Document 7422579
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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,nI
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
ab
1 ab
Solution
tan2 tan
tan tan
1 tan tan
ab
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
CD
C D
cos
2
2
D A-B
C D
By putting A+B = CCand
= D in
sin
2 we get
2 this result
the previous formula
sinC sinD 2 cos
cos C cosD 2 cos
CD
C D
cos
2
2
cos C cosD 2 sin
CD
C D
sin
2
2
Note
or
cos C cosD 2 sin
CD
DC
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
5cos 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 sinb 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