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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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
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
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.
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
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:
