Solve Graphing Trigonometric Functions

Graphs of Trigonometric Functions


The relation (variation) between the angles and the values of the trigonometric ratios at them are plotted by graphs .

Pro 1:    Graph of  y  =  sinx

x    -`pi`    `-pi/2`    0    `pi/2`    `pi`    `(3pi)/2`    `2pi`    `(5pi)/2`    `3pi`
y    0    -1    0    1    0    -1    0    1    0
Plot the graph in the coordinate plane by taking angles x in radian measure on X-axis and the values of sinx  =  y on Y-axis .

Sol :



By choosing a suitable scale , plot and join the points of  y = sin x with a smooth curve to get the graph .

This curve passes through the origin . The values of sin x vary between -1 and +1 which are respectively the minimum and the maximum . It is in the shape of a wave whose wave length is 2`pi` . This wavelength is nothing but the period .

SInce  `-1<=sinx<=1`   `AA`  x  `in`  R , the sine function is bounded . It can be proved that it is a continuous function on  R .

Pro 2:  Graph of  y  =  cosx

x    `-pi`    -`pi/2`    0    `pi/2`    `pi`    `(3pi)/2`    `2pi`    `(5pi)/2`
y    -1    0    1    0    -1    0    1    0

Sol :



By choosing a suitable scale , plot and join the points of y = cos x with a smooth curve to get the graph .

This curve does not pass through the origin . It is evident that the maximum and the minimum values are  +1 and -1 respectively .

Since  `-1<=cosx<=1`   `AA`  x  `in`  R  ,  the cosine function is bounded . It can be shown that is a contnous function and periodic with `2pi` as the period .

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Solve Graphing Trigonometric Functions : Tanx and Cotx

Pro 3:  Graph of  y  =  tanx

x    `-pi/2`    0    `pi/2`    `pi`    `(3pi)/2`
y    not defined    0    not defined    0    not defined

Sol :



The curve nearly  touches the vertical lines  at  x  =  ...........`-pi/2`  , `pi/2` , `(3pi)/2` , . .. . . . .

The curve has nreakes at  x  =  (2n + 1)`pi/2`   ,  n  `in`  Z  and passes through the origin . it is not bounded .

The tan function is periodic and `pi`  is the period of it .

Pro 4:  Graph of  y  =  cotx



The curve nearly touches the vertical lines at  x  =  . . . . . . . `-pi` , 0 , `pi`  . . . . . . .  .

The curve has breakes at  x  =  n`pi`   ,    n  `in`   Z   and does not pass through the origin . It is not bounded .

The cot function is periodic and  `pi`  is the period of it .

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Solve Graphing Trigonometric Functions : Secx and Cosecx

Pro 5:   Graph of  y = secx



The curve nearly touches the vertical lines at x  =  . . . . . . . . `-pi/2` , 0 , `pi/2` , `pi` , `(3pi)/2` . . . . . .

The curve has breakes at  x  `in`   (2n+1) `pi/2` ,  n `in`  Z . It is not bounded . The values  of  secx  lie  in  `(oo,-1]uu[1,oo)`

The secant function is periodic and 2`pi`  is the period of it .

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Pro 6:   Graph of  y  =  cosecx



The curve nearly  touches the vertical lines at  x  =  . . . . . . 0 , `pi/2` , `pi` , `(3pi)/2` , 2`pi` , . . . . . .

The curve has taken

Multiplication and Division of Whole Numbers

Division:

For numbers, the operation of assert how many times one number, the divisor, is contained in a second, the dividend. The result is known as quotient and if the divisor is not contained an integral number of times in the dividend; any number left over is called the remainder. The symbol of division is / or ÷.

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Multiplication:

The product of two numbers is known as multiplication. Here we are going to learn about some example problems of multiplication and division whole numbers.

Example Problems of Multiplication Whole Numbers:

Example 1:

12*14

Step 1: First we need to take 4 and multiply with 12 = 48.

Step 2: Now we need to take 1 and multiply with 12 = 12.

Step 3:

Step 4: Therefore the answer is 168.

Example 2:

23*42

Step 1: First we need to take 2 and multiply with 23 =46.

Step 2: Now we need to take 4 and multiply with 23 =92.

Step 3:



Step 4: Therefore the answer is 966.

Example 3:

231* 14

Step 1: First we need to take 4 and multiply with 231 = 924.

Step 2: Now we need to take 1 and multiply with 231 = 231

Step 3:



Step 4: Therefore, the answer is 3234.

These are the examples problems of multiplication of whole numbers.

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Examples of Dividing Whole Numbers:

Example 1:

Solve: `(143)/(3)`

Solution:

Step 1: Here we need to perform division operation `(143)/(3)` .

Step 2: First we need to take first two digits that is 14.  And we know that 4 times 3 = 12.

Step 3: Now we need to minus 12 from 14 = 2.

Step 4: Here we need to get the next term.

Step 5: The next term is 3. So presently we have 23.

Step 6: We know that 7 times 3 = 21.

Step 7: So the quotient is 47 and remainder is 2.



Example 2:

Solve:  `(276)/(12)`

Solution:

Step 1: Here we need to perform division operation `(276)/(12)` .

Step 2: First we need to take first two digits that is 27.  And we know that 2 times 12 = 24.

Step 3: Now we need to minus 24 from 27 = 6.

Step 4: Here we need to get the next term.

Step 5: The next term is 6. So presently we have 36.

Step 6: We know that 3 times 12 = 36.

Step 7: So the quotient is 23 and remainder is 0.



These are the examples of division of whole numbers.

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Triangles Venn Diagram

Venn diagrams are used to describe the relationship of the sets. The description of some sets is given and you are asked to draw a Venn diagram to illustrate the sets. You are well known about the triangles that they are having three sides. Based on their sides and the angles, it is classified into different types. The relationship of the triangles can also be described using Venn diagrams.

Triangles and Venn Diagrams

Venn diagrams:

Venn diagrams to determine the relationships between the sets such as subsets and intersections.

Venn diagram for A within B :



Here, all members of A belongs to B or A ⊂ B or A ∪ B = B or A ∩ B = A or n(A ∩ B) = n(A).

Venn diagram for  A overlap B:



Here, some members of A belongs to B or A ∩ B ≠ Ø or n(A ∩ B ) ≠ 0

Venn diagram for disjoint sets of A and B :



Here, no members of A belongs to B or A ∩ B = Ø or n(A ∩ B ) = 0

Triangles:

scalene triangle:

A triangle is said to be scalene triangle, if all the sides are of different lengths.Some scalene triangles are also right triangles.



Right triangle:

A triangle is said to be right triangle, if one interior angle is exactly 90°. Some right triangles are also scalene triangles.



Equilateral triangle:

A triangle is said to be equilateral triangle, if all the sides are of same length.



Isosceles triangle:

A triangle is said to be isosceles triangle, if two sides are of same length and two angles are equal.



Examples

Let U is the set of triangles, A is the set of isosceles triangles, B is the set of equilateral triangles and C is the set of right-angled triangles.

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Draw a Venn diagram.

Solution:

We have to determine the relationships between the sets.

All the equilateral triangles are said to be isosceles triangles, so B ⊂ A. (within)

Some of the right-angled triangles are said to be isosceles triangle. C ∩ A ≠ Ø (overlap)

No right-angled triangles are equilateral triangles. C ∩ B = Ø (disjoint)

Venn diagram for the triangles:



Let U is the set of triangles, S is the set of scalene triangles and R is the set of right triangles.

Draw the Venn diagram

Solution:

The relationship between scalene and right triangle to be defined.

Some of the right triangles are scalene triangles.

Integer Divide by Zero

Integer is the number which is greater than the zero or less than the zero and a number greater than zero is a positive and less than the zero is called negative. Zero has no sign like positive and negative sign. In number line two integers are same distance from the zero in opposite directions are the opposites.In this section we are going to see about any integer which is divided by zero and the integer divided the integers.


Division Property of Integers:

For integers we have some properties, let assume a, b be the two integers for that positive and negative integers. But with the zero it has some special property any integer which divided by the zero we get infinity and the zero divided by the any integer we get zero.


Let a & b be the two integers, where  `a/b` is not always an integer.

Examples: `(-4)/5` ; ` 5/ (-3)`  are not integers ;  `1/2 ` is an integer.

For any integer 'a' is not equal to zero.  `a/a ` = 1; and  `a/1` = a

Examples:
`5/5` = 1

`5/1` = 5

` 1/5`  is not  an integer

For any integer 'a' is not equal to zero.
`a/(-1)` = -a; ` a/(-a)` = -1.

For every non zero integer a; `0/a` = 0.

Examples:

` 3/ (-1)` = -3

`3/ (-3)` = -1

`1/ (-3)` is not an integer

Algebra is widely used in day to day activities watch out for my forthcoming posts on Integers and Absolute Value and greatest integer function. I am sure they will be helpful.

Problems with Integer Division:

Problem 1:
Divide +95 by 5.

Solution:
=    ` 95/5`

=19 is an integer


Problem 2:
Divide  -65 by 0.

Solution:
We know that the any integer divided by 0 is infinity by division property. Here, one twenty five by zero
= ∞

Problem 3:

Divide 106 by 0.

Solution:
We know that the any integer divided by 0  is infinity by division property. Here,thousand divided by zero.

= ∞

Problem 4:

Divide 0 by 55

Solution:
We know that the zero divided by any number is zero by division property

=0

Fraction Chart for Math

Fraction can be defined as part of something bigger than that. Suppose your class has 40 boys and 60 girls thus totaling 100 students. Boys are the part of the class and so are girls. But the class consists both boys and girls and is bigger set. In the class, we say 40 out 100 are boys. We can write this as a Fraction as below



We observe that number 40 is written at the top and then a bar below it and then number 100 below the bar. This is how the fractions are represented.

The boys in the class form a part of the class and they are 40 in number. The class is bigger and has 100 students. So, the part here is the boys and the whole is the class. So, this fraction represents that 40 boys are a part of a class of 100 students. Thus we see, fraction represents the part of a whole.

It must also be noted that there could be fractions where the part can be bigger than the whole.

Fractions Illustration

Let us illustrate how fractions can be formed as a part of the whole. Let us a draw a full circle first as below
 

Now let us divide the circle into two different parts and shade one part green as shown below



So we have total two parts and one part is green. This is represented as ½.

Now let us divide the original circle into three and shade two parts of it green as below



So we have total three parts and two parts are green. This is represented as `(2)/(3)`

 Thus fractions represent a part of the whole.

As we saw a fraction has one number at the top of a bar and one number below it. The top number is called the numerator and the bottom number is called the denominator as shown below.



The chart below shows fractions in pictorial form



Types of Fractions

• If the numerator is less than the denominator, it is called a proper fraction.

All the fractions we saw above like `(1)/(2)` , `(2)/(3)`  etc are proper fractions because the top number or numerator was smaller then the bottom number or denominator

• If the numerator is greater than the denominator, it is called an improper fraction.

Example. `(4)/(3)` , `(7)/(2)`  etc. These fractions represent where the part is greater than the whole. We will illustrate with an example. Suppose your teacher is conducting an examination for you. The examination had 25 questions and the teacher asked you to solve any 20 questions to get the maximum mark of 20. Suppose you managed to solve all the 25 correctly then your score in the test will be `(25)/(20)` . That means that you solved more than the maximum required.

• There are another kind of fractions, which has both a whole number and a proper fraction. These fractions are called mixed fractions.

Example 3 `(1)/(2)` ,4`(2)/(3)`  etc.  Thus these fractions are a mix of whole number and a proper fraction and hence they are called mixed fractions.

Exercises on Fractions

Pro 1: From the figure below, write the region shaded in the circle as a fraction of the whole circle

 i)



ii)





iii)



Ans : (i) `(2)/(10)` (ii) `(5)/(6)`       (iii) `(7)/(10)`

Pro 2: Identify the proper, improper and mixed fractions from the following

`(7)/(8)`  ,7 `(1)/(2)`   , `(16)/(2)` , `(1)/(8)` , `(4)/(3)`

Ans: Proper fractions - `(7)/(8)` ,`(1)/(8)`   - Numerator of these fractions are smaller then the denominator

Improper Fraction - `(16)/(2)` , `(4)/(3)`  - - Numerator of these fractions are bigger then the denominator

Mixed Fraction - 7 `(1)/(2)`   - This fraction has  a whole number and a proper fraction

Probability of a Intersection B

Probability of A Intersection B

Probability is the possibility of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0 (impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 6 when rolling a dice is 1/6. In this lesson we will discuss about probability problems using intersection rule.

Probability of a Intersection B – Example Problems

Example 1: A jar contains 5 red candies, 4 orange candies. If three candies are drawn at random, find the probability, that 1 is red candy and 2 are orange candies

Solution:

We have to select 3 candies, from 9 (5 + 4) candies.

n(S) = 9C3 = (9!)/(3!xx6!) = (9xx8xx7)/(3xx2xx1) = 84

Let A = Event of getting 1 red candy

B = Event of getting 2 orange candies

n(A) = 5C1 = `(5!)/(1!xx4!) ` = `5/1` = 5

n(B) = 4C2 =` (4!)/(2!xx2!) ` = `(4xx3)/(2xx1)` = 6

P(A) =` (n(A))/(n(S))` = `5/84`

P(B) = `(n(B))/(n(S))` = `6/84`

P(A intersection B) = P(A) ∙ (B) = `5/84` ∙ `6/84` = `30/7056`

P(A intersection B) = `5/1176` .



Example 2: A box contains 6 yellow marbles, 6 orange marbles. If four marbles are drawn at random, find the probability, that 2 are yellow marbles and 2 are orange marbles.

Solution:

We have to select 4 marbles, from 12 (6 + 6) marbles.

n(S) = 12C4 = `(12!)/(4!xx8!)` = `(12xx11xx10xx9)/(4xx3xx2xx1)` = 495

Let A = Event of getting 2 yellow marbles

B = Event of getting 2 orange marbles

n(A) = 6C2 = `(6!)/(2!xx4!)` = `(6xx5)/(2xx1)` = `30/2` = 15

n(B) = 6C2 = `(6!)/(2!xx4!)` = `(6xx5)/(2xx1)` = `30/2` = 15

P(A) = `(n(A))/(n(S)) ` = `15/495` = `1/33`

P(B) = `(n(B))/(n(S)) ` = `15/495` = `1/33`

P(A intersection B) = P(A) ∙ (B) = `1/33` ∙ `1/33` = `1/1089`

P(A intersection B) = `1/1089` .

Probability of a Intersection B – Practice Problems

Problem 1: A jar contains 4 lemon candies, 4 orange candies. If two candies are drawn at random, find the probability, that 1 is lemon and 1 is orange candy.

Problem 2: If P(A) = `1/5` , P(B) = `1/7` , P(A or B) = `1/9` , find (A and B)?

Answer: 1) `1/49 ` 2) `73/315`

Area of a Square Inscribed in a Circle

A square is a four sided figure; all the four sides are equal. If we situate a square inside the circle, all the four edges or vertices may touches the circle. A square is a quadrilateral, if the four sides of a square touch the circle, the four sides act as a four chords of a circle. The area of a square is measured in square units such as feet, inch, meter etc. In this article we shall discuss about the area of a square inscribed in a circle.

Area of a Square Inscribed in a Circle

If a square is inscribed in a circle and four sides of a square touches a circle.



The diagonals of the circle act as a diameter of the circle. We can find the area of a square is by:

Area of square = side times side

Area of the square = side side

= s2

Example:

Find the area of the square, sides of the square is 9ft.

Solution:

Given: Side = 8ft therefore diameter = 9ft

Area of a circle = side × side

= 9 × 9

= 81square ft.

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If the square does not touch the circle but inscribed in a circle, In such case we can also find the area of the square.

Example:

Find the area of a square inscribed in a circle, side of the square is 6ft.

Solution:

Given: side = 6t

Area of a square = side × side

= 6 × 6

= 36square ft.

The one or two sides of a square may touch the circle.

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Example for Area of a Square Inscribed in a Circle:

Example1:

Find the area of the square inscribed in a circle, sides of the square is 12m.

Solution:

Given: Side = 12

Area of a square = side × side

= 12 × 12

= 144m2.

Example2:

Find the area of the square inscribed in a circle, sides of the square is 23ft.

Solution:

Given: Side = 23ft

Area of a square = side × side

= 23 × 23

= 529square ft.