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Chapter 3 Class 11 Trigonometric Functions

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NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

• What is a positive or a negative angle
• Measuring angles in Degree , Minutes and Seconds
• Radian measure of an angle
• Converting Degree to Radians , and vice-versa
• Sign of sin, cos, tan in all 4 quadrants
• Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
• Finding Value of trigonometric functions, given angle
• Solving questions by formula like  (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula 
• Finding principal and general solutions of a trigonometric equation
• Sin and Cosine Formula with supplementary Questions

Important questions are marked, and Formula sheet is also provided. Click on an exercise or topic to begin.

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Trigonometry - Sequence of Lessons

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Trigonometry

Aug 31, 2014

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Trigonometry. BY SAI KUMAR. Maths. Trigonometry. Trigonometry is derived from Greek words trigonon (three angles) and metron ( measure). Trigonometry is the branch of mathematics which deals with triangles, particularly triangles in a plane where one angle of the triangle is 90 degrees

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Presentation Transcript

Trigonometry BY SAI KUMAR Maths

Trigonometry • Trigonometry is derived from Greek words trigonon (three angles) and metron ( measure). • Trigonometry is the branch of mathematics which deals with triangles, particularly triangles in a plane where one angle of the triangle is 90 degrees • Triangles on a sphere are also studied, in spherical trigonometry. • Trigonometry specifically deals with the relationships between the sides and the angles of triangles, that is, on the trigonometric functions, and with calculations based on these functions.

History • The origins of trigonometry can be traced to the civilizations of ancient Egypt, Mesopotamia and the Indus Valley, more than 4000 years ago. • Some experts believe that trigonometry was originally invented to calculate sundials, a traditional exercise in the oldest books • The first recorded use of trigonometry came from the Hellenistic mathematician Hipparchus circa 150 BC, who compiled a trigonometric table using the sine for solving triangles. • The Sulba Sutras written in India, between 800 BC and 500 BC, correctly compute the sine of π/4 (45°) as 1/√2 in a procedure for circling the square (the opposite of squaring the circle). • Many ancient mathematicians like Aryabhata, Brahmagupta,Ibn Yunus and Al-Kashi made significant contributions in this field(trigonometry).

Right Triangle • A triangle in which one angle is equal to 90 is called right triangle. • The side opposite to the right angle is known as hypotenuse. AB is the hypotenuse • The other two sides are known as legs. AC and BC are the legs Trigonometry deals with Right Triangles

Pythagoras Theorem • In any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of areas of the squares whose sides are the two legs. • In the figure AB2 = BC2 + AC2

Trigonometric ratios • Sine(sin) opposite side/hypotenuse • Cosine(cos) adjacent side/hypotenuse • Tangent(tan) opposite side/adjacent side • Cosecant(cosec) hypotenuse/opposite side • Secant(sec) hypotenuse/adjacent side • Cotangent(cot) adjacent side/opposite side

Values of trigonometric function of Angle A • sin = a/c • cos = b/c • tan = a/b • cosec = c/a • sec = c/b • cot = b/a

Values of Trigonometric function

Calculator • This Calculates the values of trigonometric functions of different angles. • First Enter whether you want to enter the angle in radians or in degrees. Radian gives a bit more accurate value than Degree. • Then Enter the required trigonometric function in the format given below: • Enter 1 for sin. • Enter 2 for cosine. • Enter 3 for tangent. • Enter 4 for cosecant. • Enter 5 for secant. • Enter 6 for cotangent. • Then enter the magnitude of angle.

Trigonometric identities • sin2A + cos2A = 1 • 1 + tan2A = sec2A • 1 + cot2A = cosec2A • sin(A+B) = sinAcosB + cosAsin B • cos(A+B) = cosAcosB – sinAsinB • tan(A+B) = (tanA+tanB)/(1 – tanAtan B) • sin(A-B) = sinAcosB – cosAsinB • cos(A-B)=cosAcosB+sinAsinB • tan(A-B)=(tanA-tanB)(1+tanAtanB) • sin2A =2sinAcosA • cos2A=cos2A - sin2A • tan2A=2tanA/(1-tan2A) • sin(A/2) = ±{(1-cosA)/2} • Cos(A/2)= ±{(1+cosA)/2} • Tan(A/2)= ±{(1-cosA)/(1+cosA)}

Relation between different Trigonometric Identities • Sin • Cos • Tan • Cosec • Sec • Cot

Angles of Elevation and Depression • Line of sight: The line from our eyes to the object, we are viewing. • Angle of Elevation:The angle through which our eyes move upwards to see an object above us. • Angle of depression:The angle through which our eyes move downwards to see an object below us.

Problem solved using trigonometric ratios CLICK HERE!

Applications of Trigonometry • This field of mathematics can be applied in astronomy, navigation, music theory, acoustics, optics, analysis of financial markets, electronics, probability theory, statistics, biology, medical imaging (CAT scans and ultrasound), pharmacy, chemistry, number theory (and hence cryptology), seismology, meteorology, oceanography, many physical sciences, land surveying and geodesy, architecture, phonetics, economics, electrical engineering, mechanical engineering, civil engineering, computer graphics, cartography, crystallography and game development.

Derivations • Most Derivations heavily rely on Trigonometry. Click the hyperlinks to view the derivation • A few such derivations are given below:- Parallelogram law of addition of vectors . Centripetal Acceleration . Lens Formula . Variation of Acceleration due to gravity due to rotation of earth Finding angle between resultant and the vector.

Applications of Trigonometry in Astronomy • Since ancient times trigonometry was used in astronomy. • The technique of triangulation is used to measure the distance to nearby stars. • In 240 B.C., a mathematician named Eratosthenes discovered the radius of the Earth using trigonometry and geometry. • In 2001, a group of European astronomers did an experiment that started in 1997 about the distance of Venus from the Sun. Venus was about 105,000,000 kilometers away from the Sun .

Application of Trigonometry in Architecture • Many modern buildings have beautifully curved surfaces. • Making these curves out of steel, stone, concrete or glass is extremely difficult, if not impossible. • One way around to address this problem is to piece the surface together out of many flat panels, each sitting at an angle to the one next to it, so that all together they create what looks like a curved surface. • The more regular these shapes, the easier the building process. • Regular flat shapes like squares, pentagons and hexagons, can be made out of triangles, and so trigonometry plays an important role in architecture.

Waves • The graphs of the functions sin(x) and cos(x) look like waves. Sound travels in waves, although these are not necessarily as regular as those of the sine and cosine functions. • However, a few hundred years ago, mathematicians realized that any wave at all is made up of sine and cosine waves. This fact lies at the heart of computer music. • Since a computer cannot listen to music as we do, the only way to get music into a computer is to represent it mathematically by its constituent sound waves. • This is why sound engineers, those who research and develop the newest advances in computer music technology, and sometimes even composers have to understand the basic laws of trigonometry. • Waves move across the oceans, earthquakes produce shock waves and light can be thought of as traveling in waves. This is why trigonometry is also used in oceanography, seismology, optics and many other fields like meteorology and the physical sciences.

Digital Imaging • In theory, the computer needs an infinite amount of information to do this: it needs to know the precise location and colour of each of the infinitely many points on the image to be produced. In practice, this is of course impossible, a computer can only store a finite amount of information. • To make the image as detailed and accurate as possible, computer graphic designers resort to a technique called triangulation. • As in the architecture example given, they approximate the image by a large number of triangles, so the computer only needs to store a finite amount of data. • The edges of these triangles form what looks like a wire frame of the object in the image. Using this wire frame, it is also possible to make the object move realistically. • Digital imaging is also used extensively in medicine, for example in CAT and MRI scans. Again, triangulation is used to build accurate images from a finite amount of information. • It is also used to build "maps" of things like tumors, which help decide how x-rays should be fired at it in order to destroy it.

Conclusion Trigonometry is a branch of Mathematics with several important and useful applications. Hence it attracts more and more research with several theories published year after year

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios. • sinA = 2/3 • sinQ = 11/15 • cosQ = 7/25 • secQ=13/5 • cosA = 4/5 • Tan = 5/12 • tanQ =8/15

In a ABC, right angled at B, AB+24cm,BC=7 cm. Determine • sinA, cosA • sinC, cosC

(1+sinQ) (1-sinQ) If cotQ =7/8 ,evaluate (1=cosQ) (1-cosQ)

Evaluate each of the following • sin45 sin30 + cos45 cos30 • sin60 cos30 + cos60 sin30 • cos60 cos45 – sin60 sin45 • sin²30+sin²45+sin²60+sin²90 • cos²30+cos²45+cos²60+cos²90 • tan²30+tan²60+tan²45

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• NCERT Solutions
• NCERT Class 11
• NCERT 11 Maths
• Chapter 3: Trigonometric Functions

NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are available at BYJU’S, which are prepared by our expert teachers. All these solutions are written as per the latest update on the CBSE Syllabus 2023-24 and its guidelines. BYJU’S provides step-by-step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3, Trigonometric Functions, comes under NCERT Class 11 Maths and is an important chapter for students. Though the chapter has numerous mathematical terms and formulae, BYJU’S has made NCERT Solutions for Class 11 Maths easy for the students to understand and remember with the usage of tricks.

Trigonometry has been developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is 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, the NCERT Solutions of BYJU’S help the students to attain more knowledge and score full marks in this chapter of the examination.

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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In this section, a few important terms are defined, such as principal solution and general solution of trigonometric functions, which are explained using examples too. Exercise 3.1 Solutions 7 Questions Exercise 3.2 Solutions 10 Questions Exercise 3.3 Solutions 25 Questions Exercise 3.4 Solutions 9 Questions Miscellaneous Exercise On Chapter 3 Solutions 10 Questions

Access NCERT Solutions for Class 11 Maths Chapter 3

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°

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7)

Here π radian = 180°

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

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).

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.

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

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.

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

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

θ = 10/75 radian

By further simplification

θ = 2/15 radian

(ii) l = 15 cm

θ = 15/75 radian

θ = 1/5 radian

(iii) l = 21 cm

θ = 21/75 radian

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.

2. sin x = 3/5, x lies in second quadrant.

sin x = 3/5

We can write it as

sin 2 x + cos 2 x = 1

cos 2 x = 1 – sin 2 x

3. cot x = 3/4, x lies in third quadrant.

cot x = 3/4

1 + tan 2 x = sec 2 x

1 + (4/3) 2 = sec 2 x

Substituting the values

1 + 16/9 = sec 2 x

cos 2 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

4. sec x = 13/5, x lies in fourth quadrant.

sec x = 13/5

sin 2 x = 1 – cos 2 x

sin 2 x = 1 – (5/13) 2

sin 2 x = 1 – 25/169 = 144/169

sin 2 x = ± 12/13

Here x lies in the fourth quadrant so the value of sin x will be negative

sin x = – 12/13

5. tan x = -5/12, x lies in second quadrant.

tan x = – 5/12

1 + (-5/12) 2 = sec 2 x

1 + 25/144 = sec 2 x

sec 2 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

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°

We know that values of sin x repeat after an interval of 2π or 360°

By further calculation

7. cosec (–1410°)

We know that values of cosec x repeat after an interval of 2π or 360°

= cosec 30 o = 2

We know that values of tan x repeat after an interval of π or 180°

Exercise 3.3 page: 73

Prove that:

= 1/2 + 4/4

5. Find the value of:

(i) sin 75 o

(ii) tan 15 o

(ii) tan 15°

It can be written as

= tan (45° – 30°)

Using formula

Prove the following:

= sin x cos x (tan x + cot x)

10. sin ( n  + 1) x  sin ( n  + 2) x  + cos ( n  + 1) x  cos ( n  + 2) x  = cos  x

LHS = sin ( n  + 1) x  sin ( n  + 2) x  + cos ( n  + 1) x  cos ( n  + 2) x

Using the formula

12. sin 2  6 x  – sin 2  4 x  = sin 2 x  sin 10 x

13. cos 2  2 x  – cos 2  6 x  = sin 4 x  sin 8 x

= (2 sin 4 x  cos 4 x ) (2 sin 2 x  cos 2 x )

= sin 8x sin 4x

14. sin 2x + 2sin 4x + sin 6x = 4cos 2  x sin 4x

= 2 sin 4x cos (– 2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

Taking common terms

= 2 sin 4x (cos 2x + 1)

= 2 sin 4x (2 cos 2  x – 1 + 1)

= 2 sin 4x (2 cos 2  x)

= 4cos 2  x sin 4x

15. cot 4 x  (sin 5 x  + sin 3 x ) = cot  x  (sin 5 x  – sin 3 x )

LHS = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

Hence, LHS = RHS.

22. cot  x  cot 2 x  – cot 2 x  cot 3 x  – cot 3 x  cot  x  = 1

LHS = tan 4x = tan 2(2x)

By using the formula

24. cos 4 x  = 1 – 8sin 2  x  cos 2  x

LHS = cos 4x

= cos 2(2 x )

Using the formula cos 2 A  = 1 – 2 sin 2   A

= 1 – 2 sin 2  2 x

Again by using the formula sin2 A  = 2sin  A  cos A

= 1 – 2(2 sin  x  cos  x ) 2

= 1 – 8 sin 2 x  cos 2 x

25. cos 6 x  = 32 cos 6   x  – 48 cos 4   x  + 18 cos 2   x  – 1

L.H.S. = cos 6 x

= cos 3(2 x )

Using the formula cos 3 A  = 4 cos 3   A  – 3 cos   A

= 4 cos 3  2 x  – 3 cos   2 x

Again by using formula cos 2 x  = 2 cos 2   x  – 1

= 4 [(2 cos 2   x  – 1) 3  – 3 (2 cos 2   x  – 1)

= 4 [(2 cos 2   x ) 3  – (1) 3  – 3 (2 cos 2   x ) 2  + 3 (2 cos 2   x )] – 6cos 2   x  + 3

= 4 [8cos 6 x  – 1 – 12 cos 4 x  + 6 cos 2 x ] – 6 cos 2 x  + 3

By multiplication

= 32 cos 6 x  – 4 – 48 cos 4 x  + 24 cos 2   x  – 6 cos 2 x  + 3

On further calculation

= 32 cos 6 x  – 48 cos 4 x  + 18 cos 2 x  – 1

Exercise 3.4 PAGE: 78

Find the principal and general solutions of the following equations:

1. tan x = √3

2. sec x = 2

3. cot x = – √3

4. cosec x = – 2

Find the general solution for each of the following equations:

5. cos 4x = cos 2x

6. cos 3x + cos x – cos 2x = 0

7. sin 2x + cos x = 0

sin 2x + cos x = 0

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. sec 2 2x = 1 – tan 2x

sec 2 2x = 1 – tan 2x

1 + tan 2 2x = 1 – tan 2x

tan 2 2x + tan 2x = 0

tan 2x (tan 2x + 1) = 0

tan 2x = 0 or tan 2x + 1 = 0

If tan 2x = 0

tan 2x = tan 0

2x = nπ + 0, where n ∈ Z

x = nπ/2, where n ∈ Z

tan 2x + 1 = 0

tan 2x = – 1

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

sin x + sin 3x + sin 5x = 0

(sin x + sin 5x) + sin 3x = 0

2 sin 3x cos (-2x) + sin 3x = 0

2 sin 3x cos 2x + sin 3x = 0

By taking out the common terms

sin 3x (2 cos 2x + 1) = 0

sin 3x = 0 or 2 cos 2x + 1 = 0

If sin 3x = 0

3x = nπ, where n ∈ Z

x = nπ/3, where n ∈ Z

If 2 cos 2x + 1 = 0

cos 2x = – 1/2

= – cos π/3

= cos (π – π/3)

cos 2x = cos 2π/3

Miscellaneous Exercise page: 81

2. (sin 3 x  + sin  x ) sin  x  + (cos 3 x  – cos  x ) cos  x  = 0

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

= sin 3x sin x + sin 2 x + cos 3x cos x – cos 2 x

Taking out the common terms

= cos 3x cos x + sin 3x sin x – (cos 2 x – sin 2 x)

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

= cos 2x – cos 2x

LHS = (cos x + cos y) 2 + (sin x – sin y) 2

By expanding using formula we get

= cos 2 x + cos 2 y + 2 cos x cos y + sin 2 x + sin 2 y – 2 sin x sin y

Grouping the terms

= (cos 2 x + sin 2 x) + (cos 2 y + sin 2 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)

= 2 + 2 cos (x + y)

Taking 2 as common

From the formula cos 2A = 2 cos 2 A – 1

LHS = (cos x – cos y) 2 + (sin x – sin y) 2

By expanding using formula

= cos 2 x + cos 2 y – 2 cos x cos y + sin 2 x + sin 2 y – 2 sin x sin y

= (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

From formula cos 2A = 1 – 2 sin 2 A

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

8. Find sin x/2, cos x/2 and tan x/2 in each of the following:

cos x = -3/5

From the formula

9. cos x = -1/3, x in quadrant III

10. sin x = 1/4, x in quadrant II

This chapter has 6 exercises and a miscellaneous exercise to help students understand the concepts related to Trigonometric Functions clearly. BYJU’S provides all the concepts and solutions for Chapter 3 of Class 11 Maths with clear explanations and formulas. The PDF of Maths  NCERT Solutions for Class 11 Chapter 3 includes the topics and sub-topics listed below. 3.1 Introduction The basic trigonometric ratios and identities are given here, along with the applications of trigonometric ratios in solving the word problems related to heights and distances. 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 In this section, different terms related to trigonometry are discussed, such as terminal side, initial sides, measuring an angle in degrees and radian, etc. 3.3 Trigonometric Functions 3.3.1 Sign of trigonometric functions 3.3.2 Domain and range of trigonometric functions After studying this section, students are able to understand the generalised trigonometric functions with signs. Also, they can gain knowledge on domain and range of trigonometric functions with examples. 3.4 Trigonometric Functions of Sum and Difference of Two Angles This section contains formulas related to the sum and difference of two angles in trigonometric functions.

Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Studying the Trigonometric Functions of Class 11 enables the students to understand the following:

• Introduction to Trigonometric Functions
• Positive and negative angles
• Measuring angles in radians and in degrees and conversion of one into other
• Definition of trigonometric functions with the help of unit circle
• Truth of the sin 2x + cos 2x = 1, for all x
• Signs of trigonometric functions
• Domain and range of trigonometric functions
• Graphs of Trigonometric Functions
• Expressing sin (x±y) and cos (x±y) in terms of sin x, sin y, cos x & cosy and their simple application, identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x
• The general solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a

Disclaimer – 

Dropped Topics – 

3.5 Trigonometric Equations (up to Exercise 3.4) Last five points in the Summary 3.6 Proofs and Simple Applications of Sine and Cosine Formulae

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 3

What are the topics discussed in chapter 3 of ncert solutions for class 11 maths, as per the latest update of the cbse syllabus, how many exercises are there in chapter 3 of ncert solutions for class 11 maths, what will i learn in chapter 3 trigonometric functions of ncert solutions for class 11 maths, leave a comment cancel reply.

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