1.

Find the values of sin 2A, cos 2A and tan 2A, when A lies in the
second quadrant and
1
tan A =
7
 11 6
2. If cos 2A = , show that cos A =
25 52
3. Prove the following identities:
(a) (sin  + cos )2 – (sin  – cos )2 = 2sin 2
(b) cos4  – sin4  = cos 2
(c) 2sin2  + cos2 2 + 2sin2 .cos 2 = 1
sin 3A cos 3A
(d) – =2
sin A cos A
sin3  + cos3  1 sin 2
(e) =1–
sin  + cos  2
cos   1  sin 2
f) tan 
sin   1  sin 2
(g) (2cos  + 1)(2cos  – 1)(2cos 2 – 1) = 2cos 4 + 1
(c) 2sin2  + cos2 2 + 2sin2 .cos 2 = 1
Proof:
LHS = 2sin2  + cos2 2 + 2sin2 .cos 2
= 2sin2  + 2sin2 .cos 2 + cos2 2 Rearrange
= 2sin2  (1 + cos 2) + cos2 2 Take common
= 2sin2 . 2cos2  + cos2 2 Use 1 + cos 2 = 2cos2 

= (2sin . cos )2 + cos2 2 Use 2sin .cos  = sin 2

= (sin 2)2 + cos2 2

Use sin2 A+ cos2 A = 1
= sin2 2 + cos2 2 = 1
RHS, proved.
 sin 3A cos 3A
(d) – =2
sin A cos A
Proof:

LHS = sin 3A cos 3A

–
sin A cos A
sin 3A.cos A – cos3A. sin A Take LCM
=
sin A.cos A
sin (3A – A) Use: sin A. cos B – cos A. sin B =
=
sin A.cos A sin (A – B)
sin 2A 2sin A. cos A
= = =2
sin A.cos A sin A.cos A
RHS, proved.
 sin3  + cos3  1 sin 2
(e) =1–
sin  + cos  2
Proof:
sin3  + cos3 
LHS = Use: a3 + b3 = (a + b)(a2 – ab + b2)
sin  + cos 
(sin  + cos )(sin2  – sin .cos  + cos2 )
=
(sin  + cos )

= sin2  – sin .cos  + cos2 

– 1 .2 sin .cos 
= sin  + cos 
2 2
2
=1– 1 sin 2
2 RHS, proved.
 cos   1  sin 2
f) tan 
sin   1  sin 2

cos   1  sin 2
Proof: LHS =
sin   1  sin 2
cos   sin 2   cos 2   2 sin  .cos 

sin   sin 2   cos 2   2 sin  .cos 
cos   (sin   cos  ) 2

sin   (sin   cos  ) 2
cos   (sin   cos  ) cos   sin   cos   sin 
  
sin   (sin   cos  ) sin   sin   cos   cos 

= tan  RHS, proved.

(g) (2cos  + 1)(2cos  – 1)(2cos 2 – 1) = 2cos 4 + 1
Proof: LHS = (2cos  + 1)(2cos  – 1)(2cos 2 – 1)

= (4cos2  – 1)(2cos 2 – 1) Use: (a + b)(a – b) = a2 – b2

= {2 x (2cos2 ) – 1}(2cos 2 – 1) Use: 2cos2  = 1 + cos 2

= {2 x (1 + cos 2 ) – (2cos 2 – 1)
= (2 1}
+ 2cos 2 – 1) (2cos 2 – 1)
= (2cos 2 + 1)(2cos 2 – 1) Use: (a + b)(a – b) = a2 – b2
= 4cos2 2 – 1
= 2 + 2cos 4 – 1
= 2 x (2cos2 2) – 1 Use: 2cos2 A = 1 + cos 2A
= 2cos 4 + 1
= 2 x (1 + cos 2. 2 ) – 1 A = 2
= 2 x (1 + cos 4 ) – 1 RHS, proved.
4. Prove the following identities:
1 + sin 2 + cos 2
a) = cot 
1 + sin 2  cos 2

b) (1 + sin 2 + cos 2)2 = 4cos2 (1 + sin 2)

1 1
c)  = cot A
tan 2A  tan A cot 2A  cot A
3 1
d)  =4
sin 20 o cos 20 o

e) 2  2  2  2 cos 8 A 2 cos A
f) cos3A.cos 3A + sin3 A.sin 3A = cos3 2A
 1 + sin 2 + cos 2
a) = cot 
1 + sin 2  cos 2
Proof:
1 + sin 2 + cos 2
LHS =
1 + sin 2  cos 2
2cos2  + 2sin .cos
=
2sin2  + 2sincos 
2cos(cos  + sin )
=
2sin(sin  + cos )
= cot 
= RHS, proved.
b) (1 + sin 2 + cos 2)2 = 4cos2 (1 + sin 2)

Proof:
LHS = (1 + sin 2 + cos 2)2
= {(1 + cos 2) + sin 2}2

= {2cos2  + 2sin .cos }2

= {2cos  (cos  + sin )}2
= 4cos2  (cos2  + sin2  + 2sin .cos )
= 4cos2 (1 + sin 2)
= RHS, proved.
 1 1
c)  = cot A
tan 2A  tan A cot 2A  cot A
Proof: 1 1
LHS = 
tan 2A  tan A cot 2A  cot A
1 1
= 
sin 2A sin A cos 2A cos A
 
cos 2A cos A sin 2A sin A
1 1
= 
sin 2Acos A – cos 2AsinA sin Acos 2A – sin 2Acos A
cos 2Acos A sin 2Asin A
cos 2Acos A sin 2Asin A cos 2Acos A sin 2Asin A
=  = 
sin (2A – A) sin(A – 2A) sin A  sin A
cos 2AcosA+ sin 2Asin A cos(2A – A) cos A
= = = = cot A
sin A sin A sin A
= RHS,
 3 1
d)  =4
sin 20 o cos 20 o

Proof:
Method - I
3 1
2. (
3
2
cos 20 o
1 .sin 20o
2
)
LHS =  =
sin 20o cos 20o 1
sin 2.20o
3.cos 20o  1.sin 20o 2
=
sin 20o.cos 20o 2 (sin 60o.cos 20o  cos 60o.sin 20o)
=
1
2. 1 (3.cos 20o  1.sin 20o) sin 40o
2 2
=
1 .2. sin 20o.cos 20o 4 sin (60o  20o)
=
2 sin 40o
= 4 = RHS, proved.
Proof:
Method - II
3 1
LHS = 
sin 20 o cos 20o sin (60o  20o)
=
tan 60o 1 1 1 .2. sin 20o.cos 20o
=  2 2
sin 20 o cos 20o
4 sin 40o
sin 60o 1 =
=  sin 40o
cos 60o sin 20o cos 20o
= 4 = RHS, proved.
sin 60o.cos 20o  cos 60o.sin 20o
=
cos 60o. sin 20o. cos 20o
e) 2 2 2  2 cos 8 A 2 cos A
Proof:
LHS = 2  2  2  2 cos 8A Take 2 common

 2  2  2(1  cos 8 A)

 2  2  2{1  cos 2(4 A)} Use 1 + cos 2A= 2cos2 A

 2  2  2.2 cos 4 A  2  2  2 cos 4 A Remove sq. root

2

 2  2(1  cos 4 A)  2  2.2 cos 2 2 A Repeat method

 2  2 cos 2 A  2(1  cos 2 A)

 2.2 cos 2 A 2 cos A

= RHS, proved.
 f) cos3A.cos 3A + sin3 A.sin 3A = cos3 2A
Proof: LHS = cos3A.cos 3A + sin3 A.sin 3A
1 .4. (cos3A.cos 3A + sin3 A.sin 3A)
=
4
1 cos 3A = 4 cos 3
A  3cos A
= (4cos3A.cos 3A + 4sin3 A.sin 3A)
4 4cos3 A = cos 3A + 3cos A
1
= [(cos 3A+ 3cos A).cos 3A + (3sin A – sin 3A).sin 3A]
4
1
= [cos2 3A+ 3cos A.cos 3A + 3sin A.sin 3A – sin2 3A]
4 sin 3A = 3sin A  4 sin3 A
1
= 4sin
[cos2 3A – sin2 3A+ 3(cos A.cos 3A + sin 3
=A3sin
A.sin A  sin 3A
3A)]
4
1 1 1
= [cos 2.(3A) + 3cos(A – 3A)] = [cos 3.(2A) + 3cos 2A] = .4cos3 2A
4 4 3 4
= cos 2A = RHS, proved.
 QUIZ
1. The formula of sin 2A is
A. 2sin A. cos A
B. 2cos 2A  1

C. 2tan A
1 - tan2 A
D. Both A and C

A
Answer
2. The formula of 2sin2 A is
A. cos 2A – 1
B. 1 – cos 2A

C. 2tan A
1 – tan2 A
D. 1 + cos 2A

Answer
B
3. The formula of tan 2A is

A. 2tan A
1 + tan2 A
B. 1 – cos 2A

C. 1 + cos 2A

D. 2tan A
1 – tan2 A

Answer
D
4. The formula of 3sin A is

A. 3sin A – 4sin3 A

B. 4 sin3 A + sin 3A

C. 4 sin3 A – sin 3A

D. 3sin A + 4sin3 A

Answer
B
5. The formula of 4cos3 A is

A. cos 3A + 3cos A

B. cos 3A – 3cos A

C. 4 cos3 A – cos 3A

D. 4 cos3 A + cos 3A

A
Answer
 6. The correct RHS for 2  2 cos 4A is

A. 2sin 2A

B. cos 2A

C. 2cos 2A

D. sin 2A

Answer
C
 7. The correct RHS for (2cos  +1)(2cos   1) is

A. 1 + 2cos 2

B. 2cos 2  1

C. 2cos 2 + 1

D. Both (A) and (C)

Answer
B
 3
8. If cos  = , the value of sin 3 is
5
44
A.
125
4
B.
5
44
C. 
125
4
D. 
5

AnswerA
5. Prove the following identities:
1
a ) cos   sin   (3  cos 4 )
6 6

4
 1 
b) cos   sin  cos 2  1  sin 2 
6 6 2

 4 
1
c) cos   sin  1  sin 2  sin 4 2
8 8 2

8
1
d ) sin   (3  4 cos 2  cos 4 )
4

8
1
e) cos   (3  4 cos 2  cos 4 )
4

8
f ) sin 5 16 sin 5   20 sin 3   5sin 
6. Prove the following:
a) 8cos3 10º  8sin3 20º  6cos10o + 6sin20o = 23

b) 4(cos3 20º + sin3 20º) = 3(cos20o + sin10o)

o 3 o 3 1 o 3 1
3. c) If cos 330  , show that (i) sin165  (ii) cos165 
2 2 2 2 2
6. b) 4(cos3 20º + sin3 10º) = 3(cos20o +
sin10o)
Proof: 4cos 3
A = cos 3A + 3cos A
LHS = 4(cos3 20º + sin3 10º) 4sin3=A3sin A  sin 3A
= 4cos3 20º + 4sin3 10º
= cos 3(20º) + 3 cos 20º + 3sin 10º  sin 3(10º)

= cos 60º + 3 cos 20º + 3sin 10 – sin 30o

1 o o 1
  3(cos 20  sin10 ) 
2 2
= 3(cos20o + sin10o)
= RHS, proved.
 3 3 1 3 1
3. c) If cos 330o  , show that (i) sin165o  (ii) cos165o 
2 2 2 2 2
Solution: (ii)
We have, o
o 330 
cos165 cos   42 3
 2  
Since 165 o
lies in 8 2nd
the
1  cos 330o  1  cos 
 quadrant 
cos where 3 only
2 sine
3 1
2 ratio 2
is positive 2
8
3
1
 3  2
2
3  1
2

 2

2 8

 3 1
2
2 3 2 3 1
   
4 2
2 2 
2
2 2
Proved
 REVIEW OF
Trigo. Transformation Formulae
Transformation of product into sum or difference Transformation of sum or difference into product

a) 2sin A.cos B = sin(A+B) + sin(A – B) C + D C–D

a) sin C + sin D = 2sin .cos
2 2
b) 2cos A.sin B = sin(A+B) – sin(A – B)
C + D C–D
b) sin C – sin D = 2cos .sin
c) 2cos A.cos B = cos(A + B) + cos(A – B) 2 2
C–D
d) – 2sin A.sin B = cos(A + B) – cos(A – B) c) cos C + cos D = 2cos C + D .cos
2 2
OR D–C
d) cos C – cos D = 2sin C + D .sin
2sin A.sin B = cos(A – B) – cos(A + B) 2 2

Memorize them
Next…
2. Prove that:
d) sin(45o + A) + sin(45o – A) = 2 cos A
Solution: Use: sin C + sin D
LHS = sin(45o + A) + sin(45o – A) C + D .cos C – D
=2sin
o o
(45  A)  (45  A) o o
(45  A)  (45  A) 2 2
2sin .cos
2 2 OR
90o 2A
2sin .cos
2 2 Use: 2sin A.cos B =
2sin 45o.cos A sin(A+B) + sin(A – B)
1 LHS = sin(45o + A) + sin(45o – A)
2 .cos A
2 2sin 45o.cos A
 2 cos A = RHS 1
2 .cos A
Proved
2
 2 cos A = RHS Proved
 e) 2co(45o – A).cos(45o + A) = cos 2A
Solution: Use: 2cos A.cos B
LHS = 2co(45o – A).cos(45o + A)
= cos(A + B) + cos(A – B)
= cos{(45o – A) + (45o + A)} + cos{(45o – A) – (45o + A)}

= cos90o + cos 2A

= 0 + cos 2A

= cos 2A

= RHS
Proved
  2   4 
g ) sin A  sin  A    sin  A   0
 3   3 
Solution:
 2   4 
LHS  sin A  sin  A    sin  A   Use: sin C + sin D
 3   3 
C+D C–D
=2sin .cos
  
 sin A  sin A  120o  sin A  240o  Since  = 180o 2 2

( A  120o )  ( A  240o ) ( A  120o )  ( A  240o )

 sin A  2sin .cos
2 2

2 A  360o  120o
 sin A  2sin .cos
2 2

 sin A  2sin(180o  A).cos 60o Since cos(-A) = cos A

1
 sin A  2(  sin A).  sin A  sin A  0 = RHS
2
Proved
3. Prove the following identities:
cos A  cos B AB A B
f) cot cot
cos B  cos A 2 2

Solution: C + D .cos C – D
Use: cos C + cos D = 2cos
cos A  cos B 2 2
LHS 
cos B  cos A C + D .sin D – C
and cos C – cos D = 2sin
2 2

A + B .cos A – B
2cos
2 2
=
A + B .sin A – B
2sin
2 2
A + B .cot A – B
= cot
2 2
= RHS
Proved
 sin 2 A  sin 2 B tan( A  B)
e) 
sin 2 A  sin 2 B tan( A  B)
Solution:
Use: sin C + sin D
sin 2 A  sin 2 B
LHS  C + D .cos C–D
sin 2 A  sin 2 B =2sin
2 2
2 A  2B 2 A  2B
2sin .cos Use: sin C – sin D
2 2
= C+D C–D
2 A  2B 2 A  2B =2cos .sin
2 cos .sin 2 2
2 2
2( A  B) 2( A  B)
sin .cos
2 2
=
2( A  B) 2( A  B)
cos .sin
2 2

sin( A  B) cos( A  B) tan( A  B )

= tan( A  B) cot( A  B)  = RHS
cos( A  B) sin( A  B) tan( A  B) Proved
 cos10o  sin10o o
h)  tan 35
cos10o  sin10o

Solution:
cos10o  sin10o
LHS 
cos10o  sin10o C + D .sin D – C
Use: cos C – cos D = 2sin
2 2
cos10o  sin (90o  80o ) C + D .cos C – D
= and cos C + cos D = 2cos
cos10o  sin (90o  80o ) 2 2

cos10o  cos80o

cos10o  cos80o

Do it yourself …
i ) cos 7  cos 5  cos 3  cos  4 cos  cos 2 cos 4

Solution:
LHS cos 7  cos 5  cos 3  cos  Use: cos C + cos D
(cos 7  cos  )  (cos 5  cos 3 ) C + D .cos C – D
= 2cos
2 2
7   7   5  3 5  3
2 cos cos  2 cos cos
2 2 2 2
2 cos 4 cos 3  2 cos 4 cos 

2 cos 4 (cos 3  cos  )

3   3  
2 cos 4 .2 cos cos
2 2
4 cos 4 cos 2 cos 
4 cos  cos 2 cos 4
= RHS
Proved
j ) sin 4 cos 2  cos 3 sin  sin 5 cos 

Solution:

LHS sin 4 cos 2  cos 3 sin  Use: 2sin Acos B =

sin(A + B) + sin(A – B)
1
 .(2sin 4 cos 2  2 cos 3 sin  ) &
2
1 [sin(4  2 )  sin(4  2 )  sin(3   )  sin(3   )] 2cos Asin B =

2 sin(A + B) – sin(A – B)
1
 (sin 6  sin 2  sin 4  sin 2 )
2 Use: sin C + sin D
1 .2sin 6  4 cos 6  4 C + D .cos C–D
 =2sin
2 2 2 2 2
sin 5 cos 

= RHS
Proved
4. Prove that:
cos 4 x  cos 3 x  cos 2 x
a) cot 3 x
sin 4 x  sin 3 x  sin 2 x
Solution:
cos 4 x  cos 3 x  cos 2 x
LHS 
sin 4 x  sin 3 x  sin 2 x
(cos 4 x  cos 2 x)  cos 3 x

(sin 4 x  sin 2 x)  sin 3 x
4x  2x 4x  2x
2 cos cos  cos 3 x
 2 2
4x  2x 4x  2x
2sin cos  sin 3 x
2 2
2 cos 3 x cos x  cos 3 x

2sin 3 x cos x  sin 3 x
cos 3 x(2 cos x  1)
 cot 3x = RHS
sin 3 x(2 cos x  1)
Proved
 sin(   )  2sin   sin(   )
c) tan 
cos(   )  2 cos   cos(   )
Solution:
sin(   )  2sin   sin(   )
LHS 
cos(   )  2 cos   cos(   )
[sin(   )  sin(   )]  2sin 

[cos(   )  cos(   )]  2 cos 
2sin  cos   2sin 

2 cos  cos   2 cos 
2sin  (cos   1)

2 cos  (cos   1)
tan 

= RHS
Proved
 Prove that:
i) sin5A – sin3A + 2sinAcos2A = 2sin2Acos3A
Solution:
LHS = sin5A – sin3A + 2sinAcos2A

= 2cos().sin () + 2sinAcos2A

= 2cos4A.sinA + 2sinAcos2A
= 2sinA(cos4A + cos2A)
= 2sinA(2cos ().cos())

= 4sinAcos3AcosA
= 2(2sinAcosA)cos3A
= 2sin2Acos3A
= RHS
Proved
 4f. Prove that: = tan2

=
Solution:
LHS =

= =

= = tan2

=
= R
H
Proved

= S
 4g. Prove that: = tan(
Solution:

=
=
=
=
=
= [∵ cos(-A) = cosA]
=
= tan( = RHS
Proved
 5a. Prove that:
+ - =

Solution:
LHS = + -

= {sin +

= (2sin

= (2sin

= (sin2.(sin2.

= sinsinA = =
R Proved
5c. Prove that: =
Solution:
LHS =
= [2sin2]

= [(1-cos2+120o) + 1 cos2(120o)]

= [3 – cos2+ 240o) + cos 240o)}]

= [3 – cos2.cos240o]

= [3 – cos2)] [∵ cos(240o) = cos(180o+60o) = -cos60o =

= [3 – cos2] = = RHS
Proved
5d. Prove that: = cos3Ө
Solution:
LHS
=
= cos

= cos)+ (60o – )} + cos{) (60o – )}]

= cos] = cos

= cos] [∵ cos(120
= cos
o
) = cos(180 o
- 60 o
) = -cos60 o
=

= cos = cos

= =
R
Proved
5c. Prove that: =
Solution:
LHS =
= [2sin2]

= [(1-cos2+120o) + 1 cos2(120o)]

= [3 – cos2+ 240o) + cos 240o)}]

= [3 – cos2.cos240o]

= [3 – cos2)] [∵ cos(240o) = cos(180o+60o) = -cos60o =

= [3 – cos2] = = RHS
Proved
5d. Prove that: = cos3Ө
Solution:
LHS
=
= cos

= cos)+ (60o – )} + cos{) (60o – )}]

= cos] = cos

= cos] [∵ cos(120
= cos
o
) = cos(180 o
- 60 o
) = -cos60 o
=

= cos = cos

= =
R
Proved
 7a. Prove that: (cosα +cosβ) +(sinα +sinβ) = 4cos ( )
2 2 2

LHS (cosα +cosβ)2 +(sinα +sinβ)2

Solution:

=
= (2cos 2
+ (2sin 2

= 4cos2( + 4sin2(

=4 {cos2( + sin2(

=4

=
R
H
Proved

S
 cos 2 A cos 3 A  cos 2 A cos 7 A sin 7 A  sin 3 A
Prove that: 
sin 4 A sin 3 A  sin 2 A sin 5 A sin A
Solution:
cos 2 A cos 3 A  cos 2 A cos 7 A
LHS 
sin 4 A sin 3 A  sin 2 A sin 5 A
2 cos 2 A cos 3 A  2 cos 2 A cos 7 A

2sin 4 A sin 3 A  2sin 2 A sin 5 A
cos(2 A  3 A)  cos(2 A  3 A)   cos(2 A  7 A)  cos(2 A  7 A)

cos(4 A  3 A)  cos(4 A  3 A)   cos(2 A  5 A)  cos(2 A  5 A)

cos 5 A  cos A  cos 9 A  cos 5 A sin 5 A.2sin 2 A cos 2 A

 
cos A  cos 7 A  cos 3 A  cos 7 A sin 2 A sin A

A 9A 9A  A sin(5 A  2 A)  sin(5 A  2 A)

2sin sin 
cos A  cos 9 A 2 2  sin 5 A sin 4 A sin A
 
cos A  cos 3 A 2sin A  3 A sin 3 A  A sin 2 A sin A sin 7 A  sin 3 A
2 2  = RHS
sin A
Proved
 Prove that: Cos 52o + cos 68o + cos 172o
=0
Solution:
LHS = cos 52o + cos 68o + cos 172o

52o  68o 52o  68o

2 cos cos  cos172o
2 2

2 cos 60o cos8o  cos172o

1
2. cos8o  cos(180o  8o )
2
cos8o  cos8o

0

= RHS
Proved