Circular Functions in Trig

In mathematics, the trigonometric functions (also called circular functions) are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle. The most familiar trigonometric functions are the sine, cosine, and tangent. The sine function takes an angle and tells the length of the y-component of that triangle (Source: Wikipedia)

Circular function in trig

Right triangle for all time contain a 90° angle, now denoted at C. Angles A and B might differ. Trigonometric functions identify the relations between side lengths also interior angles of a right triangle.
Illustrate the trigonometric functions used for the angle A.
Establish through either right triangle to include the angle A.
Three sides of the triangle are,
Adjacent side:
It is the side specifically in contact through both the angle we are concerned within (angle A).
Opposite side:
It is the side opposite toward the angle we are concerned within (angle A)
Hypotenuse side:
It is the side opposed the right angle. The hypotenuse is the highest part of a right-angled triangle.



Trigonometric functions:
 Sin A = opposite / hypotenuse.
Cos A = adjacent/ hypotenuse
Tan A = opposite /adjacent
Csc A =1/Sin A= hypotenuse/ opposite
Sec A =1/Cos A= hypotenuse/ adjacent
Cot A =1/Tan A adjacent / opposite
The circular function has the following equations
Let u = x + y and v = x - y. Then the three equations yields the sums otherwise differences
sin(u) + sin(v) = 2 sin(`(u + v) / 2` ) cos(`(u - v) / 2` )
cos(u) + cos(v) = 2 cos(`(u + v) / 2` ) cos(`(u - v) / 2` )
cos(v) - cos(u) = 2 sin(`(u + v) / 2` ) sin(`(u - v) / 2` )

Examples for circular function in trig

Example 1 for circular function:
Apply the tangent ratio toward solve the unidentified side of the triangle?
Given angle of triangle are 40⁰ and opposite side of triangle are 16.
Solution:
Specified to angle of triangle is 40⁰ and opposite side of triangle is 16. Find out the adjacent side
 tan 40⁰ = opposite/adjacent
tan 40⁰ x = 16
x = 16/tan 40⁰
x =16/0.8391     {since the value of tan 40 degree is 0.8391}
x=19.06
The value of adjacent side =19.06
Example 2 for circular function:
If u=30 and v=30 then find out the sin (u) + sin (v)?
Solution:
We know the formula,
sin (u) + sin(v) = 2 sin(`(u + v) / 2` ) cos (`(u - v) / 2` )
                         =2 sin(`(30+30)/2` ) cos(30-30)
                         =2 sin(60/2) cos 0
                         =2 sin30 cos 0
                         =2x(1/2)x1
 sin (u) + sin(v) =1

Triangular Pyramid Geometry

In geometry, a pyramid is a polyhedron formed by connecting a polygonal base and a point, called the apex. Each base edge and apex forms a triangle. It is a conic solid with polygonal base. A pyramid with an n-sided base will have n + 1 vertices, n + 1 faces, and 2n edge.
Source from wikipedia)

geometry triangular pyramid:

A triangular pyramid is a pyramid having a triangular base.Here we are going to study about triangular pyramid in geometry and example problems.

Formulas:

Surface area of triangular pyramid = A + `(3/2)` s*l
Here A = area of base = `1/2` * a*s
volume of triangular pyramid = `(1/6)` a*b*h

Here,

a represent the apothem

s,b is the sides of the pyramid

l is the slant height

h is the height of the pyramid.

Triangular pyramid geometry - example problems.

Example: 1

Find the surface area of the triangular pyramid with side 7 meter ,slant height is 9 meter and apothem length is 8 meter.

Solution:

We know the formula for surface area of the triangular pyramid = A + `(3/2)` sl

First we have to find the area of base A = `1/2` * a*s

Here a = 8 meter, s = 7 meter substitute the above formula we get

A = `1/2 ` * 8 * 7

A = `1/2` *56

A = 28 meter square

Now we find the surface area of the triangular pyramid

S.A = 28 + `(3/2)` * 7 * 9

= 28 + `(3/2)` *63

Simplify the above we get

= 28 + `189/2`

= 28 + 94.5

= 122.5 meter square

Therefore the surface area of the triangular pyramid = 122.5 meter square


Triangular pyramid geometry - example: 2

Find the volume of the triangular pyramid with height is 11 meter ,base is 9 meter and apothem length is 12 meter .

Solution:

We know the formula for volume = `(1/6)` abh

Here h = 11 meter, b = 9 meter, a = 12 meter substitute this value into the formula we get

Volume =  ` (1/6)` 12 * 9 * 11

Volume = `(1/6) ` 1188

Simplify the above we get

Volume = 198 meter3

Therefore the volume of the triangular pyramid is 198 meter3

Mixed Number Multiplication

Mixed number is a combination of a whole number and a proper fraction. A mixed number can be converted into an improper fraction and vice versa. All of the arithmetic operations can be performed by using the mixed numbers. For example, 2 (1)/(2) is a mixed numbers. In this example, the 2 is called as the whole number and (1)/(2) is called as the improper fraction

It will be easier to perform arithmetic operations on mixed numbers by first converting them to improper fractions

Rules for the multiplication of mixed number

There are many rules are followed for the multiplication of mixed numbers, They are given as follows,

In the first step, we have to convert the given mixed number into the improper fraction.
In the next step, we have to perform the multiplication for the numerator part
In the next step, we have to perform the multiplication in the denominator part
Then, the next step we have put this in an improper fraction.
Then the next step is simplify those improper fractions.
Then the result is converted into a mixed number.


Example problem for the mixed number multiplication

Problem 1: Multiply the given mixed numbers, 2 (4)/(6) with 1  (6)/(9) .

Solution:

Step 1: Write the mixed numbers

2(4)/(6) xx 1 (6)/(9)

Step 2: Convert the given mixed numbers into improper fractions

(16)/(6) xx (15)/(9)

Step 3: Perform the multiplication operation for the numerator terms,

16 xx 15 =240

Step 4: Perform the multiplication operation for the denominator terms,

6 xx 9 =54

Step 5: The result obtained for step 3 and 4 can be combined to an improper fractions

(240)/(54)

= (120)/(17)

This is the required solution for the multiplication of mixed numbers.

Problem 2: Multiply the given mixed numbers, 4 (2)/(7) with 7  (2)/(8) .

Solution:

Step 1: Write the mixed numbers

4 (2)/(7) xx 7 (2)/(8)

Step 2: Convert the given mixed numbers into improper fractions

(30)/(7) xx (56)/(8)

Step 3: Perform the multiplication operation for the numerator terms,

30 xx 56 = 1680

Step 4: Perform the multiplication operation for the denominator terms,

7 xx 8 = 56

Step 5: The result obtained for step 3 and 4 can be combined to an improper fractions

(1680)/(56)

= (840)/(28)

=(420)/(14)

=(210)/(7)

= 30

This is the required solution for the multiplication of mixed numbers.

Differentiate Variable

On the definition of the derivative is a dynamical diagram displaying the derivative as the slope of the tangent to a graph. The necessary preliminaries one should be familiar with are the slope of a straight line and the graph of a function.

When we differentiate any equation we denote them using different variables .these variables can be one or many depending on the equation. These variables are alphabets like x, y, t, u, v,  etc.


Examples of differentiate variables

There are many examples of differentiation. Such as:

Differentiate xx  + (cos x)1/2 with respect to x

Let y =  xx  + (cos x)1/2

We have + or – sign in between the terms, then we should not tale log.

We should write y = u + v

dy/dx = du/dx + dy/dx                                            …..(i)

consider  u = xx

taking log on both sides, we get

log u = x. log x

differentiating with respect to x, we get

1/u . du/dx = x.d/dx (log x) + log x.dx/dx

1/u. du/dx = x.1/x + log x.1

du/dx = u(1 + log x) = x2( 1 + log x)                 …..(ii)

again consider  v = (cos x)1/x

taking log on both sides, we get

log v = 1/x.log(cos x)

differentiating with respect x, we get

1/v.dv/dx = 1/x.d/dx log (cos x) + log(cos x).d/dx(1/x)

1/v.dv/dx = 1/x. 1/ cosx. (-sin x) + log(cos x). (-1/x2)

dv/dx = v[- tanx /x – 1/x2. Log (cos x)]

= (cos x)1/x[- tan x/x – 1/x2.log(cos x)]                    ……(iii)

Substituting from (ii) and (iii) in (i), we get

dy/dx = x2(1+ log x) + (cos x)1/x [-tanx/x – 1/x2. Log(cos x)]

here the variable used are x, y , u and v, and these are called the differentiate variables.


Problems on differentiate variables

Differentiate , y = (x)cos x + (sin x)tan x
y= x.sin x.log x.

Interval Notation Inequality

Interval notation inequality is a process of writting down the set of numbers. Generally interval notation inequalities is used to describe a limit of span or group of spans of numbers along a axis. However, this notation can be followed to describe any group of numbers. For the reasonable example is considered, the set of numbers that are all the numbers greater than 7. The interval notation inequality for this set, letting x be any number in the group of notation, then we it would be given in the notation as shown below,


X >7

Approaches for interval notation inequality:

This similar set possibly will be described in another type of notation is called interval notation inequality. For the collection of numbers that would be written as,

(7,+`oo` )

The interpret notation will be written as:

The span of numbers which as well as in the group is often imagined as being on a number line, normally the x-axis.The 7 on the left then the set of numbers starts at the real number which is suddenly to the right of 7 on the number line. It means we should imagine a number the tinniest bit greater than 7, and that is where the group of numbers begins from the opening stage.
So that the parenthesis to the left of 7 is called a round bracket or an exclusive bracket. Then, 7 is excluded from the group, but the numbers straightly to the right of 5 are included. Shortly put, the numbers greater than 7 are included. The group of interval notation numbers continues to include values greater than 5 all the way to a value which is infinitely greater than 7. That is, the set of all numbers continuously all the way to positive infinity. That is what the positive infinity symbol on the right means, then Infinity symbols are at all times accompanied by round brackets.


Practice problem for interval notation inequality:

Problem 1:

2 < x < 7

The entire numbers between positive two and positive seven, together with the two and the seven.  2 < x < 7,  x is less than or equal to 7 and greater than or equal to 2" { x | 2 < x < 7} "x is between 2 and 7, inclusive" [2,7]

Problem 2:

5 < x < 9

The entire numbers between positive five and positive nine, as well as the five and the nine.  5 < x < 9 x is less than or equal to nine and greater than or equal to five {x | 5 < x < 9} "x is between 5 and 9, inclusive" [5, 9].

Regression Analysis Example

In statistics, regression analysis includes any techniques for modeling and analyzing several variables, when the focus is on the relationship between a dependent variable and one or more independent variables. More specifically, regression analysis helps us understand how the typical value of the dependent variable changes when any one of the independent variables is varied, while the other independent variables are held fixed. Most commonly, regression analysis estimates the conditional expectation of the dependent variable given the independent variables — that is, the average value of the dependent variable when the independent variables are held fixed.

Source: Wikipedia.


Formulas for regression:

It is the process of finding the relationship between two variables. It is a statistical analysis which is used for the assessing the association between the two variables.

Formula for Regression:

Regression Equation:

y = a + bx

Slope:

b = (NΣXY - (ΣX) (ΣY)) / (NΣX2 - (ΣX) 2)

Intercept:

a = (ΣY - b (ΣX)) / N


Where
x and y are the variables.
b = the slope of the regression line
a = the intercept point of the regression line and the y-axis.
N = Number of values or elements
X = First Score
Y = Second Score
ΣXY = Sum of the product of first and Second Scores
ΣX = Sum of First Scores
ΣY = Sum of Second Scores
ΣX2 = Sum of square First Scores.

Using this regression formula, Find the Refression equation for the given example problem.

Example problem for regression analysis:

Some example problems for regression analysis

Example 1:

For the given set of x and y values, determine the Linear Regression and also find the slope and intercept and use this in a regression equation.
X Values Y Values
60   2.5
61   2.7
62   2.9
63   3.2
65   3.5


Solution:

Let us count the number of values.
N = 5
Determine the values for XY, X2
X Values Y Values X * Y X * X
60  2.5 150 3600
61  2.7 164.7 3721
62  2.9 179.8 3844
63  3.2 201.6 3969
65  3.5 227.5

4225

Determine the following values ΣX, ΣY, ΣXY, ΣX2.
ΣX = 311
ΣY = 14.8
ΣXY = 923.6
ΣX2 = 19359

Substitute values in the slope formula
Slope (b) = (NΣXY - (ΣX) (ΣY)) / (NΣX2 - (ΣX) 2)
= ((5)*(923.6) - (311)*(14.8)) / ((5)*(19359) - (311)2)
= (4618 – 4602.8)/ (96795 - 96721)
= 15.2 / 74
= 0.20

Substitute the values in the intercept formula given.
Intercept (a) = (ΣY - b (ΣX)) / N
= (14.8 - 0.20 (311))/5
= (14.8 – 62.2)/5
= -47.4 /5
= -9.48

Substitute Slope and intercept values in the regression equation
Regression Equation(y) = a + bx
= -9.48 + 0.20x.

Algebra is widely used in day to day activities watch out for my forthcoming posts on How to Create a Stem and Leaf Plot and Chain Rule Second Derivative. I am sure they will be helpful.

Determine the approximate value for y:

When x = 64

Substitute the x value into the regression equation

Regression Equation(y) = a + bx
= -9.48 + 0.20x.

= -9.48 + 0.20(64)

= -9.48 + 12.28
y = 3. 32.

Slope of a Line

Slope of  aline is the inclination of the lien with the horizontal. The slope of  ahorizontal line is zeroa nd that of a vertical line is infinity. Slope of a line can be positive or negative depending on the direction of inclination. By definition, slope of a line is the tangent of the angle between the line and the positive x-axis measured counter clockwise.

The slope of the line is one of the factors in the equation of a line. When the equation of the line given by the slope intercept form that is y=mx+b. here m is the slope of the line and b is the constant term . from  this equation we can find the slope of the line and the x and y intercept of the line.
The equation is y=mx+b, m is the slope. This form of a line's equation is slope intercept form, because b acts as the y-intercept of the line, it is a point  where the line meets the y-axis.


Formula for slope (slope (m) and point are known) in algebra

If the slope of a line and a point (x1, y1) on the line are both known, then the equation of the line can be found using the point-slope formula:

y- y1= m (x-x1)

Example problems:

1.what is the slope-intercept form of the line that contains one point and a slope that is  (-1, 4) with a slope 1/4.

Solution:

y - y1 = m(x - x1)

It goes through the point (-1,3) and has a slope of 1/4.  So plug in  this values into the point-slope formula gives:

y - 4 = `1/4(x - (-1))`
y - 4 = `1/4(x + 1)`
y - 4 = `x/4 + 1/4`
y = `x/4 + 1/4 + 4`
y = `x/4 + 1/4 + 16/4`
y = `x/4+ 17/4 `
The final answer  is in slope-intercept form of the given line , which is y = mx + b, where m is the slope and b is the y-intercept or constant

model problems to slope formula in algebra

write the slope-intercept  of the line that passes through (7, 0) that is perpendicular to the line x + y = 4.

Solution:

If two lines are perpendicularto each other  then their slopes are negative reciprocal to each other.  So find the slope of the line given:
x + y = 4
y = -x + 4
.

So the slope is -1.  The minus reciprocal of -1 is 1.  So put that into the point-slope formula:

y - 0 = 1(x - 7)
y = x – 7.

Multiplying Integers

In this page we are going to discuss about multiplying integers concept. The integers are formed by the natural numbers including 0 (0, 1, 2, 3, ...) together with the negatives of the non-zero natural numbers (−1, −2, −3, ...). Viewed as subset of the real numbers, they are numbers that can be written without a fractional or decimal component, and fall within the set {... −2, −1, 0, 1, 2, ...}. For example, 65, 7, and −756 are integers; 1.6 and 1½ are not integers.

Positive and Negative Integers

All the whole numbers larger than zero (1, 2, 3, 4, 5...) are positive integers and it has the positive symbol and every whole numbers lesser than zero (-1, -2, -3, -4, -5 …) are negative integers and it has the negative symbol. We need not assume about zero. Every positive integer has a negative integer. For example, -5 is the opposite of 5.

Multiplication Rules for integers

The following four kinds of rules are placed in multiplication of integers.

• Positive value x Positive value = Positive value

• Positive value x Negative value = Negative value

• Negative value x Positive value = Negative value

• Negative value x Negative value = Positive value

The first rule give details that working with the positive numbers in multiplication always provides positive answers. The second and third rule clarifies that 2 x -3 has same answer as -3 x 2, which is always true for all integers. The fourth rule is applicable only when two values of the negative symbol is multiplied, than the solution will be in positive numbers.

Multiplying integers examples

Below are the examples on multiplying integers  -

Example 1:   Find the Multiplication of the integers: 5 x 6

Solution :  Given (5) × (6)

= |5| × |6|

= 5 × 6

= 30

Example 2: Find the Multiplication of the integers: 4 x (-5)

Solution :     Given 4 x (-5)

= | 4 | × | -5 |

= 4 × -5

= -30

Example 3:  Find the Multiplication of the integers: -3 x (6)

Solution :  Given -3 x (6)

= | -3 | × | 6 |

= -3 × 6

= -18


Example 4: Find the Multiplication of the integers: -4 x (-8)

Solution :   Given (-4) x (-8)

= | -4 | × | -8 |

= -4 × -8

= 32

Conditional Probability Definition

The conditional probability of an event B in related to an event A is the probability that event B happens given that event A has already occurred. The notation for conditional probability is P(B|A), read as the probability of B given A. this formula comes from the multiplication principle and bit of algebra.

P (B | A) = P (A and B) / P (A)

The given event A has occurred; we should a reduced sample space. Instead of the entire sample space S, now we have a sample space of A since we know A has occurred.  So the old thing is about being the number in the event divided by the number in the sample space still applies.

It is the number in A and B (should be in A since A has happened) divided by the number in A. By dividing on right hand side by the number which is present in the sample space S, and then we have the probability of A and B divided by the probability of A. Another important method of conditional probabilities is given by Bayes's formula.

Conditioning on a random variables:

Conditional probability of an event a discrete random variable. Such a conditional probability is a random variable in its belongs right.

Let X is a random variable it can be equal to 0 or to 1. As above, the conditional probability of any event A given the event X = 0, and also of the conditional probability of A given the event X = 1. The former is placed by P (A|X = 0) and the latter P(A|X = 1). Define a new random variable Y, which value is P (A|X = 0) if X = 0 and P(A|X = 1) if X = 1. That is,

Y = P(A | X = 0 ) if X = 0

P(A | X = 1 ) if X = 1

variables in the conditional probability:

Random variable Y is saying that to be the conditional probability of the event A given the discrete random variable X:

Y = P ( A | X )

Conditioning, the probability of A given initial information I, P (A|I), is known as the prior probability and updating conditional probability of A, given I and the outcome of the event B, is known as the posterior probability, P (A|B,I).

Bayes's formula.

It is based on the expression P(B) = P(B|A)P(A) + P(B|Ac)P(Ac),

P(A|B) = P(B|A)P(A) / P(B|A)P(A) + P(B|Ac) P(Ac)