NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions is an important study material for students who are pursuing science or mathematics in their senior secondary education. The chapter covers the basic principles and concepts of trigonometric functions, and the solutions are structured in a concise and simple manner, making it easy for students to understand the concepts and score good marks in their exams. The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions covers various topics such as trigonometric ratios, trigonometric identities, properties of trigonometric functions, and many more.

## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

**Exercise 3.1 : Solutions of Questions on Page Number : 54**

**Q1 :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

## NCERT Solutions for Class 11 Maths Chapter 3

**Q2 :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°

(iii) (iv)

**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 =

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.

**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.**

**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“².

**Q5 :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

**Q6 :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.

**Q7 :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

**Exercise 3.2 : Solutions of Questions on Page Number : 63**

** Q1 :Find the values of other five trigonometric functions if , x lies in third quadrant.**

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

**Q2 :Find the values of other five trigonometric functions if , x lies in second quadrant.**

**Answer :
**Since x lies in the 2nd quadrant, the value of cos x will be negative

**Q3 :Find the values of other five trigonometric functions if , x lies in third quadrant.**

**Answer :
**Since x lies in the 3rd quadrant, the value of sec x will be negative.

**Q4 :Find the values of other five trigonometric functions if , x lies in fourth quadrant.**

**Answer :
**Since x lies in the 4th quadrant, the value of sin x will be negative.

**Q5 :Find the values of other five trigonometric functions if , x lies in second quadrant.**

**Answer** :

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

∴sec x =

**Q6 :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°.

**Q7 :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°.

**Q8 :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°.

**Q9 :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°.

**Q10 :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°.

**Exercise 3.3 : Solutions of Questions on Page Number : 73**

** Q1 : **

**Answer** :

L.H.S. =

**Q2 :Prove that **

**Answer :**

L.H.S. =

**Q3 :Prove that **

**Answer :**

L.H.S. =

**Q4 :Prove that **

**Answer :**

L.H.S =

**Q5 :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°)

**Q6 :Prove that: **

**Answer :
**

**Q7 :Prove that: **

**Answer :
**It is known that

∴L.H.S. =

**Q8 :Prove that **

**Answer : **

**Q9 : **

**Answer :**

L.H.S. =

**Q10 :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

4

**Q11 :Prove that **

**Answer :**

It is known that

∴L.H.S. =

**Q12 :Prove that sin ^{2} 6x – sin_{2} 4x = sin 2x sin 10x**

**Answer :**

It is known that

∴L.H.S. = sin

^{2}6x – sin

^{2}4x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

= (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.

**Q13 :Prove that cos ^{2} 2x – cos^{2} 6x = sin 4x sin 8x**

**Answer :**

It is known that

∴L.H.S. = cos

^{2}2x – cos

^{2}6x

= (cos 2x + cos 6x) (cos 2x – 6x)

= [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.

**Q14 :Prove that sin 2x + 2sin 4x + sin 6x = 4cos ^{2} x sin 4x**

**Answer :**

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

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

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

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

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

= 2 sin 4x (2 cos

^{2}x – 1 + 1)

= 2 sin 4x (2 cos

^{2}x)

= 4cos

^{2}x sin 4x

= R.H.S.

**Q15 :Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)**

**Answer :**

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

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x – sin 3x)

= 2 cos 4x. cos x

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

**Q16 :Prove that **

**Answer :
**It is known that

∴L.H.S =

**Q17 :Prove that **

**Answer :
**It is known that

∴L.H.S. =

**Q18 :Prove that **

**Answer :**

It is known that

∴L.H.S. =

**Q19 :Prove that **

**Answer :
**It is known that

∴L.H.S. =

**Q20 :Prove that **

**Answer :
**It is known that

∴L.H.S. =

**Q21 :Prove that **

**Answer :
**L.H.S. =

**Q22 :Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1**

**Answer :
**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 x cot 2x – (cot 2x cot x – 1)

= 1 = R.H.S.

**Q23 :Prove that **

**Answer :
**It is known that

∴L.H.S. = tan 4x = tan 2(2x)

**Q24 :Prove that cos 4x = 1 – 8sin ^{2} x cos^{2} x**

**Answer :**

L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin

^{2}2x [cos 2A = 1 – 2 sin

^{2}A]

= 1 – 2(2 sin x cos x)

^{2}[sin2A = 2sin A cosA]

= 1 – 8 sin

^{2}x cos

^{2}x

= R.H.S.

**Q25 :Prove that: cos 6x = 32 cos ^{6} x – 48 cos^{4} x + 18 cos^{2} x – 1**

**Answer :**

L.H.S. = cos

^{6}x

= cos 3(2x)

= 4 cos

^{3}2x – 3 cos 2x [cos 3A = 4 cos

^{3}A – 3 cos A]

= 4 [(2 cos

^{2}x – 1)

^{3}– 3 (2 cos

^{2}x – 1) [cos 2x = 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

**Exercise 3.4 : Solutions of Questions on Page Number : 78**

** Q1 :Find the principal and general solutions of the equation **

**Answer :
**Therefore, the principal solutions are x = and

Therefore, the general solution is

**Q2 :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

**Q3 :Find the principal and general solutions of the equation **

**Answer :
**Therefore, the principal solutions are x = and

Therefore, the general solution is

**Q4 :Find the general solution of cosec x = -2**

**Answer :
**cosec x= -2

Therefore, the principal solutions are x =

Therefore, the general solution is

**Q5 :Find the general solution of the equation cos 4x = cos 2x**

**Answer :
**

**Q6 :Find the general solution of the equation cos 3x + cos x – cos 2x = 0**

**Answer :
**cos 3x + cos x – cos 2x = 0

**Q7 :Find the general solution of the equation sin 2x + cos x = 0**

**Answer :
**sin 2x + cos x = 0

Therefore, the general solution is

**Q8 :Find the general solution of the equation sec ^{2 }2x = 1 – tan 2x
**

**Answer :**

sec2 2x = 1 – tan 2x

**Therefore, the general solution is**

**Q9 :Find the general solution of the equation sin x + sin 3x + sin 5x =0**

**Answer :
**sin x + sin 3x + sin 5x =0

**Therefore, the general solution is**

**Exercise Miscellaneous : Solutions of Questions on Page Number : 81**

** Q1 :Prove that: **

**Answer :**

L.H.S.

= 0 = R.H.S

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

**Answer :
**L.H.S.= (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

= RH.S.

**Q3 :Prove that: **

**Answer :**

L.H.S. =

**Q4 :Prove that: **

**Answer :**

L.H.S. =

**Q5 :Prove that: **

**Answer :**

It is known that

∴L.H.S. =

**Q6 :Prove that: **

**Answer :**

It is known that

L.H.S. =

= tan 6x

= R.H.S.

**Q7 :Prove that: **

**Answer :**

L.H.S. =

**Q8 : , x in quadrant II**

**Answer :**

Here, x is in quadrant II.

i.e.,

Therefore, are all positive.

As x is in quadrant II, cosx is negative.

∴

Thus, the respective values of are

**Q9 :Find for , x in quadrant III**

**Answer :**

Here, x is in quadrant III.

Therefore, and are negative, where as is positive.

Now,

Thus, the respective values of are

**Q10 :Find for , x in quadrant II**

**Answer :**

Here, x is in quadrant II.

Therefore,, and are all positive.

[cosx is negative in quadrant II]

Thus, the respective values of are