Solving Double Number Identities

Trigonometry is arrived from the Greek word, trigonon = triangle and metron = measure. The father of trigonometry is Hipparchus. He designed the first trigonometric table. Identity is defined as an equation that is true for all probable values of its variables. In online, many websites provide online tutoring using tutors. Double number trigonometric identities problems are easy to solve. Solving double number trigonometric identities problems are easy.  Through practice, students can learn about solving trigonometric identities. Through online, students can practice more problems on trigonometric identities. In this topic, we are going to see about; solving double number identities.

Solving Double Number Identities: - Double Number Identities

The list of double number identities are given below,

sin `2theta` = 2sin `theta` cos `theta `

cos `2theta` = 2cos2 `theta` -1

cos `2theta` = 1- 2sin2 `theta`

cos `2theta` = cos2 `theta` – sin2 `theta`


Solving Double Number Identities: - Examples

Example 1:

Evaluate sin 60 using double number identities,

Solution:

sin 2theta = 2sin theta cos theta

sin 60 = (2x30)

= 2 sin 30 cos 30

= 2 (0.5) (0.866)

= 2*0.433

= 0.866

The answer is 0.866



Example 2:

Evaluate sin 90 using double number identities,

Solution:

sin 2theta = 2sin theta cos theta

sin 90 = (2x45)

= 2 sin 45 cos 45

= 2 (0.707) (0.707)

= 2*0.499

= 1

The answer is 1



Example 3:

Evaluate cos 50 using double number identities,

Solution:

cos 2theta = cos2 theta – sin2 theta

cos 50 = cos (2x25)

= cos2 25 - sin2 25

= (0.906)2 - (0.423)2

= 0.820 - 0.179

= 0.641

The answer is 0.641

Example 4:

Evaluate cos 90 using double number identities,

Solution:

cos 2theta = cos2 theta – sin2 theta

cos 90 = cos (2x45)

= cos2 45 - sin2 45

= (0.707)2 - (0.707)2

= 0.499-0.499

= 0

The answer is 0



Example 5:

Find cos 2y if sin y = -15/16 and in 3rd quadrant

Solution:

It is given that sin 2y is in 3rd quadrant,

Use the double angle identities

cos 2theta = 1- 2sin2 theta

cos 2y = 1 -2sin2 y

= 1 – 2`(-15/16)` 2

= 1 – 2 `(225/256)`

Taking LCM, we get

= `(256-450)/256`

= `-194/256`

= -97/128

The answer is `-97/128`

Sigma Algebra Examples

In mathematics, an σ-algebra is a technological concept for a group of sets satisfy certain properties. The main advantage of σ-algebras is in the meaning of measures; particularly, an σ-algebra is the group of sets over which a measure is distinct. This concept is important in mathematical analysis as the base for probability theory, where it is construed as the group of procedures which can be allocated probabilities. Now we will see the properties and examples.
Properties - Sigma Algebra Examples

Take  A  be some set, and 2Aits power set. Then a subset Σ ⊂ 2A is known as the σ-algebra if it satisfies the following three properties:

Σ is non-empty: There is as a minimum one X ⊂ A in Σ.
Σ is closed below complementation: If X is in Σ, then so is its complement, A \ X.
Σ is closed under countable unions: If X1, X2, X3, ... are in Σ, then so is X = X1 ∪ X2 ∪ X3 ∪ … .

Eg:

Thus, if X = {w, x, y, z}, one possible sigma algebra on X is

Σ = { ∅, {w, x}, {y, z}, {w, x, y, z} }.
Examples - Sigma Algebra

Example 1

X={1,2,3,4}. What is the sigma algebra on X?

Solution:

Given set is X={1,2,3,4}

So Σ = { ∅, {1,2}, {3,4}, {1,2,3,4}}.

Example 2

What is the sigma algebra for the following set ? X={2,4,5,9,10,12}

Solution:

Given set is X={2,4,5,9,10,12}

So   Σ = { ∅, {2,4}, {5,9}, {10,12},{2,4,5,9,10,12}}.

Example 3

11x+2y+5x+12a. Simplify the given equation in basic algebra.

Solution:

The given equation is  11x+2y+5x+12a

There are two related groups are available. So join the groups.

The new equation is,

(11x+5 x)+2y+12a

Add the numbers inside the bracket. We get 16x+2y+12a.

Arrange the numbers and we get the correct format.

=12a+16x+2y

We can divide the equation by 2.

So the equation 6a+8x+y.

These are the examples for sigma algebra.

Generating Function of Exponential Distribution

In mathematics, generating function of exponential distribution is one of the most interesting distributions in probability theory and statistics. The random variable X has an exponential distribution along with the parameter λ, λ> 0. If its probability density function is given by

f(x) = lambdae^(-lambdax) ,         x >= 0

Otherwise, f(x) = 0, x < 0. The following are the moment generating functions and example problems in generating function of exponential distribution.

Generating Function of Exponential Distribution - Generating Function:

Generating function (or) Moment generating function:

Moment generating function of exponential distribution function can be referred as some random variable which contains the other definition for probability distribution. Moment generating function give the alternate way to analytical outcome compare and also it can be straightly working with the cumulative and probability density functions. Moment generating function can be denoted as Mx(t) (or) E(etx). Here we help to calculate the moment generating function for the exponential distribution function.

Mx(t) = E(etx) =int_-oo^ooe^(tx)f(x)dx

= int_0^ooe^(tx)(lambda e^(-lambdax))dx

= λ int_0^ooe^(tx)(e^(-lambdax))dx

= λ int_0^ooe^((t-lambda)x)dx

= λ [e^(((t- lambda)x)/(t- lambda))]^oo_0

= λ [e^(((t- lambda)oo)/(t- lambda)) - e^(((t- lambda)0)/(t- lambda)) ]

Here we use this eoo = 0, and also we use this e0 = 1

= λ [ 0 – 1/(t- lambda) ]

= λ [ -(1/-( lambda-t)) ]

= lambda [1/ (lambda-t)]

= lambda/( lambda-t)

Mx (t) = E (etx) = lambda/(lambda-t)

Generating Function of Exponential Distribution - Example Problem:

Example 1:

If X has probability density function f(x) = e-5x, x > 0. Determine the moment generating function E[g(x)], if g(x) = e11x/15

Solution:

Given

f(x) = e-5x

g(x) = e11x/15

E(g(x)] = E[e11x/15]

= int_0^ooe^((11x)/15)f(x)dx

=  int_0^ooe^((11x)/15) (e^(-5x))dx

= int_0^ooe^(-x(5-(11/15)))dx

= [e^((-x(5-11/15))/-(5-11/15))]^oo_0

=[(e^-oo- (e^0)/-(5-11/15)) ]

= [0+ 1/(5-11/15) ]

= [1/((75-11)/15) ]

= [1/(64/15) ]

= 15/64

E [g(x)] = 0.2344

Answer:

E [g(x)] = 0.2344

Pre Algebra Geometry

Geometry is one of a division of mathematics. Geometry is concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially the body of the practical knowledge is concerning with the lengths, areas, and volumes. In this article we shall discuss about pre algebra geometry problem. (Source: wikipedia)
Sample Problem for Pre Algebra Geometry:

Problem 1:

Find the diameter of the given circle. A given radius value of the circle is 16 cm.

Solution:

Given:

Radius of the circle is r = 16 cm

Diameter of the circle is d = 2r

d = 2 * 16

d = 32

So, the diameter of the circle is 32 cm2.

Problem 2:

Find the diameter of the circle. A given radius value of the circle is 20 cm.

Solution:

Given:

Radius of the circle is r = 20 cm

Diameter of the circle is d = 2r

d = 2 * 20

d = 40

So, the diameter of the circle is 40 cm2.

Problem 3:

A triangle has a perimeter of 34. If the two sides are equal and the third side is 4 more than the equivalent sides, what is the length of the third side?

Solution:

Let x = length of the equal side of a triangle

A formula for perimeter of triangle is

P = sum of the three sides of triangle

Plug in the values from the question

34 = x + x + x+ 4

Combine like terms

34 = 3x + 4

Isolate variable x

3x = 34 – 4

3x = 30

x = 10

The question requires length of the third side.

The length of third side is = 10 + 4 = 14

Answer: The length of third side triangle is 14.
Practice Problem for Pre Algebra Geometry:

Find the diameter of the circle. A given radius value of the circle is 14 cm.

Answer: d = 28 cm2

Find the radius of the circle. A given diameter value of the circle is 11 cm2.

Answer: 5.5 cm

Solving Word Percent Problems

In mathematics, a percentage is a way of expressing a number as a fraction of 100. It is often denoted using the percent sign, "%", or the abbreviation "pct". Percentages are used to express how large/small one quantity is, relative to another quantity. The first quantity usually represents a part of, or a change in, the second quantity, which should be greater than zero. Source: Wikipedia.
Solving Word Percent Problems

Solving percent problems - Example

Example 1: What percent of 30 is 60?

Solution:

30 × `x/100` = 60

30x = 6000

x = 200

Therefore 60 is 200 percent of 30.

Example 2: What is 45% of 250?

Solution:

`45/100`  × 250 = x

45 × 250 = 100x

x = `11250/100` = 112.5

Therefore 45 percent of 250 is 112.5.

Example 3:  25% of what is 15?

Solution:

`25/100` × (x) = 15

25x = 1500

x = 60

Therefore 25 percent of 60 is 15.

Example 4: What is the percentage increase from 14 to 24?

Solution:

Increase = 24 - 14 = 10

Percentage increase = (Change in value / Original value) × 100%

Percentage increase = `10/14` ×100% = 71.42%

Therefore 71.42% increase from 14 to 24.

Example 5: Last month, Anita earned $300. This month she earned $450. Calculate the percentage increase in her earnings.

Solution:

Increase = $450 - $300 = $150

Percentage increase = (Change in value / Original value) × 100%

Percentage increase = `150/300` × 100% = 50%

Therefore percentage increase in her earnings is 50%.

Example 6: What is the percentage decrease from 30 to 14?

Solution:

Decrease = 30 – 14 = 16

Percentage decrease = (Change in value / Original value) × 100%

Percentage decrease = `16/30` × 100% = 53.33%

Therefore 53.33% decrease from 30 to 14.

Example 7: Last month, Anita earned $400. This month she earned $340. Calculate the percentage decrease in her earnings.

Solution:

Decrease = $400 - $340 = $60

Percentage decrease = Change in value / Original value × 100%

Percentage decrease = `60/400` × 100% = 15%

The percentage decrease in her earnings is 15%.
Solving Word Percent Problems

Solving percent problems - Practice

Problem 1: What is 28% of 150?

Problem 2: Harry earned $270 in last month. He earned $370 this month. Calculate the percentage increase in his earnings.

Problem 3: Wilson earned $340 in last month. He earned $250 in this month. Calculate the percentage decrease in his earnings.

Answer: 1) 42 2) 37.03 3) 26.47

Number of Sides in a Pentagon

In geometry, a pentagon is a polygon with five sides. In a simple pentagon the sum of internal angles are about 540°. For example pentagram is a self-intersecting pentagon. Pentagon may be classified into regular and irregular. A pentagon that contains equal sides and equal internal angles are said to be regular pentagon otherwise the pentagon is irregular.pentagon


Number of Sides in a Pentagon:

The term penta indicates 5 .Hence the number of sides in a pentagon are 5 and the number of angles in a pentagon is 5.

A pentagon contains 712.694 million separate parallel lines.

Pentagon doesn’t have parallel lines.

Area of the pentagon = (perimeter of a polygon x apothem) ÷2

Where perimeter of a polygon = 5 x side.

The following are the other polygonal shapes with their sides.

Tetragon - 4 sides

Hexagon- 6 sides

Heptagon- 7 sides

Octagon- 8 sides

Nonagon Enneagon- 9 sides

Decagon-10 sides

Undecagon- 11 sides

Dodecagon- 12 sides

Properties of pentagon:

Number of diagonals:

Number of diagonals in a pentagon is 5

The number of different diagonals possible from all vertices.

Number of triangles:

Number of triangles in a pentagon is 10.

The number of triangles formed by sketching the diagonals from a given vertex.

Sum of interior angles:

Sum of interior angles of a pentagon is 540° in general 180(n–2) degrees.

Example Problem- Number of Sides in a Pentagon:

Example 1:

Find the perimeter of a regular pentagon whose side is 5ft.

Solution:’

Given that, side = 5ft.

For a regular pentagon all the sides are equal.

Therefore the perimeter of a regular pentagon = 5 x side.

= 5 x 5 =25ft.

Example 2:

Find area, from the apothem and the perimeter of a polygon is 4ft and 20ft.

Solution:

Given that, apothem = 4ft.

Perimeter = 20ft

Area of the pentagon = (perimeter of a polygon x apothem) ÷2

= (20 x 4) ÷2.

= 10 x 4.

= 40ft2

Boolean Algebra

Boolean Algebra is a branch of mathematic logics  whose use of symbols and  theory, set to represent the logical operations in the form of mathematics. This is the first logic which uses algebra and different methods for combining symbols used in proofs as well as deduction.

A Boolean Algebra is defined as:

It is  a set, having two special elements i.e, 0 and 1.
Algebra having three types of operations , which are
sum of two elements ("+"),
product sum of two elements  ("*") and
complement sum of two elements (" ' " or "prime")

these operations need to satisfy the Commutative axiom, Distributive axiom , Identity axiom (not including the boundedness identities) and Complement axiom.

These above axioms are almost equal to commutative property ,distributive property, identity property and complement property. Here we call them axioms because they are assumptions.

Boolean Algebra contains:

A  set of all propositions
The special characteristic elements - True  (1) i.e T  and False (0) i.e, F.
Three operations are
AND (product),
OR (sum) and
NOT (complement).

Laws of Boolean Algebra Axioms

To do any kind of operations using real numbers, they  depends on commutative axiom, associative axiom, and distributive axiom. In algebraic form these axioms  are expressed with letters or symbols, which are used to indicate an unknown number.

Commutative axiom

The commutative axioms explains that, numbers can be used for addition or multiplication in any manner.

Commutative axiom of Addition:

a + b = b + a        ( using addition law )

Commutative axiom of Multiplication:

a(b) = b(a)           ( using multiplication law )

Associative axiom

The associative axioms explain that, numbers which are used in addition or multiplication, also it can be grouped or regrouped in anyorder.

Associative Law of Addition:

a+(b+c) = (a+b)+c      ( using addition law )

Associative Law of Multiplication:

a(bc) = (ab)c                ( using multiplication law )

Distributive axiom

The distributive axioms are used for  both addition as well as  multiplication and state the following.

Distributive axiom for addition :

a(b + c) = ab + ac        ( using addition law )

Distributive axiom for multiplication :

(a + b)c = ac + bc             ( using multiplication law )

Identity axiom

Identity axiom for multiplication :

x · x = x                 ( using multiplication law )

Identity axiom for addition :

x + x = x                ( using addition law )



Zero Property in Boolean algebra axioms

0 · x = 0                  ( using multiplication law )

0 + x = x                   ( using addition law )

One Property  in Boolean algebra axioms

1 + x = 1                  ( using addition law )

1. x = x                     ( using multiplication law )
Examples on Boolean Algebra:

1)  `5xx(8+9)` = `5xx8 + 5xx9` ( USING DISTRIBUTIVE AXIOM  FOR MULTIPLICATION )

= `40 + 45`

=` 85`

= 5 x 8+5 x 9

2)  `3+7 = 7+3` ( BY USING COMMUTATIVE AXIOM FOR ADDITION)

3) `9xx5 = 5xx9 (` BY USING THE COMMUTATIVE AXIOM FOR MULTIPLICATION)

4)` 6xx1 = 6 ` (BY USING PROPERTY FOR ONE)
Practice Problems on Boolean Algebra:

1) Prove that  C+(A×B)=(C+A)×(C+B) by using Boolean algebra axioms

2) Prove that B+(C×A)=(B+C)×(B+A) by using Boolean algebra axioms