TRIGNOMETRY-FORMULA AND CONCEPTS BY K.H. V. AN ANGLE:An angle is the amount of rotation o f a revol ving line w.r.t a fixed straig ht line (a figure formed by two ray s having common initial point.) The two ray s 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(90 o ); 1 degree = 60 minutes (60') 1 minute = 60 seconds (60'') b)Centisimal system of angles: 1 right angle =100 grades =100 g 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 l ength is equal to the radius of t he circle. one radian is denoted by 1 c 1 radian =57 0 16 1 22'' A radian is a Constant angle. And radians = 180 0 RELATIONSHIP BETWEEN DEGREES AND RADIANS: radians =180 o 1 radian= 1 c = 180 o 1 c = 57 0 17' 45''; 1 0 = 180 o radian=0.01746 radian (approximately) Radian measure= 180 o x Degree measure i.e. To convert degrees into radians Multiply by 180 o Degree measure= 180 o 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 do es not means that π stands for 180 0 , π is real number, where as π c stands for 180 0 LENGTH OF ARC OF A CIRCLE: If an arc of length “s” su btends an angle θ radians at the center of a circle of radius 'r', then S =r θ i.e. length of arc = ra dius x angle in radians (subtended by arc) No of radians in an angle su btended by an arc of circle at the centre = arc radius = Sr1 c (1 radian) = arclengt h of magnitudeof rradiusof˚ rAREA 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 1 2 r 2 .θ RADIAN MEASURE OF SOME COMMON ANGLES: θ0 ( Degrees) 15 0 22½ 0 30 0 45 0 60 0 75 0 90 0 120 0 135 0 150 0 180 0 210 0 270 0 360 0 +ve angle -ve angle AB be the Arc, Let the length of 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 ofthe 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) T e r m i n a l s i d e (a r m ) θ Initial side(arm) θ ----r----- B Arc A D
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
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 acute angled traingle, we can write other t ratios in terms of given ratio:
Let sinθ=s=perp
hyp=
s
1
cosθ= = 1−sin2 ; tanθ=
sin
1−sin2; secθ=
1
1−sin2
;
cosecθ=1
sin ; cotθ= 1−sin2
sin
We can express sinθ in terms of other trigonometric functions by above method:
sinθ= 1−cos2 =
tan
1tan2
=1
cosec = sec2−1
sec= 1tan
2
tan
For ex. sinθ=1/3, since sine is +ve in Q1 and Q2(II quadrant), we have
cosθ= 1−1
9or - 1−
1
9ie.
2 23
or −2 2
3
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=-4
5; cosθ=
−3
5
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
sinθ
0
1
2
1
2 32 1 0 -1 0
3−1
2 2 31
2 2
cosθ
1
32
1
21
2 0 -1 0 1
31
2 2 3−1
2 2
tanθ
0
1
3 1
3 ∞
0
∞
0
2− 3
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 )
sinθ 32
1
21
2
(-1)n
0
cosθ−1
2−
1
2− 3
2 0
(-1)n
tanθ − 3-1
−1
3∞
0
e.g. cos(odd
2)=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
The above may be summed up as follows: Any angle can be expressed as n.90+θ where n is anyinteger 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 iseven 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 quadrantangle 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 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 subtractingfrom 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 θ is any angle then the angle
2- θ is its complement angle and the angl
e - 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:
VALUES OF TRIGNOMETRICAL RATIOS OF SOME IMPORTANT ANGLES:
Angle→
Ratio↓
7 1
2
0 150 180
221
2
036 0 750
sin 8−2 6−2 2
4
or 4− 6− 22 2
3−1
2 2
5−1
4
1
2 2− 2
1
4 10−2 5 31
2 2
cos 82 6−2 24
or
4 6 22 2
31
2 2
1
4 102 5
1
2 2 2
1
4 51 3−1
2 2
tan 6− 4− 3 2or
3− 2 2−1
2- 3 25−10 5
5
2−1 5−2 5 2+ 3
cot 6± 4± 3 2or
3 2 21
2+ 3 52 5 21
12
5 2- 3
sec 16−10 28 3−6 ( 6− 2 )
2−2
5 4−2 2 5−1 6 2
sin22½0 =1
2 2− 2 ;
cos22½0 =1
2 2 2 ;
tan22½0 = 2−1 ;cot22½0= 21
sin180 =1
4 5−1 =cos720 ;
cos180 =1
4 102 5 =sin720 ;
sin360 =1
4 10−2 5 =cos540 ;
cos360 =1
4 51 =sin540
tan7 ½0= 6− 4− 3 2
cot7½0= 6± 4± 3 2
sin90 = 3 5− 3− 54
cos90 = 3 5 3− 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≤1
2. 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
secA≥1 or secA≤-1 and cosecA≥1 or cosecA≤-1, In otherwords w
e never have -1<cosecA<1 and-1<secA<1
3. tanA and cotA can assume any real value.
For all values of θ, -1≤sin θ≤1 and -1≤cos θ≤1a)Max . sin θ =1; Min . sin θ =-1
b)Max . (sin θ cos θ)=Max sin2
2 =1
2; Min. (sin θ cos θ) =Min sin2
2 = -1
2
4.If y =a sinx + bcosx +c, then ∀ a , b , c∈ R , we can write y=c+ a2b2 sin(x+α)
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 =c2a
2 – b
2 2ca
c2 =a2 +b2 -2ab cosC or cos C =a2b
2 – c
2 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’:
tan B–C
2=[b –c
bc]cot
A
2or b−c
bc =tan
B−C
2
tanBC 2
tan A–B
2=[a –b
ab
]cotC
2or
a−b
ab
=
tanA− B
2
tan A B2
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=
s−b s−c
s s−a
sinB
2=
s–a s–c
accos
B
2=
s s–b
ac tan
B
2=
s−a s−c
s s−b
sinC
2=
s–a s–b
abcos
C
2=
s s–c
ab tan
C
2=
s−a s−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)
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 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 angledtriangle in which C = 900 . The side opposite to right angle Cwill 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 )
;,./ []-SWEQRTYUIXCVBNMKL ' 098
PREPARED AND DTP BY KHVASUDEVA,
LECTURER IN MATHEMATICS
14 for http://pucpcmb.wordpress.com
O X
Q
α
β
α= Angle of elevation of P
β=Angle of
Depression of Q
H
A
O
θ
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 anunreasonable 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