Trigo Identities

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TRIGONOMETRIC RATIOS & IDENTITIES

1. INTRODUCTION

The word 'Trigonometry' is derived from two Greek words

(1) Trigonon and(2) MetronThe word trigonon means a triangle and the word metron means a measurement. Hence trigonometry means thescience of measuring triangles.

2. ANGLE

Consider a ray OAuuur

. If this ray rotates about its end point O and takes the position OB , then the angle AOB has

been generated.

Vertex O

= angle

Initial side A

B

Termin

al side

An angle is considered as the figure obtained by rotating a given ray about its end - point.

The initial position OA is called the initial side and the final position OB is called terminal side of the angle. The endpoint O about which the ray rotates is called the vertex of the angle.

3. SENSE OF AN ANGLE

The sense of an angle is said to be positive or negative according as the initial side rotates in anticlockwise orclockwise direction to get to the terminal side.

O = +ve

Anticlockwise directionA

B O = ve

Clockwise directionA

B

4. RIGHT ANGLE

When two lines intersect at a point in such a way that two adjacent angles made by them are equal, then eachangle is called a right angle.

9090

OX' X

A

5. A CONSTANT NUMBER pipipipipiThe ratio of the circumference to the diameter of a circle is always equal to a constant and this constant is denotedby the Greek letter pi

TRIGONOMETRIC RATRIGONOMETRIC RATRIGONOMETRIC RATRIGONOMETRIC RATRIGONOMETRIC RATIOS & IDENTITIESTIOS & IDENTITIESTIOS & IDENTITIESTIOS & IDENTITIESTIOS & IDENTITIES


i.e. Circumference of a circle

Diameter of the circle= pi

(constant)

The constant pi is an irrational number and its approximate value is taken as 227 . The more accurate value to six

decimals places is taken as 355113 .

6. SYSTEMS OF MEASUREMENT OF AN ANGLESThere are three systems for measuring angles

6.1 Sexagesimal or English system6.2 Centesimal or French system6.3 Circular system

6.1 Sexagesimal system : The principal unit in this system is degree (). One right angle is divided into 90 equalpart and each part is called one degree (1) . One degree is divided into 60 equal parts and each part is calledone minute. Minute is denoted by (1'). One minute is equally divided into 60 equal parts and each part is calledone second (1").In Mathematical form :

One right angle = 90 (Read as 90 degrees )1 = 60' (Read as 60 minutes )1' = 60" (Read as 60 seconds )

Ex.1 40 30' is equal to

(1) o41

2

(2) 81 (3) o81

2

(4) None of these

Sol. We know that , 30' = o1

2

; 40 + o1

2

=

o812

6.2 Centesimal system : The principal unit in system is grade and is denoted by (g). One right angle is dividedinto 100 equal parts, called grades, and each grade is subdivided into 100 minutes, and each minute into 100seconds.

In Mathematical form :

One right angles = 100g (Read as 100 grades)1g = 100' (Read as 100 seconds)1' = 100" (Read as 100 seconds)

Ex.2 25' is equal to -Sol. 100' is equal to 1g

so is equal to g g1 125

100 4

=

Relation between Sexagesimal and Centesimal systems :One right angle = 90 (degree system) ..... (1)


One right angle = 100g (grade system) ..... (2)by (1) and (2)

90 = 100g or , D G90 100

=

then we can say, ; 1 = g10

9

, 1g = o9

10

Ex.3 80g is equal to

Sol. We know that 1g = o9

10

then, 80g = o9 80

10

80g = 72

6.3 Circular system : In circular system the unit of measurement is radian. One radian, written as 1C, is the measureof an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle.Consider a circle of radius r having centre of O. Let A be a point on the circle. Now cut off an arc AB whoselength is equal to the radius r of the circle.

B O rIc

rr

C

A

In the adjacent figure OA = OC = arc AC = r = radius of circle, then measurement of AOC is one radianand denoted by 1c. Thus AOC = 1c .

6.3.1 Some Important conversion

pi Radian = 180 One radian = o180

pi 6pi

Radian = 30

4pi

Radian = 45 3pi

Radian = 60 2pi

Radian = 90

23pi

Radian = 120 34pi

Radian = 13556pi

Radian = 150

76pi

Radian = 210 54pi

Radian = 22553pi

Radian = 300


Ex.4 240 is equal to

[1] C4

3

p [2] C3

4

p [3] '4

3p [4]

'34p

Sol. We know that180 = pi

240 = C

x240180

p

=

C43

p

Ans. [1]

Ex.5 The difference between two acute angle of a right angle triangle is 9p . Then the angles in degree are -

[1] 50, 30 [2] 25, 45 [3] 20, 40 [4] 35, 55Sol. In triangle ABC let C = 90

So A B =

9p = 20 .......... (i)

and sum of all the angles in ABC

A + B + C = 180

C = 90

A + B = 90 ......... (ii)Solving (i) & (ii) A = 55, B = = 35 Ans. [4]

6.3.2 Relation between systems of measurement of angles

D G 2C90 100

= =

pi

Ex.6 The length of an arc of a circle of radius 5 cm subtending a central angle measuring 15 is -

[1] 312p

cm [2] 712p

cm [3] 512p [4] None of these

Sol. Let s be the length of the arc subtending an angle at the centre of a circle of radius r.

then, = s

r

Here, r = 5 cm, and = 15 = C

15x180

p

= C

12

p

= s

r = 12

p =

s

5

= 512

pcm


7. TRIGONOMETRICAL RATIOS OR FUNCTIONSLet a line OA makes angle with a fixed line OX and AM is perpendicular from A on OX. Then in right-angledtriangle AMO, trigonometrical ratios (functions) with respect to are defined as follows :

sin = perpendicular(P)hypotenuse(H)

cos = base(B)

hypotenuse(H)

tan = perpendicular (P)

Base (B)

cosec = HP . sec =

HB , cot =

BP

Note :

(i) Since t-ratios are ratio between two sides of a right angled triangle with respect to an angle, so they arereal numbers.(ii) may be acute angle or obtuse angle or right angle.

8. RELATIONS BETWEEN TRIGONOMETRICAL RATIOS

(i) 1 1 1cosec , sec ,cotsin cos tan

= = = (ii)

sintancos

=

(iii) coscotsin

= (iv) sin

2 + cos2 = 1

(v) 1 + tan2 = sec2 (vi) 1 + cot2 = cosec2Ex.7 If cosec A + cot A = 11/2, then tan A is equal to

[1] 21/12 [2] 15/16 [3] 44/117 [4] 117/43Sol. Cosec A + cot A = 11/2

1cosecA cot A+ =

211 ....... (1)

cosec A cot A = 211 ..... (2)

(1) (2) = 2 cot A = 11 22 11- = 11722

= tan A = 44

117 Ans [3]

O B

M

H P

Y

X

A


Ex.8cos

1 tanq

q- + sin

1 cotq

q- is equal to

[1] sin cos [2] sin + cos [3] tan + cot [4] tan cot

Sol.cos

1 tanq

q- + sin

1 cotq

q-

=

cos sinsin cos1- 1cos sin

q qq qq q

+

-

=

2 2cos sincos sin cos sin

q qq q q q

-

- -

=

2 2cos sincos sin

q qq q

-

-

= cos + sin Ans [2]Ex.9 tan2 sec2 (cot2 cos2 ) equals

[1] 0 [2] 1 [3] 1 [4] 2Sol. tan2 sec2 (cot2 cos2 )

= sec2 (tan2 cot2 tan2 cos2 )

= sec2 2

22

sin1 coscos

q qq

- = sec2 (1sin2)

= sec2. cos2 = 1 Ans. [3]

9. SIGN OF TRIGONOMETRIC RATIOS(i) All ratios sin, cos, tan cot, sec and cosec are positive

in Ist quadrant.(ii) sin( or cosec) positive in IInd quadrant, rest are negative.(iii) tan( or cot) positive in IIIrd quadrant, rest are negative.(iv) cos( or sec) positive in IVth quadrant, rest are negative.

Ex.10 The value of sin and tan if cos = 1213

-

and lies in the third quadrant is -

[1] 513- and 5

12 [2] 5

12 and 5

13- [3] 1213- and

513

- [4] none of these

Sol. We have cos2 + sin2 = 1

sin = 21 cos q -

IInd Quadrant Ist Quadrant

IIIrd Quadrant IVth Quadrant

x-axis

y-axis


In the third quadrant sin is negaitve, therefore

sin = 21 cos q- sin = 2121

13

- = 5

13

then, tan = sincos

qq tan =

5 13x

13 12-

-

=

512 Ans.[1]

Ex.11 If 2p

< < pi, then 1 sin1 sin

qq

-

+ +

1 sin1 sin

qq

+

-

is equal to

[1] 2 cosec [2] 2 cosec [3] 2 sec [4] 2 sec

Sol. Exp. = 2(1 sin ) (1 sin )

1 sinq q

q

- + +

- =

2cosq

-

= 2 sec Ans.[4]

10. DOMAIN AND RANGE OF A TRIGONOMETRICAL FUNCTIONIf f : X Y is a function, defined on the set X, then the domain of the function f, written as Domain is theset of all independent variables x, for which the image f(x) is well defined element of Y, called the co-domainof f.Range of f : X Y is the set of all images f(x) which belongs to Y , i.e.Range f = {f(x) Y:x X} Y The domain and range of trigonometrical functions are tabulated as follows

Trigo. function Domain Rangesin x R, the set of all the real number 1 sin x 1

cosx R 1 cos x 1

tan x R ( )2n 1 ,n2pi

+

I R

cosecx R { }n ,npi I R { x : 1 < x < 1 }

sec x R ( )2n 1 ,n2pi

+

I R { x : 1 < x < 1 }

cot x R { }n ,npi I R


11. VARIATION OF VALUES OF TRIGONOMETRICAL RATIOS IN DIFFERENT QUADRANTS-

12. RELATION BETWEEN TRIGONOMETRICAL RATIOS AND IDENTITIES-

(1)

=cos

sintan (2)

=sincos

cot

(3) sin A cosec A = tan A cot A = cos A sec A = 1(4) sin2 + cos2 = 1 or sin2 = 1 cos2 or cos2 = 1 sin2(5) 1 + tan2 = sec2 or sec2 tan2 = 1 or sec2 1 = tan2.(6) 1 + cot2 = cosec2 or cosec2 cot2 = 1 or cosec2 1 = cot2(7) Since sin2A + cos2A = 1, hence each of sin A and cos A is numerically less than or equal to unity i.e.,

|sin A| 1 and |cos A| 1or 1 sin A 1 and 1 cos A 1

Note : The modulus of real number x is defined as |x| = x if x 0 and |x| = x if x < 0.(8) Since sec A and cosec A are respectively reciprocals of cos A and sin A, therefore the values of sec Aand cosec A are always numerically greater than or equal to unity i.e.,

sec A 1 or sec A 1

and cosec A 1 or cosec A 1

In other words, we never have

1 < cosec A < 1 and 1 < sec A < 1.

While tanA and cotA may take any real value

Sine decreases from 1 to 0

cosine decreases from 0 to 1

tangent increases from to 0

cotangent decreases from 0 to

secant increases from to 1

cosecant increases from 1 to

Sine increases from 0 to 1

cosine decreases from 1 to 0

tangent increases from 0 to

cotangent decreases from to 0

secant increases from 1 to

cosecant decreases from to 1

' Sine decreases from 0 to 1

cosine increases from 1 to 0

tangent increases from 0 to

cotangent decreases from to 0

secant decreases from 1 to

cosecant increases from to 1

IV Since increases from 1 to 0

cosine increases from 0 to 1

tangent increases from to 0

cotangent decreases from 0 to

secant decreases from to 1

cosecant decreases from 1 to

'


13. TRIGONOMETRICAL RATIOS IN TERMS OF EACH OF THE OTHER

sin cos tan cot sec cosec

sin sin 2cos1 +

2tan1

tan

+ 2cot1

1

sec

1sec2eccos

1

cos 2sin1 cos + 2tan11

+

2cot1

cot

sec1

eccos

1eccos 2

tan

2sin1sin

cos

cos1 2 tan cot1

1sec2 1eccos

12

cot 21 sin

sin

2cos

1 cos

1

tan cot 21

sec 1 2cosec 1

sec 21

1 sin 1

cos21 tan+

21 cotcot+

sec 2

cosec

cosec 1

cosec 1

sin 21

1 cos

21 tantan+

21 cot+ 2

sec

sec 1 cosec

14. TRIGONOMETRICAL RATIOS OF STANDARD ANGLES


15. TRIGONOMETRICAL RATIOS OF ALLIED ANGLES

Ex.12cos(90 )sec( ) tan(180 )

sin(360 )sec(180 )cot(90 )q q q

q q q+ - -

+ + - eqauls

[1] 2 [2] 1 [3] 1 [4] 0Sol. Given expression

( sin )(sec )( tan )(sin )( sec ) tan

q q qq q q

- -

-

= 1 Ans. [3]

Ex.13 The value of cot5 cot10 ......... cot 85 is

[1] 12 [2] 1 [3] 12 [4] 0

Sol. cot 5 cot 10 ......... cot 85= cot 5 cot 10 ........ cot(90 10) cot (90 5)= cot 5 cot 10 ...... tan 10 tan 5

= (tan 5 cot 5) (tan 10 cot 10) .......= (1) (1) (1) ........... = 1

Ex.14 Sin 10 + sin 20 + sin30 +...... + sin 360 is eqaul to[1] 1 [2] 0 [3] 1 [4] name of these

Sol. Q sin 190 = sin(180 + 10) = sin 10sin 200 = sin 20

sin 210 = sin 30

..............................

sin 360 = sin 180 = 0

\ Exp. = 0


16. GRAPH OF DIFFERENT TRIGONOMETRICAL RATIOS


17. SUM AND DIFFERENCE FORMULAE

(i) sin(A + B) = sinA cosB + cosA sin B (ii) sin(A B) = sinA cosB cosA sin B(iii) cos(A + B) = cosA cosB sinA sinB (iv) cos(A B) = cosA cosB + sinA sinB

(v) tan(A + B) =BtanAtan1BtanAtan

+ tan(A B) =

BtanAtan1BtanAtan

+

(vi) tan+

=

+pitan1tan1

4 tan +

=

pitan1tan1

4

(vii) cot(A + B) =BcotAcot1BcotAcot

+

(viii) cot(A B) = AcotBcot1BcotAcot

+

(ix) sin(A + B) sin(A B) = sin2 A sin2B = cos2B cos2A(x) cos(A + B) cos(A B) = cos2A sin2B = cos2B sin2A

(xi) sin2 = 2sin cos = )tan1(tan2

2 +

(cosA sin A)2 = 1 sin 2A

(xii) cos2 = cos2 sin2 = )tan1()tan1(

2

2

+

= 1 2 sin2 = 2 cos2 1

(xiii) tan2 = 2tan1

tan2

sin

)cos1( = tan 2

+

sin)cos1(

= cot 2

)cos1()cos1(

+

= tan2 2

)cos1()cos1(

+

= cot2 2

(xiv) A 1 cosAsin 2 2

= , cos

A 1 cosA2 2

+=

(xv) tan A 1 cos A2 1 coA

= +

(xvi) sin 3A = 3 sin A 4 sin3A or sin3 A = 14 (3 sinA sin 3A)

(xvii) cos 3A = 4cos3A 3 cosA or cos3A = 14 ( cos 3A + 3cosA )

(xix) tan 3A = 3

23 tanA tan A

1 3tan A

( A npi + pi/6 )


Ex.15 tan 20 + tan 40 + 3 tan 20 tan 40 is equal to

[1] 32

[2] 34

[3] 3 [4] 1

Sol. 3 = tan 60 = tan (40 + 20)

=

tan40 tan201 tan40 tan20

+

- +

\ 3 3 tan 40 tan 20 = tan 40 + tan 20

Hence tan 40 + tan 20 + 3 tan 40 tan 20 = 3 Ans.[3]

Ex.16 If tan A = 12 and tan B =

13 , then the value of A + B i.e. tan

1 12 + tan

1

13 is

[1] 6p [2] p [3] zero [4] 4

p

Sol. tan (A + B) = tan A tanB1 tanA tanB+

-

=

1 12 3

1 11 .2 3

+

-

=

5 / 65 / 6 = 1

\ A + B = 45 = 4p

Ans.[4]

Ex.17 If sin A = 35 , 0 < A < 2

p and cos B =

1213

-

, pi < B < 32p

then sin (A B) eqauls

[1] 1665- [2] 1665 [3]

6516 [4]

6516

-

Sol. We have : sin A = 35 , where 0 < A < 2

p

\ cos A = 21 sin A -

cos A = + 21 sin A - = 9125

- =

45 [Qcos is positive in first quadrant]


It is given that : cos B = 1213

-

and pi < B < 32p

\ sin B = 21 cos B -

sin B = 21 cos B - [Q sin is negative in the third quadrant]

sin B = 2121

13 -

- = 5

13-

Now, sin (AB) = sin A cos B cos A sin B = 3 12 4 5x x5 13 5 13- -

-

=

1665

- Ans.[1]

Ex.18sin2

1 cosq

q+ equals

[1] cot [2] tan [3] sin [4] cosec

Sol.sin2

1 cosq

q+ = 22sin cos

2cosq q

q = tan Ans. [2]

Ex.19 If sin A = 12 , then 4 cos

3 A 3 cos A is equal to (0 < A < 90)

[1] 1 [2] 0 [3] 32

[4] 12

Sol. 4 cos3 A 3 cos A = cos 3A = cos 90 = 0 [Q sin A = 12 A = 30] Ans.[2]

Ex.20 If and be between 0 and 2p

and if cos ( + ) = 1213 and sin( ) = 35 , then sin 2 is equal

[1] 1615 [2] 0 [3] 5665 [4]

6465

Sol. sin (2) = sin ( + + )= sin ( + ) cos ( ) + cos ( + ) sin ( )

=

513 .

45 +

1213 .

35 =

5665 Ans.[3]


18. FORMULAE FOR TRANSFORMATION OF SUM OR DIFFERENCE INTO PRODUCT

(i) sinC + sinD = 2sin

+

2)DC(

cos2

)DC(

(ii) sinC sinD = 2cos

+

2)DC(

sin2

)DC(

(iii) cosC + cosD = 2cos

+

2)DC(

cos2

)DC(

(iv) cosC cosD = 2sin

+

2)CD(

sin2

)DC(

(v) tanA tanB = BcosAcosBsinAcosBcosAsin

BcosBsin

AcosAsin

= BcosAcos

)BAsin( =

pi

pi+pi mB,

2nA

(vi) cotA cotB = BsinAsin)ABsin(

pi+pipi

2mB,nA

(vii) cosA sinA = 2 sin

pi A4 = 2 cos

pi A4 tanA + cotA = )AcosA(sin

1

(viii) 1 + tanA tanB = BcosAcos)BAcos(

1 tanA tanB = BcosAcos)BAcos( +

(ix) cotA tanA = 2cot2A tanA + cotA = 2cosec2A

(x) sin 2A

+ cos 2A

= Asin1+ sin 2A

cos 2A

= Asin1

Ex.21 cos 52 + cos 68 + cos 172 =

[1] 0 [2] 1 [3] 2 [4] none of theseSol. cos 52 + cos 68 + cos 172

= cos 52 + cos 68 + cos (180 8)

= 2 cos 52 68

2+

cos 68 52

2-

cos 8

= 2 cos 60 cos 8 cos 8

= cos 8 cos 8 = 0 Ans.[1]Ex.22 If sin 2 + sin 2 = 1/2, cos 2 + cos 2 = 3/2 then cos2 ( ) is equal to

[1] 3/8 [2] 5/8 [3] 3/4 [4] 5/4


Sol. Using cosine formula

2 sin ( + ) cos ( ) = 1/2 ...... (1)2 cos ( + ) cos ( ) = 3/2 ...... (2)Squaring (1) and (2) and then adding

4 cos2 ( ) = 1 94 4+ = 52

cos2 ( ) = 58 Ans.[2]

Ex.23 Sin 47 + sin 61 sin 11 sin25 is equal to

[1] sin36 [2] cos 36 [3] sin 7 [4] cos 7Sol. Given value

= (sin 47 + sin 61) (sin 11 + sin 25)= 2 sin 54 cos 7 2 sin 18 cos 7

= 2 cos 7 (sin 54 sin 18)= 2 cos 7 2 cos 36 sin 18

= 2 cos 7 2sin36cos36

cos18 x cos 36

= cos 7 2sin36cos36

cos18

= cos 7 sin72cos18 = cos 7 [Qsin 72 = cos 18] Ans.[2]

19. FORMULAE FOR TRANSFORMATION OF PRODUCT INTO SUM OR DIFFERENCE(i) 2sinA cosB = sin(A + B) + sin(A B)(ii) 2cosA sinB = sin(A + B) sin(A B)(iii) 2cosA cosB = cos(A + B) + cos(A B)(iv) 2sinA sinB = cos(A B) cos(A + B)

Ex.24 2 cos 13p

cos 913

p + cos

313

p + cos

513

p equals

[1] 0 [2] 1 [3] 2 [4] 4


Sol. LHS = 2 cos 13p

cos 913

p + cos

313

p+ cos

513

p

= cos 913 13

p p + + cos 913 13

p p - + cos

313

p + cos

513

p

= cos 1013

p + cos

813

p + cos

313

p+ cos

513

p

= cos 313

pp

- + cos513

pp

- + cos 313

p+ cos

513

p

= cos 313

p cos

513

p + cos

313

p + cos

513

p

= 0 = RHS [Q cos(pi ) = cos] Ans.[1]

Ex.25 2 sin 512

p sin 12p equals to

[1] 12 [2] 12 [3]

14 [4]

16

Sol. 2 sin A sin B = cos (A B) cos (A + B)

\ 2 sin 512

p sin 12p = cos

512 12

p p - cos

512 12

p p +

= cos 3p

cos 2p

=

12 0 =

12 Ans. [2]

20. TRIGONOMETRICAL RATIOS FOR SOME IMPORTANT ANGLES

(i) sin 17 2

=

4 2 62 2

(ii) cos 17 2

=

4 2 62 2

+ +

(iii) tan 17 2

= ( )( )3 2 2 1 (iv) sin15 = 22)13(

= cos75

(v) cos15 = 22)13( +

= sin75 (vi) tan15 = 2 3 = cot75

(vii) cot15 = 2 + 3 = tan75 (viii) sin22

21

= 2221


(ix) cos22

21

= 2221

+ (x) tan22

21

= 2 1

(xi) cot22

21

= 2 + 1 (xii) sin18 = 41 ( 5 1) = cos72

(xiii) cos18 = 41

5210 + = sin72 (xiv) sin36 = 41

2210 = cos54

(xv) cos36 = 41 ( 5 + 1) = sin54

21. FORMULAE FOR SUM OF THREE ANGLES(i) sin (A + B + C) = sinA cos B cosC + cosA sin B cos C + cos A cos B sin C sin A sin B sin C

= cos A cos B cos C ( tanA + tan B + tanC tan A tan B tan C )(ii) cos (A + B + C) = cosA cosB cosC sinA sinB cosC sinA cos B sin C cos A sin B sin C

= cos A cos B cos C (1 tan A tan B tan B tan C tan C tanA )

(iii) tan (A + B + C) = tanA tanB tanC tanA tanBtanC1 tanA tanB tanBtanC tanCtan A+ +

(iv) 4sin(60 A) sinA sin(60 + A) = sin3A 4cos(60 A) cosA cos(60 + A) = cos3A tan(60 A) tanA tan(60 + A) = tan3A

22. CONDITIONAL IDENTITIES(1) If A + B + C = 180 , then

(i) sin 2A + sin 2B + sin2C = 4 sin A sin B sin C(ii) sin 2A + sin 2B sin 2C = 4 cosA cos B sin C(iii) sin (B + C A) + sin (C + A B) + sin (A + B C) = 4 sin A sin B sin C(iv) cos 2A + cos 2B + cos 2C = 14 cos A cos B cos C(v) cos 2A + cos 2 B cos 2C = 1 4 sinA sin B cos C

(2) If A + B + C = 180, then

(i) sin A + sin B + sin C = 4cos A2 cos B2 cos

C2

(ii) sin A + sin B sin C = 4 sin A2 sin B2 cos

C2

(iii) cosA + cos B + cosC = 1 + 4 sin A2 sin B2 sin

C2

(iv) cosA + cosB cos C = 1 + 4 cos A2 cosB2 sin

C2

(v) cos A cosB cosC 2sinBsinC sinCsinA sinA sinB+ + =


(3) If A + B + C = pi , then(i) sin2A + sin2B sin2C = 2 sin A sin B cos C(ii) cos2A + cos2B + cos2C = 12 cos A cos B cos C(iii) sin2A + sin2B + sin2C = 2 + 2 cosA cos B cosC(iv) cos2A + cos2B cos2C = 12 sin A sin B cos C

(4) If A + B + C = pi , then

(i) sin2 A2

+ sin2B2 + sin

2 C2 =1 2sin

A2 sin

B2 sin

C2

(ii) 2 2 2A B C A B Ccos cos cos 2 2sin sin sin2 2 2 2 2 2+ + = +

(iii) 2 2 2A B C A B Csin sin sin 1 2cos cos sin2 2 2 2 2 2+ =

(iv) 2 2 2A B C A B Ccos cos cos 2cos cos sin2 2 2 2 2 2+ =

(5) If x + y + z = 2pi

, then

(i) sin2 x + sin2y + sin2z = 12 sin x sin y sin z(ii) cos2x + cos2y + cos2z = 2 + 2 sin x sin y sin z(iii) sin2x + sin2y + sin 2z = 4 cos x cosy cos z

(6) If A + B + C = pi , then(i) tanA + tan B + tan c = tan A tan B tan C(ii) cotB cot C + cot C cot A + cot A cot B = 1

(iii) B C C A A Btan tan tan tan tan tan 12 2 2 2 2 2+ + =

(iv) A B C A B Ccot cot cot cot cot cot2 2 2 2 2 2+ + =

(7) (a) For any angles A , B, C we have(i) sin (A + B + C) = sin A cos B cos C + cos A sin B cos C

+ cos A cos B sin C sin A sin B sin C(ii) cos (A+B+C) = cos A cos B cosC cos A sin B sin C

sin A cos B sin C sin A sin B cosC

(iii) tan (A + B + C) = tanA tanB tanC tanA tanB tanC1 tanA tanB tanB tanC tanCtan A+ +

(b) If A , B, C are the angles of a triangle, thensin(A + B + C) = sin pi = 0 andcos (A + B + C) = cos pi = 1then (a) (i) givessinA sin B sin C = sin A cos B cos C + cosA sin B cosC + cos A cos B sin Cand (a) (ii) gives1 + cos A cos B cos C = cos A sin B sin C + sin A cos B sin C + sin A sin B cos C


Ex.26 If A + B + C = pi, then sin 2A + sin 2B + sin 2C equals[1] 4 sin A sin B cos C [2] 4 sin A sin B sin C[3] 4 cos A sin B sin C [4] none of these

Sol. sin 2A + sin 2B + sin 2C

= 2 sin 2A 2B

2 + cos

2A 2B2

- + sin 2C

= 2 sin (pi c) cos (A B) + sin C sin 2C [Q A + B + C = pi, A + B = pi csin (A + B) = sin (pi C) = sin C]

= 2 sin C [cos (A B) + 2 sin C cos C]= 2 sin C [cos (A B) + cos C]= 2 sin C [cos (A B) cos (A + B)] [Q cos (A B) cos (A + B) = 2 sin A

sin B, by C & D formula]= 2 sin C [2 sin A sin B]= 4 sin A sin B sin C Ans.[2]

Ex.27 If A + B + C = pi, then cos2 A + cos2 B + cos2 C equal to[1] 2 cos A cos B cos C [2] 2 Sin A sin B sin C[3] 12cosA cos B cos C [4] 1 + 2 sin A sin B sin C

Sol. LHS = cos2 A + cos2 B + cos2 C= cos2 A +(1 sin2 B) + cos2 C = (cos2 A sin2 B) + cos2 C + 1= cos (A + B) cos (A B) + cos2 C + 1 [Q cos2A sin2 B = cos (A + B) cos(A B)]= cos (pi C) cos (A B) + cos2 C + 1 = cos C cos (A B) + cos2 C + 1= cos C [cos(A B) cos C] + 1 = cos C [(cos (AB) cos {pi(A + B)}] + 1= cos C [cos (A B) + cos (A + B)] + 1= cos C [(cos A cos B + sin A sin B) + (cos A cos B sin A sin B)] +1= cos C (2 cos A cos B) +1 = 1 2 cosA cosB cos C Ans. [3]

Ex.28 If A + B + C = pi, then tanA + tanB + tanC equals[1] cotA tanB tanC [2] tanA. cotB. tanC[3] tanA. tanB. tanC [4] None of these

Sol. A + B + C = piA + B = p C

tan (A + B) = tan (pi C)

tan A tanB

1 tanA tanB+

-

= tan C

tan A + tan B = tanC + tanC tanB tanC

tanA + tanB + tanC = tanA. tanB. tanC Ans.[3]


23. GENERAL SOLUTIONS OF TRIGONOMETRICAL EQUATIONS*(i) If sin = sin then = npi + (1)n , n Z*(ii) If cos = cos then = 2npi , n Z*(iii) If tan = tan then = npi + , n Z(iv) If sin2 = sin2(v) If cos2 = tan2 then = npi , n Z(vi) If tan2 = tan2

(vii) If sin sin

cos cos

= =

then = 2npi + , n Z

24. MOTHOD OF COMPONENDO AND DIVIDENDO

If p aq b

= , then by componendo an dividendo we can write

p q a b q p b aor

q q a b q p b a

= =

+ + + +

or p q a b q p b a

orp q a b q p b a

+ + + += =

Note :- Reference of the above formulae will be given in the solutions of problems.

25. SOME IMPORTANT RESULTS

(i) 2 2 2 2a b asinx bcos x a b + + +(ii) sin2x + cosec2

x 2

(iii) cos2x + sec2 x 2(iv) tan2x + cot2 x 2

(v) 1 sin tan sec tan1 sin 4 2+ pi

= + = +

(vi) 1 sin tan sec tan1 sin 4 2 pi

= = +

(vii) 1 cos cot cosec cot1 cos 2+

= = +

(viii) 1 cos tan cosec cot1 cos 2

= = +

(ix) cos . cos 2 . cos 22 ............ cos 2n1 = n

n

sin22 sin

; ( )n pi

(x) cosA + cos (A +B) + cos (A + 2B) + ........ + cos { A + ( n 1) B } = sin nB / 2 Bcos A (n 1)sinB / 2 2

+


26. THE GREATEST AND LEAST VALUES OF THE EXPRESSION [ a sin + b cos ]Let a = r cos ......... (1)and b = r sin ......... (2)

squaring and adding (1) and (2)then a2 + b2 = r2

or , r = 2 2a b+ a sin + b cos = r(sin cos + cos sin )= r sin ( + )But 1 sin 1so 1 sin ( + ) 1then r r sin ( + ) rhence,

2 2 2 2a b asin bcos a b + + +then the greatest and least values of

a sin + b cos are respectively 2 2 2 2a b and a b+ +

Least value of a sinx + b cos x + c is 2 2c a b + and greatest value is 2 2c a b+ +Ex.29 The maximum value of 3 sin + 4 cos is -

[1] 2 [2] 3 [3] 4 [4] 5Sol. 25 < 3 sin + 4 cos < 25 [By the standard resultes]

or 5 < 3sin + 4 cos < 5so the maximum value is 5. Ans. [4]

Ex.30 If a < 3 cos x + 5 sin (xpi/6) < b for all x, then (a, b) equals[1] ( 19, 19)- [2] (17, 17) [3] ( 21, 21)- [4] None

Sol. 12 cos x + 352

sin x = A cos x + B sin x

\ 2 2(A B )+ = 1 (76)2 = (19) Ans. [1] 27. MISCELLANEOUS POINTS(i) Some useful identities :

(a) tan (A + B + C) = tanA tanA tanB tanC1 tanA tanB

(b) tan = cot 2 cot 2(c) tan3 = tan . tan ( 60 ) . tan ( 60 + ) (d) tan (A+B) tanA tanB = tanA. tanB.tan(A+B)

(e) sin sin ( 60 ) sin (60 + ) = 1 sin34 (f) cos cos ( 60 ) cos (60 + ) = 1

cos34

(ii) Some useful series :

(a) sin + sin ( + ) + sin ( + 2) ......... + to n terms =

n 1 nsin sin

2 2; 2n

sin2

+

pi

(b) cos + cos ( + ) + cos ( + 2) + ........ + to n terms =

n 1 ncos sin

2 2; 2n

sin2

+

pi


Ex.31 The value of cos 11p

+ cos 311

p + cos

511

p + cos

711

p + cos

911

p is

[1] 0 [3] 1 [3] 12 [4] none of these

Sol. cos 11p

+ cos 311

p + cos

511

p + cos

711

p + cos

911

p

= cos 11p

+ cos 3

11 11p p + + cos

3 2.211 11

p p + + cos 3.2

11 11p p + + cos

4.211 11p p +

Use cos + cos ( + ) + cos ( + 2) + .... + cos { + (n 1) } =n

sin2

sin2

b

b cos 2 (n 1)

2a b+ -

Here = 11p

, = 211p

and n = 5 then

cos 11p

+ cos 311

p + cos

511

p + cos

711

p + cos

911

p =

5 5sin x

2 112

sin2.11

p

p =

2 2cos 4

11 112

p p +

=

5sin

11sin

11

p

p cos 511p

=

1 10sin

2 11sin

11

p

p =

sin1 112 sin

11

pp

p

-

=

12 Ans.[3]

Ex.32 The value of cos 9p

cos 29p

cos 39p

cos 49p

is

[1] 18 [2] 1

16 [3] 1

64 [4] none of these

Sol. cos 9p

cos 29p

cos 39p

cos 49p

=

2 4cos cos cos

9 9 9p p p x

3cos

9p

=

3

3sin(2 . / 9)

xcos / 32 .sin / 9

p pp

=

sin8 / 98sin / 9

pp x

12 =

18 x

12 =

116 Ans. [2]


SOLVED EXAMPLES

Ex.1 2(sin6 + cos6 ) 3 ( sin4 + cos4 ) + 1 is equal to(1) 0 (2) 1 (3) 2 (4) None of these

Sol. (1) 2 [ (sin2 + cos2 )3 3 sin2 cos2 ( sin2 + cos2 ) ] 3 [ (sin2 + cos2 )2 ] 2sin2 cos2 +1 2 [ 1 3 sin2 cos2 ] 3 [ 1 2 sin2 cos2 ] + 1 26 sin2 cos2 3 + 6 sin2 cos2 + 1 = 0

Ex.2 If 3 sin + 4 cos = 5 then the value of 4 sin 3 cos is

(1) 0 (2) 1 (3) 2 (4) 3

Sol. (1) Let 4 sin 3 cos = a

Thus we want to eliminant from both 3 sinq + 4 cos = 5 and 4 sin q 3 cos = a, i.e. squaring and

adding these equations. We get

(3 sin q + 4 cos q)2 + (4 sin 3 cos )2 = 25 + a2

9 sin2 + 16 cos2 + 24 sin cos + 16 sin2 + 9 cos2 24 cos sin = 25 + a2

9 + 16 = 25 + a2

or a2 = 0

a = 0

4 sin 3 cos = 0

Ex.3 If x = r sin cos , y = r sin sin and z = r cos . Then the value of x2 + y2 + z2 is equal to(1) 2r2 (2) r2 (3) 0 (4) none of these .

Sol. (2) Herex

2 + y2 + z2 = r2 sin2 cos2 + r2 sin2 . sin2 + r2 cos2

= r2 sin2 ( cos2 + sin2 ) + r2 cos2

= r2 sin2 + r2 cos2

= r2 ( sin2 + cos2)

= r2

x2 + y2 + z2 = r2


Ex.4 If A + B = 45, then ( 1 + tanA) ( 1 + tanB ) is equal to(1) 2 (2) 1 (3) 0 (4) 4

Sol. (1) tan (A + B) = tan A tanB1 tanA tanB+

1= tan A tanB

1 tanA tanB+

[ as A + B = 45 , tan (A + B ) = 1 ]

tan A + tan B + tan A tan B = 1

or 1 + tan A + tan B + tan A tan B = 1 + 1 [ adding 1 to both side] (1 + tanA) + tanB (1 + tan A ) = 2 ( 1 + tan A) ( 1 + tanB ) = 2

Ex.5 If cos 2x + 2 cos x = 1 then sin2x ( 2cos2x) is equal to(1) 2 (2) 1 (3) 3 (4) 4 .

Sol. (2) Here, cos 2x + 2 cos x = 1 2cos2x1 + 2 cos x 1 = 0

cos2x + cosx 1 = 0

or cos x = 1 5

2 +

, neglecting 1 5

2

as 1 cos x 1 and 1 5

2

< 1

cos2 x =

25 1 6 2 5 3 52 4

= =

sin2x (2 cos2x) = 3 5 3 51 2

2 2

=

5 1 5 1 12 2

+

=

Ex.6 3 cosec20 sec 20 is equal to

(1) 0 (2) 1 (3) 2 (4) 4

Sol. (4) 3 1 3 cos20 sin20sin20 cos20 sin20 .cos20

=

3 14 cos20 sin202 22sin20 cos20

=

( )sin60 .cos20 cos60 .sin204 .sin40

=

sin(60 20 ) sin404 4 . 4sin40 sin40

= = =


Ex.7 sin78 sin66 sin42 + sin 6 Is equal to

(1) 1/2 (2) 1/2 (3) 2 (4) 2Sol. (2) The expression

= (sin78 sin42) (sin 66 sin6)= 2cos (60) sin (18) 2 cos 36 . sin 30= sin18 cos36

=

5 1 5 1 14 4 2

+

=

Ex.8 Find set of all possible values of in [ pi , pi] such that 1 sin1 sin

+ is equal to ( sec tan ).

(1) ,2 2pi pi

(2) ,2 2pi pi

(3) ,2 2pi pi

(4) None of these

Sol. (3) Clearly / 2 pi .

as sec tan = 1 sin

cos

...(i)

and ( )2

2

1 sin1 sin 1 sin 1 sin1 sin cos coscos

= = =

+

From (i) and (ii) two expressions are equal only if cos > 0 , i.e. pi/2 < < pi / 2

1 sin1 sin

+ and sec tan are equal only

when ,2 2pi pi

Ex.9 sin 1672 + cos

1672 is equal to

(1) 1 4 2 22 + (2) 1 4 2 22

(3) 1 4 2 22 + (4) 1 4 2 22

+

Sol. (1) sin 167 2 + cos1672 =

11 sin135 12

+ = + (using cosA + sinA = 1 sin2A+ )

1 4 2 22

= + ....(i)

Ex.10 If 1 + sin 4pi

+

+ 2 cos 4pi

then the maximum value of is .

(1) 1 (2) 2 (3) 3 (4) 4


Sol. (4) We have 1 + sin 4pi

+

+ 2 cos 4pi

= 1 + 12 (cos + sin ) + 2 ( cos + sin )

= 1 + 1 22

+

(cos + sin )

= 1 + 1 2 . 2 cos

42pi

+

the maximum value of 11 2 . 2 42

+ + =

Ex.11 The value of sin 20 sin 40 sin 60 sin 80 is

(1) 3/8 (2) 1/8 (3) 3/16 (4) none of theseSol. (3) sin 20 sin 40 sin 60 sin 80

= ( ) ( )3 sin20 sin 60 20 sin 60 202

+

= ( )2 23 sin20 sin 60 sin 202 = 23 3

sin20 sin 202 4

= 33 (3sin20 4sin 20 )

8

=

3sin60

8

=

3 3 3.

8 2 16=

Ex.12 The value of 1 cos 8p +

31 cos8p +

51 cos8p +

71 cos8p + is

[1] 12 [2] cos 8p [3] 18 [4]

1 22 2+

Sol. 1 cos 8p +

31 cos8p +

31 cos8pp

+ - 1 cos

8pp

+ -

= 1 cos 8p +

31 cos8p +

31 cos8p

- 1 cos

8p

-

= 21 cos

8p

- 2 31 cos

8p

-

=

1 2 1 cos4 4

p - -

32 1 cos4p

- -


=

1 1 cos4 4

p -

31 cos4p

-

=

14

112

- 112

+ = 14

112

- =

18 Ans.[3]

Ex.12 If ABCD is a cyclic quadrilateral such that 12 tan A 5 = 0 and 5 cos B + 3 = 0 then the quadratic equationwhose roots are cos C tan D is

Sol. In a quadrilaterla no angle is greater then 180

Here tan A = 5

12 so, 0 < A < 2p

and 2p

< C < pi (Q A + C = 180)

\ tan (piC) = 512 , i.e. tan C = 5

12-

\ cos C = 1213

-

Also cos B = 35

- , so, 2p

< B < pi and 0 < D < 2p (QB + D = 180)

\ cos (pi C) = 35- , i.e., cos D = 35 \ tan D =

43

\ the required equation is x2 12 413 3

- + x +

1213

- .

43 = 0

39 x2 16 x 48 = 0 Ans.[1]

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