CBSE Class 11 Maths – Chapter 3 Trigonometric Functions- Study Materials

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CBSE Class 11 Maths Notes Chapter 3 Trigonometric Functions

Angle
Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side and the final position of ray after rotation is called terminal side of the angle. The point of rotation is called vertex. If the direction of rotation is anti-clockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative.

Measuring Angles
There are two systems of measuring angles
Sexagesimal system (degree measure): If a rotation from the initial side to terminal side is (1360)th of a revolution, the angle is said to have a measure of one degree, written as 1°.
One sixtieth of a degree is called a minute, written as 1′ and one-sixtieth of a minute is called a second, written as 1″
Thus, 1° = 60′ and 1′ = 60″

Circular system (radian measure): A radian is an angle subtended at the centre of a circle by an arc, whose length is equal to the radius of the circle. We denote 1 radian by 1°.

Relation Between Radian and Degree
We know that a complete circle subtends at its centre an angle whose measure is 2π radians as well as 360°.
2π radian = 360°.
Hence, π radian = 180°
or 1 radian = 57° 16′ 21″ (approx)
1 degree = 0.01746 radian

Six Fundamental Trigonometric Identities

  • sinx = 1cosecx
  • cos x = 1secx
  • tan x = 1cotx
  • sin2 x + cos2 x = 1
  • 1 + tan2x = sec2 x
  • 1 + cot2 x = cosec2 x

Trigonometric Functions – Class 11 Maths Notes

Trigonometric ratios are defined for acute angles as the ratio of the sides of a right angled triangle. The extension of trigonometric ratios to any angle in terms of radian measure (real number) are called trigonometric function. The signs of trigonometric function in different quadrants have been given in following table.

 IIIIIIIV
Sin x++––
Cos x+––+
Tan x+–+–
Cosec x++––
Sec x+––+
Cot x+–+–

Domain and Range of Trigonometric Functions

FunctionsDomainRange
SineR[-1, 1]
CosR[-1, 1]
TanR – {(2n + 1) π2 : n ∈ ZR
CotR – {nπ: n ∈ Z}R
SecR – {(2n + 1) π2 : n ∈ ZR – (-1, 1)
CosecR – {nπ: n ∈ Z}R – (-1, 1)

Sine, Cosine, and Tangent of Some Angles Less Than 90°

Allied or Related Angles
The angles nπ2±θ are called allied or related angle and θ ± n × (2π) are called coterminal angles. For general reduction, we have following rules, the value of trigonometric function for (nπ2±θ) is numerically equal to

  • the value of the same function, if n is an even integer with the algebraic sign of the function as per the quadrant in which angle lies.
  • the corresponding co-function of θ, if n is an odd integer with the algebraic sign of the function for the quadrant in which it lies, here sine and cosine, tan and cot, sec and cosec are cofunctions of each other.

Functions of Negative Angles

For any acute angle of θ.
We have,

  • sin(-θ) = – sinθ
  • cos (-θ) = cosθ
  • tan (-θ) = – tanθ
  • cot (-θ) = – cotθ
  • sec (-θ) = secθ
  • cosec (-θ) = – cosecθ

Some Formulae Regarding Compound Angles

An angle made up of the sum or difference of two or more angles is called compound angles. The basic results in direction are called trigonometric identities as given below:
(i) sin (x + y) = sin x cos y + cos x sin y
(ii) sin (x – y) = sin x cos y – cos x sin y
(iii) cos (x + y) = cos x cos y – sin x sin y
(iv) cos (x – y) = cos x cos y + sin x sin y

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

Transformation Formulae

  • 2 sin x cos y = sin (x + y) + sin (x – y)
  • 2 cos x sin y = sin (x + y) – sin (x – y)
  • 2 cos x cos y = cos (x + y) + cos (x – y)
  • 2 sin x sin y = cos (x – y) – cos (x + y)
  • sin x + sin y = 2 sin(x+y2) cos(x−y2)
  • sin x – sin y = 2 cos(x+y2) sin(x−y2)
  • cos x + cos y = 2 cos(x+y2) cos(x−y2)
  • cos x – cos y = -2 sin(x+y2) sin(x−y2)

Trigonometric Ratios of Multiple Angles

Product of Trigonometric Ratios

  • sin x sin (60° – x) sin (60° + x) = 14 sin 3x
  • cos x cos (60° – x) cos (60° + x) = 14 cos 3x
  • tan x tan (60° – x) tan (60° + x) = tan 3x
  • cos 36° cos 72° = 14
  • cos x . cos 2x . cos 22x . cos 23x … cos 2n-1 = sin2nx2nsinx

Sum of Trigonometric Ratio, if Angles are in A.P.

Trigonometric Equations
Equation which involves trigonometric functions of unknown angles is known as the trigonometric equation.

Solution of a Trigonometric Equation
A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation.
A trigonometric equation may have an infinite number of solutions.

Principal Solution
The solutions of a trigonometric equation for which 0 ≤ x ≤ 2π are called principal solutions.

General Solutions
A solution of a trigonometric equation, involving ‘n’ which gives all solution of a trigonometric equation is called the general solutions.

General Solutions of Trigonometric Equation

  • sin x = 0 ⇔ x = nπ, n ∈ Z
  • cos x = 0 ⇔ x = (2n + 1) π2 , n ∈ Z
  • tan x = 0 ⇔ x = nπ, n ∈ Z
  • sin x = sin y ⇔ x = nπ + (-1)n y, n ∈ Z
  • cos x = cos y ⇔ x = 2nπ ± y, n ∈ Z
  • tan x = tan y ⇔ x = nπ ± y, n ∈ Z
  • sin2 x = sin2 y ⇔ x = nπ ± y, n ∈ Z
  • cos2 x = cos2 y ⇔ x = nπ ± y, n ∈ Z
  • tan2 x = tan2 y ⇔ x = nπ ± y, n ∈ Z

Basic Rules of Triangle

In a triangle ABC, the angles are denoted by capital letters A, B and C and the lengths of sides of opposite to these angles are denoted by small letters a, b and c, respectively.
Sine Rule
sinAa=sinBb=sinCc

Cosine Rule
a2 = b2 + c2 – 2bc cos A
b2 = c2 + a2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C

Projection Rule
a = b cos C + c cos B
b = c cos A + a cos C
c = a cos B + b cos A

Trigonometric Functions Class 11 MCQs Questions with Answers

Question 1.
The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is
(a) sin (x + y)
(b) sin² (x + y)
(c) sin³ (x + y)
(d) sin4 (x + y)

Answer

Answer: (b) sin² (x + y)
Hint:
cos² x + cos² y – 2cos x × cos y × cos(x + y)
{since cos(x + y) = cos x × cos y – sin x × sin y }
= cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y)
= cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y
= cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y
= (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y
= cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y
= sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 )
= sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y
= (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y
= (sin x × cos y + sin y × cos x)²
= {sin (x + y)}²
= sin² (x + y)

Question 2.
If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is
(a) a² + b² + c²
(b) a² – b² – c²
(c) a² – b² + c²
(d) a² + b² – c²

 
Answer

Answer: (d) a² + b² – c²
Hint:
We have
(a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²

Question 3.
If cos a + 2cos b + cos c = 2 then a, b, c are in
(a) 2b = a + c
(b) b² = a × c
(c) a = b = c
(d) None of these

Answer

Answer: (a) 2b = a + c
Hint:
Given, cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)
⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2)
⇒ cos((A-C)/2) = 2sin (B/2)
⇒ cos((A-C)/2) = 2cos((A+C)/2)
⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
⇒ 2sin(A/2)sin(C/2) = sin(B/2)
⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2(s – b) = b
⇒ a + b + c – 2b = b
⇒ a + c – b = b
⇒ a + c = 2b

Question 4.
The value of cos 5π is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (c) -1
Hint:
Given, cos 5π = cos (π + 4π) = cos π = -1

Question 5.
In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals
(a) none of these
(b) c/a
(c) 1
(d) a/c

Answer

Answer: (c) 1
Hint:
Given cosec A (sin B cos C + cos B sin C)
= cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1

Question 6.
If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is
(a) 4 : (√5 – 1)
(b) 5 : 4
(c) (√5 – 1) : 4
(d) none of these

Answer

Answer: (a) 4 : (√5 – 1)
Hint:
Given, the angles of a triangle be in the ratio 1 : 4 : 5
⇒ x + 4x + 5x = 180
⇒ 10x = 180
⇒ x = 180/10
⇒ x = 18
So, the angle are: 18, 72, 90
Since a : b : c = sin A : sin B : sin C
⇒ a : b : c = sin 18 : sin 72 : sin 90
⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1
⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4
Now, c /a = 4/(√5 – 1)
⇒ c : a = 4 : (√5 – 1)

Question 7.
The value of cos 180° is
(a) 0
(b) 1
(c) -1
(d) infinite

Answer

Answer: (c) -1
Hint:
180 is a standard degree generally we all know their values but if we want to go theoretically then
cos(90 + x) = – sin(x)
So, cos 180 = cos(90 + 90)
= -sin 90
= -1 {sin 90 = 1}
So, cos 180 = -1

Question 8.
The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is
(a) 30°
(b) 90°
(c) 60°
(d) 120°

Answer

Answer: (b) 90°
Hint:
Let the lengths of the sides if ∆ABC be a, b and c
Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3
⇒ (sinA + sinB + sinC) = ( a + b + c)/2
⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2
From sin formula,Using
sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2
Now, sinB/b = 1/2
Given b = 2
So, sinB/2 = 1/2
⇒ sinB = 1
⇒ B = π/2

Question 9:
If 3 × tan(x – 15) = tan(x + 15), then the value of x is
(a) 30
(b) 45
(c) 60
(d) 90

Answer

Answer: (b) 45
Hint:
Given, 3×tan(x – 15) = tan(x + 15)
⇒ tan(x + 15)/tan(x – 15) = 3/1
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1)
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2
⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2
⇒ sin 2x/sin 30 = 2
⇒ sin 2x/(1/2) = 2
⇒ 2 × sin 2x = 2
⇒ sin 2x = 1
⇒ sin 2x = sin 90
⇒ 2x = 90
⇒ x = 45

Question 10.
If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is
(a) π/3
(b) π/2
(c) 2π/3
(d) 3π/2

Answer

Answer: (c) 2π/3
Hint:
Given, the sides of a triangle are 13, 7, 8
Since greatest side has greatest angle,
Now Cos A = (b² + c² – a²)/2bc
⇒ Cos A = (7² + 8² – 13²)/(2×7×8)
⇒ Cos A = (49 + 64 – 169)/(2×7×8)
⇒ Cos A = (113 – 169)/(2×7×8)
⇒ Cos A = -56/(2×56)
⇒ Cos A = -1/2
⇒ Cos A = Cos 2π/3
⇒ A = 2π/3
So, the greatest angle is
= 2π/3

Question 11.
The value of tan 20 × tan 40 × tan 80 is
(a) tan 30
(b) tan 60
(c) 2 tan 30
(d) 2 tan 60

Answer

Answer: (b) tan 60
Hint:
Given, tan 20 × tan 40 × tan 80
= tan 40 × tan 80 × tan 20
= [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20)
= [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20)
= [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20)
= [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20)
= [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20)
= [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20)
= [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20)
= [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}]
= (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20)
= sin 60/cos 60
= tan 60
So, tan 20 × tan 40 × tan 80 = tan 60

Question 12.
If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is
(a) 4 : (√5 – 1)
(b) 5 : 4
(c) (√5 – 1) : 4
(d) none of these

Answer

Answer: (a) 4 : (√5 – 1)
Hint:
Given, the angles of a triangle be in the ratio 1 : 4 : 5
⇒ x + 4x + 5x = 180
⇒ 10x = 180
⇒ x = 180/10
⇒ x = 18
So, the angle are: 18, 72, 90
Since a : b : c = sin A : sin B : sin C
⇒ a : b : c = sin 18 : sin 72 : sin 90
⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1
⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4
Now, c /a = 4/(√5 – 1)
⇒ c : a = 4 : (√5 – 1)

Question 13.
The general solution of √3 cos x – sin x = 1 is
(a) x = n × π + (-1)n × (π/6)
(b) x = π/3 – n × π + (-1)n × (π/6)
(c) x = π/3 + n × π + (-1)n × (π/6)
(d) x = π/3 – n × π + (π/6)

Answer

Answer: (c) x = π/3 + n × π + (-1)n × (π/6)
Hint:
√3 cos x-sin x=1
⇒ (√3/2)cos x – (1/2)sin x = 1/2
⇒ sin 60 × cos x – cos 60 × sin x = 1/2
⇒ sin (x – 60) = 1/2
⇒ sin (x – π/3) = sin 30
⇒ sin (x – π/3) = sinπ/6
⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z}
⇒ x = π/3 + n × π + (-1)n × (π/6)

Question 14.
If tan² θ = 1 – e², then the value of sec θ + tan³ θ × cosec θ is
(a) 2 – e²
(b) (2 – e²)1/2
(c) (2 – e²)²
(d) (2 – e²)3/2

Answer

Answer: (d) (2 – e²)3/2
Hint:
Given, tan² θ = 1 – e²
⇒ tan θ = √(1 – e²)
MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers 1
From the figure and Pythagorus theorem,
AC² = AB² + BC²
⇒ AC² = {√(1 – e²)}² + 12
⇒ AC² = 1 – e² + 1
⇒ AC² = 2 – e²
⇒ AC = √(2 – e²)
Now, sec θ = √(2 – e²)
cosec θ = √(2 – e²)/√(1 – e²)
and tan θ = √(1 – e²)
Given, sec θ + tan³ θ × cosec θ
= √(2 – e²) + {(1 – e²)3/2 × √(2 – e²)/√(1 – e²)}
= √(2 – e²) + {(1 – e²) × (1 – e²) × √(2 – e²)/√(1 – e²)}
= √(2 – e²) + (1 – e²) × √(2 – e²)
= √(2 – e²) × (1 + 1 – e²)
= √(2 – e²) × (2 – e²)
= (2 – e²)3/2
So, sec θ + tan³ θ × cosec θ = (2 – e²)3/2

Question 15.
The value of cos 20 + 2sin² 55 – √2 sin65 is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (b) 1
Hint:
Given, cos 20 + 2sin² 55 – √2 sin65
= cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x}
= 1 + cos 20 – cos 110 – √2 sin65
= 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula}
= 1 – 2 × sin 65 × sin (-45) – √2 sin65
= 1 + 2 × sin 65 × sin 45 – √2 sin65
= 1 + (2 × sin 65)/√2 – √2 sin65
= 1 + √2 ( sin 65 – √2 sin 65
= 1
So, cos 20 + 2sin² 55 – √2 sin65 = 1

Question 16.
If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is
(a) 2π/3
(b) π/3
(c) π/2
(d) π/6

Answer

Answer: (a) 2π/3
Hint:
Let S be the center of the circumcircle of triangle PQR.
So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii.
Thus SPQ & SPR are equilateral triangles.
⇒ ∠QSP = 60°;
Similarly ∠RQP = 60°
⇒ Angle at the center QSP = 120°
So, SRPQ is a rhombus, since all the four sides are equal.
Hence, its opposite angles are equal; so ∠P = ∠QSP = 120°

Question 17.
If cos a + 2cos b + cos c = 2 then a, b, c are in
(a) 2b = a + c
(b) b² = a × c
(c) a = b = c
(d) None of these

Answer

Answer: (a) 2b = a + c
Hint:
Given, cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2)
⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2)
⇒ cos((A-C)/2) = 2sin (B/2)
⇒ cos((A-C)/2) = 2cos((A+C)/2)
⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
⇒ 2sin(A/2)sin(C/2) = sin(B/2)
⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2(s – b) = b
⇒ a + b + c – 2b = b
⇒ a + c – b = b
⇒ a + c = 2b

Question 18.
The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is
(a) sin x
(b) sin 2x
(c) sin 3x
(d) sin 4x

Answer

Answer: (c) sin 3x
Hint:
Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3)
= 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3}
= 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2}
= 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4}
= sin x × {-sin 2x + 3 × cos 2x}
= sin x × {-sin 2x + 3 × (1 – sin 2x)}
= sin x × {-sin 2x + 3 – 3 × sin 2x}
= sin x × {3 – 4 × sin 2x}
= 3 × sin x – 4 sin 3x
= sin 3x
So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Question 19.
If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is
(a) x + y
(b) 1/x + y
(c) x + 1/y
(d) 1/x + 1/y

Answer

Answer: (d) 1/x + 1/y
Hint:
Given,
tan A – tan B = x ……………. 1
and cot B – cot A = y ……………. 2
From equation,
1/cot A – 1/cot B = x
⇒ (cot B – cot A)/(cot A × cot B) = x
⇒ y/(cot A × cot B) = x {from equation 2}
⇒ y = x × (cot A × cot B)
⇒ cot A × cot B = y/x
Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A)
⇒ cot (A – B) = (y/x + 1)/y
⇒ cot (A – B) = (y/x) × (1/y) + 1/y
⇒ cot (A – B) = 1/x + 1/y

Question 20.
The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is
(a) tan 6x
(b) 2 tan 6x
(c) 3 tan 6x
(d) 4 tan 6x

Answer

Answer: (b) 2 tan 6x
Hint:
Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x)
⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}]
⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}]
⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x)
⇒ tan 6x + tan 6x
⇒ 2 tan 6x

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