NCERT Solutions for Class 11 Science Maths Chapter 3 Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Trigonometric Functions are extremely popular among Class 11 Science students for Maths Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.

#### Page No 54:

#### Question 1:

Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47° 30' (iii) 240° (iv) 520°

#### Answer:

(i) 25°

We know that 180° = π radian

(ii) –47° 30'

–47° 30' = degree [1° = 60']

degree

Since 180° = π radian

(iii) 240°

We know that 180° = π radian

(iv) 520°

We know that 180° = π radian

#### Page No 55:

#### Question 2:

Find the degree measures corresponding to the following radian measures

.

(i) (ii) – 4 (iii) (iv)

#### Answer:

(i)

We know that π radian = 180°

(ii) – 4

We know that π radian = 180°

(iii)

We know that π radian = 180°

(iv)

We know that π radian = 180°

#### Page No 55:

#### Question 3:

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 =

In one complete revolution, the wheel turns an angle of 2π radian.

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 an angle of 12π radian.

#### Page No 55:

#### Question 4:

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.

#### 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

Therefore, forr = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36′.

#### Page No 55:

#### Question 5:

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

#### Answer:

Diameter of the circle = 40 cm

∴Radius (*r*) of the circle =

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

In ΔOAB, OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, ΔOAB is an equilateral triangle.

∴θ = 60° =

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.

Thus, the length of the minor arc of the chord is.

#### Page No 55:

#### Question 6:

If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

#### Answer:

Let
the radii of the two circles be
and.
Let an arc of length *l*
subtend an angle of 60° at the centre of the circle of radius *r*_{1},
while let an arc of length *l
*subtend
an angle of 75° at the centre of the circle of radius *r*_{2}.

Now, 60° =and 75° =

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.

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

#### Page No 55:

#### Question 7:

Find the angle in radian though 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.

It
is given that *r*
= 75 cm

(i) Here,
*l*
= 10 cm

(ii) Here,
*l *=
15 cm

(iii) Here,
*l *=
21 cm

#### Page No 63:

#### Question 1:

Find the values of other five trigonometric functions if , *x* lies in third quadrant.

#### Answer:

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

#### Page No 63:

#### Question 2:

Find the values of other five trigonometric functions if , *x* lies in second quadrant.

#### Answer:

Since
*x*
lies in the 2^{nd}
quadrant, the value of cos *x*
will be negative

#### Page No 63:

#### Question 3:

Find the values of other five trigonometric functions if , *x* lies in third quadrant.

#### Answer:

Since
*x
*lies
in the 3^{rd}
quadrant, the value of sec *x*
will be negative.

#### Page No 63:

#### Question 4:

Find the values of other five trigonometric functions if , *x* lies in fourth quadrant.

#### Answer:

Since
*x*
lies in the 4^{th}
quadrant, the value of sin *x*
will be negative.

#### Page No 63:

#### Question 5:

Find the values of other five trigonometric functions if , *x* lies in second quadrant.

#### Answer:

Since
*x*
lies in the 2^{nd}
quadrant, the value of sec *x*
will be negative.

∴sec
*x*
=

#### Page No 63:

#### Question 6:

Find the value of the trigonometric function sin 765°

#### Answer:

It
is known that the values of sin *x
*repeat
after an interval of 2π or
360°.

#### Page No 63:

#### Question 7:

Find the value of the trigonometric function cosec (–1410°)

#### Answer:

It
is known that the values of cosec *x
*repeat
after an interval of 2π or
360°.

#### Page No 63:

#### Question 8:

Find the value of the trigonometric function

#### Answer:

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

#### Page No 63:

#### Question 9:

Find the value of the trigonometric function

#### Answer:

It
is known that the values of sin *x
*repeat
after an interval of 2π or
360°.

#### Page No 63:

#### Question 10:

Find the value of the trigonometric function

#### Answer:

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

#### Page No 73:

#### Question 1:

#### Answer:

L.H.S. =

#### Page No 73:

#### Question 2:

Prove that

#### Answer:

L.H.S. =

#### Page No 73:

#### Question 3:

Prove that

#### Answer:

L.H.S. =

#### Page No 73:

#### Question 4:

Prove that

#### Answer:

L.H.S =

#### Page No 73:

#### Question 5:

Find the value of:

(i) sin 75°

(ii) tan 15°

#### Answer:

(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*]

(ii) tan 15° = tan (45° – 30°)

#### Page No 73:

#### Question 6:

Prove that:

#### Answer:

#### Page No 73:

#### Question 7:

Prove that:

#### Answer:

It is known that

∴L.H.S. =

#### Page No 73:

#### Question 8:

Prove that

#### Answer:

#### Page No 73:

#### Question 9:

#### Answer:

L.H.S. =

#### Page No 73:

#### Question 10:

Prove that
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*

#### Page No 73:

#### Question 11:

Prove that

#### Answer:

It is known that.

∴L.H.S. =

#### Page No 73:

#### Question 12:

Prove that
sin^{2} 6*x* – sin^{2} 4*x* = sin 2*x*
sin 10*x*

#### Answer:

It is known that

∴L.H.S.
= sin^{2}6*x* – sin^{2}4*x*

= (sin 6*x* + sin 4*x*) (sin 6*x* – sin 4*x*)

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

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

= sin 10*x* sin 2*x*

= R.H.S.

#### Page No 73:

#### Question 13:

Prove that
cos^{2} 2*x* – cos^{2} 6*x* = sin 4*x
*sin 8*x*

#### Answer:

It is known that

∴L.H.S.
= cos^{2} 2*x* – cos^{2} 6*x*

= (cos 2*x* + cos 6*x*) (cos 2*x *– 6*x*)

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

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

= sin 8*x* sin 4*x*

= R.H.S.

#### Page No 73:

#### Question 14:

Prove that sin 2*x* + 2sin 4*x* + sin 6*x* = 4cos^{2} *x* sin 4*x*

#### Answer:

L.H.S. = sin 2*x* + 2 sin 4*x* + sin 6*x*

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

= 2 sin 4*x* cos (– 2*x*) + 2 sin 4*x*

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

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

= 2 sin 4*x* (2 cos^{2} *x* – 1 + 1)

= 2 sin 4*x* (2 cos^{2} *x*)

= 4cos^{2} *x* sin 4*x*

= R.H.S.

#### Page No 73:

#### Question 15:

Prove that
cot 4*x* (sin 5*x* + sin 3*x*) = cot *x* (sin 5*x*
– sin 3*x*)

#### Answer:

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

= 2 cos 4*x* cos *x*

R.H.S. = cot *x* (sin 5*x* – sin 3*x*)

= 2 cos 4*x*. cos *x*

L.H.S. = R.H.S.

#### Page No 73:

#### Question 16:

Prove that

#### Answer:

It is known that

∴L.H.S =

#### Page No 73:

#### Question 17:

Prove that

#### Answer:

It is known that

∴L.H.S. =

#### Page No 73:

#### Question 18:

Prove that

#### Answer:

It is known that

∴L.H.S. =

#### Page No 73:

#### Question 19:

Prove that

#### Answer:

It is known that

∴L.H.S. =

#### Page No 73:

#### Question 20:

Prove that

#### Answer:

It is known that

∴L.H.S. =

#### Page No 73:

#### Question 21:

Prove that

#### Answer:

L.H.S. =

#### Page No 74:

#### Question 22:

Prove that
cot *x* cot 2*x* – cot 2*x* cot 3*x* –
cot 3*x* cot *x* = 1

#### Answer:

L.H.S. =
cot *x* cot 2*x* – cot 2*x* cot 3*x* –
cot 3*x* cot *x*

= cot *x* cot 2*x* – cot 3*x* (cot 2*x* +
cot *x*)

= cot *x* cot 2*x* – cot (2*x *+ *x*) (cot
2*x* + cot *x*)

= cot* x *cot 2*x *– (cot 2*x *cot *x* –
1)

= 1 = R.H.S.

#### Page No 74:

#### Question 23:

Prove that

#### Answer:

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

#### Page No 74:

#### Question 24:

Prove that
cos 4*x* = 1 – 8sin^{2 }*x *cos^{2 }*x*

#### Answer:

L.H.S. =
cos 4*x*

= cos 2(2*x*)

= 1 – 2 sin^{2} 2*x* [cos 2*A* = 1 –
2 sin^{2} *A*]

= 1 – 2(2 sin* x *cos *x*)^{2} [sin2*A*
= 2sin *A* cos*A*]

= 1 – 8 sin^{2}*x* cos^{2}*x*

= R.H.S.

#### Page No 74:

#### Question 25:

Prove
that: cos 6*x* = 32 cos^{6} *x* – 48 cos^{4}
*x* + 18 cos^{2} *x *– 1

#### Answer:

L.H.S. =
cos 6*x*

= cos 3(2*x*)

= 4 cos^{3} 2*x* – 3 cos^{ }2*x* [cos
3*A* = 4 cos^{3} *A* – 3 cos^{ }*A*]

= 4 [(2 cos^{2} *x *– 1)^{3} – 3 (2
cos^{2} *x* – 1) [cos 2*x* = 2 cos^{2}
*x *– 1]

= 4 [(2 cos^{2} *x*)^{3} – (1)^{3}
– 3 (2 cos^{2} *x*)^{2} + 3 (2 cos^{2}
*x*)] – 6cos^{2} *x* + 3

= 4 [8cos^{6}*x* – 1 – 12 cos^{4}*x*
+ 6 cos^{2}*x*] – 6 cos^{2}*x* + 3

= 32 cos^{6}*x* – 4 – 48 cos^{4}*x*
+ 24 cos^{2} *x* – 6 cos^{2}*x* + 3

= 32 cos^{6}*x *– 48 cos^{4}*x* + 18
cos^{2}*x* – 1

= R.H.S.

#### Page No 78:

#### Question 1:

Find the principal and general solutions of the equation

#### Answer:

Therefore,
the principal solutions are *x* =and.

Therefore, the general solution is

#### Page No 78:

#### Question 2:

Find the principal and general solutions of the equation

#### Answer:

Therefore,
the principal solutions are *x* =and.

Therefore,
the general solution is,
where *n* ∈ **Z**

#### Page No 78:

#### Question 3:

Find the principal and general solutions of the equation

#### Answer:

Therefore,
the principal solutions are *x* =
and.

Therefore, the general solution is

#### Page No 78:

#### Question 4:

Find the general solution of cosec *x* = –2

#### Answer:

_{cosec
}_{x}_{
= –2}

Therefore,
the principal solutions are *x*
=.

Therefore, the general solution is

#### Page No 78:

#### Question 5:

Find the general solution of the equation

#### Answer:

#### Page No 78:

#### Question 6:

Find the general solution of the equation

#### Answer:

#### Page No 78:

#### Question 7:

Find the general solution of the equation

#### Answer:

Therefore, the general solution is.

#### Page No 78:

#### Question 8:

Find the general solution of the equation

#### Answer:

Therefore, the general solution is.

#### Page No 78:

#### Question 9:

Find the general solution of the equation

#### Answer:

Therefore, the general solution is

#### Page No 81:

#### Question 1:

Prove that:

#### Answer:

L.H.S.

= 0 = R.H.S

#### Page No 81:

#### Question 2:

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

#### Answer:

L.H.S.

= (sin
3*x *+
sin *x*)
sin *x *+
(cos 3*x *–
cos *x*)
cos *x*

= RH.S.

#### Page No 82:

#### Question 3:

Prove that:

#### Answer:

L.H.S. =

#### Page No 82:

#### Question 4:

Prove that:

#### Answer:

L.H.S. =

#### Page No 82:

#### Question 5:

Prove that:

#### Answer:

It is known that.

∴L.H.S. =

#### Page No 82:

#### Question 6:

Prove that:

#### Answer:

It is known that

.

L.H.S. =

= tan 6*x*

= R.H.S.

#### Page No 82:

#### Question 7:

Prove that:

#### Answer:

L.H.S. =

#### Page No 82:

#### Question 8:

,
*x*
in quadrant II

#### Answer:

Here, *x* is in quadrant II.

i.e.,

Therefore, are all positive.

As *x* is in quadrant II, cos*x* is negative.

∴

Thus, the respective values of are.

#### Page No 82:

#### Question 9:

Find
for
,
*x* in quadrant III

#### Answer:

Here, *x*
is in quadrant III.

Therefore, and are negative, whereasis positive.

Now,

Thus, the respective values of are.

#### Page No 82:

#### Question 10:

Find
for
,
*x* in quadrant II

#### Answer:

Here, *x*
is in quadrant II.

Therefore,, and are all positive.

[cos*x*
is negative in quadrant II]

Thus, the respective values of are .

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