CBSE Class 11 Maths Revision Notes Chapter 3

One of the most crucial topics in Class 11 Mathematics is trigonometry functions. It describes how a right-angle triangle’s sides and angles relate to one another. Students looking to understand the fundamentals of trigonometric functions should refer to the Class 11 Mathematics Chapter 3 Notes.

The Chapter 3 Mathematics Class 11 Notes give students a quick overview of all the topics and formulas covered, which gives them more confidence when solving questions based on trigonometry.

For convenience, these trigonometric functions Class 11 Mathematics Notes Chapter 3 are organised systematically. It is advised that the students consult these Class 11 Mathematics Notes Chapter 3 as needed to prepare the entire chapter effectively.

The Meaning of Trigonometry

Tri Gon Metron

↓ ↓ ↓

3 sides measure

As a result, this branch of mathematics was developed in antiquity to count the three sides, three angles, and six parts of a triangle. There are many applications for time-trigonometric functions today. In a right-angled triangle, the two fundamental functions are the sine and cosine of an angle, and there are four additional derivative functions.

Basic Trigonometric Identities

(a) sin2θ + cos2 θ=1:−1⩽sinθ⩽1;−1⩽cosθ⩽1∀θ∈R

(b) sec2θ − tan2θ=1:|secθ|⩾1∀θ∈R

Trigonometric Ratios of Standard Angles:

The following can be given as the relationship between these trigonometric identities and the triangle sides:

Sine θ = Opposite/Hypotenuse
Cos θ = Adjacent/Hypotenuse
Tan θ = Opposite/Adjacent
Cot θ = Adjacent/Opposite
Cosec θ = Hypotenuse/Opposite
Sec θ = Hypotenuse/Adjacent

Trigonometric Ratios of Allied Angles

Using the trigonometric ratio of allied angles, we can calculate the trigonometric ratios of angles of any value.

Sin(–θ)=–Sinθ
Cos(–θ)=Cosθ
Tan(–θ)=–Tanθ
Sin(90o–θ)=Cosθ
Cos(90o–θ)=Sinθ
Tan(90o–θ)=Cotθ
Sin(180o–θ)=Sinθ
Cos(180o–θ)=–Cosθ
Tan(180o–θ)=–Tanθ
Sin(270o–θ)=–Cosθ
Cos(270o–θ)=–Sinθ
Tan(270o–θ)=Cotθ
Sin(90o+θ)=Cosθ
Cos(90o+θ)=–Sinθ
Tan(90o+θ)=–Cotθ
Sin(180o+θ)=–Sinθ
Cos(180o+θ)=–Cosθ
Tan(180o+θ)=Tanθ
Sin(270o+θ)=–Cosθ
Cos(270o+θ)=Sinθ
Tan(270o+θ)=–Cotθ

Trigonometric Functions of Sum or Difference of Two Angles

(a) sin(A+B)=sinA cosB+cosAsinB

(b) sin(A−B)=sinA cosB−cosAsinB

(d) cos(A−B)=cosA cosB+sinAsinB

(e) tan(A+B)=tanA+tanB÷ 1−tanAtanB

(f) tan(A−B)=tanA−tanB÷1+tanAtanB

(g) cot(A+B)=cotA cotB−1cotB÷cotB+cotA

(h)cot(A−B)=cotA cotB+1÷cotB−cotA

Multiple Angles and Half Angles

(a) sin2A=2sinAcosA

(b) sin3A=3sinA−4 sin3A

Transformation of Products into Sum or Difference of Sines & Cosines

a) 2sinAcosB=sin(A+B)+sin(A−B)

(b) 2cosAsinB=sin(A+B)−sin(A−B)

(d) 2sinAsinB=cos(A−B)−cos(A+B)

Factorisation of the Sum or Difference of Two Sines or Cosines
Important Trigonometric Ratios
Conditional Identities
Sine and Cosine Series
Graphs of Trigonometric Functions
Trigonometric Equations

Equations utilising trigonometric functions with unknown angles are known as trigonometric equations.

A solution is the value of the unknown angle that answers a trigonometric equation.

As a result, the trigonometric equation is categorised as follows and can have any number of solutions:

(i) Principal Solution

The values of sin x and cos x will repeat after an interval of two, as is known. The values of tan x will be repeated in the same way after an interval of ?.

(ii) General solution

Any trigonometric equation solution that involves the integer ‘n’ is referred to as a general solution. In addition, the set of integers is indicated by the character “Z.”

14.1 Results

Steps to Solve Trigonometric Functions:

The steps for solving trigonometric equations are as follows:

Step 1: Use the sine or cos function to decompose the trigonometric equation into a single trigonometric ratio.

Step 2: Factor the ratio into the trigonometric polynomial that is given.

Step 3: After calculating the general answer to each factor, write it down.

Note:

Unless specifically stated otherwise, zero is regarded as an integer in this chapter.
The general answer should be given unless it must fall within a specific interval or range.
The angle’s main value is denoted by α. (An angle with the least numerical value is the main value.)

Download Free Trigonometric Functions Class 11 Notes

Students in Class 11 can easily access the trigonometric functions Class 11 Mathematics Chapter 3 Notes whenever and wherever they like. Those who refer to the Class 11 Mathematics Chapter 3 Notes can regularly practise the concepts explained which will help them score better grades and make learning an enjoyable experience.

Chapter 3 Mathematics Class 11 Notes provide a concise summary of the all concepts in the chapter to help students remember them and build confidence before attempting the trigonometric exam questions. Chapter 3 Mathematics Class 11 Notes are crucial and helpful because they enable them to thoroughly review the entire chapter prior to the exam.

Class 11 Mathematics Notes Chapter 3 are meticulously prepared by subject matter experts at Extramarks while keeping in mind the updated CBSE syllabus and guidelines.

Extensive formulas and questions in each chapter of the Class 11 Mathematics textbook can be learnt and solved easily with the help of revision notes. As a result, the Trigonometric Functions Chapter 3 Mathematics Class 11 Notes are essential to helping students easily memorise the topics in this chapter.

A Few Glimpses of Class 11 Chapter 3 Trigonometric Functions

The Greek words “trignon” and “metron,” which denote “measuring the slides of a triangle,” are the roots of the English word trigonometry. This subject was initially established to address a triangle-related geometrical puzzle. Engineers, sea captains, surveyors looking for new lands, and other professionals used trigonometry. Trigonometry is currently used in many fields, including the science of seismology, designing electric currents, and estimating the heights of ocean tides.

Students must have studied the trigonometric ratios of an acute angle as a ratio of the sides of the right-angle triangle in their earlier classes, as well as the use of trigonometric ratios and trigonometric identities. They will study the properties of trigonometry functions and calculate trigonometry ratios in this chapter.

Trigonometric Functions

Trigonometry functions, also referred to as circular functions, describe how the sides and angles of a right-angle triangle relate to one another. It implies that these trigonometric functions are used to derive the relationship between the sides and angles of a right-angle triangle. The main division of trigonometric functions is into sine, cosine, and tangent angles. And from the basic trigonometric functions, the other three functions, such as cotangent, secant, and cosecant, are derived. For each of the aforementioned trigonometric functions, there is an inverse trigonometric function.

Topic and Subtopics Covered in Class 11 Chapter 3 Trigonometric Functions

The different topics and subtopics covered in Class 11 Chapter 3 Trigonometric Functions include:

3.1: Introduction to Chapter

3.2: Angles

3.2.1: Degree Measure

3.2.2: Radian Measure

3.2.3: Relation between radian and real numbers

3.2.4: Relation between degree and radian

3.3: Trigonometric Functions

3.3.1: Sign of Trigonometric Functions

3.3.2: Domain And Range of Trigonometric Functions

3.4: Trigonometric Functions of Sum and Difference of Two Angles

3.5: Trigonometric Equations

Get a quick overview of all the topics and subtopics covered in the chapter by accessing the free Trigonometric Functions Class 11 Mathematics Notes Chapter 3 right away. Students will understand all of the topics mentioned above better with the help of these revision notes because they are well-structured for simple chapter revision. Referring to these Class 11 Mathematics Notes Chapter 3 will boost their confidence and is a great tool for exam preparations.

Benefits of Extramarks Class 11 Mathematics Notes Chapter 3

The following benefits of Extramarks Class 11 Mathematics Notes Chapter 3 are described.

Class 11 Mathematics Notes Chapter 3 offers a precise summary of the chapter.
The key theorems and formulas relating to the trigonometric function can be quickly and effectively reviewed by students.
By consulting Chapter 3 Mathematics Class 11 Notes, students can save valuable time.
Written by subject matter experts, students will be able to quickly recall all the chapter’s important topics.
Students can use Class 11 Mathematics Notes Chapter 3 to thoroughly review trigonometry in the days leading up to the test.

Q.1 The minute hand of a watch is 4.2 cm long. How far does its tip move in 50 minutes? (Use π = 22/7)

Ans

In 60 min, the minute hand completes 1 revolution.
∴ In 1 minute, the minute hand turns 1/60 revolution
∴ In 50 minutes, the minute hand turns
50/60 = 5/6 revolution.
Since in 1 revolution angle made by min hand is 2π radians
∴ In 5/6 revolution, angle made by min hand is
2π × 5/6 = 5π/3 radians
Here, r =4.2 cm
θ = 5π/3
⇒ Distance covered by the tip of minute hand,
l = r θ
⇒ l = 4.2 × 5π/3
⇒ l = 4.2 × 5 × (22/7) × 1/3
⇒ l = 22 cm.

Q.2

$Prove that \cos \frac{2 π}{15} \cdot \cos \frac{4 π}{15} \cdot \cos \frac{8 π}{15} \cdot \cos \frac{14 π}{15} = \frac{1}{16}$

Ans

$We have LHS = \cos \frac{2 π}{15} \cdot \cos \frac{4 π}{15} \cdot \cos \frac{8 π}{15} \cos (π - \frac{π}{15}) = \{\cos (\frac{2 π}{15}) \cos (\frac{4 π}{15}) \cos (\frac{8 π}{15})\} (- \cos \frac{π}{15}) = - \cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15} = - cosA \cdot \cos 2 A \cdot \cos 2^{2} Acos 2^{3} A_{t} where, A = \frac{π}{15} = - [\frac{\sin 2^{4} A}{2^{4} sinA}] = \frac{- \sin 16 A}{2^{4} sinA} = \frac{- \sin (15 A + A)}{16 sinA} (∵ 15 A = π) = \frac{- \sin (π + A)}{16 sinA} = \frac{sinA}{16 sinA} = \frac{1}{16} = R . H . S$

Q.3

$Prove that \sin \frac{π}{5} \sin \frac{2 π}{5} \sin \frac{3 π}{5} \sin \frac{4 π}{5} = \frac{5}{16}$

Ans

$If A + B = π, then A = π - B \Rightarrow sinA = \sin (π - B) \Rightarrow sinA = sinB ∴ \frac{π}{5} + \frac{4 π}{5} = π \Rightarrow \sin \frac{π}{5} = \sin \frac{4 π}{5} [sinA = sinB] and \frac{2 π}{5} + \frac{3 π}{5} = π \Rightarrow \sin \frac{2 π}{5} = \sin \frac{3 π}{5} [sinA = sinB] L.H.S = \sin \frac{π}{5} \sin \frac{2 π}{5} \sin \frac{3 π}{5} \sin \frac{4 π}{5} = \sin \frac{π}{5} \sin \frac{2 π}{5} \sin \frac{2 π}{5} \sin \frac{π}{5} = {(\sin \frac{π}{5} \sin \frac{2 π}{5})}^{2} = {(\sin 36 ° \cdot \sin 72 °)}^{2} = (\sin 36 ° \cdot \cos 18 °) = {(\frac{\sqrt{10 - 2 \sqrt{5}}}{4} \cdot \frac{\sqrt{10 + 2 \sqrt{5}}}{4})}^{2} = \frac{10 - 2 \sqrt{5}}{16} \times \frac{10 + 2 \sqrt{5}}{16} = \frac{{(10)}^{2} - {(2 \sqrt{5})}^{2}}{256} = \frac{100 - 20}{256} = \frac{5}{16} = R.H.S.$

Q.4 Sketch the graph of y = cos[x – (π/4)].

Ans

We have y= cos [x – (π/4)] ⇒ y – 0 = cos [x – (π/4)] …(i)
Shifting the origin at [(π/4) , 0] we obtain
x = X + (π/4) , y = Y + 0
On substituting in (i) we get Y = cos X
Now draw the graph of y = cos x and then shift it by (π/4) to the right.
The graph is shown below:

Q.5

$Prove that \frac{\cos 8 A \cos 5 A - \cos 12 A \cos 9 A}{\sin 8 A \cos 5 A + \cos 12 A \sin 9 A} = \tan 4 A$

Ans

$L.H.S = \frac{2 \cos 8 A \cos 5 A - 2 \cos 12 A \cos 9 A}{2 \sin 8 A \cos 5 A + 2 \cos 12 A \cdot \sin 9 A} = \frac{[\cos (8 A + 5 A) + \cos (8 A - 5 A)] - [\cos (12 A - 9 A) + \cos (12 A - 9 A)]}{[\sin (8 A - 5 A) + \sin (8 A - 5 A)] + [\sin (9 A + 12 A) + \sin (9 A - 12 A)]} = \frac{[\cos 13 A + \cos 3 A] - [\cos 21 A + \cos 3 A]}{[\sin 13 A + \sin 3 A] + [\sin 21 A + \sin (- 3, A]} = \frac{\cos 13 A - \cos 21 A}{\sin 13 A + \sin 21 A} = \frac{2 \sin (\frac{13 A + 21 A}{2}) \sin (\frac{21 A - 13 A}{2})}{2 \sin (\frac{13 A + 21 A}{2}) \cos (\frac{21 A - 13 A}{2})} = \frac{\sin 17 A \cdot \sin 4 A}{\sin 17 A \cdot \cos 4 A} = \tan 4 A = R \cdot H . S$

Q.6

$Find all other trigonometrically ratios if \sin θ = \frac{- 2 \sqrt{6}}{5} and θ in third quadrant.$

Ans

$We have, \cos^{2} θ + \sin^{2} θ = 1 \Rightarrow \cos θ = \pm \sqrt{1 - \sin^{2} θ} in the third quadrant \cos θ is negative ∴ \cos θ = - \sqrt{1 - \sin^{2} θ} \Rightarrow \cos θ = - \sqrt{1 - \frac{24}{25}} = - \sqrt{\frac{1}{25}} = \frac{- 1}{5} In quadrant III tanθ is positive ∴ tanθ = \frac{\sin θ}{\cos θ} \Rightarrow \tan θ = \frac{- 2 \sqrt{6}}{5} \times \frac{- 5}{1} = 2 \sqrt{6} cosec θ = \frac{1}{sinθ} \Rightarrow cosec θ = \frac{- 5}{2 \sqrt{6}} \sec θ = \frac{1}{cosθ} \Rightarrow \sec θ = - 5 \cot θ = \frac{1}{tanθ} \Rightarrow \cot θ = \frac{1}{2 \sqrt{6}}$

Q.7 Prove that :

$\tan 3 A \cdot \tan 2 A \cdot \tan A = \tan 3 A - \tan 2 A - \tan A$

Ans

$We have, 3 A = 2 A + A \Rightarrow \tan 3 A = \tan (2 A + A) \Rightarrow \tan 3 A = \frac{\tan 2 A + \tan A}{1 - \tan 2 A \cdot \tan A} \Rightarrow \tan 3 A (1 - \tan 2 A \cdot \tan A) = \tan 2 A + \tan A \Rightarrow \tan 3 A - \tan 3 A \cdot \tan 2 A \cdot \tan A = \tan 2 A + \tan A \Rightarrow \tan 3 A - \tan 2 A - \tan A = \tan 3 A \cdot \tan 2 A \cdot \tan A$

Q.8

$Prove that : 2 \cos \frac{π}{13} \cos \frac{9 π}{13} + \cos \frac{3 π}{13} + \cos \frac{5 π}{13} = 0$

Ans

$L.H.S = 2 \cos \frac{π}{13} \cos \frac{9 π}{13} + \cos \frac{3 π}{13} + \cos \frac{5 π}{13} = \cos (\frac{9 π}{13} + \frac{π}{13}) + \cos (\frac{9 π}{13} - \frac{π}{13}) + \cos \frac{3 π}{13} + \cos \frac{5 π}{13} = \cos \frac{10 π}{13} + \cos \frac{8 π}{13} + \cos \frac{3 π}{13} + \cos \frac{5 π}{13} = \cos (π - \frac{3 π}{13}) + \cos (π - \frac{5 π}{13}) + \cos \frac{3 π}{13} + \cos \frac{5 π}{13} = - \cos \frac{3 π}{13} - \cos \frac{5 π}{13} + \cos \frac{3 π}{13} + \cos \frac{5 π}{13} = 0 = R.H.S$

Q.9

$Solve 7 \cos^{2} θ + 3 \sin^{2} θ = 4 .$

Ans

$Since, 7 \cos^{2} θ + 3 \sin^{2} θ = 4 \Rightarrow 7 (1 - \sin^{2} θ) + 3 \sin^{2} θ = 4 \Rightarrow 7 - 7 \sin^{2} θ + 3 \sin^{2} θ = 4 \Rightarrow - 4 \sin^{2} θ = 4 - 7 \Rightarrow \sin^{2} θ = \frac{3}{4} \Rightarrow \sin^{2} θ = {(\frac{\sqrt{3}}{2})}^{2} \Rightarrow \sin^{2} θ = \sin^{2} (\frac{π}{3}) \Rightarrow \frac{1 - \cos 2 θ}{2} = \frac{1 - \cos 2 (\frac{π}{3})}{2} \Rightarrow \cos 2 θ = \cos 2 (\frac{π}{3}) \Rightarrow 2 θ = 2 nπ \pm \frac{2 π}{3} \Rightarrow θ = nπ \pm \frac{π}{3}, n \in Z Thus, the general solution of given equation is θ = nπ \pm \frac{π}{3}, n \in Z .$

Q.10

$Show that : \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8 θ}}} = 2 cosθ$

Ans

$L \cdot H . S = \sqrt{2 + \sqrt{2 + \sqrt{2 (1 + \cos 8 θ)}}} = \sqrt{2 \sqrt{2 + \sqrt{(2 \times 2 \cos^{2} 4 θ)}}} (∵ 1 + \cos 8 θ = 2 \cos^{2} 4 θ) = \sqrt{2 + \sqrt{2 + \sqrt{4 \cos^{2} 4 θ}}} = \sqrt{2 + \sqrt{2 + 2 \cos 4 θ}} = \sqrt{2 + \sqrt{2 (1 + \cos 4 θ)}} = \sqrt{2 + \sqrt{2 \times 2 \cos^{2} 2 θ}} \{∵ 1 + \cos 4 θ = 2 \cos^{2} 2 θ\} = \sqrt{2 + \sqrt{4 \cos^{2} 2 θ}} = \sqrt{2 + 2 \cos 2 θ} = \sqrt{2 (1 + \cos 2 θ)} = \sqrt{2.2 \cos^{2} θ} = 2 cosθ = R \cdot H . S$

Q.11

$Prove that : \tan 6 ° \tan 42 ° \tan 66 ° \tan 78 ° = 1$

Ans

$= \frac{\sin 6 ° \sin 42 ° \sin 66 ° \sin 78 °}{\cos 6 ° \cos 42 ° \cos 66 ° \cos 78 °} = \frac{(2 \sin 66 ° \sin 6 °) (2 \sin 78 ° \sin 42 °)}{(2 \cos 66 ° \cos 6 °) (2 \cos 78 ° \cos 42 °)} = \frac{(\cos 60 ° - \cos 72 °) (\cos 36 ° - \cos 120 °)}{(\cos 60 ° + \cos 72 °) (\cos 36 ° + \cos 120 °)} = \frac{(\frac{1}{2} - \frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4} + \frac{1}{2})}{(\frac{1}{2} + \frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4} - \frac{1}{2})} = \frac{(\frac{2 - \sqrt{5} + 1}{4}) (\frac{\sqrt{5} + 1 + 2}{4})}{(\frac{2 + \sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1 - 2}{4})} = \frac{(3 - \sqrt{5}) (3 + \sqrt{5})}{(\sqrt{5} + 1) (\sqrt{5} - 1)} = \frac{9 - 5}{5 - 1} = \frac{4}{4} = 1 = RHS$

Q.12

$Prove that \cos 510 ° \cdot \cos 330 ° + \sin 390 ° \cdot \cos 120 ° = - 1$

Ans

$L.H.S. = \cos 510 ° \cos 330 ° + \sin 390 ° \cos 120 ° = \cos (360 ° + 150 °) \cos (360 ° - 30 °) + \sin (360 ° + 30 °) \cos (180 ° + 60 °) = \cos 150 ° \cos 30 ° + \sin 30 ° \times - \cos 60 ° = \cos (180 ° - 30 °) \cos 30 ° - \sin 30 ° \cos 60 ° = - \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2} = - \frac{3}{4} - \frac{1}{4} = - 1 = R.H.S.$

Q.13 Find the degree measure corresponding to the radian measure (2π/15).

Ans

We have π radian = 180^°
1 radian =(180 /π)°

$\frac{2 π}{15} radian = (\frac{2 π}{15} \times \frac{180}{π}) ° = 24 °$

Q.14 If three angles A, B, C, are in A.P. Prove that:

$cotB = \frac{sinA - sinC}{cosC - cosA}$

Ans

$R . H . S = \frac{sinA - sinC}{cosC - cosA} = \frac{2 \sin \frac{A - C}{2} \cos \frac{A + C}{2}}{2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}} = \cot (\frac{A + C}{2}) = cotB = L . H . S [\begin{matrix} ∵ A, B, C are in A.P \\ ∴ 2 B = A + C \end{matrix}]$

Q.15

$Prove that \frac{\sin 2 θ}{1 + \cos 2 θ} = tanθ$

Ans

$L.H.S = \frac{\sin 2 θ}{1 + \cos 2 θ} = \frac{2 \cdot sinθ \cdot cosθ}{2 \cos^{2} θ} = \frac{sinθ}{cosθ} = tanθ = R.H.S ∴ L . H . S = R \cdot H . S$

Q.16 (π/8) radian = ……degree.

Ans

(π/8) radian

$= (\frac{π}{8} \times \frac{180}{π}) ° = (\frac{45}{2}) ° = 22 \frac{1}{2} ° = 22 ° {(\frac{1}{2} \times 60)}^{‘} = 22 ° 30^{‘}$

Q.17 Find solution of cos x = (1/2).

Ans

$Here, cosx = \frac{1}{2} cosx = \cos (\frac{π}{3}) Thus, x = 2 nπ \pm \frac{π}{3} where n is any integer .$

Q.18

–37°30′ is = …… radian

Ans

$- 37 ° 30^{‘} = - (37 \frac{1}{2}) ° = - (\frac{75}{2}) ° = - (\frac{75}{2} \times \frac{π}{180}) radian = - (\frac{5 π}{24}) radian$

Q.19 Find solution of sin x = (√3/2).

Ans

$Here, sinx = - \frac{\sqrt{3}}{2} sinx = - \sin (\frac{π}{3}) = \sin (π + \frac{π}{3}) = \sin (\frac{4 π}{3}) Thus, x = nπ + {(- 1)}^{n} \frac{4 π}{3} where n is any integer .$

Q.20 If cos x = –(1/2), x lies in third quadrant find sin x and cot x.

Ans

$sinx = - \sqrt{1 - \cos^{2}}, (∵ x lies in third quadrant) = - \sqrt{1 - {(- \frac{1}{2})}^{2}} = - \sqrt{\frac{3}{4}} = - \frac{\sqrt{3}}{2} \cot x = \sqrt{{cosec}^{2} x - 1} = \sqrt{\frac{1}{\sin^{2} x} - 1} = \sqrt{\frac{4}{3} - 1} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$

Q.21 Show that

$\tan 3 x = \frac{3 tanx - \tan^{3} x}{1 - 3 \tan^{2} x}$

Ans

$\tan 3 x = \tan (2 x + x) = \frac{\tan 2 x + \tan x}{1 - \tan 2 x \cdot \tan x} = \frac{\frac{2 \tan x}{1 - \tan^{2} x} + \tan x}{1 - \frac{2 \tan x}{1 - \tan^{2} x} \cdot \tan x} = \frac{3 \tan x - \tan^{3} x}{1 - 3 \tan^{2} x}$

Q.22 Find the value of sin15°.

Ans

$\sin 15 ° = \sin (45 ° - 30 °) = \sin 45 ° \cos 30 ° - \cos 45 ° \sin 30 ° = \frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \frac{1}{2} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

Q.23 Find the value of

$\sin^{4} \sqrt{\frac{π}{6}} + 2 \sin^{2} \sqrt{\frac{π}{6}} \cos^{2} \sqrt{\frac{π}{6}} + \cos^{4} \sqrt{\frac{π}{6}}$

Ans

$\sin^{4} \sqrt{\frac{π}{6}} + 2 \sin^{2} \sqrt{\frac{π}{6}} \cos^{2} \sqrt{\frac{π}{6}} + \cos^{4} \sqrt{\frac{π}{6}} = {(\sin^{2} \sqrt{\frac{π}{6}})}^{2} + 2 {Sin}^{2} \sqrt{\frac{π}{6}} \cos^{2} \sqrt{\frac{π}{6}} + {(\cos^{2} \sqrt{\frac{π}{6}})}^{2} = {(\sin^{2} \sqrt{\frac{π}{6}} + \cos^{2} \sqrt{\frac{π}{6}})}^{2} = 1^{2} = 1$

Q.24 If tan (π/6) = (1/√3), then find the value of tan [π – (π/6)].

Ans

tan [π – (π/6)] = – tan(π/6) = – (1/√3)

Q.25 State sine rule.

Ans

If the sides of the triangle are a, b and c and the angles opposite those
sides are A, B and C respectively, then the law of sine states :

$\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$

Q.26 Find the Value of cos (13π/12) .

Ans

$\cos \frac{13 π}{12} = \cos (π + \frac{π}{12}) = - \cos \frac{π}{12} - \cos (\frac{π}{4} - \frac{π}{6}) = - [\cos \frac{π}{4} \cos \frac{π}{6} + \sin \frac{π}{4} \sin \frac{π}{6}] - [\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \frac{1}{2}] = - [\frac{\sqrt{3} + 1}{2 \sqrt{2}}]$

Q.27 State the laws of cosine.

Ans

The laws of cosine are:

$c^{2} = a^{2} + b^{2} - 2 ab cosC or, equivalently: a^{2} = b^{2} + c^{2} - 2 bc cosA b^{2} = c^{2} + a^{2} - 2 ca cosB$

Q.28 If the arcs of the same length in two circles subtend angles 65°and 110° at the centre, then find the ratio of their radii.

Ans

$Let r_{1} and r_{2} be the radii of the two circles. Given that θ_{1} = 45 ° = \frac{π}{180 °} \times 45 ° = \frac{π}{4} radian θ_{2} = 100 ° = \frac{π}{180 °} \times 100 ° = \frac{5 π}{9} radian Let I be the length of each of the arc. Then, I = θ_{1} r_{1} = θ_{2} r_{2} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{θ_{2}}{θ_{1}} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{\frac{5 π}{9}}{\frac{π}{4}} = \frac{5}{9} \times 4 \Rightarrow \frac{r_{1}}{r_{2}} = \frac{20}{9} Hence, r_{1} : r_{2} = 20 : 9$

Q.29 In a circle of diameter 42 cm, the length of a chord is 21 cm. Find the length of the minor arc of the chord.

Ans

$Diameter = 42 cm ∴ Radius, r = \frac{42}{2} = 21 cm Chord, AB = 21 cm \Rightarrow ∆ AOB is equilateral ∴ ∠ AOB = 60 ° = \frac{π}{180 °} \times 60 ° = \frac{π}{3} radian \Rightarrow θ = \frac{π}{3} radian, r = 21 cm ∴ The length of the minor arc, l = θ x r = \frac{π}{3} \times 21 = \frac{22}{7} \times \frac{1}{3} \times 21 cm = 22 cm$

Q.30 If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, then find the ratio of their radii.

Ans

$Let r_{1} and r_{2} be the radii of the two circles. Given that θ_{1} = 60 ° = \frac{π}{180 °} \times 60 ° = \frac{π}{3} radian θ_{2} = 75 ° = \frac{π}{180 °} \times 75 ° = \frac{5 π}{12} radian Let I be the length of each of the arc. Then, l = θ_{1} r_{1} = θ_{2} r_{2} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{θ_{2}}{θ_{1}} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{\frac{π}{3}}{\frac{5 π}{12}} = \frac{1}{3} \times \frac{12}{5} \Rightarrow \frac{r_{1}}{r_{2}} = \frac{4}{5} Hence, r_{1} : r_{2} = 4 : 5$

Q.31

$Prove that \cos 2 x = \frac{1 - \tan^{2} x}{1 + \tan^{2} x}$

Ans

$We know that \cos (x + y) = \cos x \cos y - \sin x \sin y Replacing y by x, we get \cos (x + x) = \cos x \cos x - \sin x \sin x \Rightarrow \cos 2 x = \cos^{2} x - \sin^{2} x = \frac{\cos^{2} x - \sin^{2} x}{1} = \frac{\cos^{2} x - \sin^{2} x}{\cos^{2} x + \sin^{2} x} Divide Numerator and Denominator by \cos^{2} x = \frac{1 - \frac{\sin^{2} x}{\cos^{2} x}}{1 + \frac{\sin^{2} x}{\cos^{2} x}} = \frac{1 - \tan^{2} x}{1 + \tan^{2} x}$

Q.32

$If \sqrt{7} \tan x = 1, then find the value of \frac{{cosec}^{2} x - \sec^{2} x}{{cosec}^{2} x + \sec^{2} x}$

Ans

$∵ \sqrt{7} \tan x = 1 ∴ \tan x = \frac{1}{\sqrt{7}} The value of = \frac{{cosec}^{2} x - \sec^{2} x}{{cosec}^{2} x + \sec^{2} x} = \frac{\frac{1}{\sin^{2} x} - \frac{1}{\cos^{2} x}}{\frac{1}{\sin^{2} x} + \frac{1}{\cos^{2} x}} = \frac{\cos^{2} x - \sin^{2} x}{\cos^{2} x + \sin^{2} x} Divide Numerator and Denominator by \cos^{2} x = \frac{1 - \frac{\sin^{2} x}{\cos^{2} x}}{1 + \frac{\sin^{2} x}{\cos^{2} x}} = \frac{1 - \tan^{2} x}{1 + \tan^{2} x} = \frac{1 - \frac{1}{7}}{1 + \frac{1}{7}} = \frac{\frac{6}{7}}{\frac{8}{7}} = \frac{6}{8} = \frac{3}{4}$

Q.33

$Prove that \sin 3 x = 3 sinx - 4 \sin^{3} x$

Ans

$LHS = \sin 3 x = \sin (2 x + x) = \sin 2 x \cos x + \cos 2 x \sin x = 2 \sin x \cos x \cos x + (1 - 2 \sin^{2} x) \sin x = 2 \sin x (1 - \sin^{2} x) + \sin x - 2 \sin^{3} x = 2 \sin x (1 - \sin^{3} x) + \sin x - 2 \sin^{3} x = 3 \sin x - 4 \sin^{3} x$

Q.34

$Prove that tanx = \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} .$

Ans

$RHS = \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} = \sqrt{\frac{1 - (1 - 2 \sin^{2} x)}{1 + 2 \cos^{2} x - 1}} = \sqrt{\frac{1 - 1 + 2 \sin^{2} x}{2 \cos^{2} x}} = \sqrt{\frac{2 \sin^{2} x}{2 \cos^{2} x}} = \sqrt{\frac{\sin^{2} x}{\cos^{2} x}} = \frac{sinx}{cosx} = tanx = LHS, Proved$

Q.35

$Evaluate \tan (22 \frac{1 °}{2}) .$

Ans

$We know that \tan x = \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} . . . (i) Now put x = 22 \frac{1 °}{2} = \frac{45 °}{2} in (1) we get, \tan 22 \frac{1 °}{2} = \sqrt{\frac{1 - \cos (2 \times \frac{45 °}{2})}{1 + \cos (2 \times \frac{45 °}{2})}} \tan \frac{45 °}{2} = \sqrt{\frac{1 - \cos 45 °}{1 + \cos 45 °}} = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}} = \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} = \sqrt{\frac{(\sqrt{2} - 1)}{(\sqrt{2} - 1)} \times \frac{(\sqrt{2} - 1)}{(\sqrt{2} - 1)}} = \sqrt{\frac{{(\sqrt{2} - 1)}^{2}}{2 - 1}} = (\sqrt{2} - 1)$

Q.36 Which is greater – sin1° or sin1? Justify your answer.

Ans

$First, we shall convert 1 into degree ∵ π = 180 ° ∴ 1 = (\frac{180}{π}) ° = (\frac{180}{\frac{22}{7}}) ° = (180 \times \frac{7}{22}) ° = (90 \times \frac{7}{11}) ° = (\frac{630}{11}) ° \Rightarrow 1 = 57.27 ° \Rightarrow \sin 1 = \sin 57.27 ° Hence, \sin 1 is greater than \sin 1 ° .$

Q.37

$Prove that \tan \frac{3 x}{2} = \frac{3 \tan \frac{x}{2} - \tan^{3} \frac{x}{2}}{1 - 3 \tan^{2} \frac{x}{2}}$

Ans

$LHS = \tan \frac{3 x}{2} = \tan (x + \frac{x}{2}) = \frac{\tan x + \tan \frac{x}{2}}{1 - \tan x \tan \frac{x}{2}} = \frac{(\frac{2 \tan \frac{x}{2}}{1 - \tan^{2} \frac{x}{2}}) + \tan \frac{x}{2}}{1 - (\frac{2 \tan \frac{x}{2}}{1 - \tan^{2} \frac{x}{2}}) \tan \frac{x}{2}} = \frac{\frac{2 \tan \frac{x}{2} + \tan \frac{x}{2} - \tan^{3} \frac{x}{2}}{(1 - \tan^{2} \frac{x}{2})}}{\frac{1 - \tan^{2} \frac{x}{2} - 2 \tan^{2} \frac{x}{2}}{(1 - \tan^{2} \frac{x}{2})}} = \frac{3 \tan \frac{x}{2} - \tan^{3} \frac{x}{2}}{1 - 3 \tan^{2} \frac{x}{2}} = RHS, Proved.$

Q.38 Evaluate sin18°.

Ans

Q.39 Evaluate cos72°.

Ans

$Let θ = 18 ° \Rightarrow 5 θ = 90 ° \Rightarrow 3 θ + 2 θ = 90 ° \Rightarrow 2 θ = 90 ° - 3 θ \Rightarrow \sin 2 θ = \sin (90 ° - 3 θ) \Rightarrow \sin 2 θ = \cos 3 θ \Rightarrow \cos 3 θ = \sin 2 θ \Rightarrow 4 \cos^{3} θ - 3 cosθ = 2 \sin θ \cos θ \Rightarrow 4 \cos^{3} θ - 3 cosθ - 2 \sin θ \cos θ = 0 \Rightarrow \cos θ [4 \cos^{2} θ - 3 - 2 \sin θ] = 0 \Rightarrow \cos θ [4 (1 - \sin^{2} θ) - 3 - 2 \sin θ] = 0 \Rightarrow \cos θ [4 - 4 \sin^{2} θ - 3 - 2 \sin θ] = 0 \Rightarrow \cos θ [1 - 4 \sin^{2} θ - 2 \sin θ] = 0 \Rightarrow - \cos θ [4 \sin^{2} θ + 2 \sin θ - 1] = 0 \Rightarrow \cos θ [4 \sin^{2} θ + 2 \sin θ - 1] = 0 \Rightarrow \cos θ [4 \sin^{2} θ + 2 \sin θ - 1] = 0 ∵ \cos θ \neq 0 or θ \neq 90 ° \Rightarrow 4 \sin^{2} θ + 2 \sin θ - 1 = 0 This is a quadratic equation in \sin θ with a = 4, b = 2 and c = - 1 ∴ D = b^{2} - 4 ac = 4 - 4 \times 4 \times - 1 = 4 + 16 = 20 \Rightarrow \sin θ = \frac{- b \pm \sqrt{D}}{2 a} = \frac{- 2 \pm \sqrt{20}}{2 \times 4} \Rightarrow \sin θ = \frac{- 2 \pm 2 \sqrt{5}}{2 \times 4} = \frac{- 1 \pm \sqrt{5}}{4} \Rightarrow \sin θ = \frac{- 1 + \sqrt{5}}{4} and \sin θ = \frac{- 1 - \sqrt{5}}{4} [rejected as in I quadrant all + ve] \sin 18 ° = \frac{\sqrt{5} - 1}{4} \cos 72 ° = \cos (90 ° - 18 °) \Rightarrow \cos 72 ° = \sin 18 ° \Rightarrow \cos 72 ° = \frac{\sqrt{5} - 1}{4}$

Please register to view this section

FAQs (Frequently Asked Questions)

1. What are the six trigonometric functions?

Sine, cosine, tan, cosec, sec, and cot are the six trigonometric functions.

2. What topics are covered in Chapter 3 of Mathematics 11's Trigonometry?

Chapter 3 of Class 11 Trigonometry Mathematics first provides a brief overview of the chapter before discussing the various trigonometric functions like sin, cos, and tan. This chapter covers topics like angles, radian and degree measurements, their relationships, the sign of trigonometric functions, their sums and differences, related equations, and the relationship between real numbers and radian values. Students must understand all of the topics and subtopics to do well in the exams.

CBSE Class 11 Maths Revision Notes Chapter 3

Class 11 Mathematics Revision Notes for Chapter-3 Trigonometric Functions

Download Free Trigonometric Functions Class 11 Notes

A Few Glimpses of Class 11 Chapter 3 Trigonometric Functions

Benefits of Extramarks Class 11 Mathematics Notes Chapter 3

Please register to view this section

FAQs (Frequently Asked Questions)

1. What are the six trigonometric functions?

2. What topics are covered in Chapter 3 of Mathematics 11's Trigonometry?

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions

CBSE Class 11 Maths Revision Notes Chapter 3

Class 11 Mathematics Revision Notes for Chapter-3 Trigonometric Functions

Download Free Trigonometric Functions Class 11 Notes

A Few Glimpses of Class 11 Chapter 3 Trigonometric Functions

Benefits of Extramarks Class 11 Mathematics Notes Chapter 3

Share

Please register to view this section

FAQs (Frequently Asked Questions)

1. What are the six trigonometric functions?

2. What topics are covered in Chapter 3 of Mathematics 11's Trigonometry?

Share

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions