Trignometry For Aieee Cet By Vasudeva Kh

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 figure T sid erm e( ina ar l m )

formed 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. +ve θ • Rotation can be anti clock wise or clockwise. angle • Angle is said to be +ve if rotation is anti clockwise. Initial • Angle is said to be -ve if rotation is clockwise. -ve angle side(arm)

UNITS OF MEASUREMENT OF ANGLES: b) Centisimal system of angles: a) Sexagesimal system: 1 right angle =100 grades =100g 1 grade =100 minutes =100' 1' = 100 seconds =100''

In sexagesimal system of measurement, the units of measurement are degrees, minutes and seconds. 1 right angle =90 degrees(90o); 1 degree = 60 minutes (60') 1 minute = 60 seconds (60'')

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

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=2πr C/2r =π  for any circle Circumference/diameter =π which is constant. π =3.1416(approximately)

1 radian =570 161 22'' A radian is a Constant angle. And  radians = 1800

B

D

Arc

AB be the Arc, Let the length of the arc =OA=radius

----r----- A

angle AOB =1 radian RELATIONSHIP BETWEEN DEGREES AND RADIANS: 180 o  radians =180o 1 radian= 1c =  1c = 570 17' 45''; 10 =

 radian=0.01746 radian 180 o (approximately)

 x Degree measure i.e. To convert degrees into radians Multiply by 180 o 180 o Degree measure= x Radian measure. i.e. To convert radians into degrees Multiply by  Radian measure=

 180 o 180 o 

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

LENGTH OF ARC OF A CIRCLE: If an arc of length “s” subtends an angle θ radians at the center of a circle of radius 'r', then S =r θ i.e. length of arc = radius x angle in radians (subtended by arc)

arc S = radius r

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

arc length of magnitude of r radius of r˚ AREA OF A SECTOR OF A CIRCLE:(sectorial area) 1c(1 radian) =

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

1 2

θ

r2 .θ

RADIAN MEASURE OF SOME COMMON ANGLES: θ 0 ( Degrees) θ c Radians

150

22½ 300 450

600

750

900

1200

1350

1500 1800 210 2700 3600 0

0

 12

 8

 6

 4

 3

5  12 2

2 3

1

3 4

5  6

7  3  2 6 2

SOME USEFUL FACTS ON CLOCKS: 1. Angle between two consecutive digits of a  clock is 300 or radians. 6 2. Hour hand of the clock rotates by an angle  of 300 or radians in one hour 6 1 0  and or radians in one minute. 2 360 3. Minute hand of the clock rotates by an  angle of 60 or radians in one minute. 30

DO YOU KNOW: In a regular polygon i) All the interior angles are equal ii) All the exterior angles are equal iii) 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 sides a) the sum of internal angles is (2n-4) right angles, where a rightnangle =900 b) the number of diagonals is n(n-3)/2

TRIGNOMETRIC FUNCTIONS OR RATIOS AND FUNDAMENTAL RELATIONS. 1. If θ is an acute angle of a right angled triangle OPM We define Six trigonometric ratios(t-ratios) as P opposite side adjacent side sinθ = ; cosθ = s u hypotenus hypotenus ten Opposite side o p opposite side hypotenus y H tanθ = ; cosecθ = θ adjacent side opposite side O M Adjacent side hypotenus adjacent side secθ = ; cotθ = adjacent side opposite side 2. Let θ be an angle in standard position. If P(x,y) is any point on the terminal side of θ and OP= x2  y2 =r ; then P y x y sinθ = cosθ = tanθ = r r x r r x cosecθ = secθ = cotθ = O M y x y



RELATIONS BETWEEN TRIGNOMETRIC RATIOS BASIC IDENTITTIES: a) sin2θ + cos2θ =1; DEDUCTIONS: 2 2 b) 1+ tan θ = sec θ ; cos2θ = 1 -sin2θ; sin2θ = 1- cos2θ; c) 1+ cot2θ = cosec2θ ; sec2θ -1 = tan2θ; cosec2θ -1 = cot2θ; sec2θ - tan2θ =1; cosec2θ - cot2θ =1 RECIPROCAL RELATIONS 1 1 1 cosecθ = ; secθ = cotθ = ;cosecθ.sinθ =1 ; secθ. cosθ =1 ; cotθ. tanθ =1 sinθ cosθ tanθ QUOTIENT RELATIONS sinθ 1 cosθ = tanθ = ; cotθ = cosθ tanθ sinθ SIGNS OF TRIGNOMETRIC FUNCTIONS : I

II III IV

(QUADRANT RULE) a) In First quadrant, all

sinθ

+

+

-

-

cosθ

+

-

-

+

tanθ

+

-

+

-

cosecθ +

+

-

-

c) In Third quadrant tanθ and cotθ are +ve

secθ

+

-

-

+

cotθ

+

-

+

-

d) In Fourth quadrant, only cosθ and secθ are +ve.

t-ratios are +ve. b) In Second quadrant sinθ , cosecθ are +ve.

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

2

S

0 A 0≤θ≤90

900≤θ≤1800 1800≤θ≤2700 2700≤θ≤3600 T

C

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

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 angled triangle and using the trigonometric identities given above



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



-

1−

1 9

2 2 3

ie.

or

1−

1 or 9

−2  2 according as θ ∈Q 1 or θ ∈Q2 3

We can find other ratios by forming a rightangled traingle.

3

3  , then since in Q3, sine and cosine both are negative, 2

Let tanθ=4/3,

5

4

4 −3 ; cosθ= 5 5 For acute angled traingle, we can write other t ratios in terms of given ratio:

we have sinθ=-

Let

1−s

cosθ= =

1−s

sin 

1−sin2  ; tanθ=

1

s

perp s sinθ=s= = hyp 1

2

 1−sin  2

1

; secθ=

 1−sin  2

1 ; cotθ= sin 

; cosecθ=

2

 1−sin  2

sin 

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

1−cos  = 2

tan 

 1tan

2

=



1 = cosec 

 sec −1 2

sec 

=

 1tan

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. -1≤sinA≤1 and -1≤cosA≤1 2. Since secA and cosecA are respectively reciprocals of cosA and sinA, therefore the values of secA and cosecA are always numerically greater than or equal to unity. That is secA≥1 or secA≤-1 and cosecA≥1 or cosecA≤-1, In otherwords we never have -1
  sin2  2

4.If y =a sinx + bcosx +c, then

∴ c-

 a b 2

2

≤y≤c+

⇒ r=

 a b 2

2

 a b 2

 

1 ; Min. (sin θ cos θ) =Min 2

a , b , c∈R , we can write y=c+

∀

Where a= r cos α b=r sin α

=

2

tanα = b a

sin2  2

 a b 2

1 2

sin(x+α)

2

; since -1≤sin (x+α )≤1

Hence Max. (a sinx + bcosx +c) =c+ Min (a sinx + bcosx +c)= c-

5. sin θ >cos θ >0 for

=-

 a b  a b 2

2

2

and

2

    ; 0<sin θ
4.The following formulae of Componedndo and Dividendo must be noted: If p/q =a/b then by componendo and dividendo we can write

p−q a−b = or pq ab

PERIODICITY: sin(2nΠ +α ) =sin α, cos(2nΠ +α )=cos α, tan(nΠ +α )= α ( n being any integer). All Trigonometrical functions are periodic. The period of sineθ , cosineθ, cosecθ, secθ is 2Π and that of tangent θ and cotθ is Π. sinθ is periodic with period 2 

sinkθ is periodic with period

cosθ is periodic with period 2  tanθ is periodic with period



tankθ is periodic with period

3

2 k  k

TRIGNOMETRIC RATIOS OF STANDARD and QUANDRANTAL ANGLES: Radians 0 Degrees

 6

 4

0 300

sinθ 0 cosθ

450

600

1 2

1 2

3

3

1 2

1 2

1

2

0

1 3

tanθ

 3



900

1800

2

2

2700

150

0

-1

0

0

-1

0

1

750

 3−1 2 2  31 2 2 2−3

∞ 0

5 12

 12

3600

1

∞

3

1

3 2

 2

0

 31 2 2  3−1 2 2 23

Approximate values of sinθ , cosθ and tanθ when θ is small (OUT OF SYLLUBUS) Let θ be small and measured in radian, then sinθ ≈ θ , cosθ ≈ 1; tanθ ≈ θ . These are first degree approximations. The second degree approximations are given by 1 2  , tanθ ≈ θ 2

sinθ ≈ θ ; cosθ ≈ 1-

VALUES OF T-FUNCTIONS OF SOME FREQUENTY OCCURING ANGLES. Radians 0 Degrees

3 4

2 3 1200

1350

3

1 2

sinθ

2

cosθ

−

tanθ

1 2

−

−3

e.g. cos(odd

 )=0; 2

5 6 1500

1 2



(any

 )

(-1)n 0 (-1)n

3 − 2 −

n

 2

 2

(odd )

1 2

-1

cos( odd

2n1 

0

1 3

∞ 0

)=-1, cos(even

 ) =1

cos 2n−1   =0, cos( 2n-1)  =-1, cos(2n  ) =1 2 sin(any sin sin(

 ) =0, tan(any

 5 = sin =sin 2 2 3 ) = sin 2

 ) =0 sin n

 =tan n  =0 if

n=0,1,2

9 =.......=1 2

7 11  = sin = ..........=-1 2 2

Some interesting results about allied angles: 1. cosn  `=(-1)n , sin n  =0

2)Sin(nП + θ ) =(-1)n sin θ; cos(nП + θ )=(-1)n cos θ

3) cos(

n +θ)=(-1)n+1/2 sinθ if n is odd 2

4)sin(

=(-1)n/2 sinθ if n is even

n +θ)=(-1)n-1/2 cosθ if n is odd 2 =(-1)n/2 cosθ if n is even

4

DOMAIN AND RANGE OF TRIGNOMETRIC FUNCTIONS: Function

Domain

Range

sine

R

[-1, 1]

cosine

R

[-1, 1]

tangent

R-{(2n+1)

R

 }: nε Z 2

R-{n  }; nεZ

cotangent secant

R-{(2n+1)

R

 }: nε Z 2

(- ∞ ,-1] υ [1,

R-{n  }; nεZ

cosecant

(- ∞ ,-1] υ [1,

ASTC RULE:(QUADRANT RULE):‘ASTC’ rule to remember the signs ‘allied angles’

∞ ) ∞ )

S

A

T

C

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: 90−θ 1. First decide the sign +ve or -ve depending upon the quandrant in which n.360 + θ 90+θ the angle lies using QUADRANT RULE. 180-θ 2. a) When the angle is 90+θ

or 270−θ, the trignometric ratio changes

from sine→cosine, cosine→sine, tan→cot, cot→tan, sec→cosec, cosec→sec.

180+θ 270-θ

360-θ 270+θ -θ

Hence the sine and cosine, tan &cot, sec & cosec are called co - ratios. b) When the angle is 180+θ or 360 θ , -θ, the trignometrc ratio is remains the same. i.e sin →sine, cosine→cosine , tan→tan, cot→cot, sec→sec, cosec→cosec.

ALLIED ANGLE’ FORMULAE:Trignometrc ratios of allied angles θ

sinθ

cosθ

tanθ

secθ

cosecθ

cotθ

-sinθ

cosθ

-tanθ

secθ

-cosecθ

-cotθ

900 -θ

cosθ

sinθ

cotθ

cosecθ

secθ

tanθ

0

90 + θ

cosθ

-sinθ

-cotθ

-cosecθ

secθ

-tanθ

1800 -θ

sinθ

-cosθ

-tanθ

-secθ

cosecθ

-cotθ

1800+θ

-sinθ

-cosθ

tanθ

-secθ

-cosecθ

cotθ

2700 -θ

-cosθ

-sinθ

cotθ

-cosecθ

-secθ

tanθ

2700 +θ

cosθ

-sinθ

-cotθ

-cosecθ

secθ

-tanθ

0

-sinθ

cosθ

-tanθ

secθ

-cosecθ

cotθ

-θ

360 -θ

The above may be summed up as follows: Any angle can be expressed as n.90+θ where n is any integer and θ is an angle less than 900. To get any t. ratios of this angle a) observe the quandrant n.90+θ lies and determine the sign (+ve or -ve). b) If n is odd the function will change into its co function ( i.e sine↔cosine; tan↔cot; sec↔cosec. If n is even t-ratios remains the same.(i.e sin↔sin, cos↔cos etc) ILLUSTRATION: 1. To determine sin(540-θ), we note that 5400 -θ =6 x 900 -θ is a second quadrant angle if 0<θ<900. In this quadrant , sine is positive and since the given angle contains an even multiple of  , the sine function is retained . Hence sin(540- θ ) =sin θ. 2 2. To determine cos(6300 - θ ), we note 6300 - θ =7 x 900 - θ is a third quadrant angle if 0< θ <900. In  this quadrant cosine is negative and, since the given angle contains an odd multiple of , cosine is 2 replaced by sine. Hence cos(6300 - θ ) = -sin θ. 5

Short cut: Supposing we have to find the value of t- ratio of the angle θ Step1: Find the sign of the t-ratio of θ , by finding in which quadrant the angle θ 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 θ using the following method: t-ratios of θ= t- ratio of (1800- θ ) with proper sign if θ lies in the second quandrant e.g.: cos1200 = -cos600 = -1/2 t-ratio of ( θ -180) with proper sign if θ lies in the third quandrant e.g: sin2100 = -sin300 = -1/2 t-ratio of (360- θ ) with proper sign if θ lies in the fourth quandrant e.g: cosec3000= -cosec600 = −

2 3

t-ratio of θ-n (3600 ) if θ>3600 d) If θ is greater than 3600 i.e. θ =n.3600 +α , 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 applying the above method. e.g: tan10350 =tan6750 (1035-360) =tan3150 = -tan450 =-1

COMPLIMENTARY AND SUPPLIMENTARY ANGLES:  - θ is its complement angle and the angle 2

If θ is any angle then the angle

 - θ is its

supplement angle. a) trigonometric ratio of any angle = Co-trigonometric ratio of its complement sin θ = cos(90- θ ), cos θ = sin(90- θ ), tan θ = cot(90- θ ) 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 sides are one and the same. e.g. θ and 360+ θ or θ and n.360+ θ ; - θ and 360- θ or - θ and n.360- θ are co terminal angles : a) Trig functions of θ and n.360+ θ are same b) Trig functions of -θ and n.360- θ are same . TRIGNOMETRIC RATIOS OF NEGETIVE ANGLES: For negative angles always use the following relations:

c) sin(- θ ) = -sin θ cos(- θ ) = cos θ, tan(- θ )= -tan θ , cosec(- θ )= -cosec θ ; se(- θ ) =sec θ ; ci) cot(- θ) =sec θ(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 Atan B  1 – tan A tan B

6. tan (A – B) =

tan A – tan B  1tan A tan B

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

11.Cot(A-B) =

=cos B -cos A 2

2

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

(A#nπ, B#mπ, A-B#kπ)

=cos B -sin A 2

2

2

9. tan(A+B)tan(A-B)= 10.Cot(A+B) =

cotAcotB−1 cotB−cotA

2

tan A−tan B 1 −tan 2 A . tan2 B

cotAcotB−1 cotAcotB

12. tan(A+B)=

sin AB cos  AB

13. tan(A-B)=

sin A−B cos  A−B

14.

(A#nπ, B#mπ, A+B#kπ)

tanA tanB sin  AB = tanA −tanB sin  A−B

15. 1+tanA tanB=

6

cos  A−B cosA cosB

1-tanA tanB=

cos  AB cosA cosB

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

cotA.cotB.cotC−cotA−cotB−cotC  cotAcotBcotB.cotCcotC.cotA−1 

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

sin  AB cosA. cosB

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

sin  AB cosA. cosB



22. sinA+cosA=

2 sin 4 A

sinA-cosA=

2 sin 4 −A



17.tan(Π/4 + A) =

1 tanA 1 −tanA

cosA+sinA=

2 cos 4 −A 

18.tan(Π/4 - A) =

1 −tanA 1 tanA

cosA-sinA=

2 cos 4 A 

19.cot( Π/4 + A )=

cotA −1 cotA 1

20.cot( Π/4 - A )=

cotA 1 cotA −1



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]

21. tan(A+B+C)

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

tanA tanB tanC −tanA.tanB.tanC = 1 − tanAtanBtanB.tanCtanC.tanA =



=Three cos - one cos and two sines

S 1 −S 3 1−S 2

=cosAcosBcosC[1-tanAtanB-tanBtanC-tanCtanA]

If S1 = tanA + tanB +tanC

S3 =tanA.tanB.tanC

S2 =tanAtanB +tanB.tanC +tanC.tanA

MULTIPLE ANGLE FORMULAE: T ratios of multiple angles 2tanA 2 1 −tan A

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

2

A – sin

2

DEDUCTIONS:

A

1 – 2 sin2A 2

= 2cos 2A – 1 = 3.

tan 2A =

1 −tan A 1 tan2 A

1+cos2A =2cos2A;

cos2A =

1 1cos2A  2

1-cos2A =2sin2A;

cos2A =

1 1−cos2A  2

1 −cos2A =tan2A; 1 cos2A

2 tan A 1 – tan 2 A

1 cos2A =cot2A 1 −cos2A

1+sin2A =(sinA +cosA)2 1-sin2A =(sinA -cosA)2 cotA -tanA = 2 cot2A tanA+cotA=2 cosec 2A

TRIPLE ANGLES: T - ratios of 3θ in terms of those of θ Sin 3A = 3 sin A – 4 sin3A ;

DEDUCTIONS:

cos 3A = 4 cos3A – 3 cos A ;

4 sin3A =3 sin A -Sin 3A ;

3

tan3A =

3tanA −tan A ; 1−3tan2 A

sin3A =

1 ( 3 sin A -Sin 3A ). 4

4 cos3A =3 cos A +cos 3A; cos3A =

TRIGNOMETRC RATIOS OF HALF ANGLES-t ratios of sub multiple angles

1 ( 3 cos A +cos 3A ) 4

=1-2sin2

 2   a) sinθ =2sin cos =  2 2 1 tan 2 2 2 tan

b) cosθ=cos 2

 2  =  2 1 tan 2 2 1 −tan 2

 2 c)tanθ=  1 −tan 2 2 2 tan

   -sin2 =2cos2 -1 2 2 2

7

DEDUCTIONS:



1 sin   2  = cot  1 −sin  4 2

  ; 1-cosθ=2sin2 2 2

1+cosθ=2cos 2

1 −cos  =tan2 1 cos 

1 cos   ; =cot2 2 1 −cos 



1 −sin   2  = tan − 1 sin  4 2



sin   =tan ; 2 1 cos 

 2

sin   =cot 2 1 −cos 



cos    = tan − 1 sin  4 2

;





cos    = cot  1 −sin  4 2



;



Transformation formulae:

a)

SUMS AND DIFFERENCE TO PRODUCT FORMULAE:

Formula that express sum or difference into products

2sin

Sin C + sin D =

CD C–D cos 2 2

Sin C – sin D =

2cos

CD C–D sin 2 2

CD C–D cos 2 2

Cos C – cos D =

2sin

CD D−C sin 2 2

2cos

Cos C + cos D =

−2sin

or

CD C−D sin 2 2

b) PRODUCT-TO-SUM OR DIFFERENCE FORMULAE :formula which express products 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 →

1 2

7

150

0

180

22

1 2

360

0

750

Ratio ↓ sin

 8−2 6−2 2 4

or cos

1  2−2 2

 3−1 2 2

 5−1

 31 2 2

1 1  102  5 2 2 4 2

4

1  10−2  5 4

 31 2 2

1  51  4

 3−1 2 2

 4− 6− 2 2 2

 82 6−2 2 4

or

 4 6 2 2 2

tan

6− 4−32

2-

or

3

 25−10 5

2−1

5−2  5

2+

3

5 2 5 

21



2-

3



 4−2  2

5−1

2−1

;cot22½ 0=

5

  3− 2 2−1  cot

6± 4±32

or

2+

3

  3 2 21  sec

16−10  28  3−6 (6 6−2

sin22½0 =

)

2 −

1  2−2 ; 2

2  5

tan22½0 = sin180 =

1 cos22½0 =  22 ; 2

8

1

2 5



1  5−1  =cos720 ; 4

62

21

cos180 =

1  102  5 =sin720 ; 4

cot7½0=

1  10−2  5 =cos540 ; sin360 = 4 1  51  cos360 = 4

6− 4−32

tan7 ½0=

6± 4±32  3 5−3−5

sin90 =

=sin540

4

cos90 =

 3 53−5 4

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

 

A A cos 2 2

sin

A A sin −cos 2 2

 

2

=1+sinA

so that

2

sin

=1-sinA so that

A A cos = ±1 sinA 2 2

sin

A A −cos = ±1 −sinA 2 2

By addition and subtraction, we have 2 sin

A = ±1 sinA ± ±1 −sinA ; 2 2

Using suitable signs , we can find

sin

A 2

cos

A = ±1 sinA 2

cos

,

±1 −sinA

∓

A 2

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 sin

AC 1 1 B (sum of any two) = cos (third); e.g sin =cos ) 2 2 2 2

cos

BC 1 1 A (sum of any two) = sin (third), e.g: cos =sin ) 2 2 2 2

If A is any angle of traingle and α lies between 00 and 1800 , then sinA=sin α ⇒A = α or 1800- α ; cosA=cos α ⇒Α= α; tanA=tan α ⇒Α= α

SOME IMPORTANT IDENTITTIES: If A+B+C =1800 , then 1) sin2A +sin2B+sin2C=4sinAsinBsinC i.e.

∑ sin2A

∑ cos2A

2)cos2A+cos2B+cos2C=-1-4cosAcosBcosC i.e.

B A C cos cos 2 2 2

3)sinA+sinB+sinC=4cos i.e.

∑ sinA

=4cos

∑ cosA

=-1-4cosAcosBcosC VIGNAN CLASSES

B A C cos cos 2 2 2

4)cosA+cosB+cosC=1+4sin i.e

= 4sinAsinBsinC

=1+4sin

A B C sin sin 2 2 2

A B C sin sin 2 2 2

∑ tanA = tanA.tanB.tanC 6)cotB.cotC+cotC.cotA+cotA.cotB =1 i.e. ∑ cotA.cotB =1 5)tanA+tanB+tanC=tanA.tanB.tanC i.e.

7)cot

B A B C A C +cot +cot =cot cot cot 2 2 2 2 2 2 i.e.

8)tan

A

∑ cot 2

=cot

B A C cot cot 2 2 2

A B B C C A tan +tan tan +tan tan =1 i.e. 2 2 2 2 2 2

Note: If A, B, C are the angles of a traingle , then sin(A+B+C) =sinП=0,

cos(A+B+C) =cos П= -1 and tan(A+B+C) =0; 9

A

B

∑ tan 2 tan 2

=1

GRAPHS OF TRIGNOMETRIC FUNCTIONS I quadrant sinθ

increases from 0 to 1

II quadrant decreases from

III quadrant

IV quadrant

decreses from

increases from

0 to -1

-1 to 0

1 to 0 cosθ

decreases from decreases from 1 to 0

increases from increases from -1 to 0

0 to 1

0 to -1 tanθ

increases from increases from increases from increases from 0 to

cotθ

∞

−∞ to 0

decreases from decreases from ∞ to 0

0 to

−∞ to 0

∞

decreases from decreases from ∞ to 0

0 to

∞

0 to −∞ secθ

increses from 1 to

cosecθ

∞

increases from decreases from decreases from ∞ to 1 −∞ to -1 -1 to −∞

decreases from increases from increases from decreases from ∞ to 1

1 to

∞

−∞ to -1

-1 to -infinity

Graph of sinx

Graph of cosecx

Graph of cosx

Graph of secx

Graph of tanx

Graph of cotx y

f(x)=cot(x)

8 6 4 2

x -8

-6

-4

-2

2 -2 -4 -6 -8

10

4

6

8

RELATION BETWEEN THE SIDES & ANGLES OF A TRIANGLE: A traingle consists of 6 elements, three angles and three sides. The angles of traingle ABC are denoted by A,B, and C. a,b, and c are respectively the sides opposite to the angles A,B and 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 ie

a b c = = =2R Where R is sinA sinB sinC

the circum radius of circum circle that passes through the vertices of the traingle. 2

2

2

2

2

2

cos A =

 b c – a  2bc

b2 =a2 +c2 -2ac cosB or

cos B =

 c a – b  2ca

c2 =a2 +b2 -2ab cosC or

cos C =

 a2 b2 – c2  2ab

2.‘Cosine rule’: a2 =b2 +c2 -2bc cosA or

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’:

tan

tan

 B –C  b – c  A =[ ]cot 2 bc 2

B−C b−c 2 = bc BC tan 2

 

or

tan

A−B a−b 2 = ab AB tan 2

 

 A – B a – b  C =[ ]cot or 2  ab 2

tan

etc.

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

  

  

  

traingle. sin

A s – b s – c  = 2 bc

cos

A s s – a  = 2 bc

tan

A  s−b s−c = 2 s s−a 

sin

B s – a s – c  = 2 ac

cos

B s s – b  = 2 ac

tan

B  s−a s−c = 2 s s−b 

C s – a s – b  = 2 ab

cos

sin

C s s – c  = 2 ab

6. Formula that involve the Perimeter: If S=

C  s−as−b  = 2 s s−c  abc , where a+b+c is the perimeter of 2 tan

a traingle, R the radius of the circumcircle, and r the radius of the inscribed circle, then 6. Area of traingle: ∆= ∆=

 s s−a s−b s −c 

;(HERO'S FORMULA)

1 1 1 abc a.b.SinC = b.c. sinA = c.a.sinB= 2 2 2 4R 2

1 a sinB. sinC 1 a 2 sinB. sinC 1 b 2 sin.C sinA 1 c 2 sinA. sinB ∆= = = = 2 sinA 2 sinB 2 sinC 2 sinBC  DEDUCTIONS: sinA=

2 2 =  s s−a s −b  s−c bc bc

sinB=

2 ca

SinC=

2 ab

tan

s −c s −a s −b A B B C C A tan = ; tan tan = ; tan tan = . 2 2 2 2 2 2 s s s

tan

A B C B C A   tan = cot ; tan tan = cot ; 2 2 2 2 2 2 s s

tan

C A  tan = cot 2 2 s

B . 2

11

NOTE WORTHY POINTS: In a traingle ABC If cotA +cotB +cotC= traingle is equilateral

•

If sin2 A +sin2B + sin2 C =2 then traingle is equilateral

•

If cosA + cosB +cosC =3/2 then traingle is equilateral

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

•

If cotA cotB cotC>0 then traingle is acute angled traingle

If a cosA = bcosB then traingle is isosceles or rightangled

•

If in atraingle 8R2 =a2 +b2 +c2 then traingle is rightangled.

• •

3

then

a b c = = cosA cosB cosC

•

If in a traingle

•

then traingle is equilateral

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

•

SOLUTION OF TRIANGLES To solve a triangle a) when all the 3 sides are given : GIVEN

REQUIRED

a,b, c

i) Area of ∆= sinA=

 s s−a s−b s−c

, 2s = a+b+c

2 2 2 , sinB= , sin C= OR bc ac ab

iii) First, find two of the three angles by cosine formula, then the third angle is determined by using the relation A+B+C=180 0. It is advisable to find the smallest angle first. (angle opposite to the smallest side). b) When two sides and an included angle is given: GIVEN a , b and C

REQUIRED i)Area of traingle=∆=

1 a.b.SinC ; 2

tan

 A – B a – b  C =[ ]cot 2  ab 2

AB C asinC =900 ; c= 2 sinA 2 ii) Use cosine rule to find the third side. then find the smaller of the two angles by 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 be determined sine rule or cosine rule or by projection rule. c)when one side and two angles A and B are given: GIVEN a A and B

REQUIRED i) C =180-(A+B) ; b=

asinB asinC ;c= sinA 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 just cannot 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 asinC ; C=180-(A+B); c= a sinA

CASES:i)When A is acute angle and a
b sinA gives us that a

sinB>1, 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 one solution or one traingle since A is given, we can find C using A+B+C=1800 . we can find 'c' by any one of the rules. iii)When A is acute angle and a>bsinA, sinB<1, then there exist two sub cases. a) If a≥b, then A≥B, B must be acute. Thus there exists only one solution. b) if b≥a, then B≥A., there exist two values to B for which this can be true . one being acute and the other being

obtuse.

When B is determined, we can find C using A+B+C =180 , then c by any one of the rules. this case is called an ambiguous case since there exist no solution, one solution or two solutions depending on the cases. 0

Note: It is not advisable to use sine rule to find the angle in all other cases. since it always gives an ambiguous result. Use sine rule to find the angle only when it is inevitable.

12

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 following cases arise: a, b, A given i)a
No triangle

ii)a=b sinA

Right angled triangle

iii)b>a>bsinA

Two triangles

iv)a>b

one triangle

OTHER IMPORTANT FORMULA AND CONCEPTS: 1.To find the greatest and least values of the expression asinθ +bcosθ : Let a=rcosα. b=rsinα , then a2 +b2 =r2 or r=

 a b 2

2

asinθ +bcosθ = r(sinθ cos α +cosθ sin α) = rsin(θ + α )

 a b 2

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

2

≤ asinθ +bcosθ ≤

Thus the greatest and least values of asinθ +bcosθ are respectively 2

−   , minimum value of acosθ +bsecθ is 2 2 2 0

 or 2



2

and -

2

2

2

.

2

For 0 , minimum value of a sinθ + bcosecθ is 2

For

2

2

 a b

Similarly maximum value of asinθ -bcosθ is

For

 a b

 a b  a b

 ab  ab

3 , minimum value of a tanθ +bcotθ is 2 2

2. cosA.cos2A.cos4A.cos8A............cos2

n-1

A=

 ab

1 sin 2n A 2 sinA n

(Remember)

n1

OR cos θ.cos2 θ.cos22 θ.cos23 θ............cos2n θ =

sin 2 A  (Each angle being double of preceding) 2 n sinA

3. SUM OF THE SIN AND COSINE SERIES WHEN THE ANGLES ARE IN AP: sinα +sin(α+β) +sin(α +2 β) +..........n terms cosα +cos(α+β) +cos(α +2 β) +..........n terms

diff 2 diff sin 2

sin n. =

n 2 .sin or cos  sin 2

sin =

. sin or cos

[

[

1st angle last angle 2

 n−1  2

]

]

(Remember the rule)

n 2 .sin or cos  sin 2

sin =

[

 n−1 

 2

]

Note: β is not an even multiple of Π i.e. β #2n Π because in that case sum will take the form 0/0. Particular case: Both the sum will be zero if

sin

n =0 i.e. 2

n 2r  =r Π or β = or β = even multiple of 2 n

 n

then S=0 4. SOME RESULTS IN PRODUCT FORM:

1 sin3θ 4

sinθ sin(60+θ)sin(60-θ) = cosθ cos(60+θ) cos(60-θ)

=

1 cos3 θ 4

cos3A 4cosA

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

tan3A tanA

tan2A tan3A tan5A=tan5A-tan3A-tan2A

cosθ cos(120+θ) cos(120-θ)

tanx tan2x tan3x =tan3x-tan2x-tanx

tanθ tan(60+θ )tan(60-θ ) =tan3θ sin(600 -A) sin(600 +A) =

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

(Use the above formula at time of integration)

sin3A 4sinA

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

13

cos2 n 1 −cos 2 n1 

(cos α +cos β) (cos2α +cos2 β ) (cos22α +cos22 β ) .........(cos2nα +cos 2n β ) =

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

4. i) cosA ±sinA= 5. tan θ + 6.



2 sin 4 ±A

tan

=

n

2 cos −cos  

2cos2 n 1 1 2cos 1



2 cos 4 ∓A 

ii) tanA +cotA =

1 sinA.cosA

 2    + tan  =3tan3 θ ; tan θ + tan  + tan−  =3tan3 θ 3 3 3 3

  



n

2 2 2 2............  22cos2 

=2cos θ

∀ n∈N

HEIGHTS AND DISTANCES-VIGNAN CLASSES ANGLE OF ELEVATION AND ANGLE OF DEPRESSION Suppose a st.line OX is drawn in the horizontal direction. Then the angle XOP where P is a point (or the position of the object to be observed from the point O of observation ) above OX is called Angle of Elevation of P as seen from O. O Similarly, Angle XOQ where Q is below OX, is called angle of depression of Q as seen from O. OX is the horizontal line and OP and OQ are called α= Angle of elevation of P line of sights

Properties used for solving problems related to Heights and Distances.

P

α

X

β

β=Angle of

Q Depression of Q 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 will observe that each of the angles PQA,PQB,..PQC,.... is a right angle. In other words PA is perpendicular to 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 =Ө, Where ABC is right angled triangle in which C = 900 . The side opposite to right angleC will be denoted by H(Hypotenus),

H

O

θ

A the side opposite (opposite side) to angle θ is denoted by O, the side containing angle θ (other than H)(Adjacent side) will be denoted by A Then from the figure it is clear that O=A(tanθ ) or A = O(cotθ ) i.e. Opposite = Adj(tanθ ) or Adj=opposite (cotθ ). Also O=H(sinθ ) or A =H(cosθ ) i.e opposite =Hyp( sinθ ) or Adjacent =Hyp(cosθ )

ASWEQRTYUIXCVBNMKL;,./'[]-098 PREPARED AND DTP BY KHV, LECTURER IN MATHEMATICS

THE SPIRIT OF MATHEMATICS The 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

14

15