Trigonometric Identities Sum Help

Introduction to trigonometric identities sum help:

Trigonometry is arrived from the Greek word, trigonon = triangle and metron = measure. The father of trigonometry is Hipparchus. He designed the first trigonometric table. Trigonometry has wide range of applications in many fields like science, technology, astronomy etc. Identity is defined as an equation that is true for all probable values of its variables. Online help is one of the comfortable method of getting help from anywhere around the globe. Through online study, students can get about trigonometric identities sum. In this topic, we are going to see about, trigonometric identities sum help. I like to share this Pythagorean Trigonometric Identities with you all through my article.

Trigonometric Identities Sum Help - Trigonometric Identities:

The list of trigonometric identities sum are shown below,

Sum or difference of two angles:

sin (a ± b ) = sin a cos b ± cos a sin b

cos(a ± b) = cos a cos b ± sin a sin b

tan(a ± b) = `(tan a +- tan b)/ (1 +- tan a tan b)`

Sum and product formulas:

sin a + sin b = `2sin((a+b)/2)cos((a-b)/2)`

sin a - sin b = `2cos((a+b)/2) sin((a-b)/2)`

cos a + cos b = `2cos((a+b)/2) cos((a-b)/2)`

cos a – cos b = `-2sin((a+b)/2) sin((a-b)/2)`

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Trigonometric Identities Sum Help: - Examples

Example 1:

Evaluate Sin 153

Solution:

Sin 153 = Sin (90+63)

= sin 90 cos 63 + cos 90 sin 63

= 1(0.454) + 0(0.891)

= 0.454 + 0

= 0.454

The answer is 0.454

Example 2:

Evaluate Cos 128

Solution:

Cos 128 = Cos (90 + 38)

= cos 90 cos 38 – sin 90 sin 38

= 0(0.788) – 1(0.616)

= 0 – 0.616

= -0.616

The answer is -0.616

Example 3:

Evaluate tan 38

Solution:

Tan 38 = Tan (45 - 7)

= `(tan 45 - tan 7)/(1+tan 45*tan 7)`

= `(1-0.123)/(1+(1*0.123))`

= `0.877/(1+0.123)`

=` 0.877/ 1.123`

= 0.781

The answer is 0.781

Example 4

Evaluate Cos 124

Solution:

Cos 124 = cos (90 + 34)

= cos 90 cos 34 - sin 90 sin 34

= 0(0.829) - 1(0.559)

= 0 – 0-0.559

= -0.559

The answer is -0.559

Example 5:

Evaluate, sin 50 - sin 40

Solution:

sin a - sin b = `2cos((a+b)/2)sin((a-b)/2)`

sin 50 - sin 40 = `2 cos((50+40)/2)sin((50-40)/2)`

= `2 cos (90/2) sin(10)/2`

= 2 cos 45 sin 5

= 2 (0.707)(0.087)

= 2 * 0.062

= 0.124

The answer is 0.124

Graphs and Histograms

Graphs are pictorial representation of data.Histogram is graphical representation of data in statistics.

Introduction to graphs and histograms:

A histogram graph is representation of a frequency distribution as a graph . The graph  consists of rectangles constructed with class intervals as bases and heights proportional to corresponding frequencies such that there is no gap between any two successive rectangles.

A histogram deals with continuous type of data.

Different types of histograms are:

Histogram of continuous grouped frequency distribution with equal class intervals.

Histogram of a continuous grouped frequency distribution with unequal class intervals.

Histogram when mid-points are given

Histogram for grouped frequency with inclusive classes(discontinuous class-intervals)

Steps for Construction of Histogram:

Choose a suitable scale on the x-axis and represent the class-limits on it

Choose a suitable scale on the y-axis and represent the corresponding frequencies on it

Draw rectangles with class intervals as bases and the respective frequencies as heights

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Example Histogram with Equal Class Interval:

Ex : 1 Constuct a histogram to represent the following data:

Marks
Number of

students

0-10    4
10-20    7
20-30    12
30-40    20
40-50    9
50-60    2
Sol:

Choose a scale

Step 1:Along x-axis 1 cm = 10 marks

Step 2:Along y-axis 1 cm = 5 students

Step3:Starting from 0, mark 10,20,30,40,50,60,70 on x-axis and 4,7,12,20,9,2 on the y-axis

Step4:Then we draw the rectangles with class intervals as bases and corresponding frequencies as heights.

Steps to Draw Histogram of Unequal Class Interval:
Choose a suitable scale on the x-axis and represent the class-limits on it
Determine a class interval which has the minimum class size. Let the minimum class size be h
Find the adjusted frequency of each class by using the formula :
Adjusted frequency of the class = `h/("Class - size of the class)` x frequency of the class

Choose suitable scale on the y-axis and represent the corresponding adjusted frequencies on it
Draw the rectangles , the width of rectangles will be according to class limit

Direction Vector of a Line

Introduction to direction vector of a line:

The direction vector is a vector point of direction and it indicates the direction of line. The direction vector of a line is based on real values of line segment equation. In math, the vector points are used in Euclidean space. Now we are going to see about direction vector of a line.

Explanation for Direction Vector of a Line
Direction vector:

The line is represented as `vecAB` and the arrow mark symbol is indicating the direction of line. The direction vectors are determined from line equation.

In Euclidean space, the direction vector of line D is determined by using line equation form real numbers that is the line equation form is ax + by + c = 0. Here a, b, and c are real numbers and the direction vector of line D is (-b, a). We can also consider the multiples of (-b, a) as direction vectors. Understanding empirical probability is always challenging for me but thanks to all math help websites to help me out.

More about Direction of Vector of Line

Example problems for direction vector of a line:

Problem 1: Find out the direction vector of a line segment D from line form.

4x + 3y + 1 = 0.

Solution:

The given line equation form is 4x + 3y + 1 = 0.

The line segment D has two end points AB.

The direction vector `vecAB` is determined from line form.

`vecAB` = (-3, 4).

Therefore, the direction vector of line is (-3, 4), (9, 16)…

Problem 2: Find out the direction vector of a line segment D from line form.

x - 2y + 4 = 0.

Solution:

The given line equation form is x - 2y + 4 = 0.

The line segment D has two end points AB.

The direction vector `vecAB` is determined from line form.

`vecAB` = (2, 1).

Therefore, the direction vector of line is (2, 1), (4,1)…

Exercise problems for direction vector of line:

1. Find out the direction vector of line D from 3x + 4y + 2 = 0.

Solution: The direction vector `vecAB` is (-4, 3).

2. Find out the direction vector of line D from 6x - 3y - 5 = 0.

Solution: The direction vector `vecAB` is (3, 6).

Number Theory Problems and Solutions

Introduction to number theory  problems and solutions

Number theory is the branch of pure mathematics that concerned with the properties of numbers in general, and integers in particular, as well as the wider classes of problems that arise from their study. Number theory may be subdivided into the several fields, according to the methods used and the type of questions investigated and let we about number theory problems and solutions. (Source – Wikipedia).

Number Theory Problems and Solutions
Example 1:

Find the solutions of consecutive numbers and the product of the number is 143
Solution:-

Let as assume the two consecutive numbers be x, x +1.

Where the product is 143 so,

X * (x + 1) = 143

x2 + x = 143

x2 + x – 143 = 0

x2 + 13x – 11x – 143 = 0

(x + 13) (x - 11) = 0

(x + 13) = 0 (or) (x - 11) = 0

x = -13 (or) x = 11

-13 is not possible to get 143

Therefore we take x = 11

So, x + 1 = 11 + 1 = 12

The consecutive numbers is 11 and 12.

Example 2:

Verify the given sequence described by an = 12n2 + 1 and A.P.?

Solution:

an = 12n2 + 1

a1 = 12(1)2 + 1 = 13,

a2 = 12(2)2 + 1 = 49

a3 = 12(3)2 + 1 = 109,

a4 = 12(4)2 + 1 = 193

The  number theory problems solutions in sequence is 13, 49, 109, 193...

Here, 49 – 13 = 36

Example 3 :

Find the missed term in the given numbers 31, 29, ____, 25.
The above stated number is in descending order.

The known difference between the number 29 – 31 = 2.

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More Number Theory Problems and Solutions

Example 1:

Identify the type of numbers -7, 3 ÷ 4, `sqrt(8)`

Solution:

-7 - Integer

3 ÷ 4 - Fraction

`sqrt(8)` - Irrational Number

Example 2:

Mention the values of the consecutive numbers where the sum of the two numbers is 91.

Solution: Let we mention the two consecutive numbers be x, x+1.

Where the sum is 91 so,

x + x + 1 = 91

2x + 1 = 91

2x = 90

x = 45

Therefore, x + 1 = 45 + 1 = 46

So the  number theory problems solutions of an  consecutive numbers is 45 and 46.

The above problems are stated in number theory  problems and solutions

Tenth Grade Math Problems

Introduction to tenth grade math problems:

Now in this article we are going to see about tenth grade math problems. Tenth grade math topics are linear algebra, calculus, integral, slope equation, logarithmic, line equation, geometry, trigonometry function, trigonometry identities, word problems, and quadratic equations. Students know the basic differentiation and integration problems in this grade before entering into their high school. We solve more problems in tenth grade mathematics. In this article, we solve some problems used in tenth grade mathematics.

Example Problems for Tenth Grade Math

Tenth grade math example problem 1:

Solve the first order linear equation y = x + 7 and 6x + y = 70

Solution:

Given first order linear equation is y = x + 7 and 6x + y = 70

Here,

y = x + 7 -------- (1)

6x + y = 70 ------------ (2)

Substitute the equation 1 in equation 2, we get

x + 6x + 7 = 70

After simplification, we get

7x + 7 = 70

Subtract 7 on both the sides, we get

7x = 63

Divide by 7 on both the sides, we get

x = 9

Substitute the value of x in equation 1, we get

y = (9) + 7

y = 16

The answer is x = 9 and y = 16

Answer:

The final answer is x = 9 and y = 16.

Tenth grade math example problem 2:

Find the second order value of the function y = 4x3 + 2x2 + 98x

Solution:

The given function is y = 4x3 + 2x2 + 98x

First find the first order of derivative value of the function y, we get

`(dy / dx)` = (4 * 3)x2 + (2 * 2)x + (98 * 1)

= 12x2 + 4x + 98

Second order derivative of the function y is given as,

`((d^2y) / (dx^2))` = (12 * 2)x + (4 * 1)

= 24x + 4

Answer:

The final answer is 24x + 4


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Tenth Grade Math Example Problem 3:

A train travelling in a speed of 23km/h. In 3 hours how long train travel from the starting point.

Solution:

From the given,

Distance = ?

Time = 3 hours

Speed = `23((km)/(h))`

We know the formula,

Speed = `((D)/(t))`

Here, D = Distance and t = Time

Rearrange the above formula, we get

Distance = Speed * Time

Substitute the given values, we get

= `23((km)/(h))` * 3hours

= 69km

Therefore, speed of the train is 69km

Answer:

The final answer is 69km