NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 3
Chapter Name : Trigonometric Functions

Exercise 3.1

Q1 Find the radian measures corresponding to the following degree measures:
∘ ∘ ′ ∘ ∘
(i) 25 (ii) − 47 30 (iii) 240 (iv) 520

∘
(i) 25

∘
We know that 180 = π radian

∘ π 5π
∴ 25 = × 25 radian = radian
180 36
∘ ′
(ii) − 47 30

∘ ′ 1
−47 30 = −47
2

−95
=
2

Since 180 ∘
= π radian
∘
(iii) 240

∘
We know that 180 = π radian

∘ π 4
∴ 240 = × 240 radian = π radian
180 3
∘
(iv) 520

∘
We know that 180 = π radian

∘ π 26π
∴ 520 = × 520 radian = radian
180 9

Page : 54 , Block Name : Exercise 3.1

Q2 Find the degree measures corresponding to the following radian measures (Use π = 22

7
) .
11 5π 7π
(i) (ii) − 4 (iii) (iv)
16 3 6

11
(i)
16

∘
We know that π radian = 180

11 180 11 45 × 11
∴ radain = × degree = deg ree
16 π 16 π × 4
45 × 11 × 7 315
= deg ree = degree
22 × 4 8
3
= 39 deg ree
8

∘ 3×60 ∘ ∘
= 39 + min utes [1 = 60 ]
8

∘ ′ 1
= 39 + 22 + min utes
2

∘ ′ ′′ ′ ′′
= 39 22 30 [1 = 60 ]

(ii) − 4

∘
We know that π radian = 180

180 180×7(−4)
−4 radian = × (−4) deg ree = degree
π 22

−2520 1
= deg ree = −229 degree
11 11
∘ 1×60 ∘ ∘
= −229 + min utes [1 = 60 ]
11
∘ 12 n ′ ′′
= −229 5 27 [1 = 60 ]
5π
(iii)
3

∘
We know that π radlan = 180

5π 180 5π ∘
∴ radian = × degree = 300
3 π 3

7π
(iv)
6

∘
We know that π radian = 180

7π 180 7π ∘
∴ radian = × = 210
6 π 6

Page : 55 , Block Name : Exercise 3.1

Q3 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one
second?

Answer. Number of revolutions made by the wheel in 1 minute = 360
∴ Number of revolutions made by the wheel in 1 second =
360
= 6
60

Hence, In 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., 12 π radian
Thus, In one second, the wheel turns and angle of 12 π radian.

Page : 55 , Block Name : Exercise 3.1

Q4 Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an
arc of length 22 cm (Use π =
22
)
7

Answer. We know that in a circle of radius r unit, if an arc of length l unit subtends and angle θ
radian at the centre, then
1
θ =
r

Therefore, for r = 100 cm, l = 22 cm, we have

22 180 22 180 × 7 × 22
θ = radian = × deg ree = deg ree
100 π 100 22 × 100

126 3
∘ ∘ ∘ ′
= deg ree = 12 deg ree = 12 36 [1 = 60 ]
10 5

Thus, the required angle is 12 36 .
∘ ′

Page : 55 , Block Name : Exercise 3.1

Q5 In a circle of diameter 40 cm, the length of a circle is 20 cm. Find the length of minor arc of the
chord.

Answer. Diameter of the circle = 40 cm
Radius of the circle = cm = 20cm 40

2

Let AB be a chord (length = 20 cm) of the circle.

In △OAB, OA = OB = Radius of circle = 20cm

Also, AB = 20cm

Thus, △OAB is an equilateral triangle.
∘ π
∴ θ = 60 =
3

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the
centre, then θ =
l

r

π ˜
AB 20π
= ˜ =
⇒ AB cm
3 20 3

Thus, the length of the minor arc of the chord is .
20π
cm
3

Page : 55 , Block Name : Exercise 3.1

Q6 In the two circles , arcs of the same length subtend angles 60 at the centre, nd the
∘ ∘
and 75

ratio of their radii.

Answer. Let the radii of the two circles be r and r . Let an arc of length l subtend an angle of 60 at
1 2
∘

the centre of the circle of radius r , while let an arc of length l subtend an angle of 60 at the centre
1
∘

of the circle of radius r . 2

Now, 60 = radian
∘ π ∘ 5π
and 75 =
3 12

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the
centre, then θ = or l = rθ l

r
r1 π r2 5π
∴ l = and l =
3 12

r1 π r2 5π
⇒ =
3 12

r2 5
⇒ r1 =
4

r1 5
⇒ =
r2 4

Thus, the ratio of the radii is 5 : 4.

Page : 55 , Block Name : Exercise 3.1

Q7 Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip
describes an arc of length.
(i) 10 cm
(ii) 15 cm
(iii) 21 cm

Answer. We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian
at the centre, then θ = .
l

r

It is given that r = 75 cm
(i) Here, l = 10 cm
radian
10 2
θ = radian =
75 15

(ii) Here, l = 15 cm
θ =
15

75
radian =
1

5
radian
(iii) Here, l = 21 cm
θ =
21

75
radian =
7

25
radian

Page : 55 , Block Name : Exercise 3.1

Exercise 3.2

Q1 Find the values of other ve trigonometric functions in cos x = − 1

2
, x lies in third quadrant.

1
cos x = −
2

1 1
∴ sec x = = = −2
cos x 1
(− )
2

2 2
sin x + cos x = 1

2 2
⇒ sin x = 1 − cos x

2
2 1
⇒ sin x = 1 − (− )
2

2 1 3
⇒ sin x = 1 − =
4 4

√3
⇒ sin x = ±
2

Since x lies in the 3rd quadrant, the value of sin x will be negative.

√3
∴ sin x = −
2

1 1 2
cosecx = = = −
sin x √3
√3
(− )
2

√3 √3
(− ) (− )
2 2
sin x
tan x = = = = √3
cos x cos x 1
(− )
2

1 1
cot x = =
tan x
√3

Page : 63 , Block Name : Exercise 3.2

Q2 Find the values of other ve trigonometric functions in sin x = 3

5
, x lies in second quadrant.

3
sin x =
5

1 1 5
cosecx = = =
sin x 3 3
( )
5

2 2
sin x + cos x = 1

2 2
⇒ cos x = 1 − sin x

2
2 3
⇒ cos x = 1 − ( )
5

2 9
⇒ cos x = 1 −
25

2 16
⇒ cos x =
25

4
⇒ cos x = ±
5

Since x lies in the 2nd quadrant, the value of cos x will be negative
4
∴ cos x = −
5

1 1 5
sec x = = = −
cos x 4 4
(− )
5

(
3
) (
3
) (−
4
)
.
5 5 5
sin x 3
tan x = = = = = −
cos x cos x cos x 4 4
(− )
5

1 4
cot x = = −
tan x 3

Page : 63 , Block Name : Exercise 3.2

Q3 Find the values of other ve trigonometric functions in cot x = 3

4
, x lies in third quadrant.

3
cot x =
4

1 1 4
tan x = = =
cot x 3 3
( )
4

2 2
1 + tan x = sec x

2
4 2
⇒ 1 + ( ) = sec x
3

16 2
⇒ 1 + = sec x
9

25 2
⇒ = sec x
9

5
⇒ sec x = ±
3

Since, x lies in the 3rd quadrant, the value of sec x will be negative.
5
∴ sec x = −
3

1 1 3
cos x = = = −
sec x 5 5
(− )
3

sin x
tan x =
cos x

4 sin x
⇒ =
3 −3
( )
5

4 −3 4
⇒ sin x = ( ) × ( ) = −
.
3 5 5

1 5
cosecx = = −
sin x 4

Page : 63 , Block Name : Exercise 3.2

Q4 Find the values of other ve trigonometric functions in sec x = 13

5
, lies in fourth quadrant.

13
sec x =
5

1 1 5
cos x = = =
sec x 13 13
( )
5

2 2
sin x + cos x = 1

2 2
⇒ sin x = 1 − cos x
2
2 5
⇒ sin x = 1 − ( )
13

2 25 144
⇒ sin x = 1 − =
169 169

12
⇒ sin x = ±
13

Since x lies in the 4th quadrant, the value of sin x will be negative.
12
∴ sin x = −
13

1 1 13
cosecx = = = −
sin x 12 12
(− )
13

−12 −12
( ) ( )
13 13
sin x 12
tan x = = = = −
cos x cos x 5 5
( )
13

1 1 12
cot x = = = −
tan x 12 5
(− )
5

Page : 63 , Block Name : Exercise 3.2

Q5 Find the values of other ve trigonometric functions in tan x = − 5

12
, x lies in second quadrant.

5
tan x = −
12

1 1 12
cot x = = = −
tan x 5 5
(− )
12

2 2
1 + tan x = sec x

2
5 2
⇒ 1 + (− ) = sec x
12

25 2
⇒ 1 + = sec x
144

169 2
⇒ = sec x
144

13
⇒ sec x = ±
12

Since x lies in the 2nd quadrant, the value of sec x will be negative.
13
∴ sec x = −
12

1 1 12
cos x = = = −
sec x 13 13
(− )
12

sin x
tan x =
cos x

5 sin x
⇒ − =
12 12
(− )
13

5 12 5
⇒ sin x = (− ) × (− ) =
12 13 13

1 1 13
.
cosecx = = =
sin x 5 5
( )
13

Page : 63 , Block Name : Exercise 3.2

Q6 Find the values of the trigonometric functions in sin 765 .
∘

Answer. It is known that the values of sin x repeat after an interval of 2π or 360^deg.
∘ ∘ ∘ ∘ 1
∴ sin 765 = sin(2 × 360 + 45 ) = sin 45 =
√2

Page : 63 , Block Name : Exercise 3.2

Q7 Find the values of the trigonometric functions in cosec (−1410 ).
∘

Answer. It is known that the values of cosec x repeat after an interval of 2π or 360^∘.
∘ ∘ ∘
∴ cosec (−1410 ) = cosec (−1410 + 4 × 360 )

.
∘ ∘
= cosec (−1410 + 1440 )
∘
= cosec30 = 2

Page : 63 , Block Name : Exercise 3.2

Q8 Find the values of the trigonometric functions in tan .
19π

3

Answer. It is known that the values of tan x repeat after an interval of π or 180 .
∘

.
19π 1 π π ∘
∴ tan = tan 6 π = tan(6π + ) = tan = tan 60 = √3
3 3 3 3

Page : 63 , Block Name : Exercise 3.2

Q9 Find the values of the trigonometric functions in sin(− .
11π
)
3

Answer. It is known that the values of sin x repeat after an interval of π or 360 . ∘

11π 11π π √3
∴ sin(− ) = sin(− + 2 × 2π) = sin( ) =
3 3 3 2

Page : 63 , Block Name : Exercise 3.2

Q10 Find the values of the trigonometric functions in cot(− .
15π
)
4

Answer. It is known that the values of cot x repeat after an interval of π or 180 . ∘

∴ cot(−
15π
) = cot(−
15π
+ 4π) = cot
π
= 1 .
4 4 4

Page : 63 , Block Name : Exercise 3.2

Exercise 3.3

Q1 Prove that: sin 2 π 2 π 2 π 1
+ cos − tan = −
6 3 4 2

2 π 2 π 2 π
L. H.S. = sin + cos − tan
6 3 4

2 2
1 1 2
= ( ) + ( ) − (1)
2 2

1 1 1
= + − 1 = −
4 4 2

= R.H.S.

Page : 73 , Block Name : Exercise 3.3

Q2 Prove that: 2 sin 2 π

6
+ cosec
2 7π

6
cos
2 π

3
=
3

2

2 π 2 7π 2 π
L.H.S. = 2 sin + cosec cos
6 6 3

2 2
1 2 π 1
= 2( ) + cos ec (π + )( )
2 6 2

2
1 π 1
= 2 × + (−cosec ) ( )
4 6 4

1 4 1 3
= + = + 1 =
2 4 2 2

= R.H.S.

Page : 73 , Block Name : Exercise 3.3

Q3 Prove that: cot 2 π

6
+ cosec
5π

6
+ 3 tan
2 π

6
= 6

2 π 5π 2 π
L.H.S. = cot + cosec + 3 tan
6 6 6

2
2 π 1
= (√3) + cos cc(π − ) + 3( )
6 √3

π 1
= 3 + cos ec + 3 ×
6 3

= 3 + 2 + 1 = 6

= RH S

Page : 73 , Block Name : Exercise 3.3

Q4 Prove that: 2 sin 2 3π

4
+ 2 cos
2 π

4
+ 2 sec
2 π

3
= 10

Answer. L. H. S = 2 sin 2 3π

4
+ 2 cos
2 π

4
+ 2 sec
2 π

3
2 2
π 1 2
= 2{sin(π − )} + 2( ) + 2(2)
4 √2

2
π 1
= 2{sin } + 2 × + 8
4 2

2
1
= 2( ) + 1 + 8
√2

= 1 + 1 + 8

= 10

= R.H.S

Page : 73 , Block Name : Exercise 3.3

Q5 Find the value of :
(i) sin 75
∘

(ii) tan 15 ∘

∘ ∘ ∘
(i) sin 75 = sin(45 + 30 )

∘ ∘ ∘ ∘
= sin 45 cos 30 + cos 45 sin 30

[sin(x + y) = sin x cos y + cos x sin y]

1 √3 1 1
= ( )( ) + ( )( )
√2 2 √2 2

√3 1 √3+1
= + =
2√2 2√2 2√2

(ii) tan 15
∘ ∘ ∘
= tan(45 − 30 )
∘ ∘
tan 45 −tan 30 tan x−tan y
= ∘ ∘
[tan(x − y) = ]
1+tan 45 tan 30 1+tan x tan y

1
1−
√3 √3−1
= =
1 √3
1+1( )
√3 √3

2
√3−1 (√3−1) 3+1−2√3
= = =
√3+1 (√3+1)(√3−1) (√3)2 −(1)2

4−2√3
= = 2 − √3
3−1

Page : 73 , Block Name : Exercise 3.3

Q6 Prove the following: cos( π
− x) cos(
π
− y) − sin(
π
− x) sin(
π
− y) = sin(x + y)
4 4 4 4

π π π π
cos( − x) cos( − y) − sin( − x) sin( − y)
4 4 4 4

1 π π 1 π π
= [2 cos( − x) cos( − y)] + [−2 sin( − x) sin( − y)]
2 4 4 2 4 4

1 π π π π
= [cos{( − x) + ( − y)} + cos{( − x) − ( − y)}]
2 4 4 4 4

1 π π π π
+ [cos{( − x) + ( − y)} − cos{( − x) − ( − y)}]
2 4 4 4 4

∵ 2 cos A cos B = cos(A + B) + cos(A − B)
[ ]
−2 sin A sin B = cos(A + B) − cos(A − B)

1 π π
= 2 × [cos{( − x) + ( − y)}]
2 4 4
π
= cos[ − (x + y)]
2

= sin(x + y)

= R. H. S

Page : 73 , Block Name : Exercise 3.3

π
tan( +x) 2

Q7 Prove the following:
4 1+tan x
= ( )
π 1−tan x
tan( −x)
4

Answer. It is known that
tan A+tan B tan A−tan B
tan(A + B) = and tan(A − B) =
1−tan A tan B 1+tan A tan B

π
tan +tan x
4
π ( π
) 1+tan x
tan( +x) 1−tan tan x ( ) 2
4 4 1−tan x 1+tan x
= π
= = ( ) = R. H. S.
π tan −tan x 1−tan x 1−tan x
tan( −x) 4 ( )
4 ( ) 1+tan x
π
1+tan tan x
4

Page : 73 , Block Name : Exercise 3.3

cos(π+x) cos(−x)
Q8 Prove the following: π
= cot
2
x
sin(π−x) cos( +x)
2

cos(π + x) cos(−x)
L.H.S. =
π
sin(π − x) cos( + x)
2

[− cos x][cos x]
=
(sin x)(− sin x]

2
− cos x
=
2
− cot x
2
= cot x

= R. H . S

Page : 73 , Block Name : Exercise 3.3

Q9 Prove the following: cos( 3π
+ x) cos(2π + x) [cot(
3π
− x) + cot(2π + x)] = 1
2 2

Answer. L.H.S = cos( 3π
+ x) cos(2π + x) [cot(
3π
− x) + cot(2π + x)]
2 2

= sin x cos x[tan x + cot x]

sin x cos x
= sin x cos x ( + )
cos x sin x

2 2
sin x+cos x
= (sin x cos x) [ ]
sin x cos x

= 1 = R. H. S.

Page : 73 , Block Name : Exercise 3.3

Q10 Prove the following: sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos x

Answer. L.H.S = sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x

1
= [2 sin(n + 1)x sin(n + 2)x + 2 cos(n + 1)x cos(n + 2)x]
2

cos{(n + 1)x − (n + 2)x} − cos{(n + 1)x + (n + 2)x}
1
= [ ]
2
+ cos{(n + 1)x + (n + 2)x} + cos{(n + 1)x − (n + 2)x}

∵ −2 sin A sin B = cos(A + B) − cos(A − B)
[ ]
2 cos A cos B = cos(A + B) + cos(A − B)

1
= × 2 cos{(n + 1)x − (n + 2)x}
2

= cos(−x) = cos x = R. H S

Page : 73 , Block Name : Exercise 3.3

Q11 Prove the following: cos( 3π

4
+ x) − cos(
3π

4
− x) = −√2 sin x

Answer. It is known that
A+B A−B
cos A − cos B = −2 sin( ) ⋅ sin( )
2 2

∴ = cos( 3π

4
+ x) − cos(
3π

4
− x)

3π 3π 3π 3π
( +x)+( −x) ( +x)−( −x)
4 4 4 4

= −2 sin{ } ⋅ sin{ }
2 2

3π
= −2 sin( ) sin x
4

π
= −2 sin(π − ) sin x
4

π
= −2 sin sin x
4

1
= −2 × × sin x
√2

= −√2 sin x

= R. H . S

Page : 73 , Block Name : Exercise 3.3

Q12 Prove the following: sin 6x − sin 4x = sin 2x sin 10x
2 2

Answer. It is known that
A+B A−B A+B A−B
sin A + sin B = 2 sin( ) cos( ), sin A − sin B = 2 cos( ) sin( )
2 2 2 2

2 2
∴ L. H. S. = sin 6x − sin 4x

= (sin 6x + sin 4x)(sin 6x − sin 4x)

6x+4x 6x−4x 6x+4x 6x−4x
= [2 sin( ) cos( )] [2 cos( ) ⋅ sin( )]
2 2 2 2

= (2 sin 5x cos x)(2 cos 5x sin x)

= (2 sin 5x cos 5x)(2 sin x cos x)

= sin 10x sin 2x

= R. H . S.

Page : 73 , Block Name : Exercise 3.3

Q13 Prove the following: cos 2x − cos 6x = sin 4x sin 8x
2 2

Answer. It is known that

A+B A−B A+B A−B
cos A + cos B = 2 cos( ) cos( ), cos A − cos B = −2 sin( ) sin( )
2 2 2 2

2 2
∴ L. H ⋅ S, = cos 2x − cos 6x

= (cos 2x + cos 6x)(cos 2x − 6x)

2x+6x 2x−6x 2x+6x (2x−6x)
= [2 cos( ) cos( )] [−2 sin( ) sin ]
2 2 2 2

= [2 cos 4x cos(−2x)][−2 sin 4x sin(−2x)]

= [2 cos 4x cos 2x][−2 sin 4x(− sin 2x)]

= (2 sin 4x cos 4x)(2 sin 2x cos 2x)

= sin 8x sin 4x

= R. H. S.

Page : 73 , Block Name : Exercise 3.3

Q14 Prove the following: sin 2x + 2 sin 4x + sin 6x = 4 cos x sin 4x 2

LH. S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

2x+6x 2x−6x
= [2 sin( )( )] + 2 sin 4x
2 2

A+B A−B
[∵ sin A + sin B = 2 sin( ) cos( )]
2 2

= 2 sin 4x cos(−2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

= 2 sin 4x(cos 2x + 1)

2
= 2 sin 4x (2 cos x − 1 + 1)

2
= 2 sin 4x (2 cos x)

2
= 4 cos x sin 4x

= R. H . S

Page : 73 , Block Name : Exercise 3.3

Q15 Prove the following: cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

L. H.S = cot 4x(sin 5x + sin 3x)

cot 4x 5x+3x 5x−3x
= [2 sin( ) cos( )]
sin 4x 2 2

A+B A−B
[∵ sin A + sin B = 2 sin( ) cos( )]
2 2

cos 4x
= ( ) [2 sin 4x cos x]
sin 4x

= 2 cos 4x cos x

R.4.5. = cot x(sin 5x − sin 3x)

cos x 5x+3x 5x−3x
= [2 cos( ) sin( )]
sin x 2 2

A+B A−B
[∵ sin A − sin B = 2 cos( ) sin( )]
2 2

cos x
= [2 cos 4x sin x]
sin x

= 2 cos 4x ⋅ cos x

L ⋅ H ⋅ S. = R. H . S

Page : 73 , Block Name : Exercise 3.3

Q16 Prove the following:
cos 9x−cos 5x sin 2x
= −
sin 17x−sin 3x cos 10x

Answer. It is known that
A+B A−B A+B A−B
cos A − cos B = −2 sin( ) sin( ), sin A − sin B = 2 cos( ) sin( )
2 2 2 2

cos 9x−cos 5x
∴ L. H . S =
sin 17x−sin 3x
9x+5x 9x−5x
−2 sin( )⋅sin( )
2 2

=
17x+3x 17x−3x
2 cos( )⋅sin( )
2 2

−2 sin 7x ⋅ sin 2x
=
2 cos 10x ⋅ sin 7x

sin 2x
= −
cos 10x

= R. H . S

Page : 73 , Block Name : Exercise 3.3

Q17 Prove the following: sin 5x+sin 3x

cos 5x+cos 3x
= tan 4x

Answer. It is known that
A+B A−B A+B A−B
sin A + sin B = 2 sin( ) cos( ), cos A + cos B = 2 cos( ) cos( )
2 2 2 2

sin 5x+sin 3x
∴ L ⋅ H ⋅ S. =
cos 5x+cos 3x

5x+3x 5x−3x
2 sin( ) ⋅ cos( )
2 2

5x+3x 5x−3x
2 cos( ) ⋅ cos( )
2 2

2 sin 4x⋅cos x
=
2 cos 4x⋅cos x

sin 4x
=
cos 4x

= tan 4x = R.H.S

Page : 73 , Block Name : Exercise 3.3

Q18 Prove the following:
sin x−sin y x−y
= tan
cos x+cos y 2

Answer. It is known that
A+B A−B A+B A−B
sin A − sin B = 2 cos( ) sin( ), cos A + cos B = 2 cos( ) cos( )
2 2 2 2

==
sin x−sin y
∴
cos x+cos y

x+y x−y
2 cos( )⋅sin( )
2 2
=
x+y x−y
2 cos( )⋅cos( )
2 2

x − y
sin( )
2

x−y
sin( )
2

=
x−y
cos( )
2

x − y
=tan( ) = R. H . S
2

Page : 73 , Block Name : Exercise 3.3

Q19 Prove the following:
sin x+sin 3x
= tan 2x
cos x+cos 3x

Answer. It is known that
A+B A−B A+B A−B
sin A + sin B = 2 sin( ) cos( ), cos A + cos B = 2 cos( ) cos( )
2 2 2 2

sin x+sin 3x
∴ L. H . S. =
cos x+cos 3x

x+3x x−3x
2 sin( ) cos( )
2 2

=
x+3x x−3x
2 cos( ) cos( )
2 2

sin 2x
=
cos 2x

= tan 2x

= R. H S

Page : 73 , Block Name : Exercise 3.3

Q20 Prove the following:
sin x−sin 3x
= 2 sin x
2 2
sin x−cos x

Answer. It is known that
A+B A−B 2 2
sin A − sin B = 2 cos( ) sin( ), cos A − sin A = cos 2A
2 2

sin x−sin 3x
∴ L ⋅ H ⋅ S = 2
sin x−cos2 x

x + 3x x − 3x
2 cos( ) sin( )
2 2

2 cos 2x sin(−x)
=
− cos 2x

= −2 × (− sin x)

= 2 sin x = R. H . S

Page : 73 , Block Name : Exercise 3.3

Q21 Prove the following:
cos 4x+cos 3x+cos 2x
= cot 3x
sin 4x+sin 3x+sin 2x

Answer. L.H.S =
cos 4x+cos 3x+cos 2x

sin 4x+sin 3x+sin 2x

(cos 4x + cos 2x) + cos 3x
=
(sin 4x + sin 2x) + sin 3x

4x + 2x 4x − 2x
2 cos( ) cos( ) + cos 3x
2 2

2 sin(4x + 2x 4x − 2x
= ) cos( ) + sin 3x
2 2

A+B A−B A+B A−B
[∵ cos A + cos B = 2 cos( ) cos( ), sin A + sin B = 2 sin( ) cos( )]
2 2 2 2

2 cos 3x cos x + cos 3x
=
2 sin 3x cos x + sin 3x

cos 3x(2 cos x + 1)
=
sin 3x(2 cos x + 1)

= cot 3x = R. H . S

Page : 73 , Block Name : Exercise 3.3

Q22 Prove the following: cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1

L. H.S. = cot x cot 2x − cot 2x cot 3x − cot 3x cot x

= cot x cot 2x − cot 3x(cot 2x + cot x)

= cot x cot 2x − cot(2x + x)(cot 2x + cot x)

cot 2x cot x−1
= cot x cot 2x − [ ] (cot 2x + cot x)
cot x+cot 2x

cot A cot B−1
[∵ cot(A + B) = ]
cot A+cot B

= cot x cot 2x − (cot 2x cot x − 1)

= 1 = R. H . S

Page : 74 , Block Name : Exercise 3.3

2
4 tan x(1−tan x)
Q23 Prove the following: tan 4x = 2 4
1−6 tan x+tan x

Answer. It is known that
2 tan A
tan 2A = 2
1−tan A

∴ L H.S. = tan 4x = tan 2(2x)

2 tan 2x
=
2
1−tan (2x)

2 tan x
2( )
1−tan2 x
=
2
2 tan x
1−( )
1−tan2 x

4 tan x
( )
1−tan2 x
=
2
4 tan x
1− ]
2
(1−tan2 x)

4 tan x
( )
2
1−tan x
= 2
⎡ (1−tan2 x) −4 tan2 x ⎤

2
⎣ 2
(1−tan x) ⎦

2
4 tan x (1 − tan x)
=
4 2 2
1 + tan x − 2 tan x − 4 tan x
2
4 tan x (1 − tan x)
= = RH S
2 4
1 − 6 tan x + tan x

Page : 74 , Block Name : Exercise 3.3

Q24 Prove the following: cos 4x = 1 − 8 sin x cos x 2 2

L.H.S. = cos 4x

= cos 2(2x)

2 2
= 1 − 2 sin 2x [cos 2A = 1 − 2 sin A]

2
2
= 1 − 2(2 sin x cos x) [sin 2A = 2 sin A cos A]

2 2
= 1 − 8 sin x cos x

= R.H.S.

Page : 74 , Block Name : Exercise 3.3

Q25 Prove the following: cos 6x = 32 cos x − 48 cos x + 18 cos x − 1 6 4 2

L. H.S. = cos 6x

= cos 3(2x)

3 3
= 4 cos 2x − 3 cos 2x [cos 3A = 4 cos A − 3 cos A]

3
2 2 2
= 4 [(2 cos x − 1) − 3 (2 cos x − 1) [cos 2x = 2 cos x − 1]

3 2
2 3 2 2 2
= 4 [(2 cos x) − (1) − 3(2 cos x) + 3 (2 cos x)] − 6 cos x + 3

6 4 2 2
= 4 [8 cos x − 1 − 12 cos x + 6 cos x] − 6 cos x + 3

6 4 2 2
= 32 cos x − 4 − 48 cos x + 24 cos x − 6 cos x + 3

6 4 2
= 32 cos x − 48 cos x + 18 cos x − 1

= R. H. S

Page : 74 , Block Name : Exercise 3.3

Exercise 3.4

Q1 Find the principal and general solutions of the following equations:
tan x = √3

Answer. tan x = √3
It is known that tan π

3
= √3 and tan (
4π

3
) = tan(π +
π

3
) = tan
π

3
= √3

Therefore, the principal solutions are x =
π 4π
and
3 3

Now, tan x = tan π

3
π
⇒ x = nπ + , where n ∈ Z
3

Therefore, the general solution is x = nπ + .
π
, where n ∈ Z
3

Page : 78 , Block Name : Exercise 3.4

Q2 Find the principal and general solutions of the following equations: sec x = 2

Answer. Sec x = 2
It is known that sec π

3
= 2 and sec
5π

3
= sec(2π −
π

3
) = sec
π

3
= 2

Therefore, the principal solutions are x = .
π 5π
and
3 3
π
Now 7 sec x = sec
3

π
⇒ cos x = cos
3

π
⇒ x = 2nπ ± , where n ∈ Z
3

Therefore, the general solution is x = 2nπ ±
π
, where n ∈ Z
3

Page : 78 , Block Name : Exercise 3.4

Q3 Find the principal and general solutions of the following equations: cot x = −√3

Answer. Cot x = −√3
It is known that cot = √3
π

6

π π π π
∴ cot(π − ) = − cot = −√3 and cot(2π − ) = − cot = − √3
6 6 6 6

5π 11π
i.e., cot = −√3 and cot = − √3
6 6

Therefore, the principal solutions are x =
5π 11π
and
6 6

Now, cot x = cot 5π

6

5π 1
⇒ tan x = tan [cot x = ]
6 tan x

5π
⇒ x = nπ + , where n ∈ Z
6

Therefore, the general solution is x = nπ + 5π

6
, where n ∈ Z .

Page : 78 , Block Name : Exercise 3.4

Q4 Find the principal and general solutions of the following equations: cosec x = – 2

Answer. Cosec x = -2
It is known that
π
cosec = 2
6

π π π π
∴ cos cc(π + ) = − cos cc = −2 and cosec (2π − ) = − cos cc = −2
6 6 6 6

7π 11π
cosec = −2 and cosec = −2
6 6

Therefore, the principal solutions are x =
7π 11π
and
6 6

Now, cosec x = cos ec 7π

6

7π 1
⇒ sin x = sin [cosecx = ]
6 sin x

n 7π
⇒ x = nπ + (−1) , where n ∈ Z
6

Therefore, the general solution is x = nπ + (−1) n 7π

6
, where n ∈ Z .

Page : 78 , Block Name : Exercise 3.4

Q5 Find the general solution for each of the following equations: cos 4 x = cos 2 x

cos 4x = cos 2x

⇒ cos 4x − cos 2x = 0

4x+2x 4x−2x
⇒ −2 sin( ) sin( ) = 0
2 2

A+B A−B
[∵ cos A − cos B = −2 sin( ) sin( )]
2 2

⇒ sin 3x sin x = 0

⇒ sin 3x = 0 or sin x = 0

∴ 3x = nπ
nπ
⇒ x = or x = nπ, where n ∈ Z
3

Page : 78 , Block Name : Exercise 3.4

Q6 Find the general solution for each of the following equations: cos 3x + cos x – cos 2x = 0

cos 3x + cos x − cos 2x = 0

3x+x 3x−x A+B A−B
⇒ 2 cos( ) cos( ) − cos 2x = 0 [cos A + cos B = 2 cos( ) cos( )]
2 2 2 2

⇒ 2 cos 2x cos x − cos 2x = 0

⇒ cos 2x(2 cos x − 1) = 0

⇒ cos 2x = 0 or 2 cos x − 1 = 0

1
⇒ cos 2x = 0 or cos x =
2
π π
∴ 2x = (2n + 1) or cos x = cos , where n ∈ Z
2 3

π π
⇒ x = (2n + 1) or x = 2nπ ± , where n ∈ Z
4 3

Page : 78 , Block Name : Exercise 3.4

Q7 Find the general solution for each of the following equations: sin 2x + cos x = 0

sin 2x + cos x = 0

⇒ 2 sin x cos x + cos x = 0

⇒ cos x(2 sin x + 1) = 0

⇒ cos x = 0 or 2 sin x + 1 = 0
π
Now, cos x = 0 ⇒ cos x = (2n + 1) , where n ∈ Z
2

2 sin x + 1 = 0

−1 π π π 7π
⇒ sin x = = − sin = sin(π + ) = sin(π + ) = sin
2 6 6 6 6

n 7π
⇒ x = nπ + (−1) , where n ∈ Z
6

Therefore, the general solution is (2n + 1) π
or nπ + (−1)
n 7π

6
,n ∈ Z .
2

Page : 78 , Block Name : Exercise 3.4

Q8 Find the general solution for each of the following equations: sec 2x = 1 − tan 2x 2

2
sec 2x = 1 − tan 2x

2
⇒ 1 + tan 2x = 1 − tan 2x

2
⇒ tan 2x + tan 2x = 0

⇒ tan 2x(tan 2x + 1) = 0

⇒ tan 2x = 0 or tan 2x + 1 = 0

Now, tan 2x = 0

⇒ tan 2x = tan 0

⇒ 2x = nπ + 0, where n ∈ Z
nπ
⇒ x = , where n ∈ Z
2

tan 2x + 1 = 0

π π 3π
⇒ tan 2x = −1 = − tan = tan(π − ) = tan
4 4 4

3π
⇒ 2x = nπ + , where n ∈ Z
4

nπ 3π
⇒ x = + , where n ∈ Z
2 8

Therefore, the general solution is nπ

2
or
nπ

2
+
3π

8
,n ∈ Z .

Page : 78 , Block Name : Exercise 3.4

Q9 Find the general solution for each of the following equations: sin x + sin 3x + sin 5x = 0

Answer. sin x + sin 3x + sin 5x = 0
(sin x + sin 5x) + sin 3x = 0

x+5x x−5x A+B A−B
⇒ [2 sin( ) cos( )] + sin 3x = 0 [sin A + sin B = 2 sin( ) cos( )]
2 2 2 2

⇒ 2 sin 3x cos(−2x) + sin 3x = 0

⇒ 2 sin 3x cos 2x + sin 3x = 0

⇒ sin 3x(2 cos 2x + 1) = 0

⇒ sin 3x = 0 or 2 cos 2x + 1 = 0

Now, sin 3x = 0 ⇒ 3x = nπ, where n ∈ Z
nπ
i.c., x = , where n ∈ Z
3

2 cos 2x + 1 = 0

−1 π π
⇒ cos 2x = = − cos = cos(π − )
2 3 3

2π
⇒ cos 2x = cos
3

2π
⇒ 2x = 2nπ ± , where n ∈ Z
3

π
⇒ x = nπ ± , where n ∈ Z
3

Therefore, the general solution is
nπ π
or nπ ± ,n ∈ Z
3 3

Page : 78 , Block Name : Exercise 3.4

Miscellaneous Exercise

Q1 Prove that: 2 cos
π 9π 3π 5π
cos + cos + cos = 0
13 13 13 13

L.H.S.

π 9π 5π 5π
= 2 cos cos + cos + cos
13 13 13 13

3π 5π 3π 5π
+ − x+y x−y
π 9π 13 13 13 13
= 2 cos cos + 2 cos( ) cos( ) [cos x + cos y = 2 cos( ) cos( )]
13 13 2 2 2 2

π 9π 4π −π
= 2 cos cos + 2 cos cos( )
13 13 13 13

π 9π 4π π
= 2 cos cos + 2 cos cos
13 13 13 13

π 9π 4π
= 2 cos [cos + cos ]
13 13 13

9π 4π
−
π 9π 4π 13 13
= 2 cos [2 cos( + ) cos( )]
13 13 13 2

π π 5π
= 2 cos [2 cos cos ]
13 2 26

π 5π
= 2 cos × 2 × 0 × cos
13 26

= 0 = R. H. S

Page : 81 , Block Name : Miscellaneous Exercise

Q2 Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

L.H.S.

= (sin 3x + sin x) sin x + (cos 3x − cos x) cos x
2 2
= sin 3x sin x + sin x + cos 3x cos x − cos x

2 2
= cos 3x cos x + sin 3x sin x − (cos x − sin x)

= cos(3x − x) − cos 2x [cos(A − B) = cos A cos B + sin A sin B]

= cos 2x − cos 2x

= 0

= R. H . S

Page : 81 , Block Name : Miscellaneous Exercise

Q3 Prove that: (cos x + cos y)
x+y
2 2 2
+ (sin x − sin y) = 4 cos
2

2 2
L.H.S. = (cos x + cos y) + (sin x − sin y)

2 2 2 2
= cos x + cos y + 2 cos x cos y + sin x + sin y − 2 sin x sin y

2 2 2 2
= (cos x + sin x) + (cos y + sin y) + 2(cos x cos y − sin x sin y)

= 1 + 1 + 2 cos(x + y)
[cos(A + B) = (cos A cos B − sin A sin B)]
= 2 + 2 cos(x + y)

= 2[1 + cos(x + y)]

x + y
2 2
= 2 [1 + 2 cos ( ) − 1] [cos 2A = 2 cos A − 1]
2

x + y
2
= 4 cos ( ) = R. H. S
2

Page : 82 , Block Name : Miscellaneous Exercise

Q4 Prove that: (cos x − cos y) 2 x−y
2 2
+ (sin x − sin y) = 4 sin
2

Answer. L.H.S = (cos x − cos y) 2
+ (sin x − sin y)
2

2 2 2 2
= cos x + cos y − 2 cos x cos y + sin x + sin y − 2 sin x sin y

2 2 2 2
= (cos x + sin x) + (cos y + sin y) − 2[cos x cos y + sin x sin y]

= 1 + 1 − 2[cos(x − y)] [cos(A − B) = cos A cos B + sin A sin B]

= 2[1 − cos(x − y)]
x−y
2 2
= 2 [1 − {1 − 2 sin ( )}] [cos 2A = 1 − 2 sin A]
2

2 x−y
= 4 sin ( ) = R. H . S
2

Page : 82 , Block Name : Miscellaneous Exercise

Q5 Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Answer. It is known that sin A + sin B = 2 sin(
A+B A−B
) ⋅ cos( )
2 2

L.H.S. = sin x + sin 3x + sin 5x + sin 7x

= (sin x + sin 5x) + (sin 3x + sin 7x)

x+5x x−5x 3x+7x 3x−7x
= 2 sin( ) ⋅ cos( ) + 2 sin( ) cos( )
2 2 2 2

= 2 sin 3x cos(−2x) + 2 sin 5x cos(−2x)

= 2 sin 3x cos 2x + 2 sin 5x cos 2x

= 2 cos 2x[sin 3x + sin 5x]

3x+5x 3x−5x
= 2 cos 2x [2 sin( ) ⋅ cos( )]
2 2

= 2 cos 2x[2 sin 4x ⋅ cos(−x)]

= 4 cos 2x sin 4x cos x = R.H.S.

Page : 82 , Block Name : Miscellaneous Exercise

(sin 7x+sin 5x)+(sin 9x+sin 3x)
Q6 Prove that: = tan 6x
(cos 7x+cos 5x)+(cos 9x+cos 3x)

Answer. It is known that
A+B A−B A+B A−B
sin A + sin B = 2 sin( ) ⋅ cos( ), cos A + cos B = 2 cos( ) ⋅ cos( )
2 2 2 2

(sin 7x+sin 5x)+(sin 9x+sin 3x)
L.H.S =
(cos 7x+cos 5x)+(cos 9x+cos 3x)

7x+5x 7x−5x 9x+3x 9x−3x
[2 sin( )⋅cos( )]+[2 sin( )⋅cos( )]
2 2 2 2

=
7x+5x 7x−5x 9x+3x 9x−3x
[2 cos( )−cos( )]+[2 cos( )⋅cos( )]
2 2 2 2

= [2 sin 6x ⋅ cos x] + [2 sin 6x ⋅ cos 3x]

= [2 cos 6x ⋅ cos x] + [2 cos 6x ⋅ cos 3x]

2 sin 6x[cos x+cos 3x]
=
2 cos 6x[cos x+cos 3x]

= tan 6x

= R.H.S.

Page : 82 , Block Name : Miscellaneous Exercise

Q7 Prove that: sin 3x + sin 2x − sin x = 4 sin x cos
x 3x
cos
2 2

Answer. L.H.S = sin 3x + sin 2x − sin x
= sin 3x + (sin 2x − sin x)

2x+x 2x−x A+B A−B
= sin 3x + [2 cos( ) sin( )] [sin A − sin B = 2 cos( ) sin( )]
2 2 2 2

3x x
= sin 3x + [2 cos( ) sin( )]
2 2

3x x
= sin 3x + 2 cos sin
2 2
3x 3x 3x x
= 2 sin ⋅ cos + 2 cos sin [sin 2A = 2 sin A ⋅ cos B]
2 2 2 2

3x 3x x
= 2 cos( ) [sin( ) + sin( )]
2 2 2

3x x 3x x
( )+( ) ( )−( )
2 2 2 2 A+B A−B
3x
= 2 cos( ) [2 sin{ } cos{ }] [[sin A + sin B = 2 sin( ) cos( )]
2 2 2 2 2

3x x
= 2 cos( ) ⋅ 2 sin x cos( )
2 2

x 3x
= 4 sin x cos( ) cos( ) = R. H . S
2 2

Page : 82 , Block Name : Miscellaneous Exercise

Q8 Find sin in tan x = −
x x x ′ 4
, cos and tan , x in quadrant I I
2 2 2 3

Answer. Here, x is in quadrant II.
π
< x < π
2

π x π
⇒ < <
4 2 2

Therefore, sin are all positive.
x x x
, cos and tan
2 2 2

It is given that tan x = −
4

3

2
2 2 −4 16 25
sec x = 1 + tan x = 1 + ( ) = 1 + =
3 9 9

2 9
∴ cos x =
25

3
⇒ cos x = ±
5

As x is in quadrant II, cos x is negative.
−3
cos x =
5

2 x
Now, cos x = 2 cos − 1
2

−3 2 x
⇒ = 2 cos − 1
5 2

2 x 3
⇒ 2 cos = 1 −
2 5

2 x 2
⇒ 2 cos =
2 5

2 x 1
⇒ cos =
2 5
x 1 x
⇒ cos = ∵ cos is positive ]
2 √5 2

x √5
∴ cos =
2 5

2 x 2 x
sin + cos = 1
2 2

2
2 x 1
⇒ sin + ( ) = 1
2 √5

2 x 1 4
⇒ sin = 1 − =
2 5 5

x 2 x
⇒ sin = [∵ sin is positive ]
2 √5 2

x 2√5
∴ sin =
2 5

2
x
sin ( )
x √5
2
tan = x
= = 2
2 cos 1
2 ( )
√5

2√5 √5
Thus, the respective values of sin
x x x
, cos and tan are , , and 2
2 2 2 5 5

Page : 82 , Block Name : Miscellaneous Exercise

Q9 Find sin in cos x = − in quadrant III.
x x x 1
, cos and tan ,x
2 2 2 3

Answer. Here, x is in quadrant III.
3π
i.e., π < x <
2

π x 3π
⇒ < <
2 2 4

Therefore,
is positive.
x x x
cos and tan are negative, whereas sin
2 2 2

It is given that cos x = − . 1

3
2 x
cos x = 1 − 2 sin
2

2 x 1−cos x
⇒ sin =
2 2

1 1
1−(− ) (1+ )
x 3 3 3
2 2
⇒ sin = = = =
2 2 2 2 3

x √2 x
⇒ sin = [∵ sin is positive ]
2 √3 2

x √2 √3 √6
∴ sin = × =
2 √3 √3 3

2 x
cos x = 2 cos − 1
2
1 3−1 2
1+(− ) ( ) ( )
1+cos x 3 3 3
2 x 1
⇒ cos = = = = =
2 2 2 2 2 3

x 1
⇒ cos = −
2 √3

x 1 √3 −√3
∴ cos = − × =
2 √3 √3 3

√2
x ( )
sin √3
x 2
tan = x
= = − √2
2 cos −1
2 ( )
√3

√6 −√3
Thus, the respective values of sin .
x x x
, cos and tan are , , and − √2
2 2 2 3 3

Page : 82 , Block Name : Miscellaneous Exercise

Q10 Find sin in sin x = is in quadrant II.
x x x 1
, cos and tan ,x
2 2 2 4

Answer. Here, x is in quadrant II.
π
i.e., < x < π
2

π x π
⇒ < <
4 2 2

Therefore,
and tan are all positive.
x x x
sin , cos
2 2 2

It is given that sin x = 1

4
2
2 2 1 1 15
cos x = 1 − sin x = 1 − ( ) = 1 − =
4 16 16

√15
⇒ cos x = − [cos x is negative in quadrant II ]
4
√15
1−(− )
2 x 1−cos x 4 4+√15
sin = = =
2 2 2 8

x 4+√15 x
⇒ sin = √ [∵ sin is positive ]
2 8 2

4+√15 2
= √ ×
8 2

8+2√15
= √
16

√8+2√15

=
4
√15
1+(− )
2 x 1+cos x 4 4−√15
cos = = =
2 2 2 8

x 4−√15 x
⇒ cos = √ [∵ cos is positive ]
2 8 2

4−√15 2
= √ ×
8 2

8−2√15
= √
16

√8−2√15

=
4
√
8+2√15
x (
sin 4 √8+2√15
x 2
tan = x = ) =
2 cos √
8−2√15 √8−2√15
2
4

8+2√15 8+2√15
= √ ×
8−2√15 8+2√15

(8+2√15)2 8+2√15
= √ = = 4 + √15
64−60 2

√8+2√15 √8−2√15

Thus, the respective values of sin x

2
, cos
x

2
and tan
x

2
are
4
,
4
, and 4 + √15.

Page : 82 , Block Name : Miscellaneous Exercise