Adding Cosine Functions

Trigonometry is one of the part of mathematics managing with angles, triangles and trigonometry functions such as sine, cosine, and tangent which are abbreviated as sin, cos, and tan respectively. There are several formulas for adding trigonometric functions such as sine, cosine, and tangents. The formulas for adding cosine functions are used to find the ratios when two angles are given. Those formulas for adding cosine functions are shown below.


cosine (A + B) = cosine A cosine B − sine A sine B.
Proof for Adding Cosine Functions:

To prove: cosine (A + B) = cosine A cosine B − sine A sine B.

Proof: We know that, cosine (A − B) = cosine A cosine B + sine A sine B.

Replacing B with (−B), these identities becomes,

sine (A − (− B)) = cosine A cosine (− B) + sine A sine (− B)

Since we know that, cosine (−A) = cosine A, and

sine (−A) = − sine A.

cosine (A + B) = cosine A cosine B − sine A sine B.

Hence proved that, sine (A – B) = sine A cosine B – cosine A sine B.
Solved Examples for Adding Cosine Functions:

Example 1: If sineA = 14, sineB = 6, cosineA = 12, and cosineB = 8, then find cosine(A + B).

Solution: Given that, sineA = 14, sineB = 6, cosineA = 12, and cosineB = 8.

We know that, cosine (A + B) = cosineA cosineB − sineA sineB,

=> cosine (A + B) = (12 × 8) − (14 × 6),

=> cosine (A + B) = 96 − 84,

=> cosine (A + B) = 12.

Answer: cosine (A + B) = 12.

Example 2: If sineA = 5, sineB = 9, cosineA = 17, and cosineB = 3,  then find cosine(A + B).

Solution: Given that, sineA = 5, sineB = 9, cosineA = 17, and cosineB = 3.

We know that, cosine (A + B) = cosineA cosineB − sineA sineB,

=> cosine (A + B) = (17 × 3) − (5 × 9),

=> cosine (A + B) = 51 − 45,

=> cosine (A + B) = 6.

Answer: cosine (A + B) = 6.

Rules for Division

In mathematics, especially in elementary arithmetic, division (÷) is the arithmetic operation that is the inverse of multiplication.

Understanding Rules of Divisibility is always challenging for me but thanks to all math help websites to help me out

Specifically, if c times b equals a, written:

C = b x a

where b is not zero, then a divided by b equals c, written:

a/b = c

For instance,

6/3 = 2

since

2 x 3 = 6

In the above expression, a is called the dividend, b the divisor and c the quotient.
Basic Rules for Division:

The basic rules for division are given as below:

Rule 1:

Division of 2 numbers with the same sign should be ‘+’ ve (positive sign) sign.

‘+’ve (Positive number) ÷’+’ve ( positive number) = ‘+’ve ( positive number)
‘-‘ve (Negative number) ÷ ‘-‘ ve (negative number) = =’+’ ve (positive number)

Rule 2:

Division of 2 numbers with different signs should be ‘-’ve negative

‘+’ve (Positive number) ÷(‘-‘ve negative number) = ‘-‘ve (negative number)
‘-‘ve (Negative number) ÷ ‘+’ve (positive number) = ‘-‘ve (negative number)

Examples for division:

Examples:

54 ÷ 9 = 6 (same signs)

(-32) ÷ (-2) = 16 (same signs)

10 ÷ (-2) = -5 (different signs)

(-12) ÷ 12 = -1 (different signs)


Rules for Division on Dividing Variable:

In this operation division represents dividing the variables based on the presence of values.

Example:

Solve the following division: 49 ÷ p, given that p = -7

Solution:

49 ÷ p

Substitute p = -7,

= 49 ÷ (-7)

= -7 (dissimilar signs).
Division on Dividing Decimal on Rules for Division:

Rules for decimal division solving problems,

Rule 1:

To create the decimal divisor as whole number by changing the decimal point to the right side.

Rule 2:

To change the same decimal point in the dividend to the right side to create as whole number

Rule 3:

After divide the new dividend or whole number by new divisor or whole number


Example:

Divide the following decimal function:  24.24 ÷ 0.20

Solution:

24.24 ÷ 0.20 = 24.24 / 0.20

= 242.4 / 2 (Take the decimal divisor as whole number)

= 2424 / 20 (Take the decimal dividend as whole number)

= 121.2 (Divide the new dividend by new divisor).

The Quotient

Quotients are one of the basic part of division operator in mathematics. A quotient is the result of a division. For example, when dividing 6 by 3, the quotient is 2, while 6 is called the dividend, and 3 the divisor. The quotient can also be expressed as the number of times the divisor divides into the dividend.

A relationship between dividend, quotient, divisor and remainder is :

Dividend = Quotient x Divisor + Remainder
Example Problems:

Example: Solve Quotient solver in 325 ÷ 15 by long division method.

Solution: Step 1:

-------                   15 ) 325

In the above equation 15 is divisor and 325 is dividend. In the divisor has two decimal numbers put the value dividend of 32.

Step 2:

2
-------
15 ) 325
30
---------
25

In 15 x 2 = 30 the divisor number 15 is multiplied with 2 to get an answer 30. In 30 is less than from 32.So use the value then subtract the value and get 25.

Step 3:

21
-------
15) 325
30
-------
25
15
--------
10

In 15 x 1 = 15 the divisor number 15 is multiplied with 1 to get an answer 15. In 15 is less than from 32.So use the value then subtract the value and get 10.

Step 4:

21.6
-------
15| 325
30
-------
25
15
---------
100
90
---------
100

The value 10 has no more value in the right side. So put 0 to get 100 and put decimal point on the quotient. Then normal divison 15 x 6 = 90 the divisor number 15 is multiplied with 6 to get an answer 90. In 90 is less than from 100.So use the value then subtract the value again we get 10.

Step 4:

21.66
--------
15| 325
30
--------
25
15
---------
100
90
----------
100
90
-----------
10 (continued)

Repeat the step again the value 10 have no more value in the right side. So put 0 to get 100 and put decimal point on the quotient. Then normal divison 15 x 6 = 90 the divisor number 15 is multiplied with 6 to get an answer 90. In 90 is less than from 100.So use the value then subtract the value again we get 10. Finally we get an answer 21.66.
Practice Problems:

Problem 1: Solve quotient solver in 425÷15

Answer: - 17

Problem 2: Solve quotient solver in 22÷11

Answer: - 2

Problem 3: Solve quotient solver in 58÷4

Answer: - 14.5

Problem 4: Solve quotient solver in 63÷2

Answer:  31.5

Problem 5: Solve quotient solver in 110÷5

Answer: - 22

Gaussian Function Equation

Gaussian function is defined as a function of the form. The graph of a Gaussian is a attribute symmetric "bell curve" shape that promptly falls off towards plus/minus infinity. Height of the curve's peak is a. Position of the centre of the peak is b. width of the bell is c. This function is commonly used in statistics, heat equations and diffusion equations and to define the Weierstrass transform. This function arises by the exponential and quadratic functions.In this article, we shall discuss about Gaussian function equation.

Gaussian Function Formulas:

 ( x) = a e^(-(x-a)^2/(2c^2))

here, a, b, c > 0 and  e = 2.718(approximate).

Gaussian integral is      int_(-oo)^oo e^(-x^2) dx = sqrtpi

nt_(-oo)^oo a e^(-(x-a)^2/(2c^2)) dx = a c sqrt (2pi)
Definition of Gaussian Function Equation :

Probability function of the normal distribution is,

f(x) = 1/(sigma sqrt(2pi)) e^ [(- (x-mu)^2)/(2sigma^2)] .

Full width at half maximum is finding the points x0 . So, we must solve

^ [(- (x_0-mu)^2)/(2sigma^2)] . = (1/2) (xmax).

But  f(xmax)  = mu

^ [(- (x_0-mu)^2)/(2sigma^2)] . = (1/2) (mu. = (1/2)

Now   solving the above equation ,        ^ [(- (x_0-mu)^2)/(2sigma^2)] . = 2^(-1)

(- (x_0-mu)^2)/(2sigma^2)] . = - ln2

(- (x_0-mu)^2)] . = - 2sigma^2 ln2

 (x_0-mu)^2] . = 2sigma^2 ln2 .

 (x_0-mu)] . = +-sqrt(2sigma^2 ln2)             .

_0 . = +-sqrt(2sigma^2 ln2) + mu             .

Therefore full width at half maximum is FWHM = 2 sqrt(2 ln2) sigma ~~ 2.35        .

Gaussian function equation for two dimensions:


The bi variate normal distribution is sigma = sigma_x = sigma_y

So,  f(x, y) = 1/(sigma^2 sqrt(2pi)) e^ [(- ((x-mu_x)^2 + (y - mu_y)^2))/(2sigma^2)] .

in elliptical function  sigma_x != sigma_y

f(x, y) = 1/(sigma_x sigma_y sqrt(2pi)) e^ - [(x-mu_x)^2/(2(sigma_x)^2) + (y - mu_y)^2/(2(sigma_y)^2)] .

Gaussian function

Gaussian function also used to find   A(x)   =  ^[(-x^2)/(2sigma^2)]

Instrument function is  I(k) = ^(-2pi^2 k^2 sigma^2) sigma sqrt(pi/2)[erf((a - 2pi i k sigma^2)/(sigmasqrt2)) + erf((a+2pi i k sigma^2)/(sigmasqrt2))]   .

I max    =   sigma sqrt (2pi) erf(a/(sigmasqrt2)) .

As  a -> oo  instrument  equation reduced as

im_(a->oo) I(k) = sigma sqrt(2pi)e^(-2pi^2 k^2 sigma^2).

Inequalities Lesson Plan

Inequalities are the main concepts in mathematical algebra.Usually inequalities are the combination of variables and constants  greater than and less than sign.  For example 2y -10 < 20 is a less than inequality. In this article we are going to study about inequalities lesson plan through suitable example problems.
Brief Summary about Inequalities Lesson Plan


General form of an inequality:
Inequalities take the following form,

     ax +by <= c

     Here x and y are variables and then ‘a’ and ‘b’ are coefficients of x and y and then c is the constant term.  <= is denote the inequality sign.

     The following signs are used to represent the inequalities.  They are

    < = less than
    > = Greater than
    <= less than or equal to
    >= Greater than or equal to

Examples on Solving Inequalities Lesson Plan

Example 1:

     Solve the following inequality 5p + 15 > 45

Solution:

     The given inequality is 5p + 15 > 45

     Now we have to subtract the value 15 on both sides, then we have to get the following inequality,

    5p + 15 -15 > 45 - 15

     That is 5p > 30

     Dividing the value 5 on both sides, then we have to the following inequality,

     That is p>6

     That is p value is greater than 6.

Example 2:

     Solving the following inequality 2r + 18 `<=` 36 and represent it on the number line

Solution:

     The given inequality is 2r + 18 `<=` 36

     Now we have to subtract the value 18 on both sides, then we have to get the following inequality,

     2r + 18 -18 `<=` 36 -18

     That is 2r `<= ` 18

     Dividing the value 2 on both sides, then we have to the following inequality,

     That is r`<=` 9

     That is r value is less than or equal to 9.

     That is r=9 and r is less than 9.


Representation of number line:

 Inequalities on a number line

     First we have to draw the number line and mark the values.  Here r is less than or equal to 9 so we can first shade the value 9 and then draw the arrow in the direction of negative direction.  Because it is the less then (<) value.

    These are the important problems in the inequality lesson plan.

Equation Of A Line

Equation of a line is the equation satisfied by all the points which lie on the line.

 So, we can call every point on the line as a solution of the equation.

 As there are infinitely many points on a line, the equation of a line has infinitely many solutions.

 For every value of x, you can find a value for y by substituting the x value in the equation of the line.

 That is, consider the line whose equation is 2x + y = 4.

 We can find the value of y, when x equals 2.

 Substituting x = 2 in 2x + y = 3, we have

 2(2) + y = 3

 ? 4 + y = 3

 ? y = -1

 So, (2, -1) is a point on the line 2x + y = 4.

 There are various approaches to derive equation of a line.
Point Slope Form of Equation of a Line

We can draw a line if we know its slope and a point through which it passes. And the line we get is unique. Is it? Try.

 Hence we can find the equation of a line when its slope m and a point P(x1, y1) through which passes are given.

 The equation of the line is given as (y – y1) = m(x – x1)
Slope Intercept Form of Equation of a Line :

Suppose if c is the y-intercept of the line, then (0, c) is a point on the line.

 So, the equation (y – y1) = m(x – x1) can be written as (y – c) = m(x – 0)

 That is, y = mx + c

 So, we can find the equation of the line if its slope and y-intercept are known.
Equation of a Line Passing through Two Given Points:

Equation of a  line passing through two given points is also unique.

 So, we can find the equation of a line, if we know any two points on it.

 Suppose line l passes through the points A(x1, y1) and B(x2, y2), then slope of the line l is

 m = y2 - y1 / x2 -x1

Now we know the slope and a point on the line (considering either A or B) and hence we can find the equation of the line using the formula (y – y1) = m(x – x1).

So, the equation of the line is

(y - y1) = y2 - y1 / x2 -x1  (x - x1)

OR

(y - y2) = y2 - y1 / x2 -x1  (x - x2)

That is y - y1/ x - x1 = y2 - y1 / x2 -x1

In general, we can rewrite the equation of a line in the form ax + by + c = 0, where a, b and c are real numbers.

LCM Using Prime Factorization

The expanded form of LCM is Lease common multiple which means find the common multiple with least value for a given set of numbers. This can be done by using either prime numbers or by using composite numbers or with a combination of prime numbers and composite numbers. The LCM can also be written with exponents.
Steps to Find the Lcm Using Prime Factorization:

Step 1: Write the given set of numbers.

Step 2: Take the first number from the given set.

Step 3: Write the number with the multiples of prime numbers.

Step 4: Repeat step 2 and Step 3 for all the numbers in a given set.

Step 5: Choose the highest exponent term form all and multiply it to get LCM of given numbers.
Example Problems – Lcm Using Prime Factorization:

Example 1- LCM using prime factorization:

Find the LCM of 30 and 10.

Solution:

The given numbers are 30 and 10.

First it needs to find the prime factors of 30.

30 = 2*15

     = 2 * 3  * 5

The prime factors of 30 are 21, 31 and 51.

Next it needs to find the prime factors of 10.

10 = 2 * 5.

The prime factors of 10 are 21 and 51.

The highest exponent of all terms is 21, 31 and 51.

The LCM of 30 and 10 is 2 * 3* 5 `=>` 30.

My forthcoming post is on prime factorization algorithm, adding and subtracting rational expressions with unlike denominators will give you more understanding about Algebra

Example 2- LCM using prime factorization:

Find the LCM of 42 and 60.

Solution:

The given numbers are 42 and 60.

First it needs to find the prime factors of 42.

42 = 2*21

     = 2 * 3  * 7

The prime factors of 30 are 21, 31 and 71.

Next it needs to find the prime factors of 60.

60 = 2 * 30

     = 2 * 2 * 15

     = 2* 2 * 3* 5

The prime factors of 10 is 22 , 31 and 51.

The highest exponent of all terms is 22, 31, 51 and 71.

The LCM of 30 and 10 is 22 * 3* 5* 7 `=>` 420.

Example 3- LCM using prime factorization:

Find the LCM of 50 and 100.

Solution:

The given numbers are 50 and 100.

First it needs to find the prime factors of 50.

50 = 2*25

     = 2 * 5  * 5

The prime factors of 50 are 21 and 52

Next it needs to find the prime factors of 60.

100 = 2 * 50

       = 2 * 2 * 25

       = 2* 2 * 5* 5

The prime factors of 10 is 22 and 52

The highest exponent of all terms is 22 and 52

The LCM of 30 and 10 is 22 * 52 `=>` 100