Algebra
Sequence
A sequence is an ordered set of numbers. Some sequences are simply random values;
other sequences have a definite pattern,a1 , a2 ,a3…. anis the infinite sequence.Eg3,
6,9,12is the finite sequence.
Types of sequences
Arithmetic Sequence
Geometric Sequence
Harmonic Sequence
Series
The sum of the terms of an infinite sequence is called series. The sum of a finite
sequence has defined first and last terms, whereas a series continues indefinitely.
If a1 , a2 ,a3…. an…is the infinite sequence.
Then, Sn= a1 + a2 +a3+…an=Σnk= 1is the infinite series
Arithmetic series:.
The nth term of the arithmetic series is given bytn = a + (n – 1)d
l = a + (n – 1)d
1. Arithmetic mean (A.M) = a+b2a+b2
2.If m1, m2, m3,….,mn are arithmetic means then;
m1 = a + d
m2 = a + 2d
m3 = a + 3d
………………..
mn = a + nd, where d = b−an+1b−an+1
3. Sn = n2n2 [2a +(n – 1)d]or Sn – n2n2(n + 1)
Geometric series
1.tn = arn-1
2. Geometric mean (G.M) = ab−−√ab
3. If g1, g2, g3,…..gn are geometric means then,
Gn = arn, where r = (ba)1n+1(ba)1n+1
4. Sn =a(rn−1)r−1,r≠1a(rn−1)r−1,r≠1 or, Sn = lr−ar−1,r≠1lr−ar−1,r≠1
Example 1
Find the 3 arithmetic mean between 7 and 23
a.
Here
First term (a) = 7
Last term (b) = 23
No of means = 3
Common difference (d) =b−an+1b−an+1
=23−73+123−73+1 = 4
Now, m1=a+d =11
m2=a + 2d = 15
m3=a+3d =19
Example 2
If the third and the eleventh terms of an arithmetic sequence are 8 and -8 respectively,
Find the first seven terms of sequence.
Soln
Given,
t3=8
t11=-8
we have ,
tn=a + (n-1 )d
Now,
t3= a + (3-1 )d
or , 8=a + 2d………………..1
t11= a + ( 11 -1 )d
-8 = a+ 10d………………2
Solving 1 and 2
We get a = 12 and d = -2
t1=12
t2 =12 + (2-1) *- 2= 12 -2= 10
t3 =12 + (3-1) *-2 = 12 -4 = 8
t4=12 + (4-1 )* -2 = 12 -6 = 6
t5= 12 + (5-1)*-2 = 4
t6= 12 + (6- 1 )*-2= 12-10 = 2
t7= 12 + (7- 1) *-2 = 0
Relation and functions
A relation is any set of ordered pair (x,y)such that the valueof the second coordinate `y’
depends on the value of the first coordinate `x’ then y is the dependent variable and x is
the independent variable.
Different ways of representing relationship
a. Mapping diagram
b. Set of ordered pairs
c. Description
d. Table
e. Graph
Function
A function is a relation between a set of inputs and a set of possible outputs with the
property that each input is related to exactly one output. An example is the function that
relates each real number x to its square x2.
Composite function
Let f : A → B and g: B → C be the two functions .the function gof: A→ C is called composite
functionfrom A to C
Example 3
Iff = {(1,2),(2,3) (3,4) } anf g ={ (2,a)(4,c) ,(3,b)} , then show that the composite function
gof in arrow
diagram and find the it in ordered pair form.
Soln
gof={(1, a) ,(2,b) , (3,c)}
Inverse function
If a function f(x ) =yis injective, exactly one functionwill exist such that , otherwise no
such function will exist .The functionis called the inverse function ofbecause it
"reverses" ; that is to say .
Types of function
Onto function
A function f from a set X to a set Y is onto, if every element in Y has a corresponding
element in X such that f(x) = y.
Into function
A function f from a set X to a set Y is into, if element in Y is proper subset of X
One to one on to function
A function f from a set X to a set Y is One to one on to function, if every element in Y
uniquely assign to element of X.
One to one into function
A function f from a set X to a set Y is One to one into function, if element in Y not
necessarily assign to element of X.
Many to one on to function
A function f from a set X to a set Y is Many to one on to function, if element in Y has more
than one element in x mapped to them.
Many to one into function
A function f from a set X to a set Y is Many to one into function, if element in Y has at
least one element which is not mapped to the element of X.
Domain and Range
The domain is the set of all first elements of ordered pairs (x-coordinates)
The range is the set of all second elements of ordered pairs (y-coordinates).
Rational and irrational numbers
A rational number is the set of fractional numbers. It is denoted by Q. we can define
Q = {x: x =PqPq p,qϵ Z and q ≠ O}
An irrational number cannot be written as a fraction. An irrational number has endless
non-repeating digits to the right of the decimal point.
A surd is an irrational number, a number which cannot be expressed as a fraction or as
a terminating or recurring decimal. It is left as a square root. It can be written asa√nan
Laws of radicals:
(a√n)n=a(an)n=a
a√nan × a√nan =a√nanb
a√nb√nanbn=ab√nabn
ma√n−−−√man=a√mnamn
Surds having the same order and radicals can be added or subtracted.
Eg5√ 11 + 2√ 11 =7 √ 11
Rationalization
When the surd is multiplied by its conjugate we get the rational numbers
Eg – 11−3√11−3
$ = {\rm{\: }}\frac{1}{{1 - \sqrt 3 }}* \frac{{1 + \sqrt 3 }}{{1 + {\rm{\: }}\sqrt 3 }}{\rm{\: }}$
= 1+3√1−31+31−3
= - 1212- 3√232
Polynomial:
A polynomial is an expression consisting of variables and coefficients which only
employs the operations of addition, subtraction, multiplication, and non-negative
integer exponents. An example of a polynomial of a single variable x is x2 −x + 3. An
example in three variables is x3 + 2xy2z2 − yz + 1.
Degree of polynomial: The degree of the polynomialof one variable is the value of the
largest exponent of the variable.
The Remainder theorem
If P(x) of degree n ≥ 1 divided by x-r , the remainder is P (r).
Consider P(x) = (x − r) q(x) + R
Note that if we let x = r, the expression becomes
P(r) = (r − r) q(r) + R
Simplifying gives:
P(r) = R
This leads us to the Remainder Theorem which states:
If a polynomial P(x) is divided by (x − r) and a remainder R is obtained, then P(r) = R.
Example 4
x3+px2+qx+5x3+px2+qx+5Leaves remainder 1 when divided by x+ 2 and 16 when
divided by x -1 , find p and q
Soln
letP(x) = x3+px2+qx+5x3+px2+qx+5, R = 1 and Q(x) = x+2
P(-2) =-8 + 2p -2q +5
1 = 2p -2q -3
4 = 2p -2q
2 = p- q ……………………………………….1
P(x) = x3+px2+qx+5x3+px2+qx+5, R = 16 and Q(x) = x-1
P(1) = 1 + p+q+5
16 = 6+p +q
10= p + q ……………………2
Solving 1 and 2 we getp = 6 and q =4
The Factor theorem
If P(x) of degree n ≥ 1 has remainder P(r) = 0, then (x-r) ids the factors of P(x)
The factor theorem is a theorem linking factors and zeros of a polynomial. It is commonly
applied to factorizing and finding the roots of polynomial equations. The theorem states
that is a factor of a polynomial P(x) if; that is, r is a root of P(x).
Example5
Let P(x) = x3 –kx2 –x - 6 = 0
D(x) = (x -2)
If the D(x) = (x -2) is the factor of the given polynomial the remainder is Zero
By synthetic division
As the remainder is zero (x-2) divides P(x) perfectly and therefore it is one of the factors
of P(x)
i.e4K+4= 0
K = -1
Matrix
A matrix is a rectangular array of numbers, symbols, or expressions, arranged
in rows and columns. The individual items in a matrix are called its elements or entries.
Two matrices can be added or subtracted element by element if have the same number
of rows and the same number of columns.
Order of matrix
The number of rows and columns that a matrix has is called its order or its dimension
Example[1658−29][15−2689]
The dimension of matrix is read as "two by three" because there are two rows and three
columns.
Order or dimension = number of rows × number of columns
Components of matrix
Let M =[1658−29][15−2689]
The numbers 1,5, -2,6,8,9 are the components of the matrix .It can be also written as
M =$\left[ {\begin{array}{*{20}{c}}{{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}&{{{\rm{a}}_{13}}}\\
{{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}}&{{{\rm{a}}_{23}}}\end{array}} \right]$
Where a11 denotes first row first column a11 =1 similarly a12 = 5and so on.
Types of matrix
Row matrix
Matrix having only one row is called row matrix. The order of such matrix isof 1× n
Example M =[5, 6,8] is the matrix of order 1× 3
Column matrix
Matrix containing only one column is called column matrix. The order of such matrix is of
n ×1
Example M =⎛⎝⎜56−9⎞⎠⎟(56−9)is the matrix of order 1× 3
Null matrix or Zero matrix
Matrix having all elements zero is called zero or null matrix.
M =⎛⎝⎜000⎞⎠⎟3∗1(000)3∗1
M =[000000]2∗3[000000]2∗3
Square matrix
A square matrix is a matrix with an equal number of rows and columns.
Example: M = ⎛⎝⎜56−446987−6−99⎞⎠⎟3∗3(56769−6−448−99)3∗3
M is a square matrix of order 3 × 3
Diagonal matrix
A diagonal matrix is a square matrix that has all its elements zero except for those in the
diagonal from top left to bottom right.
Example: M = ⎛⎝⎜50009000−9⎞⎠⎟3∗3(50009000−9)3∗3
Scalar matrix
A diagonal matrix having same main diagonal element is called scalar matrix
Example: M = ⎛⎝⎜500050005⎞⎠⎟(500050005)
Unit matrix
A unit matrix is a diagonal matrix whose elements in the diagonal are all ones.
I=(1001)2∗2(1001)2∗2
I = ⎛⎝⎜100010001⎞⎠⎟3∗3(100010001)3∗3
I is a unit matrix.
Symmetric matrix:
A square matrix which do not change its row and column are interchanged is called
symmetric matrix.
S = ⎛⎝⎜256584649⎞⎠⎟3∗3(256584649)3∗3
Transpose of matrix
The transpose of a matrix is a matrix which is obtained by interchanging rows into column
and column into rows.It is denoted by A’ or AT
Example1
M =[1658−29][15−2689] transpose of M= T
T =⎛⎝⎜15−2689⎞⎠⎟(1658−29)
Multiplication of matrices
Rules:
1. The number of columns in the 1st one equals the number of rows in the 2nd one
2. Multiply the elements of each row of the first matrix by the elements of each column in
the second matrix.
3. Add the products.
Example 2
$\left(
{\begin{array}{*{20}{c}}{\rm{a}}&{\rm{x}}\end{array}}
\begin{array}{*{20}{c}}
\right)\left(
{\rm{p}}&2\\3&{\rm{q}}\end{array}} \right)$= ( a.p + 3x , 2a+ xq )
Properties of the matrices multiplication
Associative property: A (BC) = A (BC)
Distributive property: A (B +C) = AB + BC
Identity property: AI = A = IA
Singular and non – singular matrices
Singular matrix is square matrix whose determinant is equal to Zero
Example
A =(5252)(5522)is the singular matrix
|A| = 10 -10 = 0
Non - Singular matrix is also square matrix whose determinant is not equal to zero.
A =(6252)(6522)is the non singular matrix
|A| = 12- 10 = 2
Determinant of 2 × 2 matrix
If A =(acbd)(abcd)be 2× 2then its determinant
|A |= ad -bc
Inverse matrix
For a matrix A its inverse B exist when AB = BA = I exists.
A =$\left( {{\rm{\: }}\begin{array}{*{20}{c}}{\rm{a}}&{\rm{b}}\\{\rm{c}}&{\rm{d}}
\end{array}{\rm{\: }}} \right)B=B=\left({{\rm{\:}}
egfhefgh
{\rm{\: }}} \right)$
The components of the inverse matrix can be obtained by
e=dad−bcdad−bc
g =−cad−bc−cad−bc
f =−bad−bc−bad−bc
h= aad−bcaad−bc
We have matrix equation
AX = B
If A-1exists if, |A| ≠ o
⇒A-1(AX) = A-1 B
⇒I.X = A-1 B
⇒ X= A-1 B
By using this equation we can solve simultaneous equations
Example3
Find the value of x and y
(5yx7)(1−2)=(19−4)(5xy7)(1−2)=(19−4)
(5−2xy−14)(5−2xy−14)=(19−4)(19−4)
Now,
5-2x = 19
∴ x = -7
And
y-14 = -4
∴ y = 10
Co-ordinate Geometry
Equation of straight line in slope intercepts form:
Let P(x, y) be the any point line AB which makes a positive angle θ with x-axis.
From figure
∠ ACD= ∠ABE = θ
Slope of line AB = tanθ = m
PD = y- c
CD = OE =x
Tanθ =PDCDPDCD
m =y−cxy−cx
or y= mx+ cis the requiredequation of AB
Angle betweenlines y = m1x+ c2andy =m2 x+ c1
Tanθ = ±m1−m21+m1.m1±m1−m21+m1.m1
Condition for parallelism
If two lines are parallel to each other then m1 = m2.
Condition for perpendicularity
If two lines are perpendicular to each other then m1.m 2 = -1
The equation that is perpendicular to the equation ax + by + c = 0 is bx – ay + k = 0.
The equation that is parallel to the equation ax + by + c = 0 is ax + by + K = 0.
Example 1
Find the equation of the straight line passing through the points of intersection of lines xy+ 2=0 and x+2y -6= 0 and parallel to line 2x+y= 4
Given lines are
x - y +2=0………..1
x + 2y - 6 = 0………………….2
Solving 1 and 2
x = 2323and y = 8383
∴ Point of intersection (x, y) = (2/3, 8/3 )
The equation of given lineis 2x + y = 4 forslope of this line y = -2x + 4
Comparing this equation with y = mx + cm = -2
The equation of the line passing through the point (2/3, 8/3) and slope m = -2
Is (y – 83)=−2(x−23)83)=−2(x−23)
or,(3y−8)=−2(3x−2)or,(3y−8)=−2(3x−2)
Or, 3y- 8= -6x+ 4
Or, 6x + 3y= 12
Or , 2x+ y = 4is the required equation for the straight line.
Equation of pair of straight lines
Theequations of a1x + b1y = 0 and a2x + b2y =0 represent the straight lines passing through
the origin. Multiplying we get
(a1x + b1y) (a2x + b2y) =0
Or, a1a2x2 + (a2b1 + b2a1) xy + b1 b2y2 =0
or ax2 + 2hxy + by2 =0 represents the second degree homogeneous equation in x and y
The angle between the two lines of a equation ax2 + 2hxy + by2 = 0 is tan?
= ±2h2−ab√a+b.±2h2−aba+b.
If the lines of the equation ax2 + 2hxy + by2 = 0 are perpendicular then they should fulfill
the condition a + b = 0 and for parallelism they should fulfill the condition h2 = ab
Circle
The equation of a circle with centre (0,0) and radius r is x2 + y2 = r2.
The equation of a circle with centre (h,k) and radius r is (x-h)2 + (y-k)2 = r2.
The equation of circle with end – points (x1,y1) and(x2, y2) is:
(x – x1)(x – x2) + (y – y1)(y – y2) = 0
If the two circle touch externallyr1+r2= C1C2
Example2
Find the equation center, radius and equation of the circlepassing through the following
points
( 3,-1 )((6,2 )( 3,5 )
let o ( h,k ) be the centre of the circle passing through the point (3,-1 ) , (6,2 ) (3,5 )
A ( 3,-1 )B (6,2 )C ( 3,5 )centre of the circle O
OA = (h−3)2+(k+1)2−−−−−−−−−−−−−−−−√(h−3)2+(k+1)2
OA =h2−6h+9+k2+2k+1−−−−−−−−−−−−−−−−−−−−−−√h2−6h+9+k2+2k+1
OA2=h2−6h+9+k2+2k+1OA2=h2−6h+9+k2+2k+1
OA2=h2−6h+9+k2+2k+1OA2=h2−6h+9+k2+2k+1
OA2=h2+k2−6h+2k+10OA2=h2+k2−6h+2k+10
OB = (h−6)2+(k−2)2−−−−−−−−−−−−−−−−√(h−6)2+(k−2)2
OB =h2−12h+36+k2−4k+4−−−−−−−−−−−−−−−−−−−−−−−√h2−12h+36+k2−4k+4
OB2=h2−12h+36+k2−4k+4OB2=h2−12h+36+k2−4k+4
OB2=h2−12h+36+k2−4k+4OB2=h2−12h+36+k2−4k+4
OB2=h2+k2−12h−4k+40OB2=h2+k2−12h−4k+40
OC = (h−3)2+(k−5)2−−−−−−−−−−−−−−−−√(h−3)2+(k−5)2
OC=h2−6h+9+k2−10k+25−−−−−−−−−−−−−−−−−−−−−−−√h2−6h+9+k2−10k+25
OC2=h2−6h+9+k2−10k+25OC2=h2−6h+9+k2−10k+25
OC2=h2−6h+9+k2−10k+25OC2=h2−6h+9+k2−10k+25
OC2=h2+k2−6h−10k+34OC2=h2+k2−6h−10k+34
As OA = OB = OC
OA2=OB2=OC2OA2=OB2=OC2
Taking two ratio
OA2=OB2OA2=OB2
h2+k2−6h+2k+10=h2+k2−6h+2k+10=h2+k2−12h−4k+40h2+k2−12h−4k+40
6h+6k-30 = 0………………..1
Taking other two ratio
OB2=OC2OB2=OC2
h2+k2−12h−4k+40=h2+k2−12h−4k+40=h2+k2−6h−10k+34h2+k2−6h−10k+34
12h -6k -6 = 0 ……………….2
Solving 1 and 2 we get,
(h, k) = (2 , 3 )
For radius
(3−2)2+(−1+3)2=r2(3−2)2+(−1+3)2=r2
(1)2+(2)2=r2(1)2+(2)2=r2
∴ r2 = 5 unit
Equation of circle is(x−2)2+(y−3)2=10is(x−2)2+(y−3)2=10
Example 3
Prove that the following points are concyclic .
(1,0) ,(2,-7),(8,1 ), (9, -6)
The equation of the circle is given by(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
If it passes through point (1,0) and 2,-7 )
Then
(1−h)2+(0−k)2=r2(1−h)2+(0−k)2=r2 …………….1
(2−h)2+(7+k)2=r2(2−h)2+(7+k)2=r2……………2
Equating 1 and 2
(1−h)2+(0−k)2=(2−h)2+(7+k)2(1−h)2+(0−k)2=(2−h)2+(7+k)2
1-2h = 4-4h +49 +14k
h-7k -26 = 0…………………3
Alsoif it pases through the point (8,1 )
(8−h)2+(1−k)2=r2(8−h)2+(1−k)2=r2……………4
From 2 and 3
(8−h)2+(1−k)2=(1−h)2+(0−k)2(8−h)2+(1−k)2=(1−h)2+(0−k)2
64−16h+1−2k=1−2h64−16h+1−2k=1−2h
14 h +2k -64 = 0
7 h + k -32 =0
Solving 3 and 4
We get (h, k ) = (5 ,-3)
Putting this value in 1 we get
(1−5)2+(0+3)2=r2(1−5)2+(0+3)2=r2
R= 5 unit
The equation of the circle is given by(x−5)2+(y+3)2=25(x−5)2+(y+3)2=25
If it pae htrought point(9 -6 )
Then ,
(9−5)2+(−6+3)2=25(9−5)2+(−6+3)2=25
25 =25
Which I true
∴ Given 4 point lies on the circle and they are concyclic
Trigonometry
Trigonometrical ratios
SinA = phph, cosA = bhbh, tanA = pbpb, cotA = bpbp, SecA = hbhb and cosecA = hp.hp.
Where, p = perpendicular, b = base and h = hypotenuse and h2 = p2 + b2.
2. SinA * cosecA = 1, cosA * secA = 1, tanA * cotA = 1.
tanA = sinAcosAsinAcosA, cotA = cosAsinAcosAsinA
3. sin2A + cos2A = 1, sec2A – tan2A = 1, cosec2A – cot2A = 1
4. The trigonometric values of different degree(0° - 180°) are listed below:
Angles
Sin
Ratios
00
Cos
Tan
Cosec
Sec
Cot
0
1
0
300
1212
3√232
13√13
2
23√23
3√3
45
12√12 12√12
1
2√2
2√2
1
600
3√232 1212
3√3
23√23
2
13√13
900
1
0
1
0
1
1200 3√232
−12−12 −3√−3
23√23
−2−2
135
−12√−12 -1
2√2
−2√−2 -1
0
12√12
1500 1212
−3√2−32 −13√−13 2
0
−13√−13
−23√−23 −3√−3
1800 0
-1
0
-1
5.With angle is (-A),
Sin(-A) = -sinA, cos(-A) = cosA. Tan(-A) = - tanA, etc.
6. When angle is (90° - A) and (90° + A)
Sin(90° - A) = cosA sin(90°+A) = cosA
Cos(90° - A) = sinA cos(90° + A) = -sinA
Tan(90° - A) = cotA tan(90°+A) = -cotA etc.
7. With angle (180° - A) and (180° + A)
Sin(180° - A) = sinA sin(180°+A) = -sinA
Cos(180° - A) = -cosA cos(180° + A) = -cosA
Tan(180° - A) = -tanA tan(180°+A) = tanA etc.
8. When angle is (270° - A) and (270° + A)
Sin(270° - A) = -cosA sin(270°+A) = -cosA
Cos(270° - A) = -sinA cos(270° + A) = sinA
Tan(270° - A) = cotA tan(270°+A) = -cotA etc.
9. With angle (360° - A) and (360° + A)
Sin(360° - A) = -sinA sin(360°+A) = sinA
Cos(360° - A) = cosA cos(360° + A) = cosA
Tan(360° - A) = -tanA tan(360°+A) = tanA etc.
Trigonometric Ratios of compound angles:
1. sin(A + B) = sinA.cosB + cosA.sinB
2. cos(A + B) = cosA.cosB – sinA.sinB
3. Tan (A + B) = tanA+tanB1−tanA.tanBtanA+tanB1−tanA.tanB
4. sin (A – B) = sinA.cosB - cosA.sinB
5. cos(A – B) = cosA.cosB + sinA.sinB
6. tan(A – B) = tanA−tanB1+tanA.tanBtanA−tanB1+tanA.tanB
Trigonometric ratios of multiple angles:
1.
(i)sin2A = 2sinA.cosA
(ii)cos2A = cosA2A – sin2A = 2cos2A – 1 = 1 – 2sin2A
(iii)tan2A = 2tanA1−tan2A2tanA1−tan2A
(iv)sin2A = 2tanA1+tan2A2tanA1+tan2A
(v)cos2A = 1−tan2A1+tan2A1−tan2A1+tan2A
2.
(i) sin3A = 3sinA – 4sin3A
(ii) cos3A = 4cos3A – 3cosA
(iii) tan3A = 3tanA−tan3A1−3tan2A3tanA−tan3A1−3tan2A
Trigonometric ratios of Sub-multiple angles:
1.
(i)sinA = 2sinA2A2.cosA2A2.
(ii)cosA = cosA2A2A2 – sin2A2A2. = 2cos2A2A2. – 1 = 1 – 2sin2A2A2.
(iii)tanA = 2tanA2.1−tan2A2.2tanA2.1−tan2A2.
(iv) sinA = 3sinA3A3 – 4sin3A3A3
(v) cosA = 4cos3A3A3 – 3cosA3A3
(vi) tanA = 3tanA3−tan3A31−3tan2A33tanA3−tan3A31−3tan2A3
Transformation of Trigonometric Formulae:
1.2sinA.cosB = sin(A + B) + sin(A – B)
2. 2cosA.sinB = sin(A+B) – sin(A – B)
3. 2cosA.cosB = cos(A + B) + cos(A – B)
4. 2sinA.sinB = cos(A + B) – cos(A + B)
5. sinC + sinD = 2sin C+D2.C+D2. cos C−D2C−D2
6. sinC – sinD = 2cos C+D2.C+D2. sin C−D2C−D2
7. cosC + cosD = 2cos C+D2C+D2. cos C−D2C−D2
8. cosC – cosD = -2sinC+D2C+D2.sin C−D2C−D2
Where, A = C+D2C+D2 and B = C−D2C−D2
Conditional Trigonometric Identities:
If A + B + C = π
1.sinA = sin[ π– (B + C)] = sin(B + C)
2. cosA = cos[π – (B + C)] = -cos(B + C)
3. tanA = tan[π – (B + C)] = -tan(B + C)etc.
If, A2A2+ B2B2 + C2C2 = π2π2
1.sinA2A2= sin[π2π2 – (B2B2 + C2C2)] = cos(B2B2 + C2C2)
2. cosA2A2 = cos[π2π2 – (B2B2 + C2C2))] = sin(B2B2 + C2C2))
3. tanA2A2 = tan[π2π2 – (B2B2 + C2C2))] = cot(B2B2 + C2C2))etc.
Example 1
Prove:
cos2(A – 120°) + cos2A + cos2(A – 120°) = 3232
L.H.S. = cos2(A – 120°) + cos2A + cos2(A – 120°)
= {cos2(A – 120°)}2 + cos2 A + {cos2(A + 120°)}2
= (cosA.cos120°+sinA.sin120°) + cos2A + (cosA.cos120°– sinA.sin120°)
= (−12cosA+3√2sinA)2(−12cosA+32sinA)2+
+ (−12cosA+3√2sinA)2(−12cosA+32sinA)2
= 1414cos2A – 2.1212.3√232 cosA.sinA + 3434sin2A +
2.1212.3√232 cosA.sinA +3434sin2A
= 1414cos2 A + 3434 sin2A + cos2A + 1414 cos2A + 3434 sin2A
= 2424 cos2A + 6464sin2A + cos2A.
= 1414 (2cos2A + 6sin2A + 4cos2A) = 1414(6cos2A + 6sin2A)
= 6464(cos2A + sin2A) = 3232 = R.H.S.
cos2A
cos2A
+ 1414cos2A
Example 2
If A + B+ C = π prove that cos2A + cos2B + cos2C = 1 – 2cosA.cosB.cosC
Soln:
L.H.S. =cos2A + cos2B + cos2C
= 1+cos2A21+cos2A2 + 1+cos2B21+cos2B2 + cos2C.
= 12+12+1212+12+12 (cos2A + cos2B) + cos2C.
= 1 + 1212. 2 cos 2A+2B22A+2B2. Cos 2A−2B22A−2B2 + cos2C.
+
= 1 + cos(A + B).cos(A – B) + cos2C.
= 1 – cosC.cos(A – B) + cos2C[A + B = π – C, So, cos(A+B) = cos(π – C) = – cosC]
= 1 – cosC {cos(A – B) – cosC}
= 1 – cosC {cos(A – B) + cos(A + B)}
= 1 – cosC.(cosA.cosB + sinA.sinB + cosA.cosB – sinA.sinB)
= 1 – cosC.(2cosA.cosB) = 1 – 2cosA.cosB.cosC = R.H.S.
Example 3
Solve:
2 sin2x – sinx = 0
Soln:
Here, 2 sin2x = sinx
Or, 4sinx cosx – sinx = 0
Or, sinx (4cosx – 1) = 0
Either, sinx = 0
So, x = nπ
Or, 4cosx – 1 = 0
Or, cosx = 1414
Or, cosx = cosα [cosx = 1414]
So, x = 2nπ ±± α.
So, x = 2nπ ±± cos–11414 n ԑ Z.
Height and Distance
Angle of elevation: The angle of elevation is defined as angle between the line of sight
and horizontal line made by the observer when the observer observes the object above
the horizontal line.
Example 4
A man observes the top of a pole 52cm height situated in front of him and finds the angle
of elevation to be 30° .If the distance between man and pole is 86m .Find the height of
that man.
Soln
Tan30° =EDADEDAD=ED86ED86
13√13=ED86ED86
Or, ED = 863√863
Or, ED = 49.65cm
Height of man (DC) = EC – ED = 50- 47.65 =2.35 m
Example 5
The angle of elevation from the roof of a house to the top of atop of tree is found to be
30°. If the height of the house and tree are 8 m and 20m respectively, find the distance
between the house and the tree.
Soln
ED = EC - DC = 20- 8 = 12 cm
Tan 30°=EDADEDAD
13√13=12AD12AD
AD =20.78 cm
∴ Distance between tree and house is20.78 cm
Vector
Cartesian form of vector in two dimensions
Vectors i and j are vectors of length 1 in the directions OX and OY respectively.
The vectorOR⃗is xi. The vectorOS⃗is yj.. The vectorOP⃗is the sum ofOR⃗and OS⃗, that is
OP⃗=xi+ yj
The modulus of the vector is positive number that measures the length of the vector in
terms of components
OP ⃗= x2+y2−−−−−−√x2+y2
Addition of vectors
if two vectors such as AB and BC are representing the two sides of a triangle ABC, then
the third side AC closing the other side of the triangle in opposite direction represents the
sum of two vectors both in magnitude and vectors.
AC⃗= AB⃗+ BC⃗
Polygon law of vector addition
For the addition of more than three vectors polygon law of vectors is used.
Figure:
R⃗ =A⃗+ B⃗+ C⃗+ D⃗
The resultant is the vector sum of two or more vectors. It is the result of adding two or
more vectors together. If displacement vectors A, B, C,D are added together, the result
will be vector R as shown in the diagram, vector R can be determined.
Scalar product of two vectors
Let a⃗ = ( a1,a2) andb⃗ = ( b1 , b2) be any two vectors thenthe scalar product of two vectors
ordot product of two vectors is given by
a⃗ . b⃗ = a1.b1 + a2 b2
It can be also written as
a⃗ .b⃗ = |a⃗| |b⃗|cosθ = abcosθ
Geometrical interpretation of Scalar product of two vectors
Here, OA ⃗ = a⃗ , OB⃗= b⃗ and θ is the angle between
a⃗. b⃗ = |a⃗| |b⃗|cosθ = abcosθ=OA OB cosθ= OA*OE= magnitude of a⃗ × projection of b⃗
on a⃗
Condition of perpendicularity
a⃗. b⃗ = a1.b1 + a2 b2+ a3b3=abcos 90° = 0
Angle between the vectors
If a⃗ = ( a1,a2) andb⃗ = ( b1 , b2) be any two vectors then,
cosθ=a1.b1+a2b2aba1.b1+a2b2ab=a⃗ .b⃗ aba→.b→ab
If a⃗ a→ =
x1\veci\veci +
y1j⃗ j→ and b⃗ b→ =
x2\veci\veci +
y2j⃗ j→ then, a⃗ .b⃗ a→.b→ = x1.x2 + y1.y2. where, i⃗ .i⃗ i→.i→ =j⃗ .j⃗ j→.j→ = 1
and i⃗ .j⃗ i→.j→ = j⃗ .i⃗ j→.i→ = 0
-|a⃗ a→| = x21+y21−−−−−−√x12+y12 and |b⃗ b→|= x22+y22−−−−−−√x22+y22 and
cos? = x1.x2+y1.y2∣∣a⃗ ∣∣∣∣b⃗ ∣∣x1.x2+y1.y2|a→||b→|
Vector Geometry
If the vector of A is a⃗ a→ and the vector of B is b⃗ b→. Then the mid-point of AB is M
so, m⃗ m→= a⃗ +b⃗ 2a→+b→2
If P divides line AB internally in ratio m:n then p⃗ =na⃗ +mb⃗ m+np→=na→+mb→m+n and
when it divides the line externally p⃗ =mb⃗ −na⃗ m−np→=mb→−na→m−n
Position vector of the centroid of the given triangle is given byr⃗a⃗ +b⃗ +c⃗ 3a→+b→+c→3
Example 1
Find the value ofKif p ⃗= k i⃗+3 j⃗= x1i⃗ + y1 j⃗, q⃗ = 9 i⃗+ 27 j⃗=x2i⃗ + y2 j⃗
p⃗= k i⃗+3 j⃗= x1i⃗ + y1 j⃗ = (k, 3 )
q⃗ = 9 i⃗+ 27 j⃗=x2i⃗ + y2 j⃗ = (9, -27)
if the p⃗ and q⃗parallelto each other p⃗ = a q⃗
(k3)=a(k3)=a(927)(927)
Equating the corresponding term
3 =27 a
∴ a = 1919
Andk = 9 a
Or, k = 1
Example 2
Find the position vector of the centroid of the triangle ABC
Letp⃗be the position vector of the centroid of the triangle and a⃗, b⃗ and c⃗ bet the position
vector of the A, B and C respectively.
a⃗ = (−10)(−10)b⃗ =(23)(23)c⃗=(−2−7)(−2−7)
Then,
p⃗= 13(a⃗ +b⃗ +c⃗ )13(a→+b→+c→)
= 1313[(−10)(−10)+(23)(23)+ (−2−7)(−2−7)
=13(−1+2−20+3−7)13(−1+2−20+3−7)
= 13(−1−4)13(−1−4) = (−13−43)(−13−43)= −13i+−43j−13i+−43j
Transformation
Transformation
Transformation is the movement of an object in plane to bring a change. The original
shape of an object is called pre- image and the final image after transformation is called
image.There are basic four methods for changing geometrical figures they are:
a. Reflection
Reflection is a mirror image of an object or basically it is called flip of an object.
P(x,y) →P’ (x ,-y)
P(2,3)→P’ (2,-3)
Transformation matrix =(100−1)(100−1)
Reflection in y-axis
Reflection in y= x
P(x,y) →P’ ( y ,x)
P(-3,2 )→P’ (2,-3)
Transformation matrix =(0110)(0110)
Reflection in y= -x
P(x,y) →P’ ( -y ,-x)
P(3,2 )→P’ ( -2,-3)
Transformation matrix =(01−10)(0−110)
b. Translation
Translation is a term used in geometry to describe a function that moves an object a
certain distance. The object is not altered in any other way. It is not rotated, reflected or
re-sized. In a translation, every point of the object must be moved in the same direction
and for the same distance.
If there co-ordinates in plane is translated by the vector(ab)(ab)then
P(x,y)→ P’ (x+ a ,y+b)
If the translation T1 =(ab)(ab)is followed by T2 =(cd)(cd)the combined translation
T2OT1=(a+cb+d)(a+cb+d)
P(x,y)→ P’ (x+ a+c ,y+b+d)
c.Rotation
In this transformation the pre-image is transformed by certain angle like 90°, 180° about
the origin or centre (h,k)
Rotation of 90° anti-clockwise about the origin
P(x,y) →P(-y,x)
Transformation matrix =(01−10)(0−110)
Rotation of 90° clockwise about the origin
P(x,y) →P(y,-x)
Transformation matrix =(0−110)(01−10)
Rotation of 180° clockwise about the origin
P(x,y) →P( -x, -y )
Transformation matrix =(−100−1)(−100−1)
d. Enlargement
Enlargement changes the size of an object .For every enlargement scalar factor kand
centre should be defined.
If k>1, then the image is larger than the object.
If k = ± 1, then the image and the object is the same size.
If 0<k<1, then the image is smaller than the object.
If O is an centre of enlargement and k is a scalar factor then , it is [ written as [ 0,K]
P(x,y)→ P(kx,ky)
Transformation matrix =(K00K)(K00K)
The combination of the enlargement of the enlargement E1 [O, k1] followed by the another
enlargement E2 [0,k2 ] is given E2 OE1 =[ O, K1 K2 ]
If the centre is other than origin, say, (a,b) and scalar factork
P(x,y)→ P(k(x-a)+a ,k(y-b) +b)
Example1
A triangle with vertices A(1,2), B(4, -1),C(2,5)is reflected successively in the lines X =5
and y= -2Find by stating the co-ordinates
Soln
A(1,2)→reflectiononx=5→reflectiononx=5A’(2*5 – 1, 2 )= A’(9,2)
B(4, -1)→reflectiononx=5→reflectiononx=5 B’(2*5- 4, -1) =B’(6, -1)
C(2,5)→reflectiononx=5→reflectiononx=5 C ‘(2*5-2, 5) = C’(8,5)
Now,
A’(9,2) →reflectionony=−2→reflectionony=−2 A’’(9 ,2 * -2- 2 ) = A’’(9, -6)
B’(6, -1)→reflectionony=−2→reflectionony=−2 B’’(6 , 2*-2+1 ) = B’’(6,-3)
C’(8,5) →reflectionony=−2→reflectionony=−2 C’’(8, 2*-2 -5) = C’’(8, -9)
Example 2
IfX(2,3) Y(4,5) and Z(6,2) are vertices of Δ XYZfind the image of Δ XYZunderthe
translation vector XY⃗ XY→followed by YZ⃗ YZ→
Soln
Translation vector XY−→−XY→= OY−→−OY→ -OX−→−OX→
= (4, 5)- (2,3)
= (2, 2)
||y translation vectorYZ−→−YZ→ = OZ−→−OZ→- OY−→−OY→
= (6, 2) – (4, 5)
= (2,-3)
The translation vector XY ⃗ followed by YZ ⃗ is given by T =(2, 2)+(2,-3)= (4 , -1)
Example 3
Using the matrix method show that a reflection in the line y= x followed by a reflection in
y=0 is equivalent to the negative quarter turn about origin.
Soln
The matrix representing the reflection in y = x
T =(0110)(0110)
Let us consider a point P(x,y) as an object.
Pre- multiplying by the given matrix M
(0110)(xy)(0110)(xy)
= (yx)(yx)
Hence P(x,y) à P’(y , x)
If it is followed by the reflection y = 0 (reflection on x-axis)
Then,
Matrix representing the reflection on x-axis T =(100−1)(100−1)
(100−1)(yx)(100−1)(yx)
=(y−x)=(y−x)s
∴ P’(y,x)= P’’(y, -x)
P(x, y)= P’’(y,-x)
The matrix representing the negative quarter turn about the origin(01−10)(0−110)
If p(x,y) be the object then,
(01−10)(xy)=(−yx)(0−110)(xy)=(−yx)
Hence
P(x,y)à P’(-y , x)
Statistics
The measures of the data that shows the range or spread of the data from the central
value is called measure of the variability and dispersion
Range, mean deviation, quartile deviation and standard deviation are the four measures
of dispersion.
Range
The difference between the highest and lowest value of any set of data is called range of
the data.
R = L- S
Coefficient of range=L−SL+SL−SL+S
For example, let us consider the following data set:
23, 65,82,59,55,25,85,56
Largest value (L) =85
Smallest value (S) =23
R = L- S85 -23 = 62
Mean
A mean is defined as the average of the numbers. The data may be individual, discrete
and continuous .The method of calculating the means depend upon the nature of data. It
is denoted byx̅
For individual data:x̅ =x1+x2+x3+x4……xnNx1+x2+x3+x4……xnN=ΣxNΣxN
For
discrete
data:x̅
=f1x1+f2x2+f3x3+x4……fnxnf1+f2+f3…………..fnf1x1+f2x2+f3x3+x4……fnxnf1+f2+f3…………..fn=
ΣfxN=ΣfxN
For
continuous
data:f1x1+f2x2+f3x3+x4……fnxnf1+f2+f3…………..fnf1x1+f2x2+f3x3+x4……fnxnf1+f2+f3…………..f
n=ΣfxNΣfxN
Mean deviation
The mean deviation is the mean of the absolute deviationsof a set of data about the
mean. For a sample size N, the mean deviation is defined by
M.D =1N∑i=1N|xi−x¯|1N∑i=1N⁡|xi−x¯|
Where x̅ is the mean of the distribution
The mean deviation for a discrete and continuous distribution defined by
M.D =1N∑i=1Nf|xi−x¯|1N∑i=1N⁡f|xi−x¯|
Coefficient of mean deviation =M.DX¯M.DX¯
Standard deviation
The Standard Deviation is a measure of how spreads out numbers are. It is denoted by
σ. it is the square root of the Variance.
For individual series
σ =Σd2N−−−√Σd2Nwhere d = |xi−x¯||xi−x¯|where I = 1…N
For discrete and continuous series
σ =Σfd2N−−−−√Σfd2Nwhere d = |xi−x¯||xi−x¯|where I = 1…N for discrete and d
= |m−x¯||m−x¯|m = mid value
Coefficient of variance= σx¯σx¯
Median
Median is the mid –value of the data.
For the individual and discrete data
Median = value of N+12N+12 item if the number of the terms is odd
= average value ofN2N2andN+22andN+22item if the number of the terms is even
For continuous data
Median = L + N2−c.ffN2−c.ff *i
N= total number of items
L = lower limit of median class
C.F = C.F of the preceding the median class
F= frequency of median class
I = class interval
Example 1
Find the mean deviation and median from the median and its coefficient of the data from
given below.
x
f
4
6
6
4
8
5
10
3
12
2
14
1
16
4
a.
x
f
c.f
|D|
f|D|
Md
4
6
6
4
24
6
4
10
2
8
8
5
15
0
0
10
3
18
2
6
12
2
20
4
8
Value of
N+12thitemN+12thitem
= 13th term
=8
14
1
21
6
6
16
4
25
8
32
N=25
84
Mean deviation (M.D) =Σf|D|NΣf|D|N= 3.36
Coefficient of mean deviation from the median is given by coefficient of M.D
= M.D(Md)MdM.D(Md)Md= 3.368=0.423.368=0.42
Example 2
Find the standard deviation and coefficient of standard deviation from the following data.
mean(X̅) d = X – X̅
x
14
66.28
d2d2
-52.3 2735.29
46
-20.3
412.09
58
-8.3
68.89
76
9.7
94.09
82
15.7
246.49
90
23.7
561.69
98
31.7 1004.89
ΣX=
464
5123.43
Standard deviation =Σd2N−−−√Σd2N=5123.437−−−−−√5123.437= 27.05
Coefficient of variation= σX¯σX¯* 100 =27.0566.2827.0566.28 *100 = 40