Trigonometric Functions of Sum

trigonometric functions of sum and difference of two angles:

The trigonometric functions of sum and difference of two angles A are B are listed below:
Sin (A+B) = Sin (A) Cos (B) + Cos (A) Sin (B) .................. (1)
Sin (A-B) = Sin (A) Cos (B) - Cos (A) Sin (B) .................... (2)
Cos (A+B) = Cos (A) Cos (B) - Sin (A) Sin (B) ................... (3)
Cos (A-B) = Cos (A) Cos (B) + Sin (A) Sin (B) ................... (4)
Adding (1) and (2) we get, 
Sin (A+B) + Sin (A-B) = 2 Sin (A) Cos (B)
Subtracting (1) and (2), we get,
Sin (A+B) - Sin (A-B) = 2 Cos (A) Sin (B)
Similarly, adding (3) and (4), we get,
Cos (A+B) + Cos (A-B) = 2 Cos (A) Cos (B)
Subtracting (3) and (4), we get,
Cos (A+B) - Cos (A-B) = - 2 sin (A) sin (B)
These can be written in the forms
Sin (A+B) + Sin (A-B) = 2 Sin (A) Cos (B) ............................ (5)
Sin (A+B) - Sin (A-B) = 2 Cos (A) Sin (B) ............................. (6)
Cos (A+B) + Cos (A-B) = 2 Cos (A) Cos (B) ........................ (7)
Cos (A - B) - Cos (A+B) = 2 sin (A) sin (B) ........................... (8)
Note: The order on the right-hand side of (8) must be carefully noted
The sum and difference of two angles A and B of a Tangent function are given below:
Tan (A+B) = [Tan (A) + Tan (B)] / [1 - Tan (A) Tan (B)] .................. (9)
Tan (A-B) = [Tan (A) - Tan (B)] / [1 + Tan (A) Tan (B)] ................... (10)

Example for trigonometric functions of sum and difference of two angles

 Express as the sum of two trigonometrical ratios 
sin (5θ) cos (3θ)
Sol : Using the Identity:  2 Sin (A) Cos (B) = Sin (A+B) + Sin (A-B)  
On substitution, we get,
sin (5θ) cos (3θ) = (1/2) {Sin (5θ+3θ) + Sin (5θ-3θ)  
= (1/2) {Sin (8θ) + Sin (2θ) 

More examples for trigonometric functions of sum and difference of two angles

Change in to sum of trigonometric functions: Sin (70^o) Sin (20^o)
Sol: Using,   2 sin (A) sin (B) = Cos (A - B) - Cos (A+B)
Sin (70^o) Sin (20^o) = (1/2) { Cos (70^o - 20^o) - Cos (70^o+20^o) }
                            = (1/2) { Cos (50^o) - Cos (90^o) }
= (1/2) {Cos (50^o) }           Since Cos(90^o) = 0

Square Pyramid

Pyramid is one where the surface of outer area looks triangular and it converges at that point. They are congruent .Square pyramid is a pyramid in which the base of pyramid looks to be square. The most popular form of square pyramid is Johnson square pyramid. If the apex is perpendicularly on top of the center of the square, it will have C4v symmetry.

what is square pyramid

Formulae for square pyramid:

Volume of a square pyramid V = `1/3` (Ah)

= `1/3` a2 h     (Area of square = a2)

Height of square pyramid h = (3 * V) / a2

Where, A -----> area of base of a pyramid

V -----> Volume of a pyramid

a ----->  Side length

what is square pyramid

Surface area of a square pyramid = side2 + 2 x side x length

what is square pyramid

Example problems for square pyramid:

ExampleExample 1:

Find the surface area of a square based pyramid with the given sides 6 cm, height 7 cm, and the slant height 8 cm.

Solution:

Formula:

Surface area of a square pyramid = side2 + 2 x side x length

Given:

Side = 6 cm, length = slant height = 8cm

Surface area             = (6)2 + 2 x 6 x 8

= 36 + 96

= 132

The Surface area of square pyramid is 132 square cm.

ExampleExample 2:

Find the volume of square pyramid for a given dimensions base = 5cm, height = 6cm.

Solution:

Formula:

Volume of the pyramid = `1/3` * Area of base * height.

Given:

Base = 5cm, height = 6cm.

Area of base = b2

Volume = `1/3` * 52 * 6 cm3

= `1/3` * (150) cm3

= 50 cm3

The Final answer is 50 cm3.

ExampleExample 3:

Find the volume of the square pyramid whose base area is 16 cm and height is 12 cm.

Solution:

Formula:

Volume of the pyramid = `1/3` * B*h

Given:

Base area is 16 cm and height is 12 cm.

Volume of the pyramid = `1/3` * 16 * 12

= `1/3` * 192

= `192/3`

= 64 cm3

Practice problems:

Problem 1:

Find the volume of square pyramid for a given dimensions base = 2cm, height =3cm.

Solution:

Volume of a square pyramid is 4cm3.

Problem 2:

Find the volume of the square pyramid whose base area is 15 cm and height is 13 cm.

Solution:

Volume of a square pyramid is 65cm3.

Decimal in Words

In mathematics, a decimal in words is nothing but, it is only represent the decimal numbers. The heart of our number system is considering by place value. Many ways can be represent the decimal numbers. But the important one is place value. The numbers in a base-10 numeral system is specified as decimal notation. A dot with a decimal number, like to present in 41.602. Decimal powers in decimals, (1, 10, 100, and 1000) and secondary symbols for half these values (5, 50, and 500) are contained as roman numerals.

How to write Decimal System in words

Below are examples which will show you how to write decimals in words:

1. Write the decimals in words for given decimal number, 5.32

Solution:

Given the decimal number is 5.32,

The (1.32) decimals in words is write,

Five and thirty two hundredths.

2. Write the decimals in words for given decimal number, 65.44

Solution:

Given the decimal number is 65.44,

The (65.44) decimals in words is write,

Sixty five and forty four hundredths.

3. Write the decimals in words for given decimal number, 24.65

Solution:

Given the decimal number is 24.65,

The (24.65) decimals in words is write,

Twenty four and sixty five hundredths.

4. Write the decimals in words for given decimal number, 36.875

Solution:

Given the decimal number is 36.875,

The (36.875) decimals in words is write,

Thirty six and eight hundred seventy five thousandths.

5. Write the decimals in words for given decimal number, 0.02

Solution:

Given the decimal number is 0.02,

The (0.02) decimals in words is write,

Two hundredths.

Solved Examples

1. Write the decimals in words for given decimal number, 6000.32

Solution:

Given the decimal number is 6000.32,

The (6000.32) decimals in words is write,

Six thousand and thirty two hundredths.

2. Write the decimals in words for given decimal number, 475.00055

Solution:

Given the decimal number is 475.00055,

The (475.00055) decimals in words is write,

Four hundred seventy five and fifty five hundred thousandths.

3. Write the decimals in words for given decimal number, 98.84

Solution:

Given the decimal number is 98.84,

The (98.84) decimals in words is write,

Ninety eight and eighty four hundredths.

4. Write the decimals in words for given decimal number, 637.674.

Solution:

Given the decimal number is 637.674,

The (637.674) decimals in words is write,

Six hundred thirty seven and six hundred seventy four thousandths.

5. Write the decimals in words for given decimal number, 9.69

Solution:

Given the decimal number is 9.69,

The (9.69) decimals in words is write,

Nine and sixty nine hundredths.

Solve Equation Numerically

In mathematics, to solve an equation is to find what values (numbers, functions, sets, etc.) fulfill a condition stated in the form of an equation (two expressions related by equality). These expressions contain one or more unknowns, which are free variables for which values are sought that cause the condition to be fulfilled. To be precise, what are sought are often not necessarily actual values, but, more in general, mathematical expressions (Source from Wikipedia). Here are going to solve equation numerically and its examples.

Solving equation numerically - Example problems:

Solve equation numerically - Example: 1

To solve the following equation,

2x+3+3x =8

Solution:

Add the like terms

Here the like terms is 2x and 3x

2x+3x=5x

Therefore the equation is

5x+3=8

Add both sides -3

5x+3-3 =8-3

5x=5

Divide both sides 5

There fore the value of x=1

Solve equation numerically -  Example: 2

Solve the following equation by substitution method

x-y =3

2x+4=6

Solution:

x-y =3…………….1

2x+4=6……………2

From the equation 1 can we write

x=y+3……………..3

Substitute x value in equation 1

Therefore

2x+4=6

2(y+3) =6

2y+6=6

Add both sides -6

2y+6-6=6-6

2y=0

y=0

Substitute y=0 in equation 3

x=y+3

x=0+3

x=3

Therefore the solution is x=3, y=0

Solve equation numerically - Example:3

To solve the following expression,

5x+3 = 2x+1

Move the 3 to the right hand side by subtracting 3 from both sides, like this:

From the left hand side:
3 - 3 = 0

The answer is 5x

From the right hand side:
1 - 3 = -2

The answer is -2+2x

Now, the equation reads:

5x = -2+2x

Move the 2x to the left hand side by subtracting 2x from both sides, like this:

From the left hand side:

5x - 2x = 3x

The answer is 3x

From the right hand side:

2x - 2x = 0

The answer is -2

Now, the equation reads:

3x = -2

To isolate the x, we have to divide both sides of the equation by the other variables around the x on the left side of the equation.

The solution to your equation is:

x =-2/3

Geometry Practice Trapezoids

Geometry is the mathematics that deals with  measurement, properties and relationships of lines, points, angles, surfaces, and solids. In geometry trapezoid is a shape that has a pair of opposite sides of parallel, the oter two sides would not be parallel. The shape of the trapezoid differs but it should satisfy the property of pair of opposite sides are parallel. Let us take several geometry practice trapezoid probblems.

Trapezoid

Geometry practice trapezoids:

Problem for Geometry practice trapezoids

Example 1 :

Find the area of the trapezoid of base b1 = 5cm, b2 = 6 cm and height of the trapezoid is 7 cm.

Trapezoid

Solution:

The b1= 5cm, b2 = 6 cm and height h = 7 cm.
The area of a trapezoid is one-half the height h times the sum of the base lengths b1 and b2 .

Area of the trapezoid = `1/2 ` (b1 +b2)* h square units.

= `1/2` (5 + 6) * 7

= `1/2` * 11 * 7

= `77/2`

= 38.5 cm2

Area of the trapezoid is equal to 38.5 cm2

Example 2:

Find the area of the trapezoid of base b1 = 4cm, b2 = 7 cm and height of the trapezoid is 7 cm.

Trapezoid

Solution:

The b1= 4cm, b2 = 7 cm and height h = 7 cm.
The area of a trapezoid is one-half the height h times the sum of the base lengths b1 and b2 .

Area of the trapezoid = `1/2 ` (b1 +b2)* h square units.

= `1/2` (4 + 7) * 7

= `1/2` * 11 * 7

= `77/2`

= 38.5 cm2

Area of the trapezoid is equal to 38.5 cm2

More problems on geometry practice trapezoids:

Example 3:

Find the perimeter of the trapezoid  if the four sides of the trapezoid are given as 5 cm, 7 cm, 4 cm, 6 cm.

Trapezoid
TRAPEZOID

solution:

Given the four sides of the trapezoid as  a = 5cm, b = 7cm, c = 4cm, d = 6cm.

The perimeter of the trapezoid (P)= a+b+c+d square units

= 5 + 7 + 4 +6

= 22 cm2

Example 4:

Find the perimeter of the trapezoid  if the four sides of the trapezoid are given as 6 cm, 7 cm, 4 cm, 5 cm.

Trapezoid
TRAPEZOID

solution:

Given the four sides of the trapezoid as  a = 6cm, b = 7cm, c = 4cm, d = 5cm.

The perimeter of the trapezoid (P)= a+b+c+d square units

= 6 + 7 + 4 + 5

= 22 cm2

Learn Online the Chain Rule

Rule for solving the derivative of a composition of two functions.

If y is a function of   u   and   u is a function of   x, then y is a function of x.

The chain rule notify us how to find the derivative of  y   with respect to x

Online:

Through online to help you learn better and faster. In fact they not only help you with accepting Math concepts better but also help you with your Math homework and assignments.

Chain rule formula:

f (x) = g (h(x))

f’(x) = g’ (h(x) h’(x)

(dy) / (dx) = (dy)/(du) xx (du)/(dx)

Learn online the Chain Rule - Examples:

Learn online the Chain Rule - Example 1:

Differentiate y = (3x + 1)2

D (3x + 1)2 = 2(3x +1) 2-1 D (3x + 1)

= 2 (3x + 1) (3)

= 6 (3x + 1).

Learn online the Chain Rule - Example 2:

Differentiate y = (1 -4x + 7x5) 30

The outer is the 30th power and the inner is (1 – 4x +7x5). Differentiate the 30th power first, leaving (1 – 4x +7x5) unchanged. Then differentiate (1 – 4x +7x5). Thus,

D ( 1– 4x + 7x5) = 30 ( 1-4x+7x5) 30-1 D( 1-4x+7x5)

= 30 (1-4x+7x5) 20-1 (-4+35x4)

= 30(35x4 – 4)(1-4x+7x5)29.

= (4x+x-5)1/2

Learn online the Chain Rule - Example 3:

Differentiate the following:

Y = sin(x2+3)

Let u = x2 +3 so y = sin u

(du)/(dx) =2x         (dy)/(du) = cos u

(dy)/(dx)  = cos ux 2x

(dy)/(dx) = 2xcos(x2+3)

Example 4:

Y = ln (3x3-4x+2)

U = 3x3 - 4x + 2         y = ln u

(du)/(dx) = 9x2 -4          (dy)/(du) = 1/u

(dy)/(dx)  = 1/u xx (9x2 – 4)

(dy)/(dx) = (9x^2 - 4)/(3x^3 - 4x + 2)

Note that (dy)/ (dx) = (u')/u

In general if

Y = ln f(x) then (dy)/(dx) = (f'(x))/f(x)

This is a useful result to remember

Example 5:

Differentiate the following:

Y = sin 2 4x

Now sin 2 4x = (sin 4x)2

Let u = sin 4x                  y = u2

(du)/(dx) = 4 cos 4x       (dy)/(du) =2u

(dy)/(dx) = 2u x 4cos 4x

(dy)/(dx)  = 8 sin 4x cos 4x

Learn online the Chain Rule - Practice Problems:

Practice problem 1:

Differentiate y = sqrt (3e^x + 4)

Answer:

(3e^x)/ (2(sqrt(3e^x + 4)))

Practice problem 2

Differentiate  y = (3x2+1)2

Answer:

36x3 + 12x

Inequality Notation

Inequalities are algebraic equations and it variable x cannot be equal. Inequalities have the different values. In general, inequality notation is used to describe intervals. The inequality notations are,

Greater than

Less than

= Greater than or equal to

= Less than or equal to

The inequality also has chained inequality notation abc which means ab and bc.

Inequality Notation Properties:

The inequality notation “a  b" is read as "a is less than b"
The inequality notation "a  b" is read as "a is greater than b"


These notations define as the sense of the inequality.

(a) If the signs of inequality point in the same direction the same logic is used

(b) If the signs of inequality point in the opposite direction then it is opposite logic/sense.

Inequality Notation for intervals:

Example:

Use inequality notation to illustrate each set.

All the Y in the interval (-4, 8)
X is nonnegative


Solution:

The inequality notation is,

-4  y  8,

0  x

Example problems on inequality:

Example 1:

Solve the inequality 4x+ 2  3

Solution:

4x+ 2  3

Subtract 2 from both sides

4x+2 -2  3-2

4x  1

Divide both sides by 4

4x/4  1/4

x1/4

This means that ANY number greater than ¼ will make the equation 4x+ 2  3 true!

Example 2:

Solve the inequality y + 12  24

Solution:
Simply subtract 12 from each sides:

y + 12 -12   24 -12

y  12


This means that ANY number less than 12 will make the equation y + 12  24 true!

Examples 3:

Solve the inequality       x-35 and 2x+14 16

Solution:

Step 1: Solve the first inequality

X-35

Add 3 on both sides,

x-3 + 3 5+3

x8

Step 2: solve the second inequality

2x+14 16

Subtract 14 from both sides,

2x+14-14  16 -14

2x12

Divide by 2 on both sides

x6

The final solution is: x8 and x6

This means that 6x8.