CLASS 11 MATH CHAPTER 3 NCERT SOLUTION
TRIGNOMETRIC FUNCTIONS
Exercise 3.1 Page No: 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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