powerpoint presentation on trigonometry for class 11

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Introduction to Trigonometry

Jul 22, 2014

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Introduction to Trigonometry. Basic Trigonometric Functions. What is Trigonometry?. The study of triangles Relationship between sides and angles of a right triangle What is a right triangle? A triangle with a 90 ⁰ angle. 90°. Review Right Triangles.

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Introduction to Trigonometry Basic Trigonometric Functions

What is Trigonometry? • The study of triangles • Relationship between sides and angles of a right triangle • What is a right triangle? A triangle with a 90⁰ angle 90°

Review Right Triangles • In relation to angle a, the sides of the triangle are: • hypotenuse- always longest side and side across from right angle (90⁰) • adjacent - side closest to angle a • opposite - side opposite toangle a a hypotenuse adjacent 90⁰ opposite

Review Right Triangles Label the sides for angle b: • hypotenuse • adjacent • opposite ? hypotenuse ? opposite 90° b ? adjacent

Trigonometric Functions Ratios of the sides in relation to angle a: • sine • cosine • tangent a hypotenuse adjacent 90° opposite

Trigonometric Functions:SINE abbreviation: sin sin(a)= • 0 ≤ sin ≤ 1 • Example: sin(60°)= = ~.866 • Ratio of opposite side to hypotenuse for 60° angle is to 2 (.866 to 1) opposite hypotenuse a hypotenuse 2 adjacent 90° opposite

Trigonometric Functions:COSINE abbreviation: cos cos(a)= • 0 ≤ cos ≤ 1 • Example: cos(60°)= = .5 • Ratio of adjacent side to hypotenuse for 60° angle is 1 to 2 (half) adjacent hypotenuse a 1 hypotenuse 2 adjacent 90° opposite

Trigonometric Functions:TANGENT abbreviation: tan tan(a)= • 0 ≤ tan ≤ • Example: tan(60°)= = ~1.732 • Ratio of opposite side to adjacent side for 60° angle is to 1 (1.732 to 1) opposite opposite adjacent adjacent a ∞ hypotenuse adjacent 90° opposite

Trigonometric Functions: SOH SOH – CAH – TOA CAH TOA REMEMBER: Sine = Cosine = Tangent = Opposite Hypotenuse a Adjacent Hypotenuse hypotenuse Opposite adjacent Adjacent 90° opposite

Using Trigonometric Functions: For any right triangle: • calculate other sides if one side and angle known • calculate angle if two sides known 90°

Calculating Sides:One Side and Angle Known What is known? • angle (50°) and adjacent side (2) Solving for hypotenuse: Which function uses adjacent and hypotenuse? 50° hypotenuse 2 COSINE 90° opposite

Calculating Sides:One Side and Angle Known What is known? • angle (50°) and adjacent side (2) Solving for hypotenuse: cos(50°)= = hypotenuse = ~3.111 50° 2 ~0.643 3.111 hypotenuse hypotenuse 2 90° opposite

Calculating Sides:One Side and Angle Known Now we know: • angle (50°) and hypotenuse (3.111) Solving for opposite: Which function uses oppositeand hypotenuse? 50° 3.111 2 SINE 90° opposite

Calculating Sides:One Side and Angle Known Now we know: • angle (50°) and hypotenuse(3.111) Solving for opposite: sin(50°)= = opposite= ~2.384 50° opposite ~.766 3.111 3.111 2 90° 2.384 opposite

Calculating Angle:Two Sides Known What is known? • adjacent (3) and opposite (5) Solving for angle (a): Which function uses adjacent and opposite? a hypotenuse 3 TANGENT 90° 5

Calculating Angle:Two Sides Known What is known? • adjacent (3) and opposite (5) Solving for angle (a): tan(a)= = * need to use inverse tan → tan-1(.6) = a= ~30.964° a 30.964° 3 .6 5 hypotenuse 3 90° 5

TEKS Reference §111.35. Precalculus (c)  Knowledge and skills. (3)  The student uses functions and their properties, tools and technology, to model and solve meaningful problems. The student is expected to: (A)  investigate properties of trigonometric and polynomial functions;

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Trigonometry - Sequence of Lessons

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TRIGONOMETRY

By : Rushikesh Reddy

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.

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.

Many ancient mathematicians like Aryabhata, Brahmagupta, Ibn Yunus and Al-Kashi made significant contributions in this field(trigonometry).

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.

AC is the hypotenuse

The other two sides are known as legs.

AB and BC are the legs

Trigonometry deals with Right Triangles

Right Triangle

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 figureAC2 = AB2 + BC2

Pythagoras Theorem

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

Trigonometric Ratios

Sin = AB/AC

Cos = BC/AC

Tan = AB/BC

Cosec = AC/AB

Sec = AC/BC

Cot = AC/AB

Value of Trigonometric Functions for Angle C

0 30 45 60 90

Sine 0 0.5 1/2 3/2 1

Cosine 1 3/2 1/2 1/2 0

Tangent 0 1/ 3 1 3 Not defined

Cosecant Not defined 2 2 2/ 3 1

Secant 1 2/ 3 2 2 Not defined

Cotangent Not defined 3 1 1/ 3 0

Values of Trigonometric Functions

sin2 + cos2 = 1 1 + tan2 = sec2 1 + cot2 = cosec2 sin(/2) = ±[(1-cos )/2] Cos(/2)= ±[(1+cos)/2] Tan(/2)= ±[(1-cos)/(1+cos)]

Trigonometric Identities

There are two Systems of measurements of angles ie., Degree and Radian.

Conversion of degree to radian: Radian= Degree×(л/180) Conversion of radian to degree: Degree=Radian×(180/л)

Conversion of Angles

Angles in Standard Position

sin (A+B) = sin A cos B + cos A sin B sin (A-B) = sin A cos B - cos A sin B cos (A+B) = cos A cos B - sin A sin B cos(A-B) = cos A cos B + sin A sin B tan (A+B) = [tan A + tan B] / [1 - tan A tan

B] tan (A-B) = [tan A - tan B] / [1 + tan A tan

A-B Formula

sin C - sin D = 2 cos (C+D)/2 sin (C-D)/2 sin C + sin D = 2 sin (C+D)/2 cos (C-D)/2 cos C - cos D = 2 sin (C+D)/2 sin (C-D)/2 cos C + cos D = 2 cos (C+D)/2 cos (C-

C-D Formula

sin 2θ = 2 sin θ cos θ cos 2 θ = cos2 θ - sin2 θ tan 2 θ = 2 tan θ / (1 - tan2 θ) tan (θ /2) = sin θ / (1 + cos θ) sin (- θ)=-sin θ cos(- θ)=cos θ tan(- θ)=-tan θ

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, etc

Applications of Trigonometry

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 .

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Trigonometry PPT Maths Class 11

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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 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]

= [2 cos 4 x  cos 2 x ] [–2 sin 4 x  (–sin 2 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

= 2 [1 + cos (x + y)]

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

= 1 + 1 – 2 [cos (x – y)]

= 2 [1 – cos (x – y)]

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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