NCERT Solutions for Class 11 Maths Chapter 3

Trigonometric Functions Class 11

Chapter 3 Trigonometric Functions Exercise 3.1, 3.2, 3.3, 3.4, miscellaneous Solutions

 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

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°

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

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

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 =

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

.

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 radiusr1, while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r2.

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

.

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.

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.

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 isknown 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

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

Q11 :

Prove that

Answer :

It is known that                                                                                              .

∴L.H.S. =

Q12 :

Prove that sin2  6x - sin2 4x = sin 2x sin 10x

Answer :

It is known

that

∴L.H.S. = sin26x – sin24x

= (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

Q13 :

Prove that cos2 2x - cos2  6x = sin 4x sin 8x

Answer :

It is known

that

∴L.H.S. = cos2 2x – cos2 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

Q14 :

Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

Answer :

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

= 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 cos2 x – 1 + 1)

= 2 sin 4x (2 cos2 x)

= 4cos2  x sin 4x

Q15 :

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

Answer :

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x – sin 3x)

= 2 cos 4x. cos x

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)

Q23 :

Prove that

Answer :

It is known that                                           .

∴L.H.S. = tan 4x = tan 2(2x)

Q24 :

Prove that cos 4x = 1 - 8sin2 x cos2  x

Answer :

L.H.S. = cos 4x

= cos 2(2x)

= 1 - 2 sin2 2x [cos 2A = 1 - 2 sin2 A]

= 1 - 2(2 sin x cos x)2 [sin2A = 2sin A cosA]

= 1 - 8 sin2x cos2x

Q25 :

Prove that: cos 6x = 32 cos6  x - 48 cos4  x + 18 cos2  x - 1

Answer :

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos3 2x - 3 cos 2x [cos 3A = 4 cos3 A - 3 cos A]

= 4 [(2 cos2 x - 1)3 - 3 (2 cos2 x - 1) [cos 2x = 2 cos2 x - 1]

= 4 [(2 cos2 x)3 - (1)3 - 3 (2 cos2 x)2 + 3 (2 cos2 x)] - 6cos2 x + 3

= 4 [8cos6x - 1 - 12 cos4x + 6 cos2x] - 6 cos2x + 3

= 32 cos6x - 4 - 48 cos4x + 24 cos2 x - 6 cos2x + 3

= 32 cos6x - 48 cos4x + 18 cos2x - 1

 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         .

Q2 :

Find the principal and general solutions of the equation

Answer :

Therefore, the principal solutions are x =      and         .

Q3 :

Find the principal and general solutions of the equation

Answer :

Therefore, the principal solutions are x =  and  .

Q4 :

Find the general solution of cosec x = -2

Answer :

cosec x= –2

Therefore, the principal solutions are x =                            .

Q5 :

Find the general solution of the equation

Answer :

Q6 :

Find the general solution of the equation

Answer :

Q7 :

Find the general solution of the equation

Answer :

Q8 :

Find the general solution of the equation

Answer :

Q9 :

Find the general solution of the equation

Answer :

 Exercise Miscellaneous : Solutions of Questions on Page Number : 81                                                                                              

Q1 :

Prove that:

Answer :

L.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

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

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.

∴

Q9 :

Find                                                      for                          , x in quadrant III

Answer :

Here, x is in quadrant III.

Therefore,                and                are negative, whereas is positive.

Now,

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                                                         .