NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions are available at Goyanka Maths Study, which are prepared by Balkishan Agrawal. All these solutions are written as per the latest guidelines of CBSE. We provides step by step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3 Trigonometric Functions of NCERT Class 11 Maths is an important chapter for students. Though the chapter has more mathematical terms and formulae, We have made NCERT Solutions for Class 11 Maths easy for the students to understand and remember them, using tricks.
Trigonometry is developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is being used in many areas such as finding the heights of tides in the ocean, designing electronic circuits, etc., In the earlier classes, students have already studied trigonometric identities and applications of trigonometric ratios. In Class 11, trigonometric ratios are generalised to trigonometric function and their properties. However, these NCERT Solutions of GoyankaMathsStudy help the students to attain more knowledge and score full marks in this chapter.

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Access NCERT Solutions for Class 11 Maths Chapter 3

Exercise 3.1 page: 54

1. Find the radian measures corresponding to the following degree measures:

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

Solution:

(iv) 520°

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7)

(i) 11/16

(ii) -4

(iii) 5π/3

(iv) 7π/6

Solution:

(i) 11/16

Here π radian = 180°

(ii) -4

Here π radian = 180°

(iii) 5π/3

Here π radian = 180°

We get

= 300o

(iv) 7π/6

Here π radian = 180°

We get

= 210o

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

Solution:

It is given that

No. of revolutions made by the wheel in

1 minute = 360

1 second = 360/60 = 6

We know that

The wheel turns an angle of 2π radian in one complete revolution.

In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian

Therefore, in one second, the wheel turns an angle of 12π radian.

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 (Use π = 22/7).

Solution:

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.

Solution:

The dimensions of the circle are

Diameter = 40 cm

Radius = 40/2 = 20 cm

Consider AB be as the chord of the circle i.e. length = 20 cm

In ΔOAB,

Radius of circle = OA = OB = 20 cm

Similarly AB = 20 cm

Hence, ΔOAB is an equilateral triangle.

θ = 60° = π/3 radian

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre

We get θ = 1/r

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

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

Solution:

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

Solution:

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r

We know that r = 75 cm

(i) l = 10 cm

So we get

θ = 10/75 radian

By further simplification

θ = 2/15 radian

(ii) l = 15 cm

So we get

θ = 15/75 radian

By further simplification

θ = 1/5 radian

(iii) l = 21 cm

So we get

θ = 21/75 radian

By further simplification

θ = 7/25 radian

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Exercise 3.2 page: 63

Find the values of other five trigonometric functions in Exercises 1 to 5.

1. cos x = -1/2, x lies in third quadrant.

Solution:

2. sin x = 3/5, x lies in second quadrant.

Solution:

It is given that

sin x = 3/5

We can write it as

We know that

sin2 x + cos2 x = 1

We can write it as

cos2 x = 1 – sin2 x

3. cot x = 3/4, x lies in third quadrant.

Solution:

It is given that

cot x = 3/4

We can write it as

We know that

1 + tan2 x = sec2 x

We can write it as

1 + (4/3)2 = sec2 x

Substituting the values

1 + 16/9 = sec2 x

cos2 x = 25/9

sec x = ± 5/3

Here x lies in the third quadrant so the value of sec x will be negative

sec x = – 5/3

We can write it as

4. sec x = 13/5, x lies in fourth quadrant.

Solution:

It is given that

sec x = 13/5

We can write it as

We know that

sin2 x + cos2 x = 1

We can write it as

sin2 x = 1 – cos2 x

Substituting the values

sin2 x = 1 – (5/13)2

sin2 x = 1 – 25/169 = 144/169

sin2 x = ± 12/13

Here x lies in the fourth quadrant so the value of sin x will be negative

sin x = – 12/13

We can write it as

5. tan x = -5/12, x lies in second quadrant.

Solution:

It is given that

tan x = – 5/12

We can write it as

We know that

1 + tan2 x = sec2 x

We can write it as

1 + (-5/12)2 = sec2 x

Substituting the values

1 + 25/144 = sec2 x

sec2 x = 169/144

sec x = ± 13/12

Here x lies in the second quadrant so the value of sec x will be negative

sec x = – 13/12

We can write it as

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°

Solution:

We know that values of sin x repeat after an interval of 2π or 360°

So we get

By further calculation

= sin 45o

= 1/ √ 2

7. cosec (–1410°)

Solution:

We know that values of cosec x repeat after an interval of 2π or 360°

So we get

By further calculation

= cosec 30o = 2

8. 

Solution:

We know that values of tan x repeat after an interval of π or 180°

So we get

By further calculation

We get

= tan 60o

= √3

9. 

Solution:

We know that values of sin x repeat after an interval of 2π or 360°

So we get

By further calculation

10. 

Solution:

We know that values of tan x repeat after an interval of π or 180°

So we get

By further calculation

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Exercise 3.3 page: 73

Prove that:

1.

Solution:

2.

Solution:

Here

= 1/2 + 4/4

= 1/2 + 1

= 3/2

= RHS

3.

Solution:

4.

Solution:

5. Find the value of:

(i) sin 75o

(ii) tan 15o

Solution:

(ii) tan 15°

It can be written as

= tan (45° – 30°)

Using formula

Prove the following:

6.

Solution:

7.

Solution:

8.

Solution:

9.

Solution:

Consider

It can be written as

= sin x cos x (tan x + cot x)

So we get

10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Solution:

LHS = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x

11.

Solution:

Consider

Using the formula

12. sin2 6x – sin2 4x = sin 2x sin 10x

Solution:

13. cos2 2x – cos2 6x = sin 4x sin 8x

Solution:

We get

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

It can be written as

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

So we get

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= RHS

14. sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

Solution:

By further simplification

= 2 sin 4x cos (– 2x) + 2 sin 4x

It can be written as

= 2 sin 4x cos 2x + 2 sin 4x

Taking common terms

= 2 sin 4x (cos 2x + 1)

Using the formula

= 2 sin 4x (2 cos2 x – 1 + 1)

We get

= 2 sin 4x (2 cos2 x)

= 4cos2 x sin 4x

= R.H.S.

15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution:

Consider

LHS = cot 4x (sin 5x + sin 3x)

It can be written as

Using the formula

= 2 cos 4x cos x

Hence, LHS = RHS.

16.

Solution:

Consider

Using the formula

17.

Solution:

18.

Solution:

19.

Solution:

20.

Solution:

21.

Solution:

22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution:

23.

Solution:

Consider

LHS = tan 4x = tan 2(2x)

By using the formula

24. cos 4x = 1 – 8sin2 x cos2 x

Solution:

Consider

LHS = cos 4x

We can write it as

= cos 2(2x)

Using the formula cos 2A = 1 – 2 sin2 A

= 1 – 2 sin2 2x

Again by using the formula sin2A = 2sin A cos A

= 1 – 2(2 sin x cos x) 2

So we get

= 1 – 8 sin2x cos2x

= R.H.S.

25. cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1

Solution:

Consider

L.H.S. = cos 6x

It can be written as

= cos 3(2x)

Using the formula cos 3A = 4 cos3 A – 3 cos A

= 4 cos3 2x – 3 cos 2x

Again by using formula cos 2x = 2 cos2 x – 1

= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1)

By further simplification

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

We get

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

By multiplication

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

On further calculation

= 32 cos6x – 48 cos4x + 18 cos2x – 1

= R.H.S.

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Exercise 3.4 PAGE: 78

Find the principal and general solutions of the following equations:

1. tan x = √3

Solution:

2. sec x = 2

Solution:

3. cot x = – √3

Solution:

4. cosec x = – 2

Solution:

Find the general solution for each of the following equations:

5. cos 4x = cos 2x

Solution:

6. cos 3x + cos x – cos 2x = 0

Solution:

7. sin 2x + cos x = 0

Solution:

It is given that

sin 2x + cos x = 0

We can write it as

2 sin x cos x + cos x = 0

cos x (2 sin x + 1) = 0

cos x = 0 or 2 sin x + 1 = 0

Let cos x = 0

8. sec2 2x = 1 – tan 2x

Solution:

It is given that

sec2 2x = 1 – tan 2x

We can write it as

1 + tan2 2x = 1 – tan 2x

tan2 2x + tan 2x = 0

Taking common terms

tan 2x (tan 2x + 1) = 0

Here

tan 2x = 0 or tan 2x + 1 = 0

If tan 2x = 0

tan 2x = tan 0

We get

2x = nπ + 0, where n ∈ Z

x = nπ/2, where n ∈ Z

tan 2x + 1 = 0

We can write it as

tan 2x = – 1

So we get

Here

2x = nπ + 3π/4, where n ∈ Z

x = nπ/2 + 3π/8, where n ∈ Z

Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.

9. sin x + sin 3x + sin 5x = 0

Solution:

It is given that

sin x + sin 3x + sin 5x = 0

We can write it as

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

Using the formula

By further calculation

2 sin 3x cos (-2x) + sin 3x = 0

It can be written as

2 sin 3x cos 2x + sin 3x = 0

By taking out the common terms

sin 3x (2 cos 2x + 1) = 0

Here

sin 3x = 0 or 2 cos 2x + 1 = 0

If sin 3x = 0

3x = nπ, where n ∈ Z

We get

x = nπ/3, where n ∈ Z

If 2 cos 2x + 1 = 0

cos 2x = – 1/2

By further simplification

= – cos π/3

= cos (π – π/3)

So we get

cos 2x = cos 2π/3

Here

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Miscellaneous Exercise page: 81

Prove that:

1.

Solution:

We get

= 0

= RHS

2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Solution:

Consider

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

By further calculation

= sin 3x sin x + sin2 x + cos 3x cos x – cos2 x

Taking out the common terms

= cos 3x cos x + sin 3x sin x – (cos2 x – sin2 x)

Using the formula

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

So we get

= cos 2x – cos 2x

= 0

= RHS

3.

Solution:

Consider

LHS = (cos x + cos y) 2 + (sin x – sin y) 2

By expanding using formula we get

= cos2 x + cos2 y + 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y

Grouping the terms

= (cos2 x + sin2 x) + (cos2 y + sin2 y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – sin A sin B)

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

By further calculation

= 2 + 2 cos (x + y)

Taking 2 as common

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

From the formula cos 2A = 2 cos2 A – 1

4.

Solution:

LHS = (cos x – cos y) 2 + (sin x – sin y) 2

By expanding using formula

= cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y – 2 sin x sin y

Grouping the terms

= (cos2 x + sin2 x) + (cos2 y + sin2 y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

= 1 + 1 – 2 [cos (x – y)]

By further calculation

= 2 [1 – cos (x – y)]

From formula cos 2A = 1 – 2 sin2 A

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Solution:

6.

Solution:

7.

Solution:

8. Find sin x/2, cos x/2 and tan x/2 in each of the following:

Solution:

cos x = -3/5

From the formula

9. cos x = -1/3, x in quadrant III

Solution:

10. sin x = 1/4, x in quadrant II

Solution:

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