Get Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi and English Medium. Trigonometric Functions Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Trigonometric Functions All Exercises Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers.

Section Name | Topic Name |

3.1 | Introduction |

3.2 | Angles |

3.3 | Trigonometric Functions |

3.4 | Trigonometric Functions of Sum and Difference of Two Angles |

3.5 | Trigonometric Equations |

3.6 | Summary |

**Contents**show

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

**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°

**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.

**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′.

**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.

**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.

**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.

**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

**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.

**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.

**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* =

**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°.

**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°.

**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°.

**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°.

**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. =

**Question 2:**

Prove that

**Answer:**

L.H.S. =

**Question 3:**

Prove that

**Answer:**

L.H.S. =

**Question 4:**

Prove that

**Answer:**

L.H.S =

**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°)

**Question 6:**

Prove that:

**Answer:**

**Question 7:**

Prove that:

**Answer:**

It is known that

∴L.H.S. =

**Question 8:**

Prove that

**Answer:**

**Question 9:**

**Answer:**

L.H.S. =

**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*

**Question 11:**

Prove that

**Answer:**

It is known that.

∴L.H.S. =

**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.

**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.

**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 (â€“ 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* â€“ 1 + 1)

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

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

= R.H.S.

**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* â€“ sin 3*x*)

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

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

**Question 16:**

Prove that

**Answer:**

It is known that

∴L.H.S =

**Question 17:**

Prove that

**Answer:**

It is known that

∴L.H.S. =

**Question 18:**

Prove that

**Answer:**

It is known that

∴L.H.S. =

**Question 19:**

Prove that

**Answer:**

It is known that

∴L.H.S. =

**Question 20:**

Prove that

**Answer:**

It is known that

∴L.H.S. =

**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.

**Question 23:**

Prove that

**Answer:**

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

**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.

**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

**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**

**Question 3:**

Find the principal and general solutions of the equation

**Answer:**

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

Therefore, the general solution is

**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

**Question 5:**

Find the general solution of the equation

**Answer:**

**Question 6:**

Find the general solution of the equation

**Answer:**

**Question 7:**

Find the general solution of the equation

**Answer:**

Therefore, the general solution is.

**Question 8:**

Find the general solution of the equation

**Answer:**

Therefore, the general solution is.

**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

**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. =

**Question 4:**

Prove that:

**Answer:**

L.H.S. =

**Question 5:**

Prove that:

**Answer:**

It is known that.

∴L.H.S. =

**Question 6:**

Prove that:

**Answer:**

It is known that

.

L.H.S. =

= tan 6*x*

= R.H.S.

**Question 7:**

Prove that:

**Answer:**

L.H.S. =

**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.

**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.

**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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