Precalculus Trigonometry

Precalculus is an advanced form of  the secondary schools algebra, is a foundational mathematical discipline. It is also called Introduction to the Analysis. In many school, precalculus is actually two separate courses: Algebra and Trigonometry.

Trigonometry is a branch of a mathematics that studies triangles, particularly right triangles. Trigonometry deal  with the relationships between the sides and the angles of triangles, and with trigonometric functions, which describe those relationships and angles in general, and the motion of waves such as sound and light waves.                                           (Source.Wikipedia)
Precalculus with Trigonometry Identities:

sin2x+ cos2 x≡ 1

1 + tan2x≡ sec2 x

1 + cot2x ≡ cosec2 x

sin2x≡ 1 – cos2x

cos2x≡ 1 – sin2x

tan2x ≡ sec2x − 1

sec2x − tan2x ≡ 1

cot2x ≡ cosec2x − 1

cosec2x − cot2x ≡ 1
precalculus trigonometry ratios:

sin (90° − x) = cosx

cos (90° − x) = sin x

tan (90° − x) = cot x

cot (90° − x) = tan x

sec (90° − x) = cosecx

cosec (90° − x) = sec x
Examples for Precalculus with Trigonometry:

Example 1:

Find x° if cosec x° = sec 55°.

Solution:

Since cosec x° =sec(90°− x°),we have sec(90°− x°)= sec 45°. ∴ 90°− x°=45°.

∴ x° = 90°− 45°= 55°.

Example 2:

Evaluate sin 30° tan 60° sec 30°

Solution:

sec 30° = sec(90°−60°) = cosec 30°=1/sin 30°

∴sin 30° tan 60° sec 70° = sin 30° tan 60° cosec 30°

=sin 30° × 3× 1/sin 30°  = 3

Examples for trigonometry identities:

Problem 1:

Prove that sin4a + cos4 a= 1 – 2sin2a cos2a.

Solution:

LHS = sin4a + cos4a = (sin2a) 2 + (cos2a) 2

= [sin2a + cos2a] 2 – 2 (sin2a) (cos2a)                 ( a2 + b2 = (a + b)2 – 2ab)

= (1)2 – 2sin2a cos2a

= 1 – 2sin2a cos2a

= RHS

Hence proved

Problem 2:

Prove that sin4b − cos4b = sin2b − cos2b

Solution:

LHS = sin4b − cos4b = (sin2b) 2 – (cos2b) 2

= (sin2b + cos2b) (sin2 b − cos2b) = (1) (sin2b − cos2b)

= sin2b − cos2b = RHS

Problem 3:

Prove that (sec c+ cos c ) (sec c − cosc ) = tan2c + sin2c.

Solution:

LHS = (sec c+ cosc ) (sec c − cosc ) = sec2c − cos2c

= (1 + tan2c) – cos2c= tan2c + (1 – cos2c)

= tan2c + sin2c = RHS

Points Lines and Planes

Points:

In math points are used to mark the location.

Lines:

A line can be formed by connected set of infinitely points. The line is extended in both the directions. The symbol `harr` is used to represent the line in math.  Lines are denoted by letters like A, B, C, D.

Planes:

A plane can be formed by an infinite group of points forming a related flat surface expanding infinitely distant in every directions.


Points - Points Lines and Planes:

Points:

this diagram shows the structure of point

In math point is used to represent the correct location or position or place. In math point has been represented by capital letters like A, B, C and D. Mark a point on a sheet of paper as like above diagram.

Properties of points:

Point has no length
Point has no breath
Point has no thickness
Point has no starting point
Point has no end point

Lines - Points Lines and Planes:

Lines:

A lines can be formed by connected set of infinitely points. The line is extended in both the directions.

Draw a line:

With the help of pencil and scale we can draw a line in the plains.

this diagram shows the structure of line

Parallel lines:

When two lines are placed on a plane in same distance without intersection is called as parallel line. The term parallel in math can be represented as ||. For example PQ || to YZ. he above statement shows that the line PQ is parallel to the line YZ. The line PQ are parallel to the line YZ, if they don’t have any common point or center point.

The above diagram shows the structure of parallel line.

Properties of lines:

Line has infinite length
Line has zero width
Line has zero height

Planes - Points Lines and Planes:

Plains:

Plane consists of four sides. Four sides closed part is called as planes.



Properties of planes:

Plane has infinite or unlimited length
Plane has unlimited or infinite width.
Plane has zero thickness (zero height)
Plane consists four sides
Planes are marked by letters like P, Q, R and S.

Segments of Equal Length

The segments are one of the basic concepts of geometry in math subject. The segment is a small piece of a full figure. Segments are two types. There are line segment and circle segment. The line segments are the small part on a large and straight infinity line, which is having both sides of end points. length is the linear extent or measurements of one end point to another one end point, usually being a long dimensions. Here in this article we are going to explain about segments and equal length of segments.


General Definition for Segments:

The segments are the small part or small distance of a long and straight infinity line, which is having two end points on both sides of an infinity line.
A straight line that is joining with two end points, those are coordinates without extending the line after that the end point. One end point to another one end point distance is called as length.


Example figure for “General line segments”:

Line segment

The above example segment figure is, an infinity line is xy, and it is having a segment AB. It is called segment` bar (AB).`

Examples for Segments of Equal Length:

The following given example line segment figure, is example for equal length of line segments. Here an infinity line is AB, and it is having 4 line segments are CE, ED, DE and EC.

Line segment length

It is called the names of line segment `bar (CE), bar (ED), bar (DE) and bar (EC)` .
Those all segments are equal lengths.

Another example for segments of equal length: (square)

Line segment of equal length



Explanatory equal length of segments:

Segment is nothing but the line segments. That is having an end point on all directions.

This above figure represented the Closed and regular polygon (square) figure.
Because of 4 end points are there in the figures, and those all points are joining with each other. And makes the closed figure that is made up of line segments.
This above figure having the 4 end points like A, B, C, and D and alla are equal lengths.
Those 4 points are makes the many line segments like, those line segment names for line segment AB, BC, CD, and DA. Those 4 are original and the reverses of above line segments are BA, CB, DC, and AD.
So those 4 line segments are named as` bar (AB), bar (BC), bar (CD), bar (DA)` , and then reverse 4 are named as `bar (BA), bar (CB), bar (DC), bar (AD).`
This is a square figure, so all the sides are equal, so same like that all the segments length are also equal.
It is having the line segments` bar (AB), bar (BC), bar (CD), bar (DA)` , and etc…

These all are the important notes in the examples of segments. And the above explanations and examples very are useful and easy for understand the equal length of segments.

Simplify and Write in Degrees

In math, degree means a position of a scale or amount of quality. It is mainly used measure the position of the angle. There are different types of angles like as acute angle, obtuse angle, right angle, straight angle, reflex angle and complete angle.

For example,

35^@ this is a way to write in degrees.
How to Learn Simplify and Write in Degrees:-

In the following example to explain the how to learn simplify and write in degrees:

Convert pi/18 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/18 xx (180^@)/pi (Multiplying the both values)
= (180^@)/18 (Cancel the pi value then divide the both numerator and denominator value)
= 10 degrees

Finally we get an answer as 10 degrees.


Example Problems for Simplify and Write in Degrees:-

Problem 1:-

Simplify pi/9 and write in degrees

Solution:

Given: pi/9

Convert pi/9 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/9 xx (180^@)/pi (Multiplying the both values)
= (180^@)/9 (Cancel the pi value then divide the both numerator and denominator value)
= 20^@ degrees

Finally we get an answer as 20^@ degrees.


Problem 2:-

Simplify pi/30 and write in degrees

Solution:

Given: pi/30

Convert pi/30 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/30 xx (180^@)/pi (Multiplying the both values)
= (180^@)/30 (Cancel the pi value then divide the both numerator and denominator value)
= 60^@ degrees

Finally we get an answer as 60^@ degrees.


Problem 3:-

Simplify pi/36 and write in degrees

Solution:

Given: pi/36

Convert pi/36 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/36 xx (180^@)/pi (Multiplying the both values)
= (180^@)/36 (Cancel the pi value then divide the both numerator and denominator value)
= 5^@ degrees

Finally we get an answer as 5^@ degrees.



Problem 4:-

Simplify pi/60 and write in degrees

Solution:

Given: pi/60

Convert pi/60 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/60 xx (180^@)/pi (Multiplying the both values)
= (180^@)/60 (Cancel the pi value then divide the both numerator and denominator value)
= 30^@ degrees

Finally we get an answer as 30^@ degrees.

R Bar Statistics

The r- bar Statistics is the science of creating ordered use of arithmetic data between the groups of individuals. The r-bar statistics deals with all features of this, jointly with not only the group, study and understanding of such data, but also the planning of the collection of data. The r-bar is the average range of set observations. The r-bar statistics is mainly used to find the average range between the groups of data series. Range measures variability of process.


Formula to Find R Bar Statistics:

Range is defined as the difference between the Maximum value to the Minimum value.

Range=Maximum value – Minimum value

R-Bar defined as the average range of set of observations

R-bar (Average range) is defined as
R-Bar =` ("Range1" + "Range2") / 2`
Coefficient of range is defined as
Coefficient of range = `("MaxValue"** "MinValue") / ("MaxValue" **" MinValue")`
Example Problems for Solving R Bar Statistics:

Example1:

Find out the r-bar (average range) for the following datas?

Series1:  2, 5,9,14,17,23,39.

Series2: 13,17,23,28,35,45,49

Solution:

Range for series1 is

Here the maximum value=39, minimum value=2

Range=Maximum value – Minimum value

=39-2

=37

Range for Series2 is

Here the maximum value=49, minimum value=13

=49-13

=36

R-bar = `("Range1"+"Range2")/2`

= `(37+36)/2`

R-bar =36.5

Example2:

Find out the r-bar (average range) for the following datas?

Series1:  10, 13, 15, 19,21,25,28

Series2: 13,17,19,22,25,27,29

Solution:

Range for series1 is

Here the maximum value=28, minimum value=10

Range=Maximum value – Minimum value

=28-10

=18

Range for Series2 is

Here the maximum value=29, minimum value=13

=29-13

=16.

R-bar = `("Range1"+"Range2")/2`

= `(18+16)/2`

R-bar =17

Practice problems for solving r bar statistics:

Problem1:

Find out the r-bar (average range) for the following datas?

series1: 12, 23,32,40,55,65,78,90.

series2: 10, 20, 36, 92, 95,40,50,56

Answer: 62

Problem2:

Find out the r-bar (average range) for the following datas?

series1: 50        56        80        70        60        60        88.

series2:   36        92        95        40        50       56        80

Answer: 48.5

Typical Relationship Problems

The following are the typical relationship problems. These problems are solved by using logic.

Problem: - Pointing out to a photograph, a man tells his friend, “She is the daughter of the only son of my father’s wife.” How is the girl in the photograph related to the man?

Solution: - The relations may be analysed as follows:

Father’s wife – Mother; Mother’s only son – Himself.

So, the girl is man’s daughter.

Problem: - Pointing to a man on the stage, Rita said, “He is the brother of the daughter of the wife of my husband.” How is the man on the stage related to Rita?

Solution: - The relations may be analysed as follows:

Wife of husband – Herself; Brother of daughter- Son.

So, the man is Rita’s son.

Problem: - A man pointing to a photograph says, “The lady in the photograph is my nephew’s maternal grandmother.” How is the lady in the photograph related to the man’s sister who has no other sister?

Solution: - Clearly, the lady is the grandmother of man’s sister’s son i.e., the mother of the mother of man’s sister’s son i.e., the mother of man’s sister.

So, the lady is man’s mother.

Problem: - Pointing out to a lady, a girl said, “She is the daughter-in-law of the grandmother of my father’s only son.” How is the lady related to the girl?

Solution: - Father’s only son- My brother; Grandmother of my brother – My grandmother;

Daughter-in-law of my grandmother – My mother.

So, the lady is girl’s mother.
Some Unsolved Typical Relationship Problems

Problem: - Pointing to a lady, a man said, “The son of her only brother is the brother of my wife.” How is the lady related to the man?           (Answer: Sister of father-in-law)

Problem: - Introducing a man to her husband, a woman said, “His brother’s father is the only son of my grandfather.” How is the woman related to this man? (Answer: Sister)

Problem: - A, B, C, D, E, F and G are members of a family consisting of four adults and three children, two of whom, F and G are girls. A and D are brothers and A is a doctor. E is an engineer married to one of the brothers and has two children. B is married to D and G is their child. How is C related to A?

(Answer: C is A’s son.)

Boolean Algebra Complement

Boolean algebra operators are AND, OR, NOT and equal. The process of NOT operator is found the complement or opposite. Through the truth table we can learn Boolean algebra operation easily. Truth table is connected with the Boolean algebra. In the truth table the expression ‘true is represented as ‘T and the expression ‘false is represented as ‘F. Let us see Boolean algebra complement and their symbols in this article.
Boolean Algebra Complement:

Only one operand is present in the NOT operator. For example ~X here the symbol ~ is used to denote expression opposite or complement. The most important operation of operator NOT is complement or opposite.

Some rules in Boolean algebra complement:

Complement of PQ:

(PQ) = P + Q

Complement of P Q

(P Q) =P + Q

Complement of P + Q + R

(P + Q + R) = P * Q * R

Complement of PQR:

(PQR) = P + Q + R

Easy way to remember this Complement rule is put * for + and put + for *

For example:

Complement of PQ = > (PQ) = P + Q
Proof for Boolean Algebra Complement:

Prove:

Complement of P Q `!=` PQ

Proof:

Given

(P Q) `!=` PQ

LHS:

P Q

Let us consider P=1 and Q =0

F = P Q

F = 1 0

F= 0 1

F = 0

Take the complement of F

We get

F =1

RHS:

PQ

P=1 and Q =0

F = 1 0

F =0

LHS is not equal to RHS therefore (P Q) `!=` PQ

Prove:

Complement of P Q = P + Q

Proof:

Given

(P Q) = P + Q

LHS:

P Q

Let us consider P=1 and Q =0

F = P Q

F = 1 0

F= 0 1

F =0

Take the complement of F

We get

F =1

RHS:

P+Q

P=1 and Q =0

F = 1 + 0

F =1

LHS is equal to RHS therefore (P Q) = P + Q.

Prove:

(PQ) = P + Q

Proof:

Given:

(PQ) = P + Q

LHS:

(PQ)

Let us consider P=1 and Q =0

F = (1 0)

F = (0)

F =1

RHS:

P + Q

Let us consider P=1 and Q =0

F =1 + 0

F =0 + 1

F = 1

LHS is equal to RHS therefore (PQ) = P + Q