Trigonometric ratios and identities 1

Mathematics

Trigonometric ratios and Identities

Session 1

Topics

Transformation of AnglesCompound Angles

Definition and Domain and Range of Trigonometric Function

Measurement of Angles

Measurement of Angles

O A

BOA InitialRay

OB Ter minal Ray

Angle is considered as the figure obtained by rotating initial ray about its end point.

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Measure and Sign of an Angle

Measure of an Angle :-Amount of rotation from initial side to terminal side.Sign of an Angle :-

O A

B

Rotation anticlockwise – Angle positive

B’

Rotation clockwise – Angle negative

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Right Angle

O

Y

X

Revolving ray describes one – quarter of a circle then we say that measure of angle is right angle

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Angle < Right angle Acute AngleAngle > Right angle Obtuse Angle

Quadrants

O

Y

Y’

X’ X

II Quadrant( , )

I Quadrant( , )

IV Quadrant( , )

III Quadrant( , )

X’OX – x - axis

Y’OY – y - axis

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System of Measurement of Angle

Measurement of Angle

Sexagesimal System or British System

Centesimal System or French System

Circular System or Radian Measure

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System of Measurement of Angles

Sexagesimal System (British System)

1 right angle = 90 degrees (=90o)1 degree = 60 minutes (=60’)1 minute = 60 seconds (=60”)

Centesimal System (French System)1 right angle = 100 grades (=100g)1 grade = 100 minutes (=100’)1 minute = 100 Seconds (=100”)

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Is 1 minute of sexagesimal1 minute of centesimal ?

=

NO

System of Measurement of Angle

Circular System

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O

r

r

r

A

B

1c

If OA = OB = arc ABcThen AOB 1radian( 1 )

System of Measurement of Angle

Circular System

O A

C

B1c

AOC arc ACAOB arcACB

1radian r2right angles r

2right angles radian

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180 radian

Relation Between Degree Grade And Radian Measure of An Angle

0 gD G 2C

90 100

OR

0 gD G C

180 200

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Illustrative ProblemFind the grade and radian measures of the angle 5o37’30”Solution

o' o30 30 130" 60 60 60 120

o37and37' 60

o oo 37 1 455 37'30" 5 60 120 8

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We know thatD G 2C

90 100

Illustrative Problem Find the grade and radian measures of the angle 5o37’30”

g10G D9

g g

g10 45 225 12.5 Ans9 8 18

c

and R D180

c45 radian Ans180 8 32

Solution

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Relation Between Angle Subtended by an Arc At The Center of Circle

O A

C

1c

B

Arc AC = r and Arc ACB = AOC arcACAOB arc ACB

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1radian r

r

Illustrative Problem A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 meters when it has traced out 72o at the center. Find the length of rope. [ Take = 22/7 approx.].Solution

P A

B

72o

Arc AB = 88 m and AP = ?c

o 272 72 rad180 5

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arc ABr AP

2 88 22AP 70m [ approx.]5 AP 7

Definition of Trigonometric Ratios

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2 2r x y

xO

Y

X

P (x,y)

M

yr

ysin rxcos rytan x

xcot yrsec x

rcosec y

Some Basic Identities

sin cosec 1 ; n ,n I

2 2sin cos 1 2 2sec tan 1

2 2cosec cot 1

sintan ; 2n 1 ,n Icos 2

tan cot 1 ; 2n 1 ; n ,n I2

coscot ; n ,n Isin

cos sec 1 ; 2n 1 ,n I2

Illustrative Problem

Solution3sec tan .cosec

3 cosecsec 1 tan sec

3sec 1 tan cot 2sec 1 tan

2 21 tan 1 tan 3

2 21 tan

3

2 22 e Proved

J0032 2I f tan 1 e ,provethat

3

3 2 2sec tan .cosec 2 e

0 2

Signs of Trigonometric Function In All Quadrants In First Quadrant

xO

Y

X

P (x,y)

M

yr

Here x >0, y>0, 2 2r x y >0

ysin 0r

xcos 0r

ytan 0x

xcot 0y

rsec 0x

rcosec 0y

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Signs of Trigonometric Function In All Quadrants In Second Quadrant

Here x <0, y>0, 2 2r x y >0

ysin 0r

rcosec 0y

XX’

Y

Y’

P (x,y)

x

y r

xcos 0r

ytan 0x

xcot 0y

rsec 0x

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Signs of Trigonometric Function In All Quadrants In Third Quadrant

Here x <0, y<0, 2 2r x y >0

rcosec 0y

rsec 0x

X’ X

P (x,y)

O

Y’

YM

ysin 0r

xcos 0r

ytan 0x

xcot 0y

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Signs of Trigonometric Function In All Quadrants In Fourth Quadrant

Here x >0, y<0, 2 2r x y >0ysin 0r

XO

P (x,y)Y’

M

xcos 0r

ytan 0x

xcot 0y

rsec 0x

rcosec 0y

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Signs of Trigonometric Function In All Quadrants

I QuadrantAll Positive

II Quadrantsin & cosec are Positive

III Quadranttan & cot are Positive

IV Quadrantcos & sec are Positive

X

Y’

X’

Y

O

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ASTC :- All Sin Tan Cos

Illustrative Problem

lies in secondIf cot = 12 ,5

quadrant, find the values of other five trigonometric functionSolution

12 5cot tan5 12 2 2 2 169sec 1 tan sec 144

13 13sec sec liesinsecondquadrant12 12

12Whichgivescos 13

13cosec 5

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Method : 1

5 12 5Thensin tan cos 12 13 13

Illustrative Problem

lies in secondIf cot = 12 ,5

quadrant, find the values of other five trigonometric function

Solution

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Method : 2

Y

XX’Y’

P (-12,5)

-12

5 r

Here x = -12, y = 5 and r = 13

y 5sin r 13

x 12cos r 13

y 5tan x 12

r 13sec x 12

r 13cosec y 5

Functions

Domain Range

sin R [-1,1]cos R [-1,1]

sec R : (2n 1) 2

R-(-1,1)

cosec R : n R-(-1,1)

tan R : (2n 1) 2

R

cot R : n R

Domain and Range of Trigonometric Function

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Illustrative problemProve that

22 (x y)sin 4xy

is possible for real values of x and y only when x=ySolution

2 2(x y) 1 x y 4xy4xy

2sin 1

2 2x y 4xy 0 x y 0

But for real values of x and y 2x y is not less than zero 2x y 0 x y Pr oved

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Trigonometric Function For Allied Angles

Trig. ratio - 90o- 90o+ 180o- 180o+ 360o- 360o+

cos cos sin - sin - cos - cos cos cos

tan - tan cot - cot -tan tan - tan tan

sin - sin cos cos sin - sin - sin sin

If angle is multiple of 900 then sin cos;tan cot; sec cosecIf angle is multiple of 1800 then sin sin;cos cos; tan tan etc.

Trigonometric Function For Allied Angles

Trig. ratio - 90o- 90o+ 180o- 180o+ 360o- 360o+

sec sec cosec - cosec - sec - sec sec sec

cosec - cosec sec sec cosec -cosec - cosec cosec

cot - cot tan -tan -cot cot - cot cot

Periodicity of Trigonometric Function

Periodicity : After certain value of x the functional values repeats itself

Period of basic trigonometric functions

sin (360o+) = sin period of sin is 360o or 2cos (360o+) = cos period of cos is 360o or 2tan (180o+) = tan period of tan is 180o or

J005If f(x+T) = f(x) x,then T is called period of f(x) if T is the smallest possible positive number

Trigonometric Ratio of Compound AngleAngles of the form of A+B, A-B, A+B+C, A-B+C etc. are called compound angles(I) The Addition Formula

sin (A+B) = sinAcosB + cosAsinB cos (A+B) = cosAcosB - sinAsinB

tanA tanBtan A B 1 tanA tanB

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sin(A B)Pr oof: tan A B cos(A B)

sinA cosB cos A sinBcos A cosB sinA sinB

Trigonometric Ratio of Compound Angle

J006sinA cosB cos A sinBcos A cosB sinA sinB

r rDividing N and D by cos A cosB

We get tanA tanB

1 tanA tanB

Proved

cotBcot A 1cot A B cotB cot A

Illustrative problemFind the value of(i) sin 75o

(ii) tan 105o

Solution(i) Sin 75o = sin (45o + 30o) = sin 45o cos 30o + cos 45o sin 30o

1 3 1 1 3 12 22 2 2 2

(ii) Ans: 2 3

Trigonometric Ratio of Compound Angle

(I) The Difference Formula

sin (A - B) = sinAcosB - cosAsinB cos (A - B) = cosAcosB + sinAsinB

tanA tanBtan A B 1 tanA tanB

Note :- by replacing B to -B in addition formula we get difference formula

cotBcot A 1cot A B cot A cotB

Illustrative problem

If tan (+) = a and tan ( - ) = b

Prove that a btan2 1 ab

Solution

tan2 tan

tan tan1 tan tan

a b1 ab

Some Important Deductions

sin (A+B) sin (A-B) = sin2A - sin2B = cos2B - cos2A

cos (A+B) cos (A-B) = cos2A - sin2B = cos2B - sin2A

tanA tanB tanC tanA tanB tanCtan A B C 1 tanA tanB tanB tanC tanC tanA

To Express acos + bsin in the form kcos or sinacos +bsin

2 22 2 2 2a ba b cos sin

a b a b

2 2 2 2a bLet cos , thensin

a b a b

2 2acos bsin a b cos cos sin sin

2 2a b cos

Similarly we get acos + bsin = sin

2 2k cos , where k a b ,

Illustrative problem

7cos +24sin

Find the maximum and minimum values of 7cos + 24sinSolution

2 22 2 2 2

7 247 24 cos sin7 24 7 24

7 2425 cos sin25 25

7 24Let cos sin25 25

7cos 24sin 25 cos cos sin sin

Illustrative problem Find the maximum and minimum value of 7cos + 24sin

Solution 25cos 25cos where

1 cos 1 25 25cos 25

Max. value =25, Min. value = -25 Ans.

Transformation Formulae

Transformation of product into sum and difference 2 sinAcosB = sin(A+B) + sin(A - B) 2 cosAsinB = sin(A+B) - sin(A - B) 2 cosAcosB = cos(A+B) + cos(A - B)

Proof :- R.H.S = cos(A+B) + cos(A - B)= cosAcosB - sinAsinB+cosAcosB+sinAsinB= 2cosAcosB =L.H.S

2 sinAsinB = cos(A - B) - cos(A+B) [Note]

Transformation Formulae Transformation of sums or difference into products

C D C DsinC sinD 2sin cos2 2

C D C DsinC sinD 2cos sin2 2

C D C DcosC cosD 2cos cos2 2

C D C DcosC cosD 2sin sin2 2

C D D CcosC cosD 2sin sin2 2

or Note

By putting A+B = C and A-B = D in the previous formula we get this result

Illustrative problemProve thatcos8x cos6x cos 4x cot 6xsin8x sin6x sin4x

Solution(cos8x cos 4x) cos6xL.H.S (sin8x sin4x) sin6x

8x 4x 8x 4x2cos cos cos6x2 28x 4x 8x 4x2sin cos sin6x2 2

2cos6x cos2x cos6x2sin6x sin2x sin6x

2cos6x(cos2x 1)2sin6x(cos2x 1)

cot 6x Proved

Class Exercise - 1If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately)

227

A

B

E (Eye)

r

Moon

Class Exercise - 1If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately)

227

A

B

E (Eye)

r

Moon

Solution :-Let the coin is kept at a distance r from the eye to hide the moon completely. Let AB = Diameter of the coin. Then arc AB = Diameter AB = 2.2 cm

c c30 130́ 60 2 180 360

arc 2.2radius 360 r

360 2.2 360 2.2 7r 252 cm22

Class Exercise - 2

Solution :-

Prove that tan3A tan2A tanA = tan3A – tan2A – tanA.

We have 3A = 2A + Atan3A = tan(2A + A) tan3A = tan2A tanA

1– tan2A tanA

tan3A – tan3A tan2A tanA = tan2A + tanA

tan3A – tan2A – tanA = tan3A tan2A tanA (Proved)

Class Exercise - 3If sin = sin and cos = cos, then

–sin 02

(c) –cos 02

(d)

sin 02

(a) cos 02

(b)

Solution :- sin sin cos cosand

sin – sin 0 and cos – cos 0

– –2sin cos 0 and – 2sin sin 02 2 2 2

–sin 02

Class Exercise - 4Prove that

3sin20 sin40 sin60 sin80

16

LHS = sin20° sin40° sin60° sin80°Solution:-

1sin60 [2sin20 sin40 ] sin802

3 1[cos(40 – 20 ) – cos(40 20 )] sin802 2

3 [cos20 – cos60 ]sin804

3 1sin80 cos20 – sin804 2

3 2sin80 cos20 – sin808

Class Exercise - 4Prove that

3sin20 sin40 sin60 sin80

16

Solution:-

3 sin(80 20 ) sin(80 – 20 ) – sin808

3 sin100 sin60 – sin808

3 sin(180 – 80 ) sin60 – sin808

3 3sin80 – sin808 2

316

Proved.

Class Exercise - 5Prove that

n n

n

cos A cosB sin A sinBsin A – sinB cos A – cosB

A B2 cot , if n is even20, if n is odd

Solution :-n ncos A cosB sinA sinBLHS sinA – sinB cos A – cosB

n nA B A – B A B A – B2cos cos 2sin cos2 2 2 2A B A – B A B A – B2cos sin –2sin sin2 2 2 2

n nA– B A – Bcot – cot2 2

Class Exercise - 5

Solution :-

n n nA – B A – Bcot (–1) cot2 2

nn nA – B2cot , if n is evenA – Bcot 1 (–1) 22 0, if n is odd

Prove thatn n

n

cos A cosB sinA sinBsin A – sinB cos A – cosB

A B2cot , if n is even20, if n is odd

Class Exercise - 6The maximum value of 3 cosx + 4 sinx + 5 is

(d) None of these(a) 5 (b) 9(c) 7

2 23cosx 4sinx 3 4 Solution :-

2 2 2 23 4cosx sinx

3 4 3 4

3 45 cosx sinx5 5

5 cosx cos sinx sin

3 4Let cos sin5 5

Class Exercise - 6The maximum value of 3 cosx + 4 sinx + 5 isSolution :-

3 4Let cos sin5 5

5cos(x – )

–1 cos(x – ) 1

–5 5cos(x – ) 5

–5 5 5cos(x – ) 5 10

0 3cosx 4sinx 5 10

Maximum value of the given expression = 10.

Class Exercise - 7If a and b are the solutions of a cos + b sin = c, then show that

2 22 2

a – bcos( )a b

Solution :-We have … (i)acos bsin c

acos c – bsin 2 2 2a cos (c – bsin )

2 2 2 2 2a 1– sin c – 2bcsin b sin

2 2 2 2 2a b sin – 2bcsin (c – a ) 0

are roots of equatoin (i),

Class Exercise - 7If a and b are the solutions of acos + bsin

= c, then show that

2 2

2 2a – bcos( )a b

Solution :-

2 22 2

c – asin sina b

Hence

Again from (i),bsin c – acos

2 2 2b sin (c – acos )

2 2 2 2 2b (1– cos ) c a cos – 2cacos

2 2 2 2 2(a b ) cos – 2accos c – b 0

sin and sin are roots of equ. (ii).

Class Exercise - 7If a and b are the solutions of acos + bsin

= c, then show that

2 2

2 2a – bcos( )a b

Solution :- (iv) and be the roots of equation (i),

cos and cos are the roots of equation (iv).2 22 2

c – bcos cosa b

2 2 2 2 2 22 2 2 2 2 2

c – b c – a a – b–a b a b a b

cos( ) cos cos – sin sin Now

Class Exercise - 8If a seca – c tana = d and b seca + d tana = c, then

(a) a2 + b2 = c2 + d2 + cd

(c) a2 + b2 = c2 + d2

(d) ab = cd

2 22 21 1a b

c d(b)

Class Exercise - 8If a seca – c tana = d and b seca + d tana = c, then

asec – ctan d a csin– dcos cos

Solution :-

bsec dtan c Againb dsin ccos cos

b ccos – dsin ….. ii

a csin dcos ….. (I)

Squaring and adding (i) and (ii), we get

2 2 2 2 2 2 2 2a b c (sin cos ) d (cos sin )

2cd sin cos 2cd sin cos 2 2c d

Class Exercise -9

2 2A Asin – sin –8 2 8 2

The value of

(a) 2 sinA

(c) 2 cosA

1 sinA2

(b)1 cos A2(d)

Class Exercise -9

2 2A Asin – sin –8 2 8 2

Solution :-

A A A Asin – sin – –8 2 8 2 8 2 8 2

1sin sinA sinA4 2

2 2sin A – sin B sin(A B) sin(A – B)

2 2A Asin – sin –8 2 8 2

The value of

Class Exercise -10If , ,

and lie between0 and , then value of tan2 is

4cos( ) 5 5sin( – ) 13

4

(a) 1

(c) 0

(b)5633

(d)3356

Solution :- and between 0 and ,4

– – and 04 4 2

Consequently, cos and sin are positive.2sin( ) 1– cos ( )

Class Exercise -10

Solution :-16 31– 25 5

2 25 12cos( – ) 1– sin ( – ) 1– 169 13

3 5tan( ) , tan( – )4 12

tan2 tan[( ) ( – )]

3 5tan( ) tan( – ) 564 12

3 51– tan( ) tan( – ) 331– 4 12

If , , and lie between0 and , then value of tan2 is

4cos( ) 5 5sin( – ) 13

4