trignometry for aieee cet by vasudeva kh
TRANSCRIPT
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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
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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
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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
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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
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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
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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
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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
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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 ;
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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;
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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
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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
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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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