Download TRIGONOMETRIC FUNCTIONS-II

Trigonometric Functions-II
MODULE - IV
Functions
17
Notes
TRIGONOMETRIC FUNCTIONS-II
In the previous lesson, you have learnt trigonometric functions of real numbers, draw and interpret
the graphs of trigonometric functions. In this lesson we will establish addition and subtraction
formulae for cos ( A ± B ) , sin ( A ± B ) and tan ( A ± B ) . We will also state the formulae for
the multiple and sub multiples of angles and solve examples thereof. The general solutions of
simple trigonometric functions also discussed in the lesson.
OBJECTIVES
After studying this lesson, you will be able to :
•
•
x
π
write trigonometric functions of − x, , x ± y, ± x, π ± x where x, y are real nunbers;
2
2
establish the addition and subtraction formulae for :
cos (A ± B) = cos A cos B m sin A sin B,
sin (A ± B) = sin A cos B ± cos A sin B and tan ( A ± B ) =
•
•
solve problems using the addition and subtraction formulae;
state the formulae for the multiples and sub-multiples of angles such as cos2A, sin 2A, tan
2A, cos 3A, sin 3A, tan 3A, sin
•
tanA ± tanB
1 m tanAtanB
A
A
A
,cos and tan ; and
2
2
2
solve simple trigonometric equations of the type :
sinx = 0,cosx = 0 , tanx = 0,
sinx = sin α , cosx = cos α , tanx = tan α
MATHEMATICS
99
Trigonometric Functions-II
MODULE - IV
Functions
Notes
EXPECTED BACKGROUND KNOWLEDGE
•
Definition of trigonometric functions.
•
Trigonometric functions of complementary and supplementary angles.
•
Trigonometric identities.
17.1 ADDITION AND MULTIPLICATION OF
TRIGONOMETRIC FUNCTIONS
In earlier sections we have learnt about circular measure of angles, trigonometric functions,
values of trigonometric functions of specific numbers and of allied numbers.
You may now be interested to know whether with the given values of trigonometric functions of
any two numbers A and B, it is possible to find trigonometric functions of sums or differences.
You will see how trigonometric functions of sum or difference of numbers are connected with
those of individual numbers. This will help you, for instance, to find the value of trigonometric
functions of
π
5π
and
etc.
12
12
π
π π
can be expressed as −
12
4 6
5π
π π
can be expressed as +
12
4 6
How can we express
7π
in the form of addition or subtraction?
12
In this section we propose to study such type of trigonometric functions.
17.1.1 Addition Formulae
For any two numbers A and B,
cos ( A + B) = cosAcosB − s i n A s i n B
In given figure trace out
∠ SOP = A
∠ POQ = B
∠ SOR = − B
Fig. 17.1
where points P, Q, R, S lie on the unit circle.
Coordinates of P, Q, R, S will be (cos A, sin A), [cos (A + B), sin (A + B)],
[ cos ( −B) ,sin ( −B) ] , and (1, 0).
From the given figure, we have
100
MATHEMATICS
Trigonometric Functions-II
MODULE - IV
Functions
side OP = side OQ
∠ POR = ∠ QOS (each angle = ∠ B + ∠ QOR)
side OR = side OS
∆ POR ≅ ∆ QOS (by SAS)
∴ PR = QS
Notes
2
2
PR = (cosA − cosB) + (sinA − sin ( − B ))
QS = (cos ( A + B) −1) + (sin (A + B) − 0)
2
Fig. 17.2
2
Since PR 2 = QS2
∴ cos 2 A + cos 2 B − 2cosAcosB + sin 2 A+ sin 2 B+ 2 s i n A s i n B
= cos2 (A + B) +1 − 2cos ( A + B) + sin 2 (A + B )
⇒
1 + 1 − 2 ( c o s A c o s B − sinAsinB) = 1 + 1− 2cos (A + B )
⇒
cosAcosB − sinAsinB = cos (A + B)
(I)
Corollary 1
For any two numbers A and B, cos (A − B) = cos A cos B + sin A sin B
Proof : Replace B by − B in (I)
cos ( A − B) = cosAcosB + sinAsinB
[Q cos ( − B) = cosB and sin ( −B ) = − sinB ]
Corollary 2
For any two numbers A and B
sin ( A + B) = sinAcosB + cosAsinB
π

Proof : We know that cos  − A  = s i n A
2

π

sin  − A  = c o s A
2

and
∴
 π

sin ( A + B ) = cos   − (A + B)  

 2
 π


= cos   − A  − B

 2

MATHEMATICS
101
Trigonometric Functions-II
MODULE - IV
Functions
π

π

= cos  − A  cosB + sin  − A  sinB
2

2

or
Notes
sin ( A + B) = sinAcosB + cosAsinB
.....(II)
Corollary 3
For any two numbers A and B
sin ( A − B) = sin A c o s B − c o s A s i n B
Proof : Replacing B by − B in (2), we have
sin ( A + ( −B ) ) = sinAcos ( − B) + cosAsin ( − B )
or
sin ( A − B) = sinAcosB − c o s A s i n B
Example 17.1
(a)
Find the value of each of the following :
(i) sin
(b)
5π
12
(ii) cos
If sinA =
π
12
(iii) cos
7π
12
1
1
π
, sinB=
show that A + B =
10
5
4
Solution :
(a)
(i) sin
5π
π
π
π
π
π π
= sin  +  = sin ⋅ cos + cos ⋅ sin
12
4 6
4
6
4
6
=
∴
1
3
1 1
3 +1
⋅
+
⋅ =
2 2
2 2 2 2
sin
5π
3 +1
=
12
2 2
(ii)
cos
π
π π
= cos  − 
12
4 6
= cos
=
∴
cos
1
3
1 1
3 +1
⋅
+
⋅ =
2 2
2 2 2 2
π
3 +1
=
12 2 2
Observe that sin
102
π
π
π
π
⋅ cos + sin ⋅ sin
4
6
4
6
5π
π
= cos
12
12
MATHEMATICS
Trigonometric Functions-II
7π
π π
= cos  + 
(iii) cos
12
3 4
π
π
π
π
= cos ⋅ cos − sin ⋅ sin
3
4
3
4
=
∴
(b)
cos
1 1
3 1 1− 3
⋅
−
⋅
=
2 2
2
2
2 2
MODULE - IV
Functions
Notes
7π 1 − 3
=
12
2 2
sin ( A + B ) = sinAcosB + cosAsinB
1
3
1
2
=
and cosB = 1− =
10
10
5
5
Substituting all these values in (II), we get
cosA = 1 −
1 2
3 1
+
10 5
10 5
sin ( A + B) =
5
5
5
1
π
+
=
=
=sin
4
10 5
50 5 2
2
=
or
A + B=
π
4
CHECK YOUR PROGRESS 17.1
1.
(a) Find the values of each of the following :
π
π
2π
π
2π
+ cos ⋅ sin
(ii) sin ⋅ cos
12
9
9
9
9
(b) Prove the following :
(i) sin
π
 1
π
1
(i) sin  + A  = ( cosA + 3 sinA) (ii) sin  − A  = ( cosA −sinA)
6
 2
4

2
(c) If s i n A =
2.
8
5
and sinB = , find sin ( A − B )
17
13
(a) Find the value of cos
5π
.
12
(b) Prove the following :
π

(i) cos θ + sin θ = 2cos  θ− 

4
MATHEMATICS
(ii)
π

3 sin θ − cos θ = 2sin  θ− 

6
103
Trigonometric Functions-II
MODULE - IV
Functions
Notes
(iii) cos ( n + 1)A cos ( n − 1) A + sin ( n +1 ) Asin ( n −1 ) A = cos2A
π

π

π

π

(iv) cos  + A  cos  − B  + sin  + A  sin  − B  = cos ( A + B )
4

4

4

4

Corollary 4 : tan ( A + B) = tanA + tanB
1 − tanAtanB
Proof : tan ( A + B ) =
sin (A + B)
cos ( A + B)
=
sin A cosB + c o s A s i n B
cosAcosB − sin A sinB
Dividing by cos A, cos B, we have
sinAcosB c o s A s i n B
+
cosAcosB cosAcosB
(
)
tan A+ B =
cosAcosB s i n A s i n B
−
cosAcosB cosAcosB
or
tan ( A+ B) =
tanA + tanB
1 − tanAtanB
......(III)
tanA − tanB
Corollary 5 : tan ( A− B) =
1 + tanAtanB
Proof : Replacing B by − B in (III), we get the required result.
cotAcotB − 1
Corollary 6 : cot ( A+ B) =
cotB + cotA
Proof : cot ( A + B) =
cos ( A + B ) cosAcosB − sinAsinB
=
sin ( A + B ) sinAcosB + cosAsinB
Dividing by sin A sin B, we have
cot ( A+ B) =
......(IV)
cotAcotB − 1
cotB + cotA
π
 1 + tanA
Corollary 7 : tan  + A  =
4
 1 − tanA
Proof :
π
tan + tanA
π

4
tan  + A  =
4
 1 − tan π ⋅ t a n A
4
=
104
1 + tanA
π
as tan = 1
1− t a n A
4
MATHEMATICS
Trigonometric Functions-II
MODULE - IV
Functions
Similarly, it can be proved that
π
 1−tanA
tan  − A  =
4
 1 + tanA
Example 17.2 Find tan
π
12
Notes
π
π
tan − tan
π
π
π
4
6
= tan  −  =
Solution : tan
π
 4 6  1 + tan ⋅ tan π
12
4
6
1
1−
3 = 3 −1
=
1
3 +1
1 + 1.
3
=
(
(
3 − 1 ) ( 3 −1 )
3 + 1) ( 3 − 1)
= 2− 3
∴
=
4 −2 3
2
tan
π
= 2− 3
12
Example 17.3 Prove the following :
(a)
7π
7π
+ sin
36
36 = tan 4 π
7π
7π
9
cos − sin
36
36
(b)
tan 7 A − tan 4 A − tan3A = tan7A tan4A ⋅ tan3A
(c)
tan
cos
7π
π
5π
= tan + 2tan
18
9
18
Solution : (a) Dividing numerator and denominator by cos
7π
, we get
36
7π
7π
7π
+ sin
1 + tan
36
36 =
36
L.H.S. =
7π
7π
7π
cos − sin
1 −tan
36
36
36
cos
π
7π
+ tan
4
36
=
π
7π
1 − tan ⋅ tan
4
36
tan
 π 7π 
= tan  +

 4 36 
MATHEMATICS
105
Trigonometric Functions-II
MODULE - IV
Functions
= tan
16 π
4π
= tan
= R.H.S.
36
9
(b) tan 7 A = tan ( 4A + 3A ) =
Notes
tan4A + tan3A
1 − tan4A tan3A
or
tan 7 A − tan7A tan 4 A tan 3A = tan4A + tan3A
or
tan7A − tan4A −tan3A =tan7Atan4Atan3A
5π
2π
tan
+ tan
7π
5
π
2
π


18
18
tan
= tan 
+ =
5
π
2π
18
 18 18  1 − tan
⋅tan
18
18
(c)
tan
7π
7π
5π
2π
5π
2π
− tan
tan
tan
= tan
+ tan
18
18
18
18
18
18
tan
7π
π
2π
π π
= tan  −  = cot = cot
18
2 9
9
18
.....(1)
∴ (1) can be written as
tan
∴ tan
7π
2π
5π
2π
5π
π
− cot
tan
tan
= tan
+ tan
18
18
18
18
18
9
7π
π
5π
= tan + 2tan
18
9
18
CHECK YOUR PROGRESS 17.2
1. Fill in the blanks :
π

π

π π
(i) sin  + A  sin  − A  = ......... (ii) cos  +
4

4

3 4

π π
 cos  −  = .........

3 4
2. (a) Prove the following :
π

π

(i) tan  + θ  tan  − θ  = 1.
4

4

(iii) tan
cotAcotB + 1
cotB − cot A
π
π
π
π
+ tan + tan
⋅ tan = 1
12
6
12
6
(b) If tan A =
a
c
; tan B = , Prove that
b
d
tan ( A + B ) =
106
(ii) cot ( A − B ) =
ad + bc
.
bd − ac
MATHEMATICS
Trigonometric Functions-II
MODULE - IV
Functions
11π
(c) Find the value of cos
.
12
3. (a) Prove the following :
π
3π

(i) tan  + A  tan 
+ A  = −1
4

 4

(iii)
(ii)
cos θ + sin θ
π

= tan  + θ 
cos θ − sin θ
4

Notes
cos θ − sin θ
π

= tan  − θ 
cos θ + sin θ
4

17.2 TRANSFORMATION OF PRODUCTS INTO SUMS
AND VICE VERSA
17.2.1 Transformation of Products into Sums or Differences
We know that
sin ( A + B ) = sin A c o s B + cosAsinB
sin ( A − B ) = sinAcosB − c o s A s i n B
cos ( A + B) = cosAcosB − sinAsinB
cos ( A − B ) = cosAcosB + s i n A s i n B
By adding and subtracting the first two formulae, we get respectively
and
2 s i n A c o s B = sin ( A + B ) + sin ( A − B )
.....(1)
2 c o s A s i n B = sin ( A + B ) − sin ( A − B )
.....(2)
Similarly, by adding and subtracting the other two formulae, we get
and
2cosAcosB = cos ( A + B) + cos ( A −B )
....(3)
2 s i n A s i n B = cos ( A − B ) − cos ( A + B )
....(4)
We can also quote these as
2 s i n A c o s B = sin ( sum ) + sin ( difference )
2 c o s A s i n B = sin ( sum ) − sin ( difference )
2cosAcosB = cos ( sum ) + cos ( difference )
2 s i n A s i n B = cos ( difference ) − cos ( sum )
17.2.2 Transformation of Sums or Differences into Products
In the above results put
MATHEMATICS
107
Trigonometric Functions-II
MODULE - IV
Functions
and
Then A =
Notes
A+B=C
A− B = D
C+D
C−D
and B =
and (1), (2), (3) and (4) become
2
2
sinC + sin D = 2sin
C+D
C−D
cos
2
2
sinC − sin D = 2cos
C+D
C−D
sin
2
2
cosC + cosD = 2cos
C+D
C−D
cos
2
2
cosD − cosC = 2sin
C+D
C−D
sin
2
2
17.2.3 Further Applications of Addition and Subtraction Formulae
We shall prove that
(i) sin ( A + B ) sin ( A − B ) = sin 2 A − sin 2 B
(ii) cos ( A + B ) cos ( A − B ) = cos 2 A − sin 2 B or cos 2 B − sin 2 A
Proof : (i) sin ( A + B ) sin ( A − B )
= ( sin A c o s B + cosAsinB ) ( sinAcosB − cosAsinB )
= sin 2 Acos 2 B − cos 2 Asin 2 B
= sin 2 A ( 1 − sin 2 B ) − ( 1 − sin 2 A ) sin 2 B
= sin 2 A − sin 2 B
(ii) cos ( A + B ) cos ( A − B )
= ( cosAcosB − sin A s i n B ) ( cosAcosB + s i n A s i n B )
= cos 2 Acos 2 B − sin 2 Asin 2 B
= cos 2 A ( 1 − sin 2 B ) − ( 1 − cos 2 A ) sin 2 B
= cos 2 A − sin 2 B
= ( 1 − sin 2 A ) − ( 1 − cos 2 B )
= cos 2 B − sin 2 A
Example 17.4 Express the following products as a sum or difference
(i) 2sin3θ c o s 2θ
108
(ii) cos6θ cos θ
(iii) sin
5π
π
sin
12
12
MATHEMATICS
Trigonometric Functions-II
Solution :
MODULE - IV
Functions
(i) 2sin3θ cos2θ = sin ( 3θ + 2θ ) + sin ( 3θ − 2θ )
= sin5θ + sin θ
(ii) cos6θ cos θ =
(iii) sin
1
( 2cos6θ cos θ )
2
Notes
=
1
[ cos ( 6θ + θ ) + cos ( 6θ − θ ) ]
2
=
1
( cos7θ + cos5θ )
2
5π
π
1
sin
=
12
12 2
5π
π

 2sin 12 sin 12 
=
1
2

 5π − π 
 5π + π  
 cos  12  − cos  12  
=
1
2
π
π

 cos 3 − cos 2 
Example 17.5 Express the following sums as products.
(i) cos
5π
7π
+ cos
9
9
(ii) sin
5π
7π
+ cos
36
36
Solution :
(i)
cos
5π
7π
5π + 7 π
5π − 7 π
+ cos
= 2cos
cos
9
9
9×2
9×2
= 2cos
π

 π
 Q cos  − 9  = cos 9 
2π
π
cos
3
9
π
π

= 2cos  π −  cos
3
9

= −2cos
= − cos
(ii)
sin
π
9
π 1

 Q cos 3 = 2 
5π
7π
7π
 π 13π 
+ cos
= sin  −
 + cos
36
36
36 
36
2
= cos
MATHEMATICS
π
π
cos
3
9
13π
7π
+ cos
36
36
109
Trigonometric Functions-II
MODULE - IV
Functions
Notes
Example 17.6 Prove that
= 2cos
13π + 7 π
13π − 7π
cos
36 × 2
36 × 2
= 2cos
5π
π
cos
18
12
cos7A − cos9A
= tan8A
sin9A − sin7A
Solution :
7A + 9A
9A − 7A
sin
2
2
L.H.S. =
9A + 7A
9A − 7A
2cos
sin
2
2
2sin
=
Example 17.7
sin8A sinA
sin8A
=
= tan8A = R.H.S.
cos8A sin A cos8A
Prove the following :
π
π
(i) cos 2  − A  − sin 2  − B  = sin ( A + B ) cos ( A − B )
4

 4

π A
π A
1
(ii) sin 2  +  − sin 2  −  = sinA
2
8 2
8 2
Solution :
(i) Applying the formula
cos 2 A − sin 2 B = cos ( A + B ) cos ( A − B ) , we have
π
π
π

π

L.H.S. = cos  − A + − B  cos  − A − + B 
4
4
4

4

π

= cos  − ( A + B) cos  − ( A − B)
2

= sin ( A + B ) cos ( A − B ) = R.H.S.
(ii) Applying the formula
sin 2 A − sin 2 B = sin ( A + B ) sin ( A − B ) , we have
π A π A
π A π A
L.H.S. = sin  + + −  sin  + − +

8 2 8 2 
8 2 8 2 
= sin
=
110
π
sin A
4
1
sinA = R.H.S.
2
MATHEMATICS
Trigonometric Functions-II
MODULE - IV
Functions
Example 17.8 Prove that
π
2π
π
4π 1
cos cos
cos cos
=
9
9
3
9 16
Solution :
L.H.S. cos
cos
Now
π
2π
π
4π
cos
cos
cos
3 
9
9 
9
Notes
=
1 1
2π
π
4π
⋅  2cos
cos  cos
2 2
9
9
9
π 1

 Q cos 3 = 2 
=
1
4
=
1
4π 1 
4π
π
cos
+  2cos
cos 
8
9
8
9
9
=
1
4π 1 
5π
π
cos
+  cos
+ cos 
8
9
8
9
3
=
1
4π 1
5π
1
cos
+ cos
+
8
9
8
9 16
.....(1)
5π
4π 
4π

= cos  π −
= − cos

9
9 
9

.....(2)
π
π
4π

cos
+
cos
cos

3
9 
9
From (1) and (2), we get
L.H.S. =
1
= R.H.S.
16
CHECK YOUR PROGRESS 17.3
1. Express each of the following as sums or differences :
(a)
2cos3θ s i n 2θ
(c)
2cos
π
π
cos
4
12
(b)
2sin4θ sin2θ
(d)
2sin
π
π
cos
3
6
2. Express each of the following as a product :
(a)
sin6θ + s i n 4θ
(b)
sin7 θ − sin3θ
(c)
cos2θ − cos4θ
(d)
cos7 θ + cos5θ
MATHEMATICS
111
Trigonometric Functions-II
MODULE - IV 3. Prove the following :
Functions
(a) sin
5π
4π
π
+ cos
= cos
18
9
9
π
7π
− cos
9
18 = 1
(b)
7π
π
sin
− sin
18
9
(c) sin
5π
7π
π
− sin
+ sin
=0
18
18
18
(d) cos
Notes
cos
π
5π
7π
+ cos
+ cos
=0
9
9
9
4. Prove the following :
(a) sin 2 ( n + 1 ) θ − sin 2 nθ = sin ( 2n + 1 ) θ ⋅ sin θ
(b) cos β cos ( 2α − β ) = cos 2 α − sin 2 ( α − β )
(c) cos 2
π
π
3
− sin 2
=
4
12
4
π

 π

5. Show that cos 2  + θ  − sin 2  − θ  is independent of θ .
4

 4

6. Prove the following :
(a)
sin θ + sin3θ + sin5θ + s i n 7θ
= tan 4θ
cos θ + cos3θ + cos5θ + cos 7θ
(b) sin
π
5π
5π
7π
1
sin
sin
sin
=
18
6
18
18 16
2
2
(c) ( cos α + cos β ) + ( sin α + sinβ ) = 4cos 2
α−β
2
17.3 TRIGONOMETRIC FUNCTIONS OF
MULTIPLES OF ANGLES
(a) To express sin 2A in terms of sin A, cos A and tan A.
We know that
sin ( A + B ) = sin A c o s B + cosAsinB
By putting B = A, we get
sin2A = sinAcosA + cosAsinA
= 2sinAcosA
∴ sin 2A can also be written as
sin2A =
2sinAcosA
cos 2 A + sin 2 A
( Q 1 = cos 2 A + sin 2 A )
Dividing numerator and denomunator by cos 2 A , we get
112
MATHEMATICS
Trigonometric Functions-II
MODULE - IV
Functions
sin A c o s A 
2 

2 tan A
cos 2 A 

sin2A =
=
2
2
cos A sin A
1 + tan 2 A
+
cos 2 A cos2 A
(b) To express cos 2A in terms of sin A, cos A and tan A.
We know that
Notes
cos ( A + B ) = cosAcosB − s i n A s i n B
Putting B = A, we have
cos2A = cosAcosA − sin A s i n A
or
cos2A = cos 2 A − sin 2 A
Also cos2A = cos 2 A − ( 1 − cos 2 A )
= cos2 A − 1 + cos 2 A
i.e,
cos2A = 2cos 2 A − 1
⇒
cos 2 A =
1 + cos2A
2
⇒
sin 2 A =
1 − cos2A
2
Also cos2A = cos 2 A − sin 2 A
= 1 − sin 2 A − sin 2 A
i.e.,
cos2A = 1 − 2sin 2 A
∴
cos2A =
cos 2 A − sin 2 A
cos 2 A + sin 2 A
Dividing the numerator and denominator of R.H.S. by cos 2 A , we have
cos2A =
1 − tan 2 A
1 + tan 2 A
(c) To express tan 2A in terms of tan A.
tan2A = tan ( A + A ) =
=
tanA + tanA
1 − tanAtanA
2tanA
1 − tan 2 A
Thus we have derived the following formulae :
sin2A = 2 s i n A c o s A =
2tanA
1 + tan 2 A
cos2A = cos A −sin A =2cos A −1 =1 2sin
−
A
2
MATHEMATICS
2
2
2
1 − tan 2 A
=
1 + tan 2 A
113
Trigonometric Functions-II
MODULE - IV
Functions
tan2A =
2tanA
1 − tan 2 A
1 + cos2A
1 − cos2A
, sin 2 A =
2
2
π
Example 17.9 If A = , verify the following :
6
2tanA
(i) sin2A = 2 s i n A c o s A =
1 + tan 2 A
cos 2 A =
Notes
(ii) cos2A = cos 2 A − sin2A = 2cos 2 A − 1 = 1 − 2sin 2 A =
(iii) tan2A =
1 − tan 2 A
1 + tan 2 A
2tanA
1 − tan 2 A
Solution :
sin2A = sin
(i)
π
3
=
3
2
π
π
1
3
3
cos = 2 × ×
=
6
6
2
2
2
2 s i n A c o s A = 2sin
2× 1 


2tanA

3 = 2 ×3 = 3
=
=
1
1 + tan 2 A 1 + tan 2 π
2
3 4
1+
6
3
Thus, it is verified that
2tan
π
6
sin2A = 2 s i n A c o s A =
cos2A = cos
(ii)
2tanA
1 + tan 2 A
π 1
=
3 2
2
cos 2
A−
sin 2
A =
cos 2
2
π
π  3
1 = 3− 1= 1
− sin 2 = 
−

 
6
6  2 
4 4 2
2
2
2cos 2
A −1=
2cos 2
= 2×
 3
π
−1 = 2 × 
 −1
6
 2 
3
1
− 1=
4
2
1 − 2sin 2 A = 1 − 2sin
114
π
1 2
1 1
= 1 − 2 ×   = 1 − 2 × =
6
4 2
2
MATHEMATICS
Trigonometric Functions-II
π 1− 

1−
A

6 =
=
2
π
1 + tan A 1 + tan 2

6 1 + 
tan 2
1 − tan 2
1
3
1
3
2

1
1−

 =
3 = 2× 3= 1
2
1
3 4 2

1+

3

Thus, it is verified that
Notes
cos2A = cos 2 A − sin 2 A = 2cos 2 A − 1 = 1 − 2sin 2 A =
tan 2 A = tan
(iii)
π
=
3
π
6
Thus, it is verified that tan2A =
Example 17.10 Prove that
1−
A
2
1 + tan A
tan 2
3
1
2tanA
3 = 2 ×3 =
=
=
1
1 − tan 2 A 1 − tan 2 π
3 2
1−
6
3
2tan
Solution :
MODULE - IV
Functions
2×
3.
2tanA
1 − tan 2 A
sin2A
= tanA
1 + cos2A
sin2A
2sinAcosA
=
1 + cos2A
2cos 2 A
sinA
= tan A
cosA
Example 17.11 Prove that cot A − tan A = 2cot2A.
=
Solution :
cotA − tan A =
=
=
1 − tan 2 A
tan A
2 ( 1 − tan 2 A )
2tanA
=
2
2
t
a
nA 



 1 − tan 2 A 
=
2
= 2cot2A.
tan2A
Example 17.12 Evaluate cos 2
MATHEMATICS
1
− tan A
tanA
π
3π
+ cos 2
.
8
8
115
Trigonometric Functions-II
MODULE - IV
Functions
3π
Solution : cos 2 π + cos 2
=
8
8
=
1 + cos
2
=
Example 17.13 Prove that
Solution : R.H.S. = tan 

1
1+
2
Notes
π
3π
1 + cos
4 +
4
2
2 +
1
1−
2
2
( 2 + 1 ) + ( 2 − 1)
2 2
=1
cosA
π A
= tan  +  .
1 − sinA
4 2
π
A
+
tan
π A
4
2
+ =
π
4 2  1 − tan tan A
4
2
tan
A
2
1+
A
A
A
cos
cos + sin
2 =
2
2
=
A
A
A
sin
cos − sin
2
2
2
1−
A
cos
2
sin
 cos A + sin A

2
2
= 
 cos A

2

  cos A − sin A 


2
2 

A 2
− sin 
2 
 cosA − s i n A  

[Multiplying Numerator and Denominator by 
 2
2  
A
A
− sin 2
2
2
=
A
A
A
A
cos 2 + sin 2 − 2cos sin
2
2
2
2
cos 2
=
cosA
= L.H.S .
1 − sinA
Example 17.14 Prove that
( cos α − cos β )2 + ( sin α − sin β )2 = 4sin 2
116
α−β
2
MATHEMATICS
Trigonometric Functions-II
Solution : ( cos α − cos β )2 + ( sin α − sin β )2
= cos 2 α + cos 2 β − 2cos α cos β + sin 2 α + sin 2 β − 2sin α sin β
MODULE - IV
Functions
= 2 − 2 ( cos α cos β + in α sin β )
= 2 {1 − cos ( α − β )}
= 2×
2sin 2
Notes
α−β
α−β
= 4sin 2
2
2
CHECK YOUR PROGRESS 17.4
1.
If A =
π
, verify that
3
(a) sin2A = 2 s i n A c o s A =
2tanA
1 + tan 2 A
−
A
(b) cos2A = cos A −sin A =2cos A −1 =1 2sin
2
2.
2
2
Find the value of sin 2A when (assuming 0 < A <
3
12
(b) s i n A =
5
13
Find the value of cos 2A when
(a) c o s A =
3.
2
1 − tan 2 A
=
1 + tan 2 A
π
)
2
(c) tan A =
16
.
63
15
4
5
(b) s i n A = (c) tan A =
17
5
12
Find the value of tan 2A when
(a) c o s A =
4.
(a) tan A =
5.
6.
3
4
(b) tan A =
a
b
π
3π
+ sin 2
.
8
8
Prove the following :
Evaluate sin 2
(a)
1 + sin2A
π

= tan 2  + A 
1 − sin2A
4


(b)
cot 2 A + 1
= sec2A
cos2 A − 1
7.
(a) Prove that
sin2A
= cosA (b) Prove that tan A + cotA = 2cosec2A .
1 − cos2A
8.
(a) Prove that
cosA
π A
= tan  − 
1 + sinA
4 2 
2
2
(b) Prove that ( cos α + cos β ) + ( sin α − sinβ ) = 4cos 2
MATHEMATICS
α−β
2
117
Trigonometric Functions-II
MODULE - IV 17.3.1 Trigonometric Functions of 3A in Terms of those of A
Functions (a)
sin 3A in terms of sin A
Substituting 2A for B in the formula
sin ( A + B ) = sin A c o s B + cosAsinB , we get
sin ( A + 2A ) = sinAcos2A + cosAsin2A
Notes
= sinA ( 1 − 2sin 2 A ) + ( cosA × 2 s i n A c o s A )
= sinA − 2sin 3 A + 2sinA ( 1 − sin 2 A )
= sinA − 2sin 3 A + 2sinA − 2sin3 A
∴
(b)
sin3A = 3 s i n A − 4sin3 A
....(1)
cos 3A in terms of cos A
Substituting 2A for B in the formula
cos ( A + B ) = cosAcosB − s i n A s i n B, we get
cos ( A + 2A ) = cosAcos2A − sinAsin2A
= cosA ( 2cos 2 A − 1 ) − ( sin A ) × 2sinAcosA
= 2cos3 A − cosA − 2cosA ( 1 − cos 2 A )
= 2cos3 A − cosA −2cosA +2cos3 A
∴
(c)
cos3A = 4cos3 A − 3 c o s A
....(2)
tan 3A in terms of tan A
Putting B = 2A in the formula
tan ( A + B ) =
tan ( A + 2A ) =
tanA + tan B
, we get
1 − tanAtanB
tanA + tan2A
1 − tanAtan2A
2tanA
1 − tan 2 A
=
2tanA
1 − tan A ×
1 − tan 2 A
tanA +
tan A − tan3 A + 2 t a n A
1 − tan 2 A
=
1 − tan 2 A − 2tan 2 A
1 − tan 2 A
118
MATHEMATICS
Trigonometric Functions-II
3tanA − tan 3 A
=
1 − 3tan 2 A
∴
(d)
Formulae for sin 3 A and cos3 A
Q
sin3A = 3 s i n A − 4sin3 A
∴
4sin3 A = 3 s i n A − sin3A
sin3 A =
or
Similarly,
∴
MODULE - IV
....(3) Functions
Notes
3sinA − sin 3A
4
cos3A = 4cos3 A − 3 c o s A
3cosA + cos3A = 4cos 3 A
3cosA + cos3A
4
cos 3 A =
or
Thus, we have derived the following formulae :
sin3A = 3 s i n A − 4sin3 A
cos3A = 4cos3 A − 3 c o s A
Example 17.15 If
tan3A =
3tanA − tan 3 A
1 − 3tan 2 A
sin3 A =
3sinA − sin 3A
4
cos 3 A =
3cosA + cos3A
4
A =
π
, verify that
4
(i) sin3A = 3 s i n A − 4sin3 A
(iii) tan3A =
(ii) cos3A = 4cos3 A − 3 c o s A
3tanA − tan 3 A
1 − 3tan 2 A
Solution :
(i)
sin3A = sin
3π
1
=
4
2
3sinA − 4sin3 A =3sin
= 3×
MATHEMATICS
π
π
−4sin3
4
4
1 3
− 4 × 

2
 2
1
119
Trigonometric Functions-II
MODULE - IV
Functions
=
3
4
1
−
=
2 2 2
2
Thus, it is verified that sin3A = 3 s i n A − 4sin3 A
cos3A = cos
(ii)
Notes
3π
1
=−
4
2
1 3
1
4cos 3 A − 3cosA = 4 × 
 − 3×
 2
2
=
4
2 2
−
3
1
=−
2
2
Thus, it is verified that cos3A = 4cos3 A − 3 c o s A
tan3A = tan
(iii)
3π
= −1 ,
4
3tanA − tan 3 A 3 × 1 − 13
2
=
=
= −1
2
2
1 − 3tan A
1−3×1
−2
Thus, it is verified that tan3A =
Example 17.16 If cos θ =
Solution :
3tanA − tan 3 A
1 − 3tan 2 A
1
1
1
1 
 a +  , prove that cos3θ =  a 3 + 3 
2
a
2
a 
cos3θ = 4cos3 θ − 3cos θ
1
1 3
1
1
= 4   a +   − 3 ×  a + 
a 
2
a
 2
1
1
1
1  3a 3
= 4 ×  a3 +3a 2 ⋅ + 3a⋅ 2+ 3 −
−
8
a
a
a  2 2a
=
a3
1
1
1 
+ 3 =  a3 + 3 
2 2a
2
a 
Example 17.17 Prove that
π

π
 1
sin α sin  + α  sin  − α  = s i n 3α
3

3
 4
π

π

Solution : sin α sin  + α  sin  − α 
3

3

=
120
1
2π 

sin α  cos2α − cos
2
3 

MATHEMATICS
Trigonometric Functions-II
1

π 

= sin α 1 −2sin 2 α −1 2sin
− 2 
2
3 


=2
MODULE - IV
Functions
1
π


sin α  sin 2 − sin 2 α 
2
3


3
 3sin α − 4sin 3 α 1
= sin α  − sin 2 α  =
= s i n 3α
4
4
4

Notes
Example 17.18 Prove that
cos 3 Asin3A + sin 3 Acos3A =
3
sin4A
4
Solution : cos 3 Asin3A + sin 3 Acos3A
= cos3 A ( 3sinA − 4sin 3 A ) + sin3 A ( 4cos3 A − 3 c o s A )
= 3sinAcos3 A − 4sin 3 Acos3 A + 4sin 3 Acos 3 A − 3sin3 A c o s A
= 3sinAcos3 A − 3sin 3 A c o s A
= 3sinAcosA ( cos 2 A − sin 2 A ) = ( 3sinAcosA ) cos2A
=
3sin2A
× cos2A
2
=
3 sin4A 3
= sin4A .
2 2
4
Example 17.19 Prove that cos 3
Solution : L.H.S. =
π
π
3
π
π 
+ sin 3
=  cos + sin 
9
18 4 
9
18 
1
π
π 1 
π
π
3cos + cos  +  3sin
−sin 

4
9
3 4
18
6
=
3
4
π
π  11 1

 cos 9 + sin 18  + 4  2 − 2 
=
3
4
π
π

 cos 9 + sin 18  = R.H.S.
CHECK YOUR PROGRESS 17.5
1.
If A =
π
, verify that
3
(a) sin3A = 3 s i n A − 4sin3 A
MATHEMATICS
(b) cos3A = 4cos3 A − 3 c o s A
121
Trigonometric Functions-II
MODULE - IV
Functions
(c) tan3A =
3tanA − tan 3 A
1 − 3tan 2 A
2
3
(b) sinA =
p
.
q
(b) cosA =
c
.
d
2.
Find the value of sin 3A when (a) s i n A =
3.
Find the value of cos 3A when (a) c o s A = −
4.
π
π
1
Prove that cos α cos  − α  cos  + α  = cos3α.
3

3
 4
5.
(a) Prove that sin3
Notes
(b) Prove that
6.
1
3
2π
π 3
2π
π
− sin 3 =  sin
− sin 
9
9 4
9
9
sin3A cos3A
is constant.
−
sin A
cosA
(a) Prove that cot3A =
cot 3 A − 3 c o t A
3cot 2 A − 1
(b) Prove that
cos10A + cos8A + 3cos4A + 3cos2A = 8cosAcos3 3A
17.4 TRIGONOMETRIC FUNCTIONS OF
SUBMULTIPLES OF ANGLES
A A A
, ,
are called submultiples of A.
2 3 4
It has been proved that
sin 2 A =
Replacing A by
sin
1 − cos2A
1 − cos2A
1 + cos2A
, cos 2 A =
, tan 2 A =
2
1 + cos2A
2
A
A
, we easily get the following formulae for the sub-multiple
:
2
2
A
1 − cosA
A
1 + cosA
=±
, cos = ±
and
2
2
2
2
tan
A
1 − cosA
=±
2
1 + cosA
We will choose either the positive or the negative sign depending on whether corresponding
A
value of the function is positive or negative for the value of
. This will be clear from the
2
following examples
122
MATHEMATICS
Trigonometric Functions-II
Example 17.20 Find the values of cos
Solution : We use the formulae cos
cos
MODULE - IV
Functions
π
π
and cos
.
12
24
A
1 + cosA
and take the positive sign, because
=±
2
2
π
π
and cos
are both positive.
12
24
Notes
π
1 + cos
π
6
cos
=±
12
2
=
2
=
2+ 3
2×2
=
4+2 3
8
=
=
cos
3
2
1+
π
=
24
=
(
3 +1
)
2
8
(

Q 4 + 2 3 =1 +3 +2 3 = 1 + 3

)
2


3 +1
2 2
1 + cos
π
12
2

3 +1
1+

2 2 

2
=
2 2 + 3 +1
4 2
=
4+ 6+ 2
8
π
π
Example 17.21 Find the values of sin  −  and cos  −  .
 8
 8
Solution : We use the formula sin
A
1 − cosA
=±
2
2
π
and take the lower sign, i.e., negative sign, because sin  −  is negative.
 8
MATHEMATICS
123
Trigonometric Functions-II
MODULE - IV
Functions
π
1 − cos  
π
4
sin  −  = −
2
 8
=−
Notes
2 −1
2− 2
=−
2
2 2
π
1 + cos  − 
π
 4
−  = ±
8
2
=
A
2
1
1+
2
2
=
2 +1
2 2
=
2+ 2
4
=
2+ 2
2
Example 17.22 If c o s A =
(i) sin
2
2
=−
Similarly, cos 

1
1−
(ii) cos
7
3π
and
< A < 2 π , find the values of
25
2
A
2
(iii) tan
A
2
Solution : Q A lies in the 4th-quardrant, 3π < A < 2 π
2
⇒
124
3
π A
<
<π
4
2
A
A
A
> 0 , cos
< 0 , tan < 0.
2
2
2
∴
sin
∴
A
sin =
2
1 − cosA
=
2
1−
2
7
25
=
18
=
50
9
3
=
25 5
MATHEMATICS
Trigonometric Functions-II
and
1+
cos
A
1 + cosA
=−
=−
2
2
tan
1−
A
1 − cosA
=−
=−
2
1 + cosA
1+
7
25
=−
32
16
4
=−
=−
50
25
5
=−
18
9
3
=−
=−
32
16
4
2
7
25
7
25
MODULE - IV
Functions
Notes
CHECK YOUR PROGRESS 17.6
1.
If A =
π
, verify that
3
(a) sin
A
1 − cosA
=
2
2
(c) tan
A
=
2
A
1 + cosA
=
2
2
1 − cosA
1 + cosA
π
π
.
and sin
24
12
2.
Find the values of sin
3.
Determine the values of
π
8
(a) sin
(b) cos
(b) cos
π
8
(c)
tan
π
.
8
Example 17.23 Prove that following :
(a) sin
π
=
10
5 −1
π
and cos
=
4
10
π
=
5
5 +1
π
and sin =
4
5
(b) cos
Solution : (a) Let A =
2A =
∴
∴
10 + 2 5
4
10 − 2 5
4
π
π
⇒ 5A =
10
2
π
− 3A
2
π

sin2A = sin  − 3A  = cos3A
2

∴ 2 s i n A c o s A = 4cos 3 A − 3 c o s A
or
cosA  2sinA − 4cos 2 A + 3  = 0
MATHEMATICS
.....(1)
125
Trigonometric Functions-II
MODULE - IV As
Functions
cosA ≠ 0 and 1 − sin 2 A = cos 2 A
∴ (1) becomes 2sinA − 4 ( 1 − sin 2 A ) + 3 = 0
4sin 2 A + 2 s i n A − 1 = 0
Notes
−2 ±
4 + 16
−1 ± 5
=
8
4
⇒
sinA =
∴
sinA =
5 −1
4
⇒
sin
π
=
10
5 −1
4
2
Now
(b) Let
 5 −1 
π
π
cos
= 1 − sin 2
= 1− 
 =
10
10
 4 
10 + 2 5
4
π
π
, 2A =
10
5
A =
cos2A = 1 − 2sin 2 A
2
∴
 5 −1 
6 −2 5  2 +2 5
π
cos = 1 − 2 
 = 1− 2
=
5
8
 4 
 16

=
Now
sin
5 +1
4
π
π
= 1 − cos 2 =
5
5
10 − 2 5
4
Example 17.24 Prove the following :
tan α + 2tan2α + 4tan4α + 8cot8α = cot α
Solution : We have to prove that
tan α − cot α +2tan2 α +4tan4 α +8cot8 α =0
126
or
 sin α cos α 
−

 + 2 tan 2α + 4 t a n 4α + 8cot8α = 0
 cos α sin α 
or
−
or
−2
2 c o s 2α
+ 2 t a n 2 α +4 tan 4 α +8cot8 α =0
2sin α cos α
c o s 2α
+ 2 t a n 2α + 4 t a n 4α + 8cot8α = 0
s i n 2α
MATHEMATICS
Trigonometric Functions-II
or
−2cot2α + 2tan2α + 4tan4 α + 8cot8α = 0
Combining
−2cot2α + 2 t a n 2α
(1) becomes
−4cot4α + 4tan4α + 8cot8α = 0
Combining
−4cot4α + 4 t a n 4α ; L.H.S. of (2) becomes
......(1) MODULE - IV
Functions
......(2)
Notes
−8cot8α + 8cot8α = 0 = R.H.S. of (2)
17.5 TRIGONOMETRIC EQUATIONS
You are familiar with the equations like simple linear equations, quadratic equations in algebra.
You have also learnt how to solve the same.
Thus, (i) x − 3 = 0 gives one value of x as a solution.
(ii) x 2 − 9 = 0 gives two values of x.
You must have noticed, the number of values depends upon the degree of the equation.
Now we need to consider as to what will happen in case x's and y's are replaced by trigonometric
functions.
Thus solution of the equation sin θ − 1 = 0 , will give
sin θ = 1 and θ =
π 5 π 9π
,
,
,....
2 2 2
Clearly, the solution of simple equations with only finite number of values does not necessarily
hold good in case of trigonometric equations.
So, we will try to find the ways of finding solutions of such equations.
17.5.1 To find the general solution of the equation sin θ = 0
It is given that sin θ = 0
But we know that sin 0, sin π, sin2 π,....,sinnπ are equal to 0
∴
θ = nπ , n ∈ N
But we know sin ( −θ ) = − sin θ = 0
∴
sin ( −π ) , sin ( −2 π ) , sin ( −3 π ) ,....,sin ( − n π) = 0
∴
θ = nπ ,
n ∈ I.
Thus, the general solution of equations of the type sin θ = 0 is given by θ = nπ where n is an
integer.
17.5.2 To find the general solution of the equation cos θ = 0
It is given that cos θ = 0
π
But in practice we know that cos = 0 . Therefore, the first value of θ is
2
π
θ=
2
MATHEMATICS
.....(1)
127
Trigonometric Functions-II
MODULE - IV
π
π

Functions We know that cos ( π + θ) = − cos θ or cos  π + 2  = − cos 2 =0.
cos
or
3π
=0
2
In the same way, it can be found that
Notes
cos
∴
5π cos7π
9π
π
,
, cos
,....., cos ( 2n + 1 ) are all zero.
2
2
2
2
θ = ( 2n + 1 )
π
,n∈N
2
But we know that cos ( −θ ) = cos θ
 π
 3π 
 5π 
∴ cos  −  =cos  −  =cos  − 
 2
 2
 2 
∴
θ = ( 2n + 1 )

....
= cos
= 

π
0=
2 

− 1) + 
( 2n
π
, n∈I
2
Therefore, θ = ( 2n + 1 ) π is the solution of equations cos θ = 0 for all numbers whose
2
cosine is 0.
17.5.3 To find a general solution of the equation tan θ = 0
It is given that tan θ = 0
or
i.e.
sin θ
=0
cos θ
or
sin θ = 0
θ = nπ, n ∈ I.
We have consider above the general solution of trigonometric equations, where the right hand
is zero. In the following, we take up cases where right hand side is non-zero.
17.5.4 To find the general solution of the equation sin θ = sin α
It is given that sin θ = sin α
⇒
128
sin θ − sin α = 0
or
θ+α
θ−α 
2cos 
 sin 
=0
 2 
 2 
∴ Either
θ−α
θ+α
cos 
 = 0 or sin  2  = 0


 2 
MATHEMATICS
Trigonometric Functions-II
⇒
θ+α
π
= ( 2p + 1 )
or
2
2
⇒
θ = ( 2p + 1 ) π − α
or
MODULE - IV
Functions
θ−α
= qπ, p, q ∈ I
2
θ = 2qπ + α
....(1)
From (1), we get
n
θ = nπ + ( −1 ) α, n ∈ I as the geeneral solution of the equation sin θ = sin α
Notes
17.5.5 To find the general solution of the equation cos θ = cos α
It is given that, cos θ = cos α
⇒
cos θ − cos α = 0
−2sin
⇒
∴ Either,
sin
θ+α
θ−α
sin
=0
2
2
θ+α
=0
2
or
θ+α
= pπ
2
θ = 2pπ − α
⇒
⇒
sin
or
or
θ−α
=0
2
θ−α
= qπ, p,q ∈ I
2
θ = 2pπ + α
....(1)
From (1), we have
θ = 2n π ± α, n ∈I as the general solution of the equation cos θ = cos α
17.5.6 To find the general solution of the equation tan θ = tan α
It is given that, tan θ = tan α
⇒
sin θ sin α
−
=0
cos θ cos α
⇒
sin θ cos α − sin α cos θ = 0
⇒
sin ( θ − α ) = 0
⇒
⇒
Similarly, for
θ − α = nπ, n ∈ I
θ = nπ + α n ∈ I
cosec θ = cosec α , the general solution is
θ = nπ + ( − 1 ) α
n
and, for
sec θ = sec α , the general solution is
θ = 2nπ ± α
and for
cot θ = cot α
θ = nπ + α is its general solution
MATHEMATICS
129
Trigonometric Functions-II
MODULE - IV If
Functions
sin 2 θ = sin 2 α , then
1 − cos2θ 1 − cos2α
=
2
2
⇒
Notes
cos2θ = cos2α
2θ = 2nπ ± 2α , n ∈ I
⇒
⇒
θ = nπ ± α
Similarly, if cos 2 θ = cos 2 α , then
α = n π ± α, n ∈ I
Again, if tan 2 θ = tan 2 α , then
1 − tan 2 θ 1 − tan 2 α
=
1 + tan 2 θ 1 + tan 2 α
⇒
cos2θ = cos2α
⇒
2θ = 2nπ ± 2α
θ = n π ± α, n ∈ I is the general solution.
⇒
Example 17.25 Find the general solution of the following equations :
(a)
(i) sin θ =
(b)
(i) cos θ =
1
2
3
2
cot θ = − 3
Solution :
(c)
(a)
(i) sin θ =
∴
(b)
130
3
2
(ii) cos θ = −
1
2
(d)
n
π
, n ∈I
6
− 3
π
π
4π
= − sin = sin  π +  = sin
2
3
3
3

θ = nπ + ( −1 )
(i) cos θ =
4sin 2 θ = 1
1
π
= sin
2
6
θ = nπ + ( −1 )
(ii) sin θ =
∴
(ii) sin θ = −
n
4π
, n ∈I
3
3
π
= cos
2
6
MATHEMATICS
Trigonometric Functions-II
∴
(ii) cos θ = −
1
π
π
2π

= − cos = cos  π −  = cos
2
3
3
3

2π
, n∈I
3
θ = 2nπ ±
∴
Notes
cot θ = − 3
(c)
tan θ = −
1
3
= − tan
θ = nπ +
∴
(d)
MODULE - IV
Functions
π
θ = 2nπ ± , n ∈ I
6
4sin 2 θ = 1
π
π
5π

= tan  π −  = tan
6
6
6

5π
, n∈I
6
⇒
sin 2 θ =
1  1 2
π
=   = sin 2
4 2
6
π
sin θ = sin  ± 
 6
⇒
θ = nπ ±
π
, n∈I
6
Example 17.26 Solve the following :
(a)
2cos 2 θ + 3sin θ = 0
cos3x = sin2x
Solution :
(c)
(a)
2cos 2 θ + 3sin θ = 0
⇒
2 ( 1 − sin 2 θ ) + 3sin θ = 0
⇒
2sin 2 θ − 3sin θ − 2 = 0
⇒
( 2sin θ + 1 )( sin θ − 2 ) = 0
⇒
sin θ = −
1
2
(b)
cos4x = cos2x
(d)
sin2x + sin 4 x + sin6x = 0
sin θ = 2
or
Since sin θ = 2 is not possible.
∴
∴
sin θ = − sin
π
π
7π

= sin  π +  = sin
6
6
6

θ = nπ + ( −1 ) ⋅
MATHEMATICS
n
7π
,n∈I
6
131
Trigonometric Functions-II
MODULE - IV (b)
Functions i.e.,
Notes
cos4x = cos2x
cos4x − cos2x = 0
⇒
−2sin3xsinx = 0
⇒
sin3x = 0
⇒
3x = nπ
⇒
x=
nπ
3
or
sinx = 0
or
x = nπ
or
x = nπ
(c)
cos3x = sin2x
⇒
π

cos3x = cos  − 2x 
2

π

3x = 2nπ ±  − 2x 
2

⇒
n∈I
n∈I
Taking positive sign only, we have
3x = 2nπ +
π
− 2x
2
⇒
5x = 2nπ +
π
2
⇒
x=
2nπ π
+
5
10
Now taking negative sign, we have
3x = 2nπ −
x = 2nπ −
⇒
( sin6x + sin2x ) + sin4x = 0
or
or
2sin4xcos2x + sin 4 x = 0
or
sin4x [ 2cos2x + 1 ] = 0
∴
⇒
π
n∈I
2
sin2x + sin 4 x + sin6x = 0
(d)
4x = nπ
x=
132
π
+ 2x
2
nπ
4
sin4x = 0
or 2x = 2nπ ±
or
or cos2x = −
1
2π
= cos
2
3
2π
,n∈I
3
x = nπ ±
π
3
n∈I
MATHEMATICS
Trigonometric Functions-II
MODULE - IV
Functions
CHECK YOUR PROGRESS 17.7
1. Find the most general value of θ satisfying :
3
2
(i)
sin θ =
(ii)
sin θ = −
(ii) cosec θ =
3
2
(ii)
2
Notes
sin θ = −
1
2
2
3
2. Find the most general value of θ satisfying :
1
2
(ii)
sec θ = −
3
2
(iv)
sec θ = − 2
(i)
cos θ = −
(iii)
cos θ =
3. Find the most general value of θ satisfying :
tan θ = −1
(i)
tan θ =
3
(iii)
cot θ = −1
(ii)
c o s 2θ =
1
2
(iii)
tan3θ =
1
3
(v)
sin 2 θ =
3
4
(vi)
sin 2 2θ =
1
4
(viii)
cos 2 2θ =
(ii)
4. Find the most general value of θ satisfying :
1
2
(i)
s i n 2θ =
(iv)
cos3θ = −
(vii)
4cos 2 θ = 1
3
2
3
4
5. Find the general solution of the following :
(i)
2sin 2 θ + 3 cos θ + 1 = 0 (ii)
(iii)
cot θ + tan θ = 2 cosec θ
4cos 2 θ− 4sin θ =1
LET US SUM UP
l
sin ( A ± B ) = sin A c o s B ± cosAsinB ,
cos ( A ± B ) = cosAcosB m s i n A s i n B
tan ( A + B ) =
MATHEMATICS
tanA + tan B
tanA − tan B
, tan ( A − B ) =
1 − tanAtanB
1 + tanAtanB
133
Trigonometric Functions-II
MODULE - IV
Functions
cot ( A + B ) =
l
cotAcotB − 1
cotAcotB + 1
, cot ( A − B ) =
cotB + c o t A
cotB − cot A
2 s i n A c o s B = sin ( A + B ) + sin ( A − B )
2 c o s A s i n B = sin ( A + B ) − sin ( A − B )
2cosAcosB = cos ( A + B ) − cos ( A − B )
Notes
2 s i n A s i n B = cos ( A − B ) − cos ( A + B )
l
sinC + sin D = 2sin
C+D
C−D
cos
2
2
sinC − sin D = 2cos
C+D
C−D
sin
2
2
cosC + cosD = 2cos
cosC − cosD =2sin
l
C+D
C−D
cos
2
2
C+ D
D −C
sin
2
2
sin ( A + B ) ⋅ sin ( A− B ) = sin 2 A − sin 2 B
cos ( A + B ) ⋅ cos ( A− B) = cos 2 A − sin 2 B
134
2tanA
1 + tan 2 A
l
sin2A = 2 s i n A c o s A =
l
cos2A = cos 2 A −sin 2 A =2cos 2 A −1 =1 2sin
− 2A
l
tan2A =
2tanA
1 − tan 2 A
l
sin 2 A =
1 − cos2A
1 + cos2A
1 − cos2A
, cos 2 A =
, tan 2 A =
2
2
1 + cos2A
l
sin3A = 3 s i n A − 4sin3 A , cos3A = 4cos3 A − 3 c o s A
1 − tan 2 A
=
1 + tan 2 A
tan3A =
3tanA − tan 3 A
1 − 3tan 2 A
l
sin3 A =
3sinA − sin 3A
3cosA + cos3A
, cos 3 A =
4
4
l
sin
A
1 − cosA
A
1 + cosA
=±
, cos = ±
2
2
2
2
tan
A
1 − cosA
=±
2
1 + cosA
MATHEMATICS
Trigonometric Functions-II
MODULE - IV
Functions
l
π
sin
=
10
5 −1
π
, cos =
4
5
5 +1
4
l
sin θ = 0
⇒
θ = nπ ,
cos θ = 0
⇒
θ = ( 2n + 1 )
tan θ = 0
⇒
θ = nπ ,
l
sin θ = sin α
⇒
θ = nπ + ( −1) α,
n
n∈I
l
cos θ = cos α
⇒
θ = 2nπ ± α,
n∈I
l
tan θ = tan α
⇒
θ = nπ + α ,
n∈I
l
sin 2 θ = sin 2 α
⇒
θ = nπ ± α ,
n∈I
l
cos 2 θ = cos 2 α
⇒
θ = nπ ± α ,
n∈I
l
tan 2 θ = tan 2 α
⇒
θ = nπ ± α ,
n∈I
n∈I
π
,
2
n∈I
Notes
n∈I
NS
SUPPORTIVE WEB SITES
l
l
http://www.wikipedia.org
http://mathworld.wolfram.com
TERMINAL EXERCISE
1.
cos 2 B − cos 2 A
Prove that tan ( A + B ) × tan ( A − B ) = 2
cos B − sin 2 A
2.
π
Prove that cos θ − 3sin θ = 2cos  θ + 
3

3.
If A + B =
π
4
Prove that ( 1 + tanA ) (1 + tan B ) = 2 and ( cot A − 1 ) ( cosB − 1 ) = 2
4.
Prove each of the following :
(i)
sin ( A − B ) sin ( B − C ) sin ( C − A )
+
+
=0
cosAcosB
cosBcosC
cosCcosA
MATHEMATICS
135
Trigonometric Functions-II
MODULE - IV
Functions
 π

π

 2π

 2π

(ii) cos  − A  ⋅cos  + A  + cos  − A  ⋅cos  + A  = cos2A
 10

 10

 5

 5

(iii) cos
2π
4π
9π
1
⋅ cos
⋅ cos
=−
9
9
9
8
(iv) cos
13π
17 π
43 π
+ cos
+ cos
=0
45
45
45
π
π
1
(v) tan  A +  + cot  A −  =
6
6  sin2A − sin π


3
Notes
(vi)
sin θ + s i n 2θ
cos θ + sin θ
= tan θ (vii)
= tan 2θ + sec2θ
1 + cos θ + c o s 2θ
cos θ − sin θ
2
 1 − sin θ
 π θ
= tan 2  − 
(viii) 

 1 + sin θ 
4 2
π
π
3
(ix) cos 2 A + cos 2  A +  + cos 2  A −  =
3
3 2


(x)
sec8A − 1 tan8A
=
sec 4 A − 1 tan2A
(xii) sin
5.
(xi) cos
π
7π
11 π
13π 11
cos
cos
cos
=
30
30
30
30
16
π
13π
1
+ sin
=−
10
10
2
Find the general value of ' θ ' satisfying
(a) sin θ =
1
2
(b) sin θ =
3
2
1
(d) cosec θ = 2
2
Find the general value of ' θ ' satisfying
(c) sin θ = −
6.
(a) cos θ =
1
2
(b) sec θ =
2
3
− 3
(d) sec θ = −2
2
Find the most general value of ' θ ' satisfying
(c) cos θ =
7.
(a) tan θ = 1
8.
(c) cot θ = −
1
3
Find the general value of ' θ ' satisfying
(a) sin 2 θ =
136
(b) tan θ = −1
1
2
(b) 4cos 2 θ = 1
(c) 2cot 2 θ = cosec2 θ
MATHEMATICS
Trigonometric Functions-II
9.
Solve the following for θ :
(a) cospθ = cos qθ
(b) sin9θ = sin θ
MODULE - IV
Functions
(c) tan5 θ = cot θ
10.
Solve the following for θ :
Notes
(a) sinmθ + sin nθ = 0
(b) tan mθ + cotnθ = 0
(c) cos θ + cos2θ + cos3θ = 0
(d) sin θ + sin2θ + sin3θ + sin4θ = 0
MATHEMATICS
137
Trigonometric Functions-II
MODULE - IV
Functions
ANSWERS
CHECK YOUR PROGRESS 17.1
Notes
1.
(a)
2.
(a)
3 −1
2 2
(i)
(ii)
3
2
(c)
21
221
3 −1
2 2
CHECK YOUR PROGRESS 17.2
1.
(i)
cos 2 A − sin 2 A
2
(ii) −
1
4
(c) −
2.
( 3 + 1)
2 2
CHECK YOUR PROGRESS 17.3
1.
(a) sin5θ − sin θ ;
(c) cos
2.
(b) cos2θ − cos6θ
π
π
+ cos
3
6
(d) sin
π
π
+ sin
2
6
(a) 2sin5θ cos θ
(b) 2cos5θ ⋅ s i n 2θ
(c) 2sin3θ ⋅ sin θ
(d) 2cos6θ ⋅ cos θ
CHECK YOUR PROGRESS 17.4
2.
(a)
24
25
(b)
120
169
(c)
3.
(a)
161
289
(b)
−7
25
(c)
4.
(a)
24
7
(b)
2ab
b2 − a 2
5.
2016
4225
119
169
1
CHECK YOUR PROGRESS 17.5
2.
(
3pq 2 − 4p3
22
(a)
(b)
27
q3
)
3.
(a)
23
4c3 − 3cd 2
(b)
27
d3
CHECK YOUR PROGRESS 17.6
2.
138
(a)
3 −1
,
2 2
(4−
2− 6
)
2 2
MATHEMATICS
Trigonometric Functions-II
3.
(a)
2− 2
2
(b)
2+ 2
2
(c)
MODULE - IV
Functions
2 −1
CHECK YOUR PROGRESS 17.7
1.
(i) θ = nπ + ( −1)
n
(iii) θ = nπ + ( −1 )
π
, n ∈I
3
n
2.
3.
(iv) θ = nπ + ( −1 )
n
(ii) θ = 2nπ ±
π
(iii) θ = 2nπ ± , n ∈ I
6
(iv) θ = 2nπ ±
(i) θ = nπ +
(i) θ =
3π
, n∈I
4
π
, n ∈I
4
Notes
5π
, n ∈I
4
5π
, n ∈I
6
3π
, n∈I
4
π
(ii) θ = nπ + , n ∈ I
3
π
, n ∈I
4
nπ
n π
+ ( −1)
, n ∈I
2
12
(iii) θ =
nπ π
+ , n ∈I
3 18
π
3
(v) θ = nπ ± , n ∈ I
(i) θ = 2nπ ±
π
6
(ii) θ = nπ ± , n ∈ I
(iv) θ =
2nπ 5π
± , n∈I
3
18
(vi) θ =
nπ π
± , n ∈I
2 12
π
3
(viii) θ =
5π
, n ∈I
6
(ii) θ = nπ + ( −1)
(vii) θ = nπ ± , n ∈ I
5.
n
2π
,n ∈I
3
(i) θ = 2nπ ±
(iii) θ = nπ −
4.
4π
, n ∈I
3
(ii) θ = nπ + ( −1)
nπ π
± , n ∈I
2 12
n
π
, n ∈I
6
π
3
(iii) θ = 2nπ ± , n ∈ I
TERMINAL EXERCISE
5.
6.
(a) θ = nπ + ( −1)
n
(c) θ = nπ + ( −1)
n
π
, n ∈I
4
(b) θ = nπ + ( −1)
n
5π
, n ∈I
4
(d) θ = nπ + ( −1)
n
π
(a) θ = 2nπ ± , n ∈ I
3
MATHEMATICS
π
, n ∈I
3
π
, n ∈I
4
π
(b) θ = 2nπ ± , n ∈ I
6
139
Trigonometric Functions-II
MODULE - IV
Functions
(c) θ = 2nπ ±
7.
5π
,n∈I
6
(d) θ = 2nπ ±
(a) θ = nπ + π , n ∈ I
4
(b) θ = nπ +
2π
, n ∈I
3
3π
, n ∈I
4
(c) θ = nπ + 2π , n ∈ I
3
Notes
8.
π
(a) θ = nπ ± , n ∈ I
4
π
(b) θ = nπ ± , n ∈ I
3
π
(c) θ = nπ ± , n ∈ I
4
9.
(a) θ =
2nπ
, n ∈I
pm q
(c) θ = ( 2n + 1)
10.
(a) θ =
(d) θ =
140
π
,n∈I
12
( 2k + 1) π
m− n
(c) θ = ( 2n + 1)
nπ
π
or ( 2n + 1) , n ∈ I
4
10
(b) θ =
π
4
or
2kπ
, k ∈I
m+ n
or
2nπ ±
(b) θ =
( 2k + 1) π
,k∈I
2(m − n)
2π
, n ∈I
3
2nπ
π
or θ = nπ ± , n ∈ I or θ = ( 2n − 1) π, n ∈ I
5
2
MATHEMATICS