NCERT Solutions for Class 11 Maths Chapter 3
Chapter 3.1: Introduction to Trigonometry
- Definition: The study of relationships between the sides and angles of a triangle.
- Trigonometric Ratios: Sine, cosine, tangent, cotangent, secant, cosecant.
- Trigonometric Identities: Fundamental relationships between trigonometric functions.
Chapter 3.2: Trigonometric Functions
- Domain and Range: The set of possible input and output values for trigonometric functions.
- Graphs of Trigonometric Functions: Visual representation of the behavior of trigonometric functions.
- Periodic Functions: Functions that repeat their values after a fixed interval (period).
Chapter 3.3: Trigonometric Identities
- Sum and Difference Formulas: Formulas for trigonometric functions of the sum or difference of angles.
- Multiple Angle Formulas: Formulas for trigonometric functions of multiples of an angle.
Chapter 3.4: Trigonometric Equations
- Solving Trigonometric Equations: Finding the values of x that satisfy a given trigonometric equation.
- General Solution: Expressing the solution of a trigonometric equation in terms of the principal solution and the period of the function.
Key Concepts:
- Trigonometric ratios and their relationships
- Trigonometric functions and their graphs
- Trigonometric identities
- Solving trigonometric equations
This chapter provides a foundational understanding of trigonometry, which is essential for various mathematical and scientific applications.
NCERT Solutions for Class 11 Maths Chapter 3
Exercise 3.1
1. Find the radian measures corresponding to the following degree measures:
(i) 25° (ii) – 47°30′ (iii) 240° (iv) 520°
Ans :
o convert from degrees to radians, we multiply by π/180.
(i) 25° = 25 * (π/180) = 5π/36 radians
(ii) -47°30′ = -47.5° = -47.5 * (π/180) = -19π/72 radians
(iii) 240° = 240 * (π/180) = 4π/3 radians
(iv) 520° = 520 * (π/180) = 26π/9 radians
2.
Ans :
To convert from radians to degrees, we multiply by 180/π.
Given: π = 22/7
(i) 11π/16 = 11 * (22/7) * (180/π) = 11 * 22 * 180 / 7 = 5820/7 ≈ 831.43°
(ii) -4 = -4 * (180/π) ≈ -4 * 180 * 7/22 ≈ -229.09°
(iii) 5π/3 = 5 * (22/7) * (180/π) = 5 * 22 * 180 / 7 = 2828.57° ≈ 300°
(iv) 7π/6 = 7 * (22/7) * (180/π) = 7 * 22 * 180 / 7 = 2520° ≈ 210°
3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Ans :
Given:
- The wheel makes 360 revolutions in one minute.
Steps:
- Convert minutes to seconds:
- 1 minute = 60 seconds
- Calculate revolutions per second:
- Revolutions per second = 360 revolutions / 60 seconds = 6 revolutions/second
- Convert one revolution to radians:
- 1 revolution = 2π radians
- Calculate radians per second:
- Radians per second = 6 revolutions/second * 2π radians/revolution = 12π radians/second
Therefore, the wheel turns through 12π radians in one second.
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).
Ans :
- Radius of the circle (r) = 100 cm
- Arc length (s) = 22 cm
Formula:
- θ (in radians) = s / r
Calculation:
- θ = 22 cm / 100 cm = 0.22 radians
Converting radians to degrees:
- 1 radian = 180° / π
- θ (in degrees) = 0.22 * (180° / π) ≈ 0.22 * (180° * 7/22) ≈ 12.63°
Therefore, the angle subtended at the center of the circle is approximately 12.63 degrees.
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.
Ans :
Given:
- Diameter of the circle = 40 cm
- Length of the chord = 20 cm
Steps:
- Find the radius of the circle:
- Radius = Diameter / 2 = 40 cm / 2 = 20 cm
- Determine the type of triangle formed:
- Since the chord is equal to the radius, the triangle formed by the chord and the radii to its endpoints is an equilateral triangle.
- Find the central angle θ:
- Convert 60 degrees to radians: θ = 60 * (π/180) = π/3 radians
- Use the formula for arc length:
- Arc length (s) = r * θ
- s = 20 cm * (π/3) ≈ 20.94 cm
Therefore, the length of the minor arc of the chord is approximately 20.94 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.
Ans :
Given:
- Two circles with arcs of equal length.
- One arc subtends an angle of 60° at the center.
- The other arc subtends an angle of 75° at the center.
Formula:
- Arc length (s) = r * θ
- (where r is the radius and θ is the angle in radians)
Approach:
- Since the arc lengths are equal, we can set up an equation using the formula for arc length.
- Let r1 and r2 be the radii of the two circles, respectively.
- Then, we can write:
- r1 * (60° * π/180) = r2 * (75° * π/180)
Simplifying the equation:
- r1 * (π/3) = r2 * (5π/12)
- r1 / r2 = (5π/12) / (π/3)
- r1 / r2 = 5/4
radii of the two circles is 5:4.
7. Find the angle in radian through which a pendulum swings if its length is 75 cm and th e tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm
Ans :
θ = s / r
where:
- θ is the angle in radians
- s is the arc length
- r is the radius (length of the pendulum)
In this case, the radius (r) is 75 cm.
(i) When the arc length (s) is 10 cm: θ = 10 cm / 75 cm = 2/15 radians
(ii) When the arc length (s) is 15 cm: θ = 15 cm / 75 cm = 1/5 radians
(iii) When the arc length (s) is 21 cm: θ = 21 cm / 75 cm = 7/25 radians
Therefore, the angles in radians through which the pendulum swings for the given arc lengths are:
(i) 2/15 radians
(ii) 1/5 radians
(iii) 7/25 radians
Exercise 3.2
Find the values of other five trigonometric functions in Exercises 1 to 5.
1. cos x = – 1/ 2 , x lies in third quadrant.
2. sin x = 3/ 5 , x lies in second quadrant.
3. cot x = 4 /3 , x lies in third quadrant.
4. sec x = 13/ 5 , x lies in fourth quadrant.
5. tan x = – 5 /12 , x lies in second quadrant.
Ans :
1. cos x = -1/2, x lies in third quadrant.
- sin x: Since x is in the third quadrant, sin x is negative. Using the Pythagorean identity:
- sin^2 x + cos^2 x = 1
- sin^2 x + (-1/2)^2 = 1
- sin^2 x = 3/4
- sin x = -√3/2
- tan x: tan x = sin x / cos x = (-√3/2) / (-1/2) = √3
- cot x: cot x = 1 / tan x = 1 / √3 = √3/3
- sec x: sec x = 1 / cos x = 1 / (-1/2) = -2
- csc x: csc x = 1 / sin x = 1 / (-√3/2) = -2√3/3
2.
- cos x: Since x is in the second quadrant, cos x is negative. Using the Pythagorean identity:
- sin^2 x + cos^2 x = 1
- (3/5)^2 + cos^2 x = 1
- cos^2 x = 16/25
- cos x = -4/5
- tan x: tan x = sin x / cos x = (3/5) / (-4/5) = -3/4
- cot x: cot x = 1 / tan x = 1 / (-3/4) = -4/3
- sec x: sec x = 1 / cos x = 1 / (-4/5) = -5/4
- csc x: csc x = 1 / sin x = 1 / (3/5) = 5/3
3.
- tan x: tan x = 1 / cot x = 1 / (4/3) = 3/4
- sin x: Since x is in the third quadrant, sin x is negative. Using the Pythagorean identity:
- 1 + tan^2 x = sec^2 x
- 1 + (3/4)^2 = sec^2 x
- sec^2 x = 25/16
- sec x = -5/4
- cos x = 1 / sec x = -4/5
- sin x = tan x * cos x = (3/4) * (-4/5) = -3/5
- csc x: csc x = 1 / sin x = 1 / (-3/5) = -5/3
4.
- cos x: cos x = 1 / sec x = 1 / (13/5) = 5/13
- sin x: Since x is in the fourth quadrant, sin x is negative. Using the Pythagorean identity:
- sin^2 x + cos^2 x = 1
- sin^2 x + (5/13)^2 = 1
- sin^2 x = 144/169
- sin x = -12/13
- tan x: tan x = sin x / cos x = (-12/13) / (5/13) = -12/5
- cot x: cot x = 1 / tan x = 1 / (-12/5) = -5/12
- csc x: csc x = 1 / sin x = 1 / (-12/13) = -13/12
5. tan x = -5/12, x lies in second quadrant.
- cot x: cot x = 1 / tan x = 1 / (-5/12) = -12/5
- sin x: Since x is in the second quadrant, sin x is positive. Using the Pythagorean identity:
- 1 + tan^2 x = sec^2 x
- 1 + (-5/12)^2 = sec^2 x
- sec^2 x = 169/144
- sec x = -13/12
- cos x = 1 / sec x = -12/13
- sin x = tan x * cos x = (-5/12) * (-12/13) = 5/13
- csc x: csc x = 1 / sin x = 1 / (5/13) = 13/5
Find the values of the trigonometric functions in Exercises 6 to 10.
6. sin 765°
7. cosec (– 1410°)
8. tan 19π /3
9. sin (– 11π /3 )
10. cot (– 15π/ 4 )
Ans :
6. sin 765°
- 765° = 2 * 360° + 45°
- Since sin(360° + θ) = sin(θ), we have:
- sin 765° = sin 45° = √2/2
7. cosec (-1410°)
- csc(-1410°) = -csc(1410°) (since cosecant is odd)
- 1410° = 3 * 360° + 330°
- csc(-1410°) = -csc(330°) = -csc(360° – 30°) = -csc(30°) = -2
8. tan (19π/3)
- 19π/3 = 6π + π/3
- Since tan(2π + θ) = tan(θ), we have:
- tan (19π/3) = tan (π/3) = √3
9. sin (-11π/3)
- sin(-11π/3) = -sin(11π/3) (since sine is odd)
- 11π/3 = 3π + 2π/3
- sin(-11π/3) = -sin(2π/3) = -√3/2
10. cot (-15π/4)
- cot(-15π/4) = -cot(15π/4) (since cotangent is odd)
- 15π/4 = 4π + 3π/4
- cot(-15π/4) = -cot(3π/4) = 1
NCERT Solutions for Class 11 Maths Chapter 3
Exercise 3.3
Prove that:
1. sin2 π/6 + cos2 π/3 – tan2 π/4 = – ½
Ans :
Given: sin²(π/6) + cos²(π/3) – tan²(π/4) = -1/2
Solution:
Let’s calculate each term separately:
- sin²(π/6): The sine of π/6 (30 degrees) is 1/2. So, sin²(π/6) = (1/2)² = 1/4.
- cos²(π/3): The cosine of π/3 (60 degrees) is also 1/2. So, cos²(π/3) = (1/2)² = 1/4.
- tan²(π/4): The tangent of π/4 (45 degrees) is 1. So, tan²(π/4) = 1².
Substituting these values into the original equation:
(1/4) + (1/4) – 1 = -1/2 1/2 – 1
= -1/2 -1/2 = -1/2
Conclusion: The left side of the equation equals the right side. Therefore, the identity is proven:
sin²(π/6) + cos²(π/3) – tan²(π/4)
= -1/2
2. sin2 π/6 + cosec2 7π/6 cos2 π/3 = 3/2
Ans :
3. cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 = 6
Ans :
1. cot²(π/6):
- cot(π/6) = 1 / tan(π/6)
- = 1 / (1/√3)
- = √3
- cot²(π/6) = (√3)² = 3
2. cosec(5π/6):
- 5π/6 is in the second quadrant, where cosecant is positive.
- cosec(5π/6) = 1 / sin(5π/6) = 1 / (1/2) = 2
3. tan²(π/6):
- tan(π/6) = 1/√3
- tan²(π/6) = (1/√3)² = 1/3
3 + 2 + 3 * (1/3) = 6
Simplifying:
3 + 2 + 1 = 6 6 = 6
Therefore, the given identity is true.
4. 2 sin2 3π/4 + 2 cos2 π/4 + 2 sec2 π/3 = 10
Ans :
1. sin²(3π/4):
- sin(3π/4) = 1/√2
- sin²(3π/4) = (1/√2)² = 1/2
2. cos²(π/4):
- cos(π/4) = 1/√2
- cos²(π/4) = (1/√2)² = 1/2
3. sec²(π/3):
- sec(π/3) = 1 / cos(π/3) = 1 / (1/2) = 2
- sec²(π/3) = 2² = 4
2 * (1/2) + 2 * (1/2) + 2 * 4 = 10
Simplifying:
1 + 1 + 8 = 10 10 = 10
Therefore, the given identity is true.
5. Find the value of:
(i) sin 75° (ii) tan 15°
Ans :
1. sin 75°
- 75° = 150°/2
- cos 150° = -√3/2 (since 150° is in the second quadrant)
Therefore,
sin 75° = √[(1 – (-√3/2))/2] = √[(2 + √3)/4] = (√6 + √2)/4
2. tan 15°
- 15° = 30°/2
- cos 30° = √3/2
Therefore,
tan 15° = √[(1 – √3/2)/(1 + √3/2)] = √[(2 – √3)/(2 + √3)] = 2 – √3
So, the values are:
(i) sin 75° = (√6 + √2)/4 (ii) tan 15° = 2 – √3
Prove the following:
6. cos(π/4−x)cos(π/4−y)−sin(π/4−x)sin(π/4−y) = sin (x + y)
Ans :
Given: cos(π/4−x)cos(π/4−y)−sin(π/4−x)sin(π/4−y) = sin (x + y)
Proof:
We will use the following trigonometric identity:cos(A + B) = cosAcosB – sinAsinB
In our case, we have:
- A = π/4 – x
- B = π/4 – y
- cos((π/4 – x) + (π/4 – y))
Simplifying:
- cos(π/2 – (x + y))
Using the identity cos(π/2 – θ) = sin θ, we get:
- sin(x + y)
7. tan(π/4+x)/tan(π/4−x)=(1+tanx1−tanx)2
Ans :
tan(A + B) = (tan A + tan B) / (1 – tan A * tan B)
In this case, A = π/4 and B = x.
Substituting into the formula:
tan(π/4 + x) = (tan(π/4) + tan x) / (1 – tan(π/4) * tan x)
Since tan(π/4) = 1, we get:
tan(π/4 + x) = (1 + tan x) / (1 – tan x)
Now, we can rewrite the original identity as:
(tan(π/4 + x) / tan(π/4 – x)) = (1 + tan x / 1 – tan x)²
Substituting the expression for tan(π/4 + x):
((1 + tan x) / (1 – tan x)) / (tan(π/4 – x)) = (1 + tan x / 1 – tan x)²
Since tan(π/4 – x) = (1 – tan x) / (1 + tan x) (using the tangent subtraction formula), we can substitute this into the equation:
((1 + tan x) / (1 – tan x)) / ((1 – tan x) / (1 + tan x)) = (1 + tan x / 1 – tan x)²
Simplifying the left side:
(1 + tan x)² / (1 – tan x)² = (1 + tan x / 1 – tan x)²
Therefore, the identity is proven.
8. cos(π+x)cos(−x)/sin(π−x)cos(π/2+x) = cot2 x
Ans :
To prove the given identity, we will use the following trigonometric identities:
- cos(π + x) = -cos(x)
- cos(-x) = cos(x)
- sin(π – x) = sin(x)
- cos(π/2 + x) = -sin(x)
Substituting these identities into the given equation, we get:
(-cos(x)) * cos(x) / (sin(x) * (-sin(x))) = cot² x
Simplifying the expression:
cos²(x) / sin²(x) = cot² x
Since cot x = cos x / sin x, we can rewrite the left side as:
(cos x / sin x)² = cot² x
Therefore, the given identity is proven.
9.
Ans :
=1= R.H.S
10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Ans :
cos(A + B) = cos A cos B – sin A sin B
In this case, let A = (n + 1)x and B = (n + 2)x.
Substituting into the identity:
cos((n + 1)x + (n + 2)x) = cos(n + 1)x cos(n + 2)x – sin(n + 1)x sin(n + 2)x
Simplifying the left side:
cos(2n + 3)x = cos(n + 1)x cos(n + 2)x – sin(n + 1)x sin(n + 2)x
Now, we can rearrange the terms to match the given identity:
sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos(2n + 3)x
Since cos(2n + 3)x is equal to cos(x) for all integer values of n (due to the periodicity of the cosine function), we have:
sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos x
Therefore, the identity is proven.
11. cos(3π/4+x)−cos(3π/4−x)=−2–√sinx
Ans :
cos(A – B) – cos(A + B) = 2sin A sin B
let A = 3π/4 and B = x.
Substituting into the identity:
cos(3π/4 – x) – cos(3π/4 + x) = 2sin(3π/4)sin(x)
Since sin(3π/4) = √2/2, we can substitute this value into the equation:
cos(3π/4 – x) – cos(3π/4 + x) = 2(√2/2)sin(x)
Simplifying:
cos(3π/4 – x) – cos(3π/4 + x) = √2sin(x)
Therefore, the given identity is proven.
12. sin2 6x – sin2 4x = sin 2x sin 10 x
Ans :
To prove this identity, we will use the following trigonometric identity:
sin² A – sin² B = sin(A + B)sin(A – B)
Substituting into the identity:
sin²(6x) – sin²(4x) = sin(6x + 4x)sin(6x – 4x)
Simplifying:
sin²(6x) – sin²(4x) = sin(10x)sin(2x)
Therefore, the given identity is proven.
13. cos2 2x cos2 6x = sin 4x sin 8x
Ans :
sin(2A) = 2sin(A)cos(A)
We can rewrite the given identity as:
2sin(4x)cos(4x) = sin(8x)sin(2x)
Now, using the identity sin(2A) = 2sin(A)cos(A) for both terms:
2(2sin(2x)cos(2x)) = 2sin(4x)sin(2x)
Simplifying:
4sin(2x)cos(2x) = 2sin(4x)sin(2x)
Dividing both sides by 2sin(2x):
2cos(2x) = sin(4x)
Since sin(4x) = 2sin(2x)cos(2x),
the given identity is proven.
14. sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
Ans :
L.H.S.= sin 2x + 2 sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x
= [2sin(2x+6×2)cos(2x−6×2)] + 2 sin 4x
[∵ sin A + sin B = 2 sin(A+B2)cos(A−B2)
= 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)
= 4 cos2 x sin 4x
= R.H.S.
Hence proved.
NCERT Solutions for Class 11 Maths Chapter 3
15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Ans :
16. Cos9x−cos5x / sin17x−sin3x =−sin2x / cos10x
Ans :
17. Sin5x+sin3x / cos5x+cos3x = tan 4x
Ans :
sin(A) + sin(B) = 2sin((A + B)/2)cos((A – B)/2)
cos(A) + cos(B) = 2cos((A + B)/2)cos((A – B)/2)
Substituting A = 5x and B = 3x into these formulas:
sin(5x) + sin(3x) = 2sin((5x + 3x)/2)cos((5x – 3x)/2) = 2sin(4x)cos(x)
cos(5x) + cos(3x) = 2cos((5x + 3x)/2)cos((5x – 3x)/2) = 2cos(4x)cos(x)
Now, we can substitute these expressions into the given identity:
(sin(5x) + sin(3x)) / (cos(5x) + cos(3x)) = (2sin(4x)cos(x)) / (2cos(4x)cos(x))
Simplifying:
(sin(4x)cos(x)) / (cos(4x)cos(x)) = sin(4x) / cos(4x)
Since tan(x) = sin(x) / cos(x), we have:
sin(4x) / cos(4x) = tan(4x)
Therefore, the given identity is proven.
18. sinx−siny/cosx+cosy =tanx−y/2
Ans :
.
19. Sinx+sin3x / cosx+cos3x = tan 2x
Ans :
20. Sinx−sin3x / sin2x−cos2x = 2 sin x
Ans :
It is known that
sin A – sin B = 2 cos(A+B /2)
sin(A−B/ 2)cos2 A – sin2 A = cos 2A
∴ L.H.S. = sinx−sin3x / sin2x−cos2x
= 2cos(x+3x/2)sin(x−3x/2)/ 2cos2x
= – 2 × (- sin x) = 2 sin x
= R.H.S
Hence proved.
21. Cos4x+cos3x+cos2x / sin4x+sin3x+sin2x= cot 3x
Ans :
22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Ans :
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 – [cot2xcotx−1 / cotx+cot2x] (cot 2x + cot x)
[∵ cot (A + B) = cotAcotB−1 / cotA+cotB]
= cot x cot 2x – (cot 2x cot x – 1)
= 1 = R.H.S.
Hence proved.
23. tan 4x = 4tanx(1−tan2x) / 1−6tan2x+tan4x
Ans :
.
24. cos 4x = 1 – 8sin2 x cos2 x
Ans :
To prove the given identity, we will use the following trigonometric identities:
cos(2A) = 1 – 2sin²(A) sin(2A) = 2sin(A)cos(A)
Substituting A = 2x into the first identity:
cos(4x) = 1 – 2sin²(2x)
Now, using the identity sin(2A) = 2sin(A)cos(A) with A = x:
sin²(2x) = (2sin(x)cos(x))²
Substituting this into the previous equation:
cos(4x) = 1 – 2(2sin(x)cos(x))²
Simplifying:
cos(4x) = 1 – 8sin²(x)cos²(x)
Therefore, the given identity is proven.
25. cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1
Ans :
To prove this identity, we will use the double angle formula for cosine:
cos(2A) = 2cos²(A) – 1
We can start by expressing cos(6x) in terms of cos(2x) using this formula:
cos(6x) = cos(2(3x)) = 2cos²(3x) – 1
Now, we can use the double angle formula again to express cos²(3x) in terms of cos(6x):
cos²(3x) = (1 + cos(6x))/2
Substituting this into the previous equation:
cos(6x) = 2((1 + cos(6x))/2) – 1
Simplifying:
cos(6x) = 1 + cos(6x) – 1
cos(6x) = cos(6x)
This is a tautology, meaning it is always true. Therefore, the given identity is proven.
NCERT Solutions for Class 11 Maths Chapter 3
FAQ’s
What is Class 11 Maths Chapter 3 about?
Class 11 Maths Chapter 3 covers Trigonometric Functions, including angles, radian measure, identities, and domain–range concepts.
Why are trigonometric functions important in Class 11 Maths?
They form the base for higher mathematics, physics, engineering, and help understand angles, periodic motion, and real-life applications.
What are the key topics in Chapter 3 Trigonometric Functions?
The chapter includes radian and degree measures, trigonometric ratios, identities, signs of functions, and graphs of basic functions.
How do NCERT Solutions help with Chapter 3?
NCERT Solutions provide step-by-step explanations that make complex trigonometric concepts easier to understand and revise for exams.
Where can I find free study material for Class 11 Maths Chapter 3?
You can access free notes, examples, and solutions for Class 11 Maths Chapter 3 – Trigonometric Functions – on various trusted educational platforms.
