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TRIGNOMETRY-FORMULA AND CONCEPTSBY 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 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.

• 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 and seconds.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 = 180o

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 180o

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)

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

radius =Sr

1c(1 radian) =arc length of magnitude of r

radiusof r̊AREA OF A SECTOR OF A CIRCLE:(sectorial area)

The area of the sector formed by the angle θ at the center of a circle of radius r is 12 r2 .θ

RADIAN MEASURE OF SOME COMMON ANGLES: θ 0

(Degrees)150 22½

0300 450 600 750 900 1200 1350 1500 1800 210

02700 3600

θ c

Radians

1 for http://pucpcmb.wordpress.com

+ve angle

-ve angle

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

angle AOB =1 radian

VIGNAN CLASSESDo 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)

Termina

l

side(a

rm)

θ

Initial side(arm)

θ

----r-----

BArc

A

D

3 4

56

7 6

3 2

212

8

6

4

3

512

2

23

SOME USEFUL FACTS ON CLOCKS: 1. Angle between two consecutive digits of a

clock is 300 or 6

radians.

2. Hour hand of the clock rotates by an angle

of 300 or 6

radians 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 30

radians in one minute.

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

sinθ =opposite sidehypotenus ; cosθ =

adjacent sidehypotenus

tanθ =oppositesideadjacent side ; cosecθ =

hypotenusopposite side

secθ =hypotenus

adjacent side ; cotθ =adjacent sideopposite 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 c

sinθ =yr cosθ =

xr tanθ =

yx

cosecθ =ry secθ =

rx cotθ =

xy

RELATIONS BETWEEN TRIGNOMETRIC RATIOSBASIC IDENTITTIES: a) sin2θ + cos2θ =1; b) 1+ tan2θ = sec2θ ; c) 1+ cot2θ = cosec2θ ;

DEDUCTIONS:cos2θ = 1 -sin2θ; sin2θ = 1- cos2θ; sec2θ -1 = tan2θ; cosec2θ -1 = cot2θ;sec2θ - tan2θ =1; cosec2θ - cot2θ =1

RECIPROCAL RELATIONS

cosecθ =1

sinθ ; secθ =1

cosθ cotθ =1

tanθ ;cosecθ.sinθ =1 ; secθ. cosθ =1 ; cotθ. tanθ =1

QUOTIENT RELATIONS

tanθ =sinθcosθ ; cotθ =

1tanθ

=cosθsinθ

SIGNS OF TRIGNOMETRIC FUNCTIONS :

I II III IV

sinθ + + - -

cosθ + - - +

tanθ + - + -cosecθ + + - -secθ + - - +cotθ + - + -


Opposite sideHypoten

us

Adjacent sideθ

O

P

M

P(x, y)

O M

900≤θ≤1800

1800≤θ≤2700 2700≤θ≤3600

(QUADRANT RULE)

a) In First quadrant, all

t-ratios are +ve. b) In Second quadrant sinθ , cosecθ are +ve. c) In Third quadrant tanθ and cotθ are +ve

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

DO YOU KNOW: In a regular polygon i) All the interior angles are equalii) All the exterior angles are equaliii) All the sides are equaliv)Sum of all the exterior angles is 3600

v) Each exterior angle = 3600/number of exterior anglesvi)Each interior angle = 1800 -exterior anglevii) 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

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

CT

A=All are +veS=Sin & cosec are +veT=Tan & Cot are +veC=Cos & Sec are +veShort Cut to remember:ALL STUDENTS TAKE COFFEE or ALL STUDENTS TO CINEMA

00≤θ≤900

AS

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

Let sinθ=s=perphyp =

s1

cosθ= = 1−sin2 ; tanθ=sin

1−sin2; secθ=

1

1−sin2;

cosecθ=1

sin ; cotθ= 1−sin2sin

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

sinθ= 1−cos2 =tan

1tan2 =

1cosec = sec2−1

sec = 1tan2

tan

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

cosθ= 1−19

or - 1−19

ie. 2 23

or −223

according as θ ∈Q1 or θ ∈Q2

We can find other ratios by forming a rightangled traingle. Let tanθ=4/3, 3 2

,

then since in Q3, sine and cosine both are negative, we have sinθ=-45 ; cosθ=

−35

TRIGNOMETRIC RATIOS OF STANDARD and QUANDRANTAL ANGLES:

Radians 0

6

4

3

2

32

2 12

512

Degrees 0 300 450 600 900 1800 2700 3600 150 750

sinθ0

12

12

32 1 0 -1 0

3−12 2

312 2

cosθ1

32

12

12 0 -1 0 1

312 2

3−12 2

tanθ0

13 1

3 ∞0

∞0

2−3

VALUES OF T-FUNCTIONS OF SOME FREQUENTY OCCURING ANGLES.

Radians 0 2 3

34

56

2n1 2

n

Degrees 1200 1350 1500

(odd ) 2

(any )

sinθ 32

12

12

(-1)n

0

cosθ −12

−12

−32 0

(-1)n

tanθ −3-1

−13

∞0

e.g. cos(odd2 )=0; cos( odd )=-1, cos(even ) =1

cos 2n−1 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


1−s2

s

1

1−s2

1 3

4

3

5

sin 2 = sin

5 2 =sin

92 =.......=1

sin(3

2 ) = sin 7 2 = sin

112 = ..........=-1

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(n2 +θ)=(-1)n+1/2 sinθ if n is odd 4)sin(

n2 +θ)=(-1)n-1/2 cosθ if n is odd

=(-1)n/2 sinθ if n is even =(-1)n/2 cosθ if n is even

DOMAIN AND RANGE OF TRIGNOMETRIC FUNCTIONS:

Function Domain Rangesine

cosine

tangent

cotangent

secant

cosecant

R

R

R-{(2n+1) 2

}: nε Z

R-{n }; nεZ

R-{(2n+1) 2

}: nε Z

R-{n }; nεZ

[-1, 1]

[-1, 1]

R

R

(- ∞ ,-1] υ [1, ∞ )

(- ∞ ,-1] υ [1, ∞ )

ASTC RULE:(QUADRANT RULE):‘ASTC’ rule to remember the signs ‘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 which the angle lies using QUADRANT RULE.

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. 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-θ -sinθ cosθ -tanθ secθ -cosecθ -cotθ900 -θ cosθ sinθ cotθ cosecθ secθ tanθ900 + θ cosθ -sinθ -cotθ -cosecθ secθ -tanθ1800 -θ sinθ -cosθ -tanθ -secθ cosecθ -cotθ1800+θ -sinθ -cosθ tanθ -secθ -cosecθ cotθ2700 -θ -cosθ -sinθ cotθ -cosecθ -secθ tanθ2700 +θ cosθ -sinθ -cotθ -cosecθ secθ -tanθ3600 -θ -sinθ cosθ -tanθ secθ -cosecθ cotθ


S A

T C

90-θn.360 + θ90+θ

180-θ

180+θ270-θ

360-θ270+θ-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 2

, the sine function is retained . Hence sin(540- θ ) =sin θ.

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 2

, cosine is

replaced by sine. Hence cos(6300 - θ ) = -sin θ.

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 quandrante.g.: cos1200 = -cos600 = -1/2t-ratio of ( θ -180) with proper sign if θ lies in the third quandrante.g: sin2100 = -sin300 = -1/2

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

e.g: cosec3000= -cosec600 = −23

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:

If θ is any angle then the angle 2

- θ is its complement angle and 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 AtanB 1– tan A tanB

6. tan (A – B) = tan A– tanB 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)=tan2 A−tan2B

1−tan2 A . tan2B

10.Cot(A+B) =cotAcotB−1cotAcotB

(A#nπ, B#mπ, A+B#kπ)

11.Cot(A-B) =cotAcotB−1cotB−cotA

(A#nπ, B#mπ, A-B#kπ)

12. tan(A+B)=sinABcos AB

13. tan(A-B)=sinA−Bcos A−B

14. tanAtanBtanA−tanB

=sin ABsin A−B

15.1+tanA tanB=cos A−BcosAcosB

1-tanA tanB=cos ABcosAcosB

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

= sin ABcosA. cosB

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

17.tan(Π/4 + A) =1tanA1−tanA

18.tan(Π/4 - A) =1−tanA1tanA

19.cot( Π/4 + A )=cotA−1cotA1

20.cot( Π/4 - A )=cotA1cotA−1

21. tan(A+B+C)

=tanAtanBtanC−tanA.tanB.tanC

1− tanAtanBtanB.tanCtanC.tanA

=S1−S3

1−S2

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

S2 =tanAtanB +tanB.tanC +tanC.tanA

21.The cot(A+B+C) = cotA.cotB.cotC−cotA−cotB−cotC

cotAcotBcotB.cotCcotC.cotA−1

22. sinA+cosA= 2sin 4A

sinA-cosA= 2 sin 4−A

cosA+sinA= 2cos 4−A

cosA-sinA= 2 cos 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

1−tan2 A

2.cos 2A = cos 2 A – sin 2 A

= 1 – 2 sin2A

= 2cos2A – 1 =1−tan2 A1tan2 A

3. tan 2A = 2 tan A

1 – tan2 A

DEDUCTIONS:

1+cos2A =2cos2A; cos2A =121cos2A

1-cos2A =2sin2A; cos2A =121−cos2A

1−cos2A1cos2A =tan2A;

1cos2A1−cos2A =cot2A

1+sin2A =(sinA +cosA)2

1-sin2A =(sinA -cosA)2

cotA -tanA = 2 cot2A

tanA+cotA=2 cosec 2A


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

Sin 3A = 3 sin A – 4 sin3A ;

cos 3A = 4 cos3A – 3 cos A ;

tan3A = 3tanA−tan3 A

1−3tan2 A;

DEDUCTIONS:

4 sin3A =3 sin A -Sin 3A ;

sin3A =14 ( 3 sin A -Sin 3A ).

4 cos3A =3 cos A +cos 3A;

cos3A =14 ( 3 cos A +cos 3A )

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

a) sinθ =2sin2 cos

2 =

2 tan 2

1tan2 2

b) cosθ=cos2 2 -sin2

2 =2cos2 2 -1

=1-2sin22 =

1−tan2 2

1tan2 2

c)tanθ=

2 tan 2

1−tan2 2

DEDUCTIONS:

1+cosθ=2cos22 ; 1-cosθ=2sin2

2

1−cos 1cos

=tan2 2 ;

1cos 1−cos

=cot2 2

1−sin 1sin

= tan24 −2 ;

1sin 1−sin

= cot2 4 2

sin

1cos =tan

2 ; sin

1−cos =cot

2

cos

1sin = tan4 −

2 ;

cos1−sin

= cot4 2

Transformation formulae:

a) SUMS AND DIFFERENCE TO PRODUCT FORMULAE:

Formula that express sum or difference into products

Sin C + sin D = 2sin CD 2

cos C–D2 Sin C – sin D = 2cos CD

2sin C–D

2

Cos C + cos D = 2cos CD2

cos C–D2 Cos C – cos D = 2sin CD

2sin D−C

2

or −2sin CD2

sin C−D2

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)

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

sin A2cos A

2 2

=1+sinA so that sin A2cos A

2 = ±1sinA

sin A2−cos A

2 2

=1-sinA so that sin A2−cos A

2 = ±1−sinA

By addition and subtraction, we have

2 sin A2 = ±1sinA ± ±1−sinA ; 2 cos A

2 = ±1sinA ∓ ±1−sinA

Using suitable signs , we can find sin A2 , cos A

2


VALUES OF TRIGNOMETRICAL RATIOS OF SOME IMPORTANT ANGLES :

Angle→

Ratio↓

712

0 150 180 2212

0 360 750

sin 8−26−2 24

or 4−6−22 2

3−12 2

5−14

12 2−2 1

410−25 31

2 2

cos 826−2 24

or

4622 2

312 2

141025 1

2 22 1451 3−1

2 2

tan 6−4−32or

3−22−1

2- 3 25−1055

2−1 5−25 2+ 3

cot 6±4±32or

3221

2+ 3 52 5 21 1 25

2- 3

sec 16−10283−66( 6−2 ) 2− 25

4−22 5−1 62

sin22½0 =12 2−2 ;

cos22½0 =12 22 ;

tan22½0 = 2−1 ;cot22½0= 21

sin180 =145−1 =cos720 ;

cos180 =141025 =sin720 ;

sin360 =1410−25 =cos540 ;

cos360 =1451 =sin540

tan7 ½0= 6−4−32

cot7½0= 6±4±32

sin90 = 35−3−54

cos90 = 353−54

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≤12. 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<cosecA<1 and -1<secA<13. tanA and cotA can assume any real value. For all values of θ, -1≤sin θ≤1 and -1≤cos θ≤1 a)Max . sin θ =1; Min . sin θ =-1

b)Max . (sin θ cos θ)=Max sin22 =12 ; Min. (sin θ cos θ) =Min sin22 = -

12

4.If y =a sinx + bcosx +c, then ∀ a ,b , c∈R , we can write y=c+ a2b2 sin(x+α)

Where a= r cos α b=r sin α ⇒ r= a2b2 tanα = ba

; since -1≤sin (x+α )≤1

∴ c- a2b2 ≤y≤c+ a2b2 Hence Max. (a sinx + bcosx +c) =c+ a2b2 and

Min (a sinx + bcosx +c)= c- a2b2


5. sin θ >cos θ >0 for 4

2

; 0<sin θ <cos θ for 04

4.The following formulae of Componedndo and Dividendo must be noted:

If p/q =a/b then by componendo and dividendo we can writep−qpq

=a−bab

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 cosθ is periodic with period 2

tanθ is periodic with period sinkθ is periodic with period 2k

tankθ is periodic with period k

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

sin12 (sum of any two) = cos

12 (third); e.g sin

AC2 =cos

B2 )

cos12 (sum of any two) = sin

12 (third), e.g: cos

BC2 =sin

A2 )

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 = 4sinAsinBsinC

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

3)sinA+sinB+sinC=4cosA2 cos

B2 cos

C2

i.e. ∑ sinA =4cosA2 cos

B2 cos

C2

4)cosA+cosB+cosC=1+4sinA2 sin

B2 sin

C2

i.e ∑ cosA =1+4sinA2 sin

B2 sin

C2

5)tanA+tanB+tanC=tanA.tanB.tanC i.e. ∑ tanA = tanA.tanB.tanC

6)cotB.cotC+cotC.cotA+cotA.cotB =1 i.e. ∑ cotA.cotB =1

7)cotA2 +cot

B2 +cot

C2 =cot

A2 cot

B2 cot

C2

i.e. ∑ cot A2 =cot

A2 cot

B2 cot

C2

8)tanA2 tan

B2 +tan

B2 tan

C2 +tan

C2 tan

A2 =1 i.e. ∑ tan A

2tan B

2 =1

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;


GRAPHS OF TRIGNOMETRIC FUNCTIONSI quadrant II quadrant III quadrant IV quadrant

sinθ increases from 0 to 1 decreases from 1 to 0 decreses from 0 to -1 increases from -1 to 0

cosθ decreases from 1 to 0 decreases from 0 to -1 increases from -1 to 0 increases from 0 to 1

tanθ increases from 0 to ∞ increases from ∞ to 0 increases from 0 to ∞ increases from −∞ to 0

cotθ decreases from ∞ to 0 dec. from 0 to ∞ decreases from ∞ to 0 decreases from 0 to ∞secθ increses from 1 to ∞ incr. from ∞ to -1 decreases from -1 to −∞ decreases from ∞ to 1

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

Graph of sinx Graph of cosecx

Graph of cosx Graph of secx

Graph of tanx Graph of cotx


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 iea

sinA=

bsinB

=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 = b2c2– a2

2bc

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

2ca

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

2ab3.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 B–C 2

=[b –c bc

]cot A2 or b−c

bc=tan B−C

2

tan BC2

tan A–B2

=[a –b ab

]cot C2 or a−b

ab=tan A−B

2

tan AB2

etc.

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

traingle. sin A2= s –b s – c bc

cos A2= s s –a bc

tan A2= s−b s−c

s s−a

sin B2= s –a s –c ac

cos B2= s s –b ac

tan B2= s−a s−c

s s−b

sin C2= s –a s –b ab

cosC2= s s –c ab

tanC2= s−as−b

s s−c

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: ∆= s s−a s−b s−c ;(HERO'S FORMULA)

∆=12 a.b.SinC =

12 b.c. sinA =

12 c.a.sinB=

abc4R

∆= 12a2 sinB. sinC

sinA= 1

2b2 sin.C sinA

sinB= 1

2c2 sinA. sinB

sinC= 1

2a2 sinB. sinCsinBC

DEDUCTIONS:

sinA=2 bc

= 2bc s s−a s−b s−c sinB=

2 ca

SinC=2 ab

tanA2 tan

B2 =

s−cs ; tan

B2 tan

C2 =

s−as ; tan

C2 tan

A2 =

s−bs .

tanA2 tan

B2 =

s cot

C2 ; tan

B2 tan

C2 =

s cot

A2 ;

tanC2 tan

A2 =

s cot

B2 .


SOLUTION OF TRIANGLESTo solve a triangle a) when all the 3 sides are given :

GIVEN REQUIRED

a,b, c i) Area of ∆= s s−a s−b s−c , 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 is determined by using the relation A+B+C=1800. 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 REQUIRED

a , b and Ci)Area of traingle=∆=

12 a.b.SinC ; tan A–B

2=[

a –b ab

]cot C2

AB

2=900 -

C2

; c=asinCsinA

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 REQUIRED

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

asinBsinA ;c=

asinCsinA

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

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

CASES:i)When A is acute angle and a<bsinA; In this relation sinB= b sinA

a gives us that

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 =1800 , 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.

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.

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<b sinA

ii)a=b sinA

iii)b>a>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 θ +bcos θ :

Let a=rcosα. b=rsinα , then a2 +b2 =r2 or r= a2b2

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

But -1≤sin(θ + α )≤1 so that -r ≤rsin(θ + α )≤r. Hence - a2b2 ≤ asinθ +bcosθ ≤ a2b2

Thus the greatest and least values of asinθ +bcosθ are respectively a2b2 and - a2b2 .

Similarly maximum value of asinθ -bcosθ is a2b2

For 0 , minimum value of a sinθ + bcosecθ is 2 ab

For −

2

2

, minimum value of acosθ +bsecθ is 2 ab

For 02

or 3 2

, minimum value of a tanθ +bcotθ is 2 ab

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

2n sinAsin 2n A (Remember)

OR cos θ.cos2 θ.cos22 θ.cos23 θ............cos2n θ =sin 2n1 A

2n sinA(Each angle being double of preceding)

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

=sin n. diff

2

sin diff2

. sin or cos [1stanglelast angle2 ] (Remember the rule)

=sin n

2

sin2

.sin or cos [n−1 2 ] =

sin n 2

sin2

.sin or cos [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 n2

=0 i.e. n 2

=r Π or β =2rn

or β = even multiple of n

then S=04. SOME RESULTS IN PRODUCT FORM:

sinθ sin(60+θ)sin(60-θ) =14

sin3θ

cosθ cos(60+θ) cos(60-θ)

=14

cos3 θ

cosθ cos(120+θ) cos(120-θ) tanθ tan(60+θ )tan(60-θ ) =tan3θ

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

cos(600 -A) cos(600 +A)=cos3A4cosA

tan(600 -A) tan(600 +A) =tan3AtanA

tan2A tan3A tan5A=tan5A-tan3A-tan2A tanx tan2x tan3x =tan3x-tan2x-tanx (Use the above formula at time of integration) tan(x-α). tan(x+ α ) tan 2x= tan2x-tan(x+ α )-tan(x- α )

4. i) cosA ±sinA= 2 sin 4±A = 2cos

4∓A ii) tanA +cotA =

1sinA.cosA

5. tan θ + tan3 + tan2

3 =3tan3 θ ; tan θ + tan

3 + tan−

3 =3tan3 θ

6. 2222............22cos2n =2cos θ ∀n∈N


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 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.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 line of sightsProperties 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 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),

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 AThen from the figure it is clear thatO=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 KHVASUDEVA, LECTURER IN MATHEMATICS


O X

Q

α

β

α= Angle of elevation of P

β=Angle of Depression of Q

H

A

O

θ

THE SPIRIT OF MATHEMATICSThe only way to learn mathematics is to recreate it for oneself -J.L.KelleyThe objects of mathematical study are mental constructs. In order to understand these one

must study , meditate, think and work hard -SHANTHINARYANMathematical 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

an angle - puc pcmb | ಇದು ... · pdf filetrignometry-formula and concepts by k.h. v. an...

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