Unit Circle Sine Cosine

The Unit circle is used to understanding the sins and cos of angles to find 90 degree triangle. It is radius is exactly one. The center of circle is said to be origin and its perimeter comprises the set of all points that are exactly one unit from the center of the circle while placed in the plane.

It s just a circle with radius ‘one’.

The distance from the origin point(x,y) is `sqrt(X^2+Y^2)` by using Pythagorean Theorem.

Here radius is one So, The expression should becomes sqrt(x²+y²) =1

Take square on both sides then the eqn. becomes,

                      X²+y² =1

Positive angles are found using counterclockwise from the positive x axis

And negative angles are found using  anti clockwise from negative axis.

To find sine and cosine angle:

Angles are measures using counterclockwise from the positive x axis.The point where the ray at angle theta meets the unit circle is,
                                      (cos(`theta` ),Sin(`theta` )).
Here,cos(`theta` )----->Represents notation for the cosine of theta
Sine(`theta` ) is the notation of the sine of theta.
X= cos`(theta)`
Y=sin`(theta)`
The angle measured in radians is the length of the circular arc from the positive x
Axis counterclockwise until the desired angle is reached.
Since the radius of the unit circle is 1, the length of its perimeter is
2pr = 2p(1) = 2p.
Hence the angle for going exactly around the circle once is 2 radians. This
corresponds to 360 degrees.

Example problem -Unitcircle sines and cosine

Example 1: 

                 The angle q=0 radians is found by starting at the positive x-axis and going a distance 0. Hence the ray at angle q = 0 meets the unit circle at (1,0).

Hence
(cos(0), sin(0)) = (1,0),
cos(0) = 1,
sin (0)= 0.
Example 2. Without a calculator, find cos(p) and sin(p).
Solution:

The angle q= p radians is found by starting at the positive x-axis and going a distance p counterclockwise around the circle. But 2p is the total perimeter, so going a distance p arrives exactly half-way around the circle. Hence the ray at angle q = p meets the unit circle at (-1,0).

Hence
(cos(p), sin(p)) = (-1,0),
cos (p) = -1,
sin (p)= 0.

Polynomial Inequality

Inequality is a statement which means that the two things are not equal. Polynomials are the expressions which have the degree of at least 2 and more. An inequality that has polynomials is known as polynomial inequality. And polynomial inequality is not linear anymore. Solving polynomial inequality is little more complicated when comparing with linear inequalities. We will see how to solve polynomial inequality in the following sections.

Steps to solve polynomial inequality:

The following are the steps involved in solving polynomial inequality

Step 1: Make the inequality to obtain a zero on one side of the inequality

Step 2: The next step is to factor the polynomial which is on the left side of the inequality sign

Step 3: Determine the points at where the polynomial is zero.

Step 4: Graph the points we get in the previous step that is where the polynomial is zero

Step 5: Write down the answer

Example problems on solving polynomial inequality:

1)Solve the polynomial inequality x2 +15 < 8x

Solution:

Step 1: Make the inequality to obtain a zero on one side of the inequality

x2 +15 < 8x

x2-8x+15<0

Step 2: factor the polynomial x2-8x+15

x2-8x+15

x2-3x-5x+15

x(x-3) -5(x-3)

(x-5)(x-3)

Step 3: Determine the points at where the polynomial is zero.

Our polynomial is (x-5)(x-3)

When x= 5 ,

(x-5)(x-3) = (5-5)(5-3)

= 0(2)

=0

When x=3

(x-5)(x-3) = (3-5)(3-3)

= (-2)0

=0

The polynomial gets zero at x=5 and x= 3

Step 4: Graph the points we get in the previous step x=5, x=3 ( where the polynomial is zero )




Step 5: Write down the answer.

The solution of the given polynomial inequality is 3<x<5 .

2) Solve the polynomial inequality (x+2)(x+3) >0

Solution:

Step 1: Make the inequality to obtain a zero on one side of the inequality

Here in our example already there is a zero on the right side of the inequality .

Step 2: Factor the polynomial (x+2)(x+3)

No problem. It is in factored form

Step 3: Determine the points at where the polynomial (x+2)(x+3) is zero.

(x+2)(x+3)

When x= -2 ,

(x+2)(x+3) = (-2+2)(3+3)

=(0)(6)=0

When x= -3,

(x+2)(x+3) = (-2+3)(-3+3) = 0

Our polynomial get 0 at x=-2, x=-3

Step 4: Graph the points we get in the previous step x=-2, x=-3 ( where the polynomial is zero )

Algebra is widely used in day to day activities watch out for my forthcoming posts on Common Factor and solving systems of linear equations by elimination. I am sure they will be helpful.

Step 5: Write down the answer.

The solution for the given polynomial inequality is  -3<x<-2

learn expected value Binomial

The binomial distribution is discrete probability distribution. The digits of success in a sequence of n self-sufficient yes/no experiment, each one of which yields success among probability p. Such an achievement/failure experiment is called a Bernoulli experiment or Bernoulli trail. Into detail, when n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the root for the popular binomial test of statistical consequence.


Binomial Distribution:

A learn expected value Binomial is an algebraic expression contain two variables identified as x and y. direct multiplication gets quite deadly and can be slightly difficult for larger powers or more complicated expressions. The coefficients appeared during the binomial expansion are called Binomial Coefficient. They are the identical as their entries of Pascal’s triangle, and know how to be determined by a simple formula using factorials.

The formal expression of the Binomial Theorem is as follows:

`(a+b)^(n) = sum_(k=0)^n ([n],[k])a^(n-k) b^(k)`

It is often used to representation number of successes in a example of size n from a population of size N. Since the examples are not self-determining the resulting distribution is a hyper geometric distribution, not a binomial one. Though, for N much larger than n, the binomial distribution is a good quality approximation, and extensively used.

Binomial Expansion Formula

The binomial theorem formula is used to develop the binomials to some given power without direct multiplication.

n! = n (n − 1)(n − 2) ... (3)(2)(1)

Basic Binomial expansion:

(x + y)0 = 1

(x + y)1 = x + y

(x + y)2 = x2 + 2xy + y2

Examples for learn expected value binomial:

Example 1:

given the equation  (5+3y)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) 5^(2-k)3y^(k)`


=  `([2],[0]) 5^(2)3y^(0) + ([2],[1]) 5^(1)3y^(1) +([2],[2]) 5^(0)3y^(2)`

the answer is = 25 + 15y + 3y2

Example 2:

given the equation  (5-y2)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) (5x^(2-k))(-y^2)^k`


=  `([2],[0]) 5x^(2)(-y^2)^(0) + ([2],[1]) 5x^(1)(-y^2)^1 +([2],[2]) 5x^(0)(-y^2)^2`

the answer is = 5x2-5xy2+y2.


Example 3:

given the equation  (x-6y)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) (x^(2-k))(6y)^k`


=  `([2],[0]) x^(2)(6y)^(0) + ([2],[1]) x^(1)(6y)^1 +([2],[2]) x^(0)(6y)^2`

the answer is = x2+6xy+6y2.

Mathematical Expression

In mathematical, an expression is a finite combination of symbols that are well-formed according to the rules applicable in the context at hand. In mathematical, symbols can designate values, variables, operations, relations, or can constitute punctuation.

         In mathematical, an expression may be used to designate a value. In mathematical, which value might depend on values assigned to variables occurring in the expression; the determination of this value depends on the semantics attached to the symbols of the expression.

Source: Wikipedia

Basic natural law for mathematical expression:

        In elementary algebra, we list the fundamental rules and properties of pre-algebra and give examples on they may be used natural law.

        Suppose that a, b and c are variables or mathematical expressions using natural law.

1. Commutative Property of Addition In mathematical.

           a + b = b + a

 2. Commutative Property of Multiplication In mathematical.

           a * b = b * a

3. Associative Property of Addition In mathematical.

       (a + b) + c = a + (b + c)

4. Associative Property of Multiplication In mathematical.

          (a * b) * c = a * (b * c)

5. Distributive Properties of Addition Over Multiplication In mathematical.

         a * (b + c) = a * b + a * c
    and
        (a + b) * c = a * c + b * c

More about In mathematical expression:

In algebra 1 expression is a finite group of algebraic terms and mathematical symbols combined with no equal or in equality sign.

Steps for simplify the mathematical expression:

            Step 1: Group the terms containing the same variable together in algebra expressions using natural law.

            Step 2: Perform the operation inside the parentheses for the variable and other.

            Step 3: Rewrite the expressions and simplifying the algebra expressions.

            Step 4: To check the equation, if there is able to simplify the expression, then repeat the step 1 to 4.

Example problems for simplify the mathematical expression using natural law:

        In mathematical expressions, find the value of x using natural law.

              3x(1x+4x²)

Solution:

            A(B+C)=AB+AC

            Step 1: is to determine what terms represent A,B and C in the given equation.

                        A represents 3x.

                        B represents 1x.

                        C represents 4x^x.

            Step 2: is to perform the multiplication operation.

                        AB=3x(1x)=3x²

                        AC=3x(4x²)=12x³

            Step 3: is to rewrite the problem.

                        3x(1x+4x)=3x²+12x³

            Step 4: is to simplify the answer.

                        3x²+12x³=0

                                 12x³=-3x²

                                                12x=-3

                                        x=`-1/4`

                 The answer is x=`-1/4`.

Trigonometry Values

If a rotation from the initial position to the terminal position is (`1/360` )^th of the revolution, the angle is said to have a measure of one degree and written as 1°. A degree is divided into minutes, and minute is divided into seconds.

The trigonometric values that exist are based on the angle which are formed of the rotation which represents it provides numeric nalues for the angles in the functions.

Tables for trigonometry values:

The following table in trigonometry shows that the trigonometric radians for their degrees.

The following table in trigonometry shows that the trigonometric functions which provide the trigonometric values for their degrees of their angles.

By the use of this table, the given trigonometric functions with their values  are also used for finding the other trigonometric functions like secant, cosecant and cotangent.

Example problems for trigonometry values:

1.Simplify :

(i) tan 735° (ii) cos 980° (iii) sin 2460° (iv) cos (−870°)
(v) sin (−780°) (vi) cot (−855°) (vii) cosec 2040° (viii)sec (− 1305°)

Solution:
(i) tan (735°) = tan (2 × 360° + 15°) = tan 15°

(ii) cos 980° = cos (2 × 360° + 260°) = cos 260°
= cos (270° - 10°) = − sin 10°

(iii) sin (2460°) = sin (6 × 360° + 300°) = sin (300°)

= sin (360° − 60°)
= − sin 60°
= −`sqrt(3)/2`

(iv) cos (− 870°) = cos (870°) = cos (2 × 360° + 150°)

= cos 150 = cos (180° - 30°)
= − cos 30° = −`sqrt(3)/2`

(v) sin (− 780°) = − sin 780°

= − sin (2 × 360° + 60°)
= − sin 60° = −`sqrt(3)/2`

(vi) cot (− 855°) = − cot (855°) = − cot (2 × 360° + 135°)

= − cot (135°) = − cot (180° - 45°)
= cot 45° = 1

(vii) cosec (2040°) = cosec (5 × 360° + 240°) = cosec (240°)

= cosec (180° + 60°) = − cosec (60°)
= − −`2/3`

(viii) sec (− 1305°) = sec (1305°) = sec (3 × 360° + 225°)

= sec (225°) = sec (270° − 45°)
= − cosec 45° = − 2

2.Simplify :

[  cot (90° − θ ) sin (180° + θ) sec (360° − θ) ] / [ tan (180° + θ) sec (− θ) cos (90° + θ) ]

Solution:

The given expression =[ tan θ (− sin θ) (sec θ) ] / [ tan θ (sec θ) (− sin θ) ]
                                = 1

Because the terms in that fraction gets easily cancelled.

Building Complex Systems

Any systems which involves a quantity of elements, set in structures which can subsist on a lot of scales is called the complex systems. These leave during processes of modify that are not describable by a solitary regulation nor are reducible to single stage of details, these levels often include features whose appearance cannot be predicted from their present specifications. Complex Systems speculation also includes the study of the connections of the a lot of parts of the systems.

Building complex systems:

            The complex number is specify the Z = a + ib
            The imaginary number is called as the tetragon root of a negative number. It is specify the `sqrt(-a)`  a > 0 is an imaginary number.
            The real numbers are a and b. The complex number is called as the a+ib. The real part is called a and imaginary part is called b.
Types of building complex system
            The first type is equality of complex numbers. The second type is calculation of two complex numbers. The third type is negative of a complex number. The fourth type is preservative identity of the complex number. The fifth type is preservative inverse of a complex number. The sixth type is creation of two complex numbers. The last type is multiplicative individuality of complex numbers.
Properties of building complex system
            There are a lot of propertied of the building complex system.  
            The first property is the commutative law for calculation.   
            The second property is Commutative Law for increase.
            The third property is Additive character Exists.
            The fourth property is Multiplicative character Exist.
            The fifth property is Reciprocals Exist for nonzero complex numbers.
            The sixth property is Negatives be real for every complex numbers.
            The last property is Non Zero invention Law.

Example problem of building complex systems:

Problem 1 :
`sqrt(-64)`
Solution
        `sqrt(-64)`
     =  `sqrt(64(-1))`
     =`sqrt(64)sqrt(-1)`
     = 8i
Problem 2:
          `-sqrt(-81)`
Solution
     = `-sqrt(-81)` 
     = `-sqrt(81(-1))`
     = `-sqrt(81)sqrt(-1)`
     =-9i

Rules of Radicals

In math, the radical is one of the number representations which indicate the square root or nth root. The symbol for denoting the radical is "√". For example, `sqrt(a)`, `root(n)(a)` are radicals. The term inside the symbol √ is called as radicand and the 'nth root' or 'square root ' is called as index value. Radical is considered as the opposite operation of exponent. For example, 4² is an exponent which is equal to 16. Now, the radical can be written as `sqrt (16)` =` sqrt (4xx4) ` = ±4. It is also possible to write the radical as an exponent by writing the reciprocal of nth root or square root as a power of radicand. Example: `sqrt(4)` = `4^(1/2)`. The radical has some basic rules for simplifying the problems. This article helps to give the detail explanation about the radical rules with some example problems for simplifying.

I like to share this Simplifying Ratios with you all through my article. 

Explanation to rules for radicals:

Rule 1:
          If the radical has a two radicands in multiplication form, then we can distribute common the index value to each radicand and then relate them by multiplication operation.
          For example, `root(n)(ab)` is the radical which has two radicands with a common index value 'n'. So, we can distribute the index value 'n' to each radicand 'a' and 'b' and then write it as `root(n)(a)` `root(n)(b)`.

  `root(n)(ab)`  =  `root(n)(a)` `root(n)(b)`  

Rules 2:
          If the radical has a two radicands in division form then we can distribute the index value to each radicand and then relate them by division operation.
          For example, `root(n)((a/b))` is the radical which has two radicands in division form with a common index value 'n'. So, we can distribute the index value 'n' to each radicand 'a' and 'b' and then write them as `root(n)(a)` /`root(n)(b)`.

  `root(n)((a/b))` =  `(root(n)(a))` / `(root(n)(b))`

Rules 3:
          If suppose the radicand has a power equals to the index value then it results the radicand without any power.
          For example, if we have radical, `root(n)(x^n)`  then we can also write it as `x^(n/n)`. Now, we can cancel the common term 'n'. So, we get the answer as 'x'.

  `root(n)(x^n)`    = x^(n/n) = x

Rule 4:
          If we have a radical `root(n)(-x)` then we can write it as -`root(n)(x)` where 'n' is odd.
By applying the above basic rules, we can easily simplifying the radicals.

Example Problem - Rules for simplifying radicals:

Example: 1

Simplify:`sqrt(50)`

Solution:
Given: `sqrt(50)`
For simplifying the given radical `sqrt(50)` , use the rules of radicals.
Step 1:

First find the factors of the radicand 50.

Factors of 50 are 25 and 2. (50 = 25 x 2)

Now, the radical can be written as `sqrt(50)` = `sqrt(25xx2)`

Step 2:

We need to apply the rule 1 for simplifying the radical.

The rule is, `root(n)(ab)` = `root(n)(a)` `root(n)(b)`

`sqrt(25xx2)` = `sqrt(25)` `sqrt(2)` (Here, we distribute the square root to both radicands 25 and 2)

= `sqrt(5xx5)` `sqrt(2)`

The square root of 25 is 5. Because the square value of 5 is 25.(5 x 5 = 25)

`sqrt(50)` = 5 `sqrt(2)`

Answer: 5`sqrt(2)`
Example: 2

Simplify: `root(n)(125/27)`

Solution:
Given: `root(3)(125/27)`
We need to apply the rules of radicals to solve this problem.

The radial rule is,

`root(n)((a/b))` =  `(root(n)(a))` / `(root(n)(b))`

Now, the given radical `root(3)(125/27)` can be written as,

`root(3)(125/27)` =`(root(3)(125))/(root(3)(27))`

Step 2:

The cubic root of 125 is 5. Because the cubic values of 5 is 125 (5 x 5 x 5 = 125)

`root(3)(125)` = `root(3)(5xx5xx5)`

= 5

The cubic root of 27 is 3. Because the cubic values of 3 is 27(3 x 3 x 3 = 27)

`root(3)(27)` = `root(3)(3xx3xx3)`

= 3

Step 3:

Now, we get the answer as,

`root(3)(125/27)` =`(root(3)(125))/(root(3)(27))`= `5/3`

Answer: `5/3`
Example: 3

Simplify: `root(5)(25^(1/5))`

Solution:
Given: `root(5)(25^(1/5))` 
Step 1:

The rule of radical for simplifying the exponent is,

  `root(n)(x^n)`    = x^(n/n) = x

Step 2:

When we are applying the rule, we get the answer 25.

`root(5)(25^(1/5)`    = 25^(5/5) =25

Answer: 25
Example: 4

Simplify: `root(3)(-64)`

Solution:
Given: `root(3)(-64)`
Step 1:

The index value of the given radical `root(3)(-64)` is odd number 3.

So, we have to take the negative outside and then find the cubic root ot radicand 64.

`root(3)(-64)` = - `root(3)(64)`

= - 4

Answer: -4