 cos       25 cos  where     
1  cos   1
 25  25 cos   25
 Max. value =25, Min. value = -25
Ans.
Transformation Formulae
 Transformation of product into sum
and difference
 2 sinAcosB = sin(A+B) + sin(A - B)
 2 cosAsinB = sin(A+B) - sin(A - B)
 2 cosAcosB = cos(A+B) + cos(A - B)
Proof :- R.H.S = cos(A+B) + cos(A - B)
= cosAcosB - sinAsinB+cosAcosB+sinAsinB
= 2cosAcosB
=L.H.S
 2 sinAsinB = cos(A - B) - cos(A+B)
[Note]
Transformation Formulae
 Transformation of sums or difference
into products
 sinC  sinD  2 sin
CD
C D
cos
2
2
D A-B
C D
By putting A+B = CCand
= D in
sin
2 we get
2 this result
the previous formula
 sinC  sinD  2 cos
 cos C  cosD  2 cos
CD
C D
cos
2
2
 cos C  cosD  2 sin
CD
C D
sin
2
2
Note
or
 cos C  cosD  2 sin
CD
DC
sin
2
2
Illustrative problem
Prove that
cos 8x  cos 6x  cos 4x
 cot 6x
sin8x  sin6x  sin 4x
Solution
(cos 8x  cos 4x)  cos 6x
(sin8x  sin 4x)  sin6x
L.H.S 
8x  4x
8x  4x
cos
 cos 6x
2 cos 6x cos 2x  cos 6x
2
2


8x  4x
8x  4x
2 sin6x sin2x  sin6x
2 sin
cos
 sin6x
2
2
2 cos

2 cos 6x(cos 2x  1)
2 sin6x(cos 2x  1)
 cot 6x
Proved
Class Exercise - 1
If the angular diameter of the moon
be 30´, how far from the eye can a
coin of diameter 2.2 cm be kept to
hide the moon? (Take p = 22
7
approximately)
Moon
B
r
E (Eye)
A
Class Exercise - 1
If the angular diameter of the moon be 30´,
how far from the eye can a coin of diameter
22
2.2 cm be kept to hide the moon? (Take p =
7
approximately)
Solution :Let the coin is kept at a distance r
from the eye to hide the moon
completely. Let AB = Diameter of
the coin. Then arc AB = Diameter
AB = 2.2 cm
c
 
 30 
1
  30´ 



 2 180 
 60 


c
  


 360 
arc

2.2
 


radius
360
r
r
Moon
360  2.2 360  2.2  7

 252 cm

22
B
r
E (Eye)
A
Class Exercise - 2
Prove that tan3A tan2A tanA =
tan3A – tan2A – tanA.
Solution :We have 3A = 2A + A
tan3A = tan(2A + A)
 tan3A = tan2A  tan A
1 – tan2A tan A
 tan3A – tan3A tan2A tanA = tan2A + tanA
tan3A – tan2A – tanA = tan3A tan2A tanA
(Proved)
Class Exercise - 3
If sin = sinb and cos = cosb, then
b
(a) sin
0
2
(b) cos   b  0
2
–b
0
(c) sin
2
(d) cos  – b  0
2
Solution :-
sin   sin b and cos   cos b
 sin  – sin b  0 and cos – cosb  0
 2 sin
–b
b
 –b
b
cos
 0 and – 2 sin
sin
0
2
2
2
2
 sin
–b
0
2
Class Exercise - 4
Prove that
sin20 sin 40 sin60 sin80 
3
16
Solution:LHS = sin20° sin40° sin60° sin80°
1
 sin60  [2sin20 sin40]  sin80
2



3 1
 [cos(40 – 20) – cos(40  20)]  sin80
2 2
3
[cos 20 – cos 60]sin80
4
3
4
1
3


sin80

cos20

–

sin80



 8 2 sin80 cos 20 – sin80
2


Class Exercise - 4
Prove that
sin20 sin 40 sin60 sin80 
3
16
Solution:3
sin(80  20)  sin(80 – 20) – sin80

8
3

sin100  sin60 – sin80

8

3
sin(180 – 80)  sin60 – sin80

8
16
 
3
3
3

– sin80
sin80 
Proved.

8 
2

Class Exercise - 5
n
n
Prove that  cos A  cos B    sin A  sinB 
 sin A – sinB 
 cos A – cos B 
Solution :-

n A B
2
cot
, if n is even




 2 

0,
if n is odd

n
n
 cos A  cosB 
 sin A  sinB 
LHS  


 cos A – cosB 
sin
A
–
sinB




n
n
A B
A –B 
A B
A –B 


2
cos
cos
2
sin
cos




2
2
2
2

 

A

B
A
–
B
A

B
A
–
B
 2 cos

 –2 sin

sin
sin

2
2 

2
2 
n
n


 A– B 
 A – B 
 cot 
 –cot 



2
2







Class Exercise - 5
Prove
n
n
that cos A  cos B    sin A  sinB 
 sin A – sinB 
 cos A – cos B 

n A B
2 cot 
 , if n is even
 
 2 

0,
if n is odd

Solution : A –B
 A –B
n
n
n
 cot 
  (–1) cot 


2


2

A –B

n
2
cot
, if n is even

 A –B
n
n

 cot 
1  (–1)   
2


 2 

0,
if n is odd
Class Exercise - 6
The maximum value of 3 cosx + 4
sinx + 5 is
(b) 9
(a) 5
(c) 7
(d) None of these
Solution :32  42
3 cos x  4 sin x 

3
cos x 

 32  42

sin x 

32  42

4
4
3

 5  cos x  sin x 
5
5

3
4

Let cos  5  sin   5 


 5cos x cos   sinx sin 
Class Exercise - 6
The maximum value of 3 cosx + 4 sinx + 5
is
Solution :3
4

Let cos  5  sin   5 


 5 cos(x – )
–1  cos(x – )  1
 –5  5 cos(x – )  5
 –5  5  5 cos(x – )  5  10
 0  3cosx  4sinx  5  10
 Maximum value of the given expression = 10.
Class Exercise - 7
If a and b are the solutions of a
cos + b sin = c, then show that
cos(  b) 
Solution :-
a2 – b2
a2  b2
We have acos   b sin   c … (i)
 acos   c – b sin 
 a2 cos2   (c – b sin )2


 a2 1– sin2   c2 – 2bc sin   b2 sin2 


 a2  b2 sin2  – 2bc sin   (c2 – a2)  0
bare roots of equatoin (i),
Class Exercise - 7
If a and b are the solutions of acos + bsin
= c, then show that cos(  b) 
a2 – b 2
a2  b2
Solution :sin and sinb are roots of equ.
c2 – a2
Hence sin  sin b  2
a  b2
(ii).
Again from (i),b sin   c – acos 
 b2 sin2   (c – acos )2
 b2(1– cos2 )  c2  a2 cos2  – 2cacos 
 (a2  b2)cos2  – 2ac cos   c2 – b2  0
Class Exercise - 7
If a and b are the solutions of acos + bsin
= c, then show that cos(  b) 
a2 – b 2
a2  b2
Solution :- (iv)
 and b be the roots of equation (i),
cos and cosb are the roots of equation (iv).
c2 – b2
cos  cos b 
a2  b2
Now cos(  b)  cos  cos b – sin  sin b
c2 – b2 c2 – a2
a2 – b2

–

a2  b2 a2  b2 a2  b2
Class Exercise - 8
If a seca – c tana = d and b seca
+ d tana = c, then
(a) a2 + b2 = c2 + d2 + cd
(b) a2  b2 
1
c2

1
d2
(c) a2 + b2 = c2 + d2
(d) ab = cd
Class Exercise - 8
If a seca – c tana = d and b seca
+ d tana = c, then
asec  – c tan   d
Solution :a
c sin 

–
d
cos  cos 
 a  c sin   dcos  ….. (I)
Again b sec   dtan   c
b
dsin 


c
cos  cos 
 b  ccos  – dsin  ….. ii
a2  b2  c2(sin2   cos2 )  d2(cos2   sin2 )
2cd sin  cos   2cd sin  cos   c2  d2
Squaring and adding
(i) and (ii), we get
Class Exercise -9
The value of
A

2
sin2  
–
sin
2 
8
(a) 2 sinA
(b)
(c) 2 cosA
(d)
A

–
8 2 


1
2
1
2
sin A
cos A
Class Exercise -9
The value of
A
A

2 
sin2 

–
sin
–
8
2
2
8



Solution : A
 A
sin2    – sin2  – 
8 2 
8 2 
  A    A 
  A    A 
 sin      –  sin    –  – 
 8 2   8 2 
 8 2   8 2 

1
 sin  sin A 
sin A
4
2
 sin2 A – sin2 B  sin(A  B)sin(A – B)


Class Exercise -10
5
4
If cos(  b)  , sin( – b) 
,
5
13

 and b lie between0 and , then value
4
 of tan2 is
(a) 1
56
(b) 33
(c) 0
33
(d)
56
Solution :-
–

 and bbetween 0 and 4 ,



  – b  and 0    b 
4
4
2
Consequently, cos b and sinb are positive.
sin(  b)  1 – cos2(  b)
Class Exercise -10
If cos(  b) 
5
4
, sin( – b) 
,
13
5

 and b lie between0 and , then value
4
 of tan2 is
Solution :- 
1–
16
3

25
5
25
12
cos( – b)  1 – sin ( – b)  1 –

169 13
3
5
 tan(  b)  , tan( – b) 
4
12
2
tan2  tan[(  b)  ( – b)]
3 5

tan(  b)  tan( – b)
4
12  56


3 5
1 – tan(  b) tan( – b)
33
1– 
4 12