HOW TO UNDERSTAND RELATIONS AND FUNCTIONS ,INVERSE OF A FUNCTION

we are going to  discuss Relations and Functions , "How to understand  Relations and  Functions, Inverse of a Function" under the topic  Relations and Functions.

Ordered-Pair Numbers :-

Ordered-pair number is written within a set of parentheses and separated by a comma.

For example, (5, 6) is an ordered-pair number; the order is designated by the first element 5 and the second element 6. The pair (3, 6) is not the same as (6,3) because they have different order. Sets of ordered-pair numbers can represent relations or functions.

Example of ordered pair :

(3,8),(2,1),(7,6)

Relation

A relation is a  set of ordered-pair numbers.

consider the following table

________________________________________________________________________

Numbers of students      1             2          3         4           5          6

_______________________________________________________________________

Marks Obtained             96          98       97         78        77         86

_______________________________________________________________________

In the above table the numbers of students and marks obtained by them  is a relation and can be written as a set of ordered-pair numbers.

A= {(1, 96), (2, 98), (3, 97), (4, 88),(5,77),(6,86)}

When we collect all the elements written in 1st column of the ordered pairs and placed in a set then the set so formed is called  Domain of the relation.
The domain of A= {1, 2, 3, 4,5,6}

As all the elements written in 2nd column of the ordered pairs and placed in a set then the set so formed is called  Range of the relation.

The range of A = {  96,98,97,88,77,86}

we can better understand this concept with the help of this video

Function

A function is a relation in which every first element in ordered pairs have unique second element associated with them. Second  elements may or may not be same.

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Example

 {(1, 2), (2, 3), (3, 4), (4, 5),(5,6)}  is an example of function 

 { (1, 2), (2, 3), (3, 4), (4, 5),(5,6) } is a function because all the  first elements are different.

Example

{(1, 3), (3, 3), (2, 1), (4, 2)}  is an example of function 

 {(1, 3), (2, 3), (2, 1), (4, 2)}  is a function because all the first elements are different.

Example

{ (1, 6), (2, 5), (1, 9), (4, 3) }  is not an  example of function 

As in  {(1, 6), (2, 5), (1, 9), (4, 3)}  the element "1 "   appeared twice .

Example

{(2, 15), (3, 15), (4, 15), (5, 13),(6,18)}  is  an  example of function 

As in  {(2, 15), (3, 15), (4, 15), (5, 15)}   all the first elements are different.

Example

{(1, 1), (-1, 1),(2,4),(-2,4), (3, 9), (-3, 9),(4,16),(-4,16)}  is an  example of function although   the element "1" and "-1" ,"2" and "-2" , "3" and "-3" ,"4","-4" have same images. This is an example of many one function.

Question:-   Find x and y if: 

(i) (5x + 3, y) = (4x + 5,  2)

(ii) (x – y, x + y) = (8, 12)
(iii) ( 2x-y , y+5 ) = ( -2,3 )

Solution

(1)  Given  (5x + 3 , y) = (4x + 5, 2)
So By the equality of ordered pair elements,
1st element of the ordered number written on the left hand side will be equal to the 1st element of the ordered pair number written on the  right hand side . Therefore 

5x + 3 = 4x + 5   and y =  2 
5x-4x = 5 -3   and y = 2 
x = 2 and y = 2

(ii) So By the equality of ordered pair elements
x – y = 8 and  x + y = 12

Solving these two equations for x and y 

 2x =20  and    10+ y =12 

x=10   y = 2

(iii) So By the equality of ordered pair elements
2x-y  =-2  , y+5 = 3 
2x = -2+y  , y = 3-5
2x = -2+y  , y = -2
Putting the value of y in 1st Equation ,we get
2x = -2 - 2
2x = -4
x = -2
so x= -2 and y =-2

Types of Relations

A relation R in a set A is called

(i) reflexive, if (a, a) ∈ R, for every a ∈ A,

(ii) symmetric, if (a, b) ∈ R implies that (a, b) ∈ R, for all a,b ∈ A.

(iii) transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b,c ∈ A.

Equivalence Relation

A relation R in a set A is said to be an equivalence  relation if R is reflexive, symmetric and transitive.

1 ) Let B be the set of all triangles in a plane with R a relation in B given by

R = {(T1, T2) : T1 is congruent to T2}. Then R is an equivalence relation.

2 ) Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7}  by

R = {(a, b) : both a and b are either odd or even}. Then R is an equivalence

one-one Function

A function f : X → Y is defined to be one-one (or injection ), if the images of distinct elements of X under f are distinct, i.e., for every x, y ∈ X, f (x) = f (y) implies x = y. Otherwise, f is called many-one.

Onto Function

A function f : X → Y is said to be onto (or surjective), if every element of Y is the image of some element of X under f, i.e., for every y ∈ Y, there exists an

element x in X such that f (x) = y.

Example

1   Function f : R → R, given by f (x) = 2x, is one-one and Onto As all the elements  have only one and uniqe image under f.

2  Function f : N → N, given by f (x) = 2x, is one-one but not onto.Because  the elements  have only one and unique image under f Therefore it is one one function .But not all elements of N have image under f 

e. g .  1,3,5,7... are not the image of any elements of N under f so it is not onto function

Example

The function f : N → N, given by f (1) = f (2) = 1 and f (x) = x – 1,

for every x > 2, is onto but not one-one.

Solution

Since f is Not one-one, as f (1) = f (2) = 1. 

But f is Onto, as given any y ∈ N, y ≠ 1,

Choose x = y + 1 s.t.

 f (y + 1) = y + 1 – 1

f (y + 1)  = y. 

Also for 1 ∈ N, 

we are given  f (1) = 1

Inverse of a Function

A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f –1

Example

Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as below have inverses. Find f , if it exists.

(a) f = {(1, 1), (2, 2), (3, 3)}

(b) f = {(2, 2), (3, 1), (4, 1)}

(c) f = {(1, 5), (3, 4), (2, 1)}

Solution

(a) It is to  proved that  f is one-one and onto Hence f is invertible with the inverse f –1 of  f given by f –1 = {(1, 1), (2, 2), (3, 3)} = f.

(b) Since f (3) = f (4) = 1, f is not one-one, so that f is not invertible.

(c) Here  f   is one-one and onto, so that f is invertible with

 f –1 = {(5, 1), (4, 3), (1, 2)}.

Composition of Functions

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A → C given by

gof (x) = g(f (x)), ∀ x ∈ A

Example

fof(x) = (16x + 12 + 18x -12 ) / ( 24x + 18 - 24x +16)

fof(x) = (34 x ) / ( 34)

fof(x) =  x  =  I(x)

Example

Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as f (2) = 3, f (3) = 4, f (4) = f (5) = 5 and g(3) = g(4) = 7 and g(5) = g(9) = 11. Find gof = ?

Solution

We are given

 gof (2) = g (f (2)) 

               = g(3) 

               = 7

 gof (3) = g(f (3)

             = g(4)

              = 7,

gof (4) = g(f (4)) 

           = g(5) 

             = 11 

and  gof (5) = g(f (5))

                   = g (5)                     

                    = 11

So gof ={(2,7),(3,7),(4,11),(5,11)} 

Example

Conclusion

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MEMORISE A B AND C D FORMULAS IN TRIGONOMETRY IN AN EASY MANNER

Hello Friends 

welcome to this post of learning trigonometric formulas with me .Most of the Students or Mathematics Learner ,most of the time confuse to remember or memorise A  B and C  D formulas,They  mixed A B and C D formulas with each other and could not reproduce what they have learnt . So today we going to learn new techniques to learn "How to memorise AB and CD formulas" forever. Before this we must have knowledge of different trigonometric values of different angles in different quadrants.

First of all have a quick look at some of  these formulas .

To clear your  all doubts on   " How to Calculate Different Trigonometric values in different quadrants "  in an easy Method. click on the  above  links  .

 

Tricks to Learn    A  B   Formulae  For  sine  angles

When angles are added   i. e  Sin  ( A+B )  

When Angles are added and then their Trigonometric Ratios is taken , and if we have to take the  Sine of  added angles, then it can be done like this.

Start with  sine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign      Start with Cosine of angle A and multiply with Sine of angle B. i.e. start with sine and ends with sine and in middle both the terms are cosine ,and angles start  A then B again A then again B.

Sin (A+B) = Sin A Cos B + Cos A Sin B

When angles are subtracted    i. e  Sin  ( A-B )  

When Angles are subtracted and  their Trigonometric Ratios is taken , and if we have to take the  Sine of  subtracted  angles, then it can be done like this

    
Start with  sine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign    Start with Cosine of angle A and multiply with Sine of angle B. i.e  start with sine and ends with sine and in middle both the terms are cosine ,and angles start  with A then B again A then again B.

Sin (A - B) = Sin A Cos B - Cos A Sin B

Tricks to Learn    A  B   Formulae For  Cosine  angles

When angles are added   i. e  Cos  ( A+B )  

When Angles are added and then their Trigonometric Ratios is taken , and if we have to take the  Cosine  of  added angles, then it can be done like this.

Start with  cosine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  -ve sign      Start with Sine of angle A and multiply with Sine of angle B. i.e. 1^st   and 2^nd terms are   cosine and  3^rd and 4^th terms    are sine , Angles start with   A then B again A then again B.

"Here  Sum of cosine of  Two angles  is equal to difference of  product of  cosines of both the angles    and product of sine of both the angles ".

Cos (A+B) = Cos A Cos B - Sin A Sin B

When angles are subtracted    i. e  Cos  ( A-B )  

When Angles are subtracted and  then their Trigonometric Ratios is taken , and if we have to take the  cosine of  subtracted  angles, then it can be done like this.

Cos (A - B) = Cos A Cos B + Sin A Sin B

Start with  cosine of  angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign      Start with Sine of angle A and multiply with Sine of angle B. i. 1^st   and 2^nd terms are   cosine and  3^rd and 4^th  terms    are sine , and angles start with   A then B again A then again B.

"Here  Difference  of cosine of  Two angles  is equal to the  Sum  of  product of  cosines of both the angles    and product of sine of both the angles" .

Want to Learn WHAT IS SET, TYPES OF SETS ,UNION ,INTERSECTION AND VENN DIAGRAMS .

How  to  Memorise     C D   Formulae

To learn C D formulae 

Step 1 

Place 2 for all four formulae and  take Trigonometric Ratio of 1st angle for all four formulae which  is (C+D)/2 and again  trigonometric Ratio of  2nd angle which  is (C-D)/2.

Step 2.1 

For addition of Sine Formula start with sine of 1st angle as mentioned in step 1 and multiply it with cos of  2nd angle as mentioned in step 1.

Step 2.2 

For subtraction of Sine Formula start with cosine of 1st angle as mentioned in step 1 and multiply it with sine of  2nd angle as mentioned in step 1.

Step 3.1

For addition of cosine Formula start with cosine of 1st angle as mentioned in step 1 and multiply it with cosine  of  2nd angle as mentioned in step 1.

Step 3.2

For subtraction of cosine Formula start with sine of 1st angle as mentioned in step 1 and multiply it with sine   of  2nd angle as mentioned in step 1,and do not forget to multiply it with -ve sign.

or  

If you do not want to multiply it with -ve sign  ,then you can change 2nd angle (D-C)/2 instead of (C-D)/2

How to Memorise A   B and C   D  formulas  easily ,watch this video 

Want to learn MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS .


Want to learn HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES || MATRIX METHOD TO SOLVE SYSTEM LINEAR EQUATIONS .

Conclusion

Thanks for devoting your valuable time for the post Easy Tricks to Memorise A B and C D Formulae in Trigonometry and trigonometry's shortcut formulas of this blog. If you found this this blog/post of your concern, Do Follow me on my blog and share this post with your friends . We shall meet again in next post ,till then Good Bye.

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HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES

Hello Friends welcome to this post of learning trigonometric formulas in an easy way with me . AS most of the Students or Mathematics Learner ,most of the time confuse to remember or memorise value of different trigonometric angles in different quadrants and could not reproduce what they have learnt . So today we going to learn new techniques to learn "How to memorise different values of trigonometric angles in various quadrants" forever. Before this we must have knowledge of different trigonometric values of different angles in different quadrants.

When angle lies in 1st Quadrant

(1) When angle lies in 1st quadrant, then all the t- Ratios have positive values. As in 1st quadrant all the three  sides Perpendicular ,base and hypotenuse of  right angled triangle are positive.

      

(2) When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan  x to cot x , cos x to sin x ,cosec c to sec x and sec  x to cosec x.

sin (π/2 - x)     =  cos x

cos (π/2 - x)    =   sin x
tan (π/2 - x)     =  cot x
cot (π/2 - x)     =  tan x
sec (π/2 - x)     =  cosec x
cosec (π/2 - x) =  sec x

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x  to sin x  , cos x  to cos x  , tan x  to tan  x and so on.

sin (2π + x)      =    sin x
cos (2π + x)    =     cos x
tan (2π + x)      =    tan x
cot (2π + x)     =     cot x
sec (2π + x)     =     sec x
cosec (2π + x) =     cosec x

ALSO READ   MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS .

When angle lies in 2nd  Quadrant

(1)  when angle lies in 2nd quadrants ,then only two t - ratios sin x and it reciprocal cosec x shall have +ve values and remaining t-Ratios shall have -ve values. As in 2nd  quadrant two  out of the three  sides Perpendicular and  hypotenuse of  right angled triangle are positive  while base is negative. So in all those  t -ratios ,when base involves  they will be negative. So  cos x, tan x ,cot x ,sec x involve with -ve value of base therefore these t- ratios shall be negative.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x to cosec x.

sin (π/2 + x)     =  sin x

cos (π/2 + x)    = -cos x
tan (π/2 + x)     =- cot x
cot (π/2 + x)     = -tan  x
sec (π/2 + x)     =  -cosec x
cosec (π/2 + x) = sec x

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (π - x)      =   sin x
cos (π - x)    =     cos x
tan (π - x)      = - tan x
cot (π - x)     =  - cot x
sec (π - x)     =   sec x
cosec (π - x) =  cosec x

ALSO READ EASY TRICKS TO MEMORISE A B AND C D FORMULAS IN TRIGONOMETRY .

When angle lies in 3rd Quadrant

(1)  when angle lies in 3rd quadrants ,then only two t - ratios tan x and it reciprocal cot x shall have +ve values and remaining t-Ratios shall have -ve values. As in 3rd  quadrant two  out of the three  sides Perpendicular and  base of  right angled triangle are negative  and hypotenuse is positive. So in all those  t -ratios ,when one of perpendicular or base  involves  they will be negative. So  sin x, cos x , sec x ,cosec x involve with -ve value of base or perpendicular therefore these t- ratios shall be negative. And tan x and cot x involves with both -ve values of perpendicular and base so they are positive in 3rd quadrant.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x  to cosec x.

sin (3π/2 - x)     = - cos x
cos (3π/2 - x)    = -  sin x
tan (3π/2 - x)     =    cot x
cot (3π/2 - x)     =    tan x
sec (3π/2 - x)     = - cosec x
cosec (3π/2 - x) = - sec x


(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (π + x)      =   -sin x
cos (π + x)    =    - cos x
tan (π + x)      =   tan x
cot (π + x)     =    cot x
sec (π + x)     =  - sec x
cosec (π + x) =  - cosec x

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Let us understand these learning of trigonometric formulae with the help of this video

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When angle lies in 4th Quadrant

(1)  when angle lies in 4th quadrants ,then only two t - ratios cos x and it reciprocal sec x shall have +ve values and remaining t-Ratios shall have -ve values. As in 4th quadrant two out of the three sides Base and hypotenuse of right angled triangle are positive and perpendicular is negative. So in all those t -ratios ,when perpendicular involves they will be negative. So sin x, tan x ,cot x ,cosec x involve with -ve value of perpendicular therefore these t- ratios shall be negative.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x  to cosec x.

sin (3π/2 + x)     =   cos x
cos (3π/2 + x)    = - sin x
tan (3π/2 + x)     = - tan x
cot (3π/2 + x)     = - cot  x
sec (3π/2 + x)     =  - sec x

cosec (3π/2 + x) =  cosec x

So if we want to calculate sin 300° ,sin 240° and sin 330°  then it can be find out as follows

sin 300° = sin (270° + 30° ) = - cos 30° = -√3/2

sin 330° = sin (360° - 30° )  = - sin 30° = -1/2

sin 240° = sin (270° - 30° ) = - cos 30° = -√3/2

and if we want to calculate cos 300° , cos 240° and   cos 330°  then it can be find out as follows

cos 300° = cos (270° + 30° ) =  sin 30° = 1/2

cos 330° = cos (360° - 30° )  =  cos 30° = √3/2

cos 240° = cos (270° - 30° ) = - sin 30° = 1/2

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (2π-x)      =   - sin x
cos (2π-x)    =      cos x
tan (2π-x)      =  - tan x
cot (2π-x)     =   - cot x
sec (2π-x)     =     sec x
cosec (2π-x) =  -  cosec x

sin (-x)      =   - sin x
cos (-x)    =      cos x
tan (-x)      =  - tan x
cot (-x)     =   - cot x
sec (-x)     =     sec x
cosec (-x) =  -  cosec x

GENERALISATION OF THE FORMULAE

So when an angle involves integral multiple of π , i,e -3π, -2π, -π, 2π, 3π, 4π then of the T-Ratios will change , But + or - sign can be added at beginning , e. g . sin(nπ + x) may change to + sin x or - sin x ,similarly cos (nπ + x) may change to + or - cos x depending upon the quadrant in which angle lies.


So if we want to find the value of sin 1110° ,then it can be written as sin (3×360° + 30°) = sin 30° = 1/2 . (As the angle is lying in 1st Quadrant )

Similarly if we want to find the value of sin 1050° ,then it can be written as sin (3×360° - 30°) = - sin 30° = - 1/2. (As the angle is lying in 4th Quadrant )


and when an angle involves odd integral multiple of π/2 i.e. (2n+1)π/2 , i . e -7π/2 , -5π/2 , -3π/2 , π/2 , 3π/2 , 5π/2.

Conclusion

Thanks for devoting your valuable time for the post memorising different values of trigonometric angles in different quadrants of this blog. If you found this  blog/post of your concern, Do Follow me on my blog and share  post with your friends . We shall meet again in next post ,till then Good Bye.

ALSO READ HOW TO SOLVE LINEAR EQUATIONS OF TWO AND THREE VARIABLES || MATRIX METHOD TO SOLVE SYSTEM LINEAR EQUATIONS .

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