trignometry for aieee cet by vasudeva kh

8/14/2019 Trignometry for Aieee Cet by Vasudeva Kh

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VIJNAN STUDY CIRCLE-TRIGNOMETRY-FORMULA AND CONCEPTS

BY K.H. V.AN ANGLE:An angle is the amount of rotation of a revolving line w.r.t a fixed straight line (a figureformed by two rays having common initial point.) The two rays or lines are called the sides of the angle and

common initial point is called the vertex of the angle.

Rotation of the initial arm to the terminal arm generates the angle. Rotation can be anti clock wise or clockwise.

Angle is said to be +ve if rotation is anti clockwise.

Angle is said to be -ve if rotation is clockwise.

UNITS OF MEASUREMENT OF ANGLES:a) Sexagesimal system:In sexagesimal system of measurement,the units of measurement are degrees, minutes andseconds.1 right angle =90 degrees(90o);1 degree = 60 minutes (60')1 minute = 60 seconds (60'')

b)Centisimal system of angles:1 right angle =100 grades =100g

1 grade =100 minutes =100'1' = 100 seconds =100''

c) RADIAN OR CIRCULAR MEASURE : In this system units of measurement is radian.A radian is the measure of an angle subtended at the center of a circle by an arc whose length is equal

to the radius of the circle. one radian is denoted by 1c

1 radian =570 161 22''

A radian is a Constant angle. And

radians = 1800

RELATIONSHIP BETWEEN DEGREES AND RADIANS:

radians =180o 1 radian= 1c =180

o

1c = 570 17' 45''; 10 =

180o

radian=0.01746 radian

(approximately)

Radian measure=

180o

x Degree measure i.e. To convert degrees into radians Multiply by

180o

Degree measure= 180o

x Radian measure. i.e. To convert radians into degrees Multiply by 180

o

NOTE: 1. Radian is the unit to measure angle 2. It does not means that stands for 1800 , is real number,where as c stands for 1800

LENGTH OF ARC OF A CIRCLE:If an arc of length s subtends an angle radians at the center of a circle of radius 'r', then

S =r i.e. length of arc = radius x angle in radians (subtended by arc)

No of radians in an angle subtended by an arc of circle at the centre =arc

radius=

S

r

1c(1 radian) = arclength of magnitude of rradius ofr

AREA OF A SECTOR OF A CIRCLE:(sectorial area)

The area of the sector formed by the angle at the center of a circle of radius r is 1

2r2 .

RADIAN MEASURE OF SOME COMMON ANGLES:

0(Degrees)

150 220

300 450 600 750 900 1200 1350 1500 1800 2100

2700 3600

c

Radians

1

+veangle

-ve angle

AB be the Arc, Let the lengthof the arc =OA=radius

angle AOB =1 radian

VIGNAN CLASSES

Do You know?

When no unit is

mentioned with an angle,

it is understod to be in

radians. If the radius of

the circle is r and its

circumference is C then

C=2rC/2r =

for any circleCircumference/diameter

= which is constant.

=3.1416(approximately)

Termin

al

si

de(arm

)

Initialside(arm)

----r-----

B

Arc

A

D

3

4

56

7

6

3

2

212

8

6

4

3

512

2

23


2/15

SOME USEFUL FACTS ON CLOCKS:

1. Angle between two consecutive digits of a

clock is 300 or

6radians.

2. Hour hand of the clock rotates by an angle

of 300 or

6radians in one hour

and 12

0 or 360

radians in one minute.

3. Minute hand of the clock rotates by an

angle of 60 or

30radians in one minute.

TRIGNOMETRIC FUNCTIONS OR RATIOS AND FUNDAMENTAL RELATIONS.1. If is an acute angle of a right angled triangle OPM

We define Six trigonometric ratios(t-ratios) as

sin =opposite side

hypotenus; cos =

adjacent side

hypotenus

tan = opposite sideadjacent side

; cosec = hypotenusopposite side

sec =hypotenus

adjacent side; cot =

adjacent side

oppositeside

2. Let be an angle in standard position. If P(x,y) is any point on the terminal side of and

OP= x2y2 =r ; thensin =

y

rcos =

x

rtan =

y

x

cosec =r

ysec =

r

xcot =

x

y

RELATIONS BETWEEN TRIGNOMETRIC RATIOS

BASIC IDENTITTIES:

a) sin2 + cos2 =1;

b) 1+ tan2 = sec2 ;

c) 1+ cot2 = cosec2 ;

DEDUCTIONS:

cos2 = 1 -sin2; sin2 = 1- cos2;

sec2 -1 = tan2; cosec2 -1 = cot2;

sec2 - tan2 =1; cosec2 - cot2 =1

RECIPROCAL RELATIONS

cosec =1

sin; sec =

1

coscot =

1

tan;cosec.sin =1 ; sec. cos =1 ; cot. tan =1

QUOTIENT RELATIONStan =

sin

cos; cot =

1

tan=

cos

sin

SIGNS OF TRIGNOMETRIC FUNCTIONS :

I II III IV

sin + + - -

cos + - - +

tan + - + -

cosec + + - -

sec + - - +

cot + - + -

2

Opposite side

Hypote

nus

Adjacent side

O

P

M

P

O M

0900

9001800

18002700 27003600

(QUADRANT RULE)

a) In First quadrant, all

t-ratios are +ve.

b) In Second quadrantsin , cosec are +ve.

c) In Third quadrant tan and cotare +ve

d) In Fourth quadrant, only cos

and sec are +ve.

DO YOU KNOW:In a regular polygon

i) All the interior angles are equal

ii) All the exterior angles are equaliii) All the sides are equal

iv)Sum of all the exterior angles is 3600

v) Each exterior angle = 3600/number of

exterior angles

vi)Each interior angle = 1800 -exterior angle

vii) For a polygon with n sidesa) the sum of internal angles is

(2n-4) right angles, where a rightnangle

=900b) the number of diagonals is n(n-3)/2

The following approximate values are quite helpful:

2 = 1.41; 3 =1.73;

1/ 2 =0.7; 3 /2 =0.87 ;

1 /3 =0.58 2/ 3 ==1.154

AS

CT

A=All are +ve

S=Sin & cosec are +ve

T=Tan & Cot are +ve

C=Cos & Sec are +ve

Short Cut to remember:

ALL STUDENTS TAKE

COFFE


3/15

TO DETERMINE THE VALUES OF OTHER TRIGNOMETRIC RATIOS WHEN ONE

TRIGNOMETRIC RATIO IS GIVEN:

If one of the t-ratio is given , the values of other t-ratios can be obtained by constructing a right angledtriangle and using the trigonometric identities given above

For ex. sin=1/3, since sine is +ve in Q1 and Q2(II quadrant), we have cos= 11

9or

-

11

9ie. 2 23 or 223 according as Q1 or Q2

We can find other ratios by forming a rightangled traingle.

Let tan=4/3, 3

2, then since in Q3, sine and cosine both are negative,

we have sin=-4

5; cos=

3

5

For acute angled traingle, we can write other t ratios in terms of given ratio:

Let sin=s=perp

hyp

=s

1

cos= = 1sin2 ; tan=sin

1sin2; sec=

1

1sin2; cosec=

1

sin ; cot= 1sin

2

sin

We can express sin in terms of other trigonometric functions by above method:

sin= 1cos2 =tan

1tan2 =

1

cosec = sec

21

sec= 1tan

2

tan

MAXIMUM AND MININUM VALUES :

1. since sin2A+cos2A =1, hence each of sinA and cosA is numerically less than or equal to unity, that is

|sinA|1 and |cosA|1 i.e. -1sinA1 and -1cosA12. Since secA and cosecA are respectively reciprocals of cosA and sinA, therefore the values of secA andcosecA are always numerically greater than or equal to unity. That is

secA1 or secA-1 and cosecA1 or cosecA-1, In otherwords we never have -1


4/15

TRIGNOMETRIC RATIOS OF STANDARD and QUANDRANTAL ANGLES:

Radians

0

6

4

3

2

3

2

2

12

5

12

Degrees 0 300 450 600 900 1800 2700 3600 150 750

sin

0

1

2

1

232 1 0 -1 0

312 2

312 2

cos

1

32

1

21

2 0 -1 0 1

312 2

312 2

tan

0

1

3 13

0

0

23 23

Approximate values of sin , cos and tan when is small (OUT O

F SYLLUBUS)

Let be small and measured in radian, then sin , cos 1; tan .

These are first degree approximations. The second degree approximations are given by

sin ; cos 1-1

2

2, tan

VALUES OF T-FUNCTIONS OF SOME FREQUENTY OCCURING ANGLES.

Radians 0 2

3

3

4

5

62n1

2

n

Degrees 1200 1350 1500(odd )

2

(any )

sin

321

21

2

(-1)n

0

cos

1

2

1

23

2 0

(-1)n

tan 3-1

1

3

0

e.g. cos(odd

2)=0; cos( odd )=-1, cos(even ) =1

cos 2n1

2 =0, cos( 2n-1) =-1, cos(2n ) =1

sin(any ) =0, tan(any ) =0 sin n =tan n =0 if n=0,1,2

sin

2= sin

5

2=sin

9

2=.......=1

sin(3

2) = sin

7

2= sin

11

2= ..........=-1

Some interesting results about allied angles:

1. cosn `=(-1)n , sin n =0 2)Sin(n + ) =(-1)n sin ;

cos(n + )=(-1)n cos

3) cos(n

2+)=(-1)n+1/2 sin if n is odd 4)sin(

n

2+)=(-1)n-1/2 cos if n is odd

=(-1)n/2 sin if n is even =(-1)n/2 cos if n is even

4


5/15

DOMAIN AND RANGE OF TRIGNOMETRIC FUNCTIONS:

Function Domain Range

sine

cosine

tangent

cotangent

secant

cosecant

R

R

R-{(2n+1)

2}: n Z

R-{n }; nZ

R-{(2n+1)

2}: n Z

R-{n }; nZ

[-1, 1]

[-1, 1]

R

R

(- ,-1] [1, )

(- ,-1] [1, )

ASTC RULE:(QUADRANT RULE):ASTC rule to remember thesigns allied angles

A denotes all angles are positive in the I quadrant

S says that sin (and hence cosec) is positive in the II quadrant.

The rest are negative.

T means tan (and hence cot) is positive in the III quadrant. The rest are negative.

C means cos (and hence sec) is positive in the IV quadrant. The rest are negative.

The trignometric ratios of allied angles can be easily remembred from the

following clues:

1. First decide the sign +ve or -ve depending upon the quandrant in whichthe angle lies using QUADRANT RULE.

2. a) When the angle is 90+ or 270, the trignometric ratio changes

from sinecosine, cosinesine, tancot, cottan, seccosec,

cosecsec.

Hence the sine and cosine, tan &cot, sec & cosec are called co - ratios.

b) When the angle is 180+ or 360 , -, the trignometrc ratio is remains the same. i.e

sin sine, cosinecosine , tantan, cotcot, secsec, coseccosec.

ALLIED ANGLE FORMULAE:Trignometrc ratios of allied angles

sin cos tan sec cosec cot

- -sin cos -tan sec -cosec -cot

900 - cos sin cot cosec sec tan

900 + cos -sin -cot -cosec sec -tan

1800 - sin -cos -tan -sec cosec -cot

1800+ -sin -cos tan -sec -cosec cot

2700 - -cos -sin cot -cosec -sec tan

2700 + cos -sin -cot -cosec sec -tan

3600 - -sin cos -tan sec -cosec cot

The above may be summed up as follows: Any angle can be expressed as n.90+ where n is anyinteger and is an angle less than 900. To get any t. ratios of this angle

a) observe the quandrant n.90+ lies and determine the sign (+ve or -ve).

b) If n is odd the function will change into its co function ( i.e sinecosine; tancot; seccosec. If n iseven t-ratios remains the same.(i.e sinsin, coscos etc)

ILLUSTRATION: 1. To determine sin(540-), we note that 5400 - =6 x 900 - is a second quadrantangle if 0


6/15

Short cut: Supposing we have to find the value of t- ratio of the angle

Step1: Find the sign of the t-ratio of , by finding in which quadrant the angle lies. This can be done

by applying the quadrant rule, i.e. ASTC Rule.

Step 2: Find the numerical value of the t-ratio of using the following method:

t-ratios of =

t- ratio of (1800- ) with proper sign if lies in the second quandrant

e.g.: cos1200 = -cos600 = -1/2

t-ratio of ( -180) with proper sign if lies in the third quandrant

e.g: sin2100 = -sin300 = -1/2

t-ratio of (360- ) with proper sign if lies in the fourth quandrant

e.g: cosec3000= -cosec600 = 2

3

t-ratio of -n (3600 ) if >3600

d) If is greater than 3600 i.e. =n.3600 + , then remove the multiples of 3600 (i.e. go on subtracting

from 3600 till you get the angle less than 3600 ) and find the t-ratio of the remaining angle by applyingthe above method. e.g: tan10350 =tan6750 (1035-360) =tan3150 = -tan450 =-1

COMPLIMENTARY AND SUPPLIMENTARY ANGLES:

If is any angle then the angle

2- is its complement angle and the angle - is its

supplement angle.

a) trigonometric ratio of any angle = Co-trigonometric ratio of its complement

sin = cos(90- ), cos = sin(90- ), tan = cot(90- ) e.g. sin600 =cos300 , tan600 =cot300 .

b) sin of(any angle) = sin of its supplement ; cos of ( any angle) = -cos of its supplement

tan of any angle = - tan of its supplement i.e. sin 300 =sin 1500 , cos 600 =-cos 1200

CO-TERMINAL ANGLES: Two angles are said to be co terminal angles , if their terminal sidesare one and the same. e.g. and 360+ or and n.360+ ; - and 360- or - and n.360-

are co terminal angles : a) Trig functions of and n.360+ are same

b) Trig functions of - and n.360- are same .

TRIGNOMETRIC RATIOS OF NEGETIVE ANGLES:

For negative angles always use the following relations:

c) sin(- ) = -sin cos(- ) = cos , tan(- )= -tan , cosec(- )= -cosec ; se(- ) =sec ;

ci) cot(- ) =sec (V.IMP)

TRIGNOMETRICAL RATIOS FOR SUM AND DIFFERENCE:

COMPOUND ANGLE FORMULAE: (Addition and Subtraction formulae)

1. Sin (A + B) = sin A cos B + cos A sin B

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

3. Cos (A + B) = cos A cos B sin A sin B

4. cos (A B) = cos A cos B + sin A sin B

5. tan (A + B) =tan AtanB

1tan A tanB

6. tan (A B) =tan AtanB

1tan A tanB

DEDUCTIONS:

7. sin(A-B)sin(A-B) =sin2A-sin2B

=cos2B -cos2A

8. cos(A+B)cos(A-B) =cos2A-sin2B

=cos2B -sin2A

9. tan(A+B)tan(A-B)=tan

2Atan2B

1tan2A . tan2B

10.Cot(A+B) =cotAcotB1

cotAcotB

(A#n, B#m, A+B#k)

11.Cot(A-B) =

cotAcotB1

cotBcotA

(A#n, B#m, A-B#k)

12.tan(A+B)=sinAB

cos AB

13.tan(A-B)=sinAB

cos AB

14. tanAtanB

tanAtanB=

sin AB

sin AB

15.1+tanA tanB= cos ABcosAcosB

6


7/15

1-tanA tanB=cos AB

cosAcosB

16. tanA+tanB=tan(A+B)(1-tanA.tanB)

=sin AB

cosA. cosB

tanA-tanB=tan(A-B)(1-tanA.tanB)=sin AB

cosA.cosB

17.tan(/4 + A) =1tanA

1tanA

18.tan(/4 -A) =1tanA

1tanA

19.cot( /4 + A )=cotA1

cotA1

20.cot( /4 -A )=cotA1

cotA1

21. tan(A+B+C)

=tanAtanBtanCtanA.tanB.tanC

1 tanAtanBtanB.tanCtanC.tanA

=S1S31S2

If S1 = tanA + tanB +tanC S3 =tanA.tanB.tanC

S2 =tanAtanB +tanB.tanC +tanC.tanA

21.The cot(A+B+C) =

cotA.cotB.cotCcotAcotBcotC

cotAcotBcotB.cotCcotC.cotA1

22. sinA+cosA= 2sin

4A

sinA-cosA= 2sin

4A

cosA+sinA= 2cos

4A

cosA-sinA= 2cos

4A

23. sin(A+B+C)

=SinA.cosB.CosC +sinB.cosC.cosA + SinC.cosA.cosB

-sinA.sinB.sinC

=one sine and two cos - three sines= sinA.sinB.sinC [cotA.cotB-1]

24. cos(A+B+C) =cosA.CosB.cosC -sinA.sinB.cosC-sinBsinCcosA -sinCsinAcosB

=Three cos - one cos and two sines

=cosAcosBcosC[1-tanAtanB-tanBtanC-tanCtanA]

MULTIPLE ANGLE FORMULAE: T ratios of multiple angles

1.Sin 2A = 2 sin A cos A =2tanA

1tan2A

2.cos 2A = cos 2 A sin 2 A

= 1 2 sin2A

= 2cos2A 1 =1tan

2A

1tan2A

3. tan 2A =2 tan A

1tan2A

DEDUCTIONS:

1+cos2A =2cos2A; cos2A =1

21cos2A

1-cos2A =2sin2A; cos2A =1

21cos2A

1cos2A

1cos2A=tan2A;

1cos2A

1cos2A=cot2A

1+sin2A =(sinA +cosA)2

1-sin2A =(sinA -cosA)2

cotA -tanA = 2 cot2AtanA+cotA=2 cosec 2A

TRIPLE ANGLES: T - ratios of 3 in terms of those of

Sin 3A = 3 sin A 4 sin3A ;

cos 3A = 4 cos3A 3 cos A ;

tan3A =3tanAtan

3A

13tan2A

;

DEDUCTIONS:

4 sin3A =3 sin A -Sin 3A ;

sin3A =1

4( 3 sin A -Sin 3A ).

4 cos3A =3 cos A +cos 3A;

cos

3

A =

1

4 ( 3 cos A +cos 3A )

TRIGNOMETRC RATIOS OF HALFANGLES-t ratios of sub multiple angles

a) sin =2sin

2cos

2=

2tan

2

1tan2

2

b) cos=cos2

2-sin2

2=2cos2

2-1

=1-2sin2

2=

1tan2

2

1tan2

2

c)tan=

2tan

2

1tan2

2

7


8/15

DEDUCTIONS:

1+cos=2cos2

2; 1-cos=2sin2

2

1cos

1cos =tan2

2;

1cos

1cos=cot2

2

1sin

1sin = tan

2

4

2

;

1sin

1sin = cot

2 4 2

sin

1cos =tan

2;

sin

1cos =cot

2

cos

1sin = tan4 2 ;

cos

1sin = cot4 2

Transformation formulae:

a) SUMS AND DIFFERENCE TO PRODUCT FORMULAE:

Formula that express sum or difference into products

Sin C + sin D = 2sinCD

2cos

C D

2Sin C sin D = 2cos

CD

2sin

C D

2

Cos C + cos D = 2cosCD

2cos

C D

2Cos C cos D = 2sin

CD

2sin

DC

2

or 2sinCD

2sin

CD

2

b) PRODUCT-TO-SUM OR DIFFERENCE FORMULAE :formula which expressproducts as sum or Difference of sines and cosines.

2 sin A cos B = sin (sum) + sin (diff) i.e 2 sinA cosB = sin(A+B) + sin(A-B)

2 cos A sin B = sin (sum) sin (diff) i.e 2 cosA sinB = sin(A+B) - sin(A-B)

2 cos A cos B = cos (sum) + cos (diff) i.e. 2 cosA.cosB = cos(A+B)+cos(A-B)

2 sin A sin B = cos (diff) cos (sum) i.e. -2 sinA.sin B = cos(A+B)-cos(A-B)

OR 2 sinA.sin B = cos(A-B)-cos(A+B)

VALUES OF TRIGNOMETRICAL RATIOS OF SOME IMPORTANT ANGLES :

Angle

Ratio

71

20

150 18022

1

20

360 750

sin 8262 24

or 4622 2

312 2

514

1

222

1

41025 31

2 2

cos 8262 24

or

4622 2

312 2

1

41025

1

222

1

451

31

2 2

tan 6432or

3

2

2

1

2- 3 251055

21 525 2+ 3

cot 6432or

3221

2+ 3 52 5 21

12

52- 3

sec 16102836 ( 62 )2

2

5 422 51 62

sin220 =1

222 ;

cos220

=

1

2 22 ;

tan220 = 21 ;cot220= 21

sin180 =1

451 =cos720 ;

8


9/15

cos180 =1

4102 5 =sin720 ;

sin360 =1

41025 =cos540;

cos360 =1

451 =sin540

tan7 0= 6432

cot70= 6432

sin90 = 35354

cos90 = 35354

EXPRESSION FOR Sin(A/2) and cos(A/2) in terms of sinA:

sin A2 cos A2 2

=1+sinA so that sinA

2cos

A

2= 1sinA

sin A2 cos A2 2

=1-sinA so that sinA

2cos

A

2= 1sinA

By addition and subtraction, we have

2 sinA

2= 1sinA 1sinA ; 2 cos

A

2= 1sinA 1sinA

Using suitable signs , we can find sin A2

, cos A2

IDENTITTIES CONNECTED WITH TRAINGLE:

If A,B,C are angles of a traingle,

sin(sum of any two) =sin(third); e.g.:sin(B+C) =sinA;

cos(sum of any two)= -cos(third); e.g.: cos(A+B)= -cosC]

tan(sum of ny two) = -tan(third) e.g. : tan(A+B) =-tanC

sin1

2(sum of any two) = cos

1

2(third); e.g sin

AC

2=cos

B

2)

cos1

2(sum of any two) = sin

1

2(third), e.g: cos

BC

2=sin

A

2)

If A is any angle of traingle and lies between 00 and 1800 , then

sinA=sin A = or 1800- ; cosA=cos = ; tanA=tan =

SOME IMPORTANT IDENTITTIES:

If A+B+C =1800 , then

1) sin2A +sin2B+sin2C=4sinAsinBsinC i.e. sin2A = 4sinAsinBsinC2)cos2A+cos2B+cos2C=-1-4cosAcosBcosC i.e. cos2A =-1-4cosAcosBcosC

3)sinA+sinB+sinC=4cosA

2cos

B

2cos

C

2VIGNAN CLASSES

i.e. sinA =4cosA

2cos

B

2cos

C

2

4)cosA+cosB+cosC=1+4sinA

2sin

B

2sin

C

2

i.e cosA =1+4sinA

2sin

B

2sin

C

2

5)tanA+tanB+tanC=tanA.tanB.tanC i.e. tanA = tanA.tanB.tanC6)cotB.cotC+cotC.cotA+cotA.cotB =1 i.e. cotA.cotB =1

7)cotA

2+cot

B

2+cot

C

2=cot

A

2cot

B

2cot

C

2

i.e. cotA

2=cot

A

2cot

B

2cot

C

2

8)tanA

2tan

B

2+tan

B

2tan

C

2+tan

C

2tan

A

2=1 i.e. tan

A

2tan

B

2=1

Note: If A, B, C are the angles of a traingle , thensin(A+B+C) =sin=0, cos(A+B+C) =cos = -1 and tan(A+B+C) =0;

9


10/15

GRAPHS OF TRIGNOMETRIC FUNCTIONS

I quadrant II quadrant III quadrant IV quadrant

sin increases

from 0 to 1

decreasesfrom

1 to 0

decreses from

0 to -1

increases from

-1 to 0

cos decreases from

1 to 0

decreases

from

0 to -1

increases from

-1 to 0

increases from

0 to 1

tan increases from

0 to

increases from

to 0

increases from

0 to

increases from

to 0

cot decreases from

to 0

decreasesfrom

0 to

decreases from

to 0

decreases from

0 to

sec increses from

1 to

increases from

to -1

decreases from

-1 to

decreases from to 1

cosec decreases from

to 1

increases from

1 to

increases from

to -1

decreases from

-1 to -infinity

Graph of sinx Graph of cosecx

Graph of cosx Graph of secx

Graph of tanx Graph of cotx

10

f(x)=cot(x)

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

x

y


11/15

RELATION BETWEEN THE SIDES & ANGLES OF A TRIANGLE:

A traingle consists of 6 elements, three angles and three sides. The angles of traingle ABCare denoted by A,B, and C. a,b, and c are respectively the sides opposite to the angles A,Band C.

In any traingle ABC , the following results or rule hold good.

1 Sine rule: a = 2R sin A, b = 2R sin B, c = 2R sin C iea

sinA=

b

sinB=

c

sinC=2R Where R is

the circum radius of circum circle that passes through the vertices of the traingle.

2.Cosine rule: a2 =b2 +c2 -2bc cosA or cos A =b

2c

2 a

2

2bc

b2 =a2 +c2 -2ac cosB or cos B =c2a2 b2

2ca

c2 =a2 +b2 -2ab cosC or cos C =a2b2 c2

2ab

3.Projection rule:

a = b cos C + c cos B; b = c cos A + a cos C; c = a cos B + b cos A

4.Napier's formula or Law of Tangents:

tanBC

2=[b c

bc]cot

A

2or bcbc =

tanBC

2

tanBC

2

tan AB2

=[a b ab

]cot C2

or

abab =

tanAB

2

tanAB

2

etc.

5.Half-angle rule: In any traingle ABC, a+b+c =2s, where 2s is the perimeter of the

traingle. sinA

2=

s b s c

bccos

A

2=

s s a

bc tan

A

2=

sbsc

s sa

sinB

2=

s a s c

accos

B

2=

s s b

ac tan

B

2=

sasc

s sb

sin

C

2 =

s a s b

ab cos

C

2 =

s s c

ab tan

C

2 =

sasb

s sc

6. Formula that involve the Perimeter: If S=abc

2, where a+b+c is the perimeter of

a traingle, R the radius of the circumcircle, and r the radius of the inscribed circle, then

6. Area of traingle:= ssa sb sc ;(HERO'S FORMULA)

=1

2a.b.SinC =

1

2b.c. sinA =

1

2c.a.sinB=

abc

4R

=1

2

a2sinB. sinC

sinA=

1

2

b2sin.C sinA

sinB=

1

2

c2sinA. sinB

sinC=

1

2

a2sinB.sinC

sinBC

DEDUCTIONS:

sinA=2

bc=

2

bcs sa sb sc sinB=

2

caSinC=

2

ab

tanA

2tan

B

2=

sc

s; tan

B

2tan

C

2=

sa

s; tan

C

2tan

A

2=

sb

s.

tanA

2tan

B

2=

scot

C

2; tan

B

2tan

C

2=

scot

A

2;

tanC

2tan

A

2

=

scot

B

2.

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NOTE WORTHY POINTS: In a traingle ABC

If cotA +cotB +cotC= 3 thentraingle is equilateral

If sin2 A +sin2B + sin2 C =2 thentraingle is equilateral

If cosA + cosB +cosC =3/2 then traingleis equilateral

If cotA cotB cotC>0 then traingle isacute angled traingle

If cos2 A+cos2 B +cos2C =1 thentraingle is rightangled traingle

If in a trainglea

cosA=

b

cosB=

c

cosCthen traingle is equilateral

In a traingle a sinA =b sinB, thentraingle is isosceles

If a cosA = bcosB then traingle is

isosceles or rightangled If in atraingle 8R2 =a2 +b2 +c2 then

traingle is rightangled.

SOLUTION OF TRIANGLES

To solve a triangle a) when all the 3 sides are given :

GIVEN REQUIRED

a,b, c i) Area of = ssa sbsc , 2s = a+b+c

sinA=2bc , sinB=

2ac , sin C=

2ab OR

iii) First, find two of the three angles by cosine formula, then the third angle isdetermined by using the relation A+B+C=1800. It is advisable to find the smallestangle first. (angle opposite to the smallest side).

b) When two sides and an included angle is given:

GIVEN REQUIRED

a , b and Ci)Area of traingle==

1

2a.b.SinC ; tan

AB

2=[a b

ab]cot

C

2

AB2

=900 - C2

; c= asinCsinA

ii) Use cosine rule to find the third side. then find the smaller of the two anglesby cosine formula. Use A+B+C=1800 to find the third angle

iii)Use Napier's formula and find two angles, then the third side can bedetermined sine rule or cosine rule or by projection rule.

c)when one side and two angles A and B are given:

GIVEN REQUIRED

a A and Bi) C =180-(A+B) ; b=

asinB

sinA

;c=asinC

sinA

d) When two sides and an angle opposite to one of them is given.

Let us assume that a,b, and A are given. Now we are required to find c,B and C. We justcannot find c or C directly before finding B. There exist only one relation with which we can

find B i.e. by using sine Rule. sinB =b sinA

a; C=180-(A+B); c=

asinC

sinA

CASES:i)When A is acute angle and a1, which is impossible. then there exists no solution or no traingle.

ii)When A is acute angle and a=bsinA: In this case only one traingle is possible

which is rightangled at B. If a=bsinA , sinB =1, then B=900 there exist only onesolution or one traingle since A is given, we can find C using A+B+C=1800 . we canfind 'c' by any one of the rules.

iii)When A is acute angle and a>bsinA, sinB


13/15

SUMMERY:

A unique traingle exists if I)three sides are given (b+c>a etc)

ii)one side and two angles are given

iii)two sides and included angle are given

iv)But two sides and angle opposite to one of these sides are given , the followingcases arise: a, b, A given

i)aa>bsinA

iv)a>b

No triangle

Right angled triangle

Two triangles

one triangle

OTHER IMPORTANT FORMULA AND CONCEPTS:

1.To find the greatest and least values of the expression asin +bcos :

Let a=rcos. b=rsin , then a2 +b2 =r2 or r= a2b2 asin +bcos = r(sin cos +cos sin ) = rsin( + )

But -1sin( + )1 so that -r rsin( + )r. Hence -

a2

b

2 asin +bcos

a2

b

2

Thus the greatest and least values of asin +bcos are respectively a2b2 and - a2b2 .Similarly maximum value ofasin -bcos is a2b2

For 0 , minimum value of a sin + bcosec is 2 ab

For

2

2, minimum value of acos +bsec is 2 ab

For 0

2or

3

2, minimum value of a tan +bcot is 2 ab

2. cosA.cos2A.cos4A.cos8A............cos2n-1 A = 12

nsinA

sin 2nA

(Remember)

OR cos .cos2 .cos22.cos23............cos2 n =sin 2

n1A

2nsinA

(Each angle being double of preceding)

3. SUM OF THE SIN AND COSINE SERIES WHEN THE ANGLES ARE IN AP:

sin +sin(+) +sin( +2 ) +..........n terms

cos +cos(+) +cos( +2 ) +..........n terms

=

sin n.diff

2

sindiff

2

. sin or cos

[1st anglelast angle

2 ](Remember the rule)

=

sinn

2

sin

2

.sin or cos [n1 2 ] =sin

n

2

sin

2

.sin or cos [n1 2 ]Note: is not an even multiple of i.e. #2n because in that case sum will take the form 0/0. Particular

case: Both the sum will be zero if sinn

2=0 i.e.

n

2=r or =

2r

nor = even multiple of

n

then S=04. SOME RESULTS IN PRODUCT FORM:

sin sin(60+)sin(60-) =1

4sin3

cos cos(60+) cos(60-)

=1

4cos3

cos cos(120+) cos(120-)

tan tan(60+ )tan(60- ) =tan3

sin(600 -A) sin(600 +A) = sin3A4sinA

cos(600 -A) cos(600 +A)=cos3A

4cosA

tan(600 -A) tan(600 +A) =tan3A

tanA

tan2A tan3A tan5A=tan5A-tan3A-tan2A

tanx tan2x tan3x =tan3x-tan2x-tanx

(Use the above formula at time of integration)

tan(x-). tan(x+ ) tan 2x= tan2x-tan(x+ )-tan(x- )

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(cos +cos ) (cos2 +cos2 ) (cos22 +cos22 ) .........(cos2n +cos 2n ) =cos2

n1cos2n1

2n cos cos

(2cos -1)(2cos2 -1)(2cos22 -1).......(2cos2n ) =2cos2

n11

2cos 1

4. i) cosA sinA= 2sin

4 A = 2cos

4 A ii) tanA +cotA =

1

sinA.cosA

5. tan + tan

3 + tan

2

3 =3tan3 ; tan + tan

3 + tan

3 =3tan3

6. 2222............22cos2n =2cos nN

HEIGHTS AND DISTANCES-VIGNAN CLASSESANGLE OF ELEVATION AND ANGLE OF DEPRESSIONSuppose a st.line OX is drawn in the horizontal direction.

Then the angle XOP where P is a point (or the positionof the object to be observed from the point O of observation )

above OX is called Angle of Elevation of P as seen from O.Similarly, Angle XOQ where Q is below OX, is calledangle of depression of Q as seen from O.

OX is the horizontal line and OP and OQ are called

line of sights

Properties used for solving problems

related to Heights and Distances.1.Any line perpendicular to a plane is

perpendicular to every line lying in the plane.Explanation: Place your pen PQ upright on your notebook, so that its lower end Q is on the notebook.Through the point Q draw line QA,QB,QC,....... in your notebook in different directions and you willobserve that each of the angles PQA,PQB,..PQC,.... is a right angle. In other words PA is perpendicularto each of the lines QA, QB, QC, lying in the plane.

2.To express one side of a right angled triangle in terms of the other side.

Explanation: Let ABC =, Where ABC is right angledtriangle in which C = 900 . The side opposite to right angle Cwill be denoted by H(Hypotenus),

the side opposite (opposite side) to angle is denoted by O,the side containing angle (other than H)(Adjacent side) will be denoted by AThen from the figure it is clear thatO=A(tan ) or A = O(cot ) i.e. Opposite = Adj(tan ) or Adj=opposite (cot ).Also O=H(sin ) or A =H(cos ) i.e opposite =Hyp( sin ) or Adjacent =Hyp(cos )

;,./ []-SWEQRTYUIXCVBNMKL ' 098PREPARED AND DTP BY KHV,

LECTURER IN MATHEMATICS

14

O X

Q

= Angle ofelevation of P

=Angle ofDepression of Q

H

A

O

P

THE SPIRIT OF MATHEMATICSThe only way to learn mathematics is to recreate it for oneself -J.L.Kelley

The objects of mathematical study are mental constructs. In order to understand these one

must study , meditate, think and work hard -SHANTHINARYAN

Mathematical theories do not try to find out the true nature of things, that would be an

unreasonable aim for them. Their only purpose is to co-ordinate the physical laws we find

from experience but could not even state without the aid of mathematics. -A. POINCARE

Experience and intution, though usually obtained more painfully, may be doveloped by

mathematical insight. -R Aris


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15

trignometry for aieee cet by vasudeva kh

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