# How to prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically

In this post I’ll be demonstrating how one can prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically…

First of all, let me show you this diagram…

sin(A-B)=sin(A)cos(B)-cos(A)sin(B) proof

*If you click on the diagram, you will be able to see its full size version.

Now, to begin with, I will have to write about some of the properties related to the diagram…

Property 1:

Angle B + (A – B) = B + A – B = A

Therefore, angle POR = A.

Property 2:

Angle OPS = 90 degrees

Property 3:

Length OS = 1

Also note:

All angles within a triangle on a flat plane should add up to 180 degrees. If you understand this rule, you will be able to discover why the angles shown on the diagram are correct. Angles which are 90 degrees are shown on the diagram too.

PROVING THAT SIN(A-B)=SIN(A)COS(B)-COS(A)SIN(B)

Since I’ve noted down some of the important properties related to the diagram, I can now focus on demonstrating why the formula above is true. I will demonstrate why the formula above is true using mathematics and the SOH CAH TOA rule…

$\sin { \left( A-B \right) } =\frac { O }{ H } =\frac { ST }{ 1 } =ST$

But it turns out that…

$ST=PR-PQ$

Because:

$QR=ST$

Now, what is PR and what is PQ?

$\sin { \left( B \right) } =\frac { O }{ H } =\frac { PS }{ 1 } =PS\\ \\ \cos { \left( B \right) } =\frac { A }{ H } =\frac { OP }{ 1 } =OP\\ \\ \sin { \left( A \right) } =\frac { O }{ H } =\frac { PR }{ \cos { \left( B \right) } } \quad \\ \\ \therefore \quad \sin { \left( A \right) } \cos { \left( B \right) } =PR\\ \\ \cos { \left( A \right) } =\frac { A }{ H } =\frac { PQ }{ \sin { \left( B \right) } } \\ \\ \therefore \quad \cos { \left( A \right) } \sin { \left( B \right) } =PQ$

And finally, to sum it all up:

$ST=PR-PQ\\ \\ \therefore \quad \sin { \left( A-B \right) =\sin { \left( A \right) } \cos { \left( B \right) } -\cos { \left( A \right) } \sin { \left( B \right) } }$

Need a better explanation? Watch this video…

Related Videos:

Related posts:

Simple But Elegant Way To Prove That sin(A+B)=sinAcosB+cosAsinB (Edexcel Proof Simplified)

# Properties of C squared, Pythagorean Theorem

In this post, I’ll be writing about some peculiar properties of C squared in Pythagoras’ theorem.

Look at this diagram very carefully…

*What are the weird properties of C^2..? It turns out that A1=A2 and A3=A4. A2 + A4 = C^2.

It turns out out that area A1 is equal to area A2, and that area A3 is equal to area A4:

A1 = A2

A3 = A4

This can be proven because:

1. ${ A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$
2. ${ x }^{ 2 }+{ D }^{ 2 }={ B }^{ 2 }$
3. ${ \left( C-x \right) }^{ 2 }+{ D }^{ 2 }={ A }^{ 2 }$

Now, due to the above:

${ D }^{ 2 }={ B }^{ 2 }-{ x }^{ 2 }\\ \\ { D }^{ 2 }={ A }^{ 2 }-{ \left( C-x \right) }^{ 2 }\\ \\ \therefore \quad { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-{ \left( C-x \right) }^{ 2 }\\ \\ { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-\left\{ { C }^{ 2 }-2Cx+{ x }^{ 2 } \right\} \\ \\ { B }^{ 2 }-{ x }^{ 2 }={ A }^{ 2 }-{ C }^{ 2 }+2Cx-{ x }^{ 2 }\\ \\ { B }^{ 2 }={ A }^{ 2 }-{ C }^{ 2 }+2Cx\\ \\ { B }^{ 2 }={ A }^{ 2 }-\left\{ { A }^{ 2 }+{ B }^{ 2 } \right\} +2Cx\\ \\ { B }^{ 2 }={ A }^{ 2 }-{ A }^{ 2 }-{ B }^{ 2 }+2Cx\\ \\ { B }^{ 2 }=-{ B }^{ 2 }+2Cx\\ \\ 2{ B }^{ 2 }=2Cx\\ \\ \therefore \quad { B }^{ 2 }=Cx\\ \\$

But… B^2 is actually the area A1 and Cx is the area A2, which means that A1=A2.

Now, if B^2=Cx, this means that:

${ A }^{ 2 }+Cx={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }={ C }^{ 2 }-Cx\\ \\ { A }^{ 2 }=C\left( C-x \right) \\ \\$

However, A^2 is equal to the area A3, and C(C-x) is equal to the area A4 – which means that A3=A4. Hence, we’ve proven that:

A1=A2

A3=A4

Related:

2 ways to derive Pythagoras’ equation from scratch

# 2 ways to derive Pythagoras’ equation from scratch

The other day I discovered one more way to derive Pythagoras’ equation from scratch, completely by accident. I was deriving Pythagoras’ equation using the usual method, whilst navigating  a diagram similar to the one below, but without (B-A) measurements…

*Note (regarding diagram above): x+y = 90 degrees

The usual method goes like this…

The area of the largest square is:

${ \left( A+B \right) }^{ 2 }$

It is also:

$4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }$

Which means that:

${ \left( A+B \right) }^{ 2 }=4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }\\ \\ { A }^{ 2 }+2AB+{ B }^{ 2 }=2AB+{ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Now, when I added the lengths (B-A) to my diagram, which are included in the diagram above, I discovered a new way to derive Pythagoras’ equation…

I did this by focusing on the area C^2. It turns out that:

$4\cdot \frac { 1 }{ 2 } AB+{ \left( B-A \right) }^{ 2 }={ C }^{ 2 }$

And since:

${ \left( B-A \right) }^{ 2 }\\ \\ ={ \left( A+B \right) }^{ 2 }-4AB\\ \\ ={ A }^{ 2 }+2AB+{ B }^{ 2 }-4AB\\ \\ ={ B }^{ 2 }-2AB+{ A }^{ 2 }$

I was able to say that:

$4\cdot \frac { 1 }{ 2 } AB+\left\{ { B }^{ 2 }-2AB+{ A }^{ 2 } \right\} ={ C }^{ 2 }\\ \\ 2AB+{ B }^{ 2 }-2AB+{ A }^{ 2 }={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Obviously, I was quite pleased. Have you discovered other ways in which to derive Pythagoras’ equation??

Related:

Video on how to come up with Pythagoras’s equation…

How To Come Up With Pythagoras’s Equation

# How to add up all the even numbers from 0 onwards quickly

In this post, I’ll be demonstrating how you can add up all the even numbers from 0 onwards.

Adding up all the even numbers from 0 to 2:

In this diagram, we are going to say that n=2. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 2 added up, is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 4:

In this diagram, we are going to say that n=4. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 4 added up, is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 6:

In this diagram, we are going to say that n=6. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 6 added up, is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Adding up all the even numbers from 0 to 8:

In this diagram, we are going to say that n=8. The height of the rectangle is (n+2) and its length is n/2. This means that the area shaded in red, which is in fact equal to all the even numbers from 0 to 8 added up, is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

What we’ve discovered:

We’ve discovered that a simple formula can be used to add up all the even numbers from 0 to “n”, whereby “n” is an even number. This formula is:

$\left\{ \left( n+2 \right) \cdot \frac { n }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { n\left( n+2 \right) }{ 4 }$

Alternative method:

There is also an alternative formula you can use to add up even numbers, from 0 onwards. That is:

# How to add up odd numbers from 0 upwards

In this post, I’ll be demonstrating how to add up all the odd numbers from 0 to any specific odd number. To create a robust demonstration, I’ll be taking the footsteps below:

• I’ll first be showing you how to add up all the odd numbers from 0 to 1, using a diagram and formula.
• I’ll then be showing you how to add up all the odd numbers from 0 to 3, using a diagram and formula.
• I’ll also be showing you how to add up all the odd numbers from 0 to 5, using a diagram and also the same formula which was used to count up all the odd numbers from 0 to 1 and 0 to 3.
• And finally, I’ll be using similar diagrams and formulas used to count odd numbers from 0 to 1, 0 to 3 and 0 to 5 to count odd numbers from 0 to 7 and 0 to 9.

What you will find, after I complete the tasks above – is that a pattern emerges. You will notice that the formula I use to count odd numbers from 0 to n (n which is an odd number) is very robust and will allow you to count all the odd numbers from 0 to n very easily.

COUNTING ALL THE ODD NUMBERS FROM 0 to 1:

If you count all the odd numbers from 0 to 1, what you will get is obviously 1. Furthermore, what you will also get as a formula (if n=1, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 1 into n, you will get 1. 1 is the value of all the odd numbers added up from 0 to 1.

COUNTING ALL THE ODD NUMBERS FROM 0 to 3:

If you count all the odd numbers from 0 to 3, what you will get is 4. Furthermore, what you will also get as a formula (if n=3, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 3 into n, you will get 4. 4 is the value of all the odd numbers added up from 0 to 3.

COUNTING ALL THE ODD NUMBERS FROM 0 to 5:

If you count all the odd numbers from 0 to 5, what you will get is 9. Furthermore, what you will also get as a formula (if n=5, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 5 into n, you will get 9. 9 is the value of all the odd numbers added up from 0 to 5.

COUNTING ALL THE ODD NUMBERS FROM 0 to 7:

If you count all the odd numbers from 0 to 7, what you will get is 16. Furthermore, what you will also get as a formula (if n=7, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 7 into n, you will get 16. 16 is the value of all the odd numbers added up from 0 to 7.

COUNTING ALL THE ODD NUMBERS FROM 0 to 9:

If you count all the odd numbers from 0 to 9, what you will get is 25. Furthermore, what you will also get as a formula (if n=9, H=Height and L=Length) is:

$\left\{ H\cdot L \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\left\{ \left( n+1 \right) \cdot \frac { \left( n+1 \right) }{ 2 } \right\} \cdot \frac { 1 }{ 2 } \\ \\ =\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

*If you plug the value 9 into n, you will get 25. 25 is the value of all the odd numbers added up from 0 to 9.

THE FORMULA WHICH CAN BE USED TO ADD UP ALL THE ODD NUMBERS FROM 0 TO n, WHEREBY n IS AN ODD NUMBER:

If you look at each and every diagram and formula above, what you will notice is that the formula

$Formula=\frac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

will allow you to add up all the odd numbers from 0 to n, whereby n is an odd number. The diagrams above have demonstrated why this formula is robust and completely logical. If you need to add up all the odd numbers from 0 to n (n is an odd number), the formula above is one you can trust.

ALTERNATIVE METHOD:

Using the table below, we can come up with an alternative method of calculating every odd number from 0 to n (n is an odd number):

n: Sum Total Total (Exponential form)
1 1 1 1^2
3 1+3 4 2^2
5 1+3+5 9 3^2
7 1+3+5+7 16 4^2
9 1+3+5+7+9 25 5^2

It turns out that:

*Note that 2x+1 can be used to denote an odd number.

# Solving The Student Handshake Problem

The other day, a question came up on a site called Brainly.com.

It went like this…

In a cafeteria, all students shook hands with one another. There were 66 handshakes in total. How many students were in the cafeteria?

As this question is quite interesting, I’m going to explain how you can answer it, and in the process – I’ll also be revealing its answer.

Now, to answer such a question we first have to perform a few experiments and ask ourselves mini questions. The data from these experiments and mini questions will have to be recorded, so that we can spot potential patterns which may ultimately help us create a formula to solve the main problem.

EXPERIMENTS + MINI QUESTIONS

Will a pattern emerge??

1. Firstly, let’s think about how many handshakes there’d be with only one student in this cafeteria. Well, we can say 0. Why would someone shake their own hand?

2. Secondly, how many handshakes would there be if there are 2 students in this cafeteria? Well, the answer to this question is 1. These two students would be able to shake hands with one another.

3. Thirdly, how many handshakes would there be if there are 3 students in the cafeteria? Haha, now things get a little more complicated… To answer this mini question, let’s attach the variables A, B and C to these students {A, B, C}.

It turns out that:

• A can shake hands with B (A,B).
• A can shake hands with C (A,C).
• B can shake hands with C (B, C).

*Possible combinations: (A, B), (A, C) and (B, C).

So the answer to this mini question has to be 3.

4. Fourthly, how many handshakes would there be if there are 4 students in this cafeteria? To answer this question we can use the same strategy we used to answer the third question. Let’s attach the variables A, B, C and D to these students {A, B, C, D}.

It turns out that:

• A can shake hands with B (A,B).
• A can shake hands with C (A, C).
• A can shake hands with D. (A, D).
• B can shake hands with C (B, C).
• B can shake hands with D (B, D).
• C can shake hands with D (C,D).

* Possible combinations: (A, B), (A, C), (A, D), (B, C), (B, D) and (C, D).

So, the answer to this mini question would have to be 6.

5. Fifthly, how many handshakes would there be if there are 5 students in this cafeteria? Using the same strategy we used to answer mini questions 3 and 4 – we will answer this question too. Let’s attach the variables A, B, C, D and E to these students {A, B, C, D, E}.

It turns out that:

• A can shake hands with B (A, B).
• A can shake hands with C (A, C).
• A can shake hands with D (A, D).
• A can shake hands with E (A, E).
• B can shake hands with C (B, C).
• B can shake hands with D (B, D).
• B can shake hands with E (B, E).
• C can shake hands with D (C, D).
• C can shake hands with E (C, E).
• D can shake hands with E (D, E).

So, the answer to this mini question would have to be 10.

Possible combinations: (A,B), (A, C), (A, D), (A, E), (B, C), (B, D), (B, E), (C, D), (C, E) and (D, E).

CAN WE SOLVE THE MAIN PROBLEM WITH A FORMULA? WHAT PATTERN WILL DEFINE THE FORMULA?

Alright… Now that we’ve performed a few experiments and have answered a few mini questions – let’s see if we can spot a pattern in our data. If we can spot a pattern in our data, we may be able to solve the problem relating to 66 handshakes. We need to find a pattern so that we don’t have to answer the main question using brute force and hundreds, if not, thousands of calculations. Remember, solving mathematical problems is all about spotting patterns.

To spot patterns, the best tool we can use is a table. Let’s create a table which contains the information we’ve just produced, related to the mini questions…

Student(s) Handshakes Pattern (Related to handshakes)
1 0 0
2 1 1
3 3 1+2
4 6 1+2+3
5 10 1+2+3+4

Ok… Let’s look at this table carefully. It turns out that a pattern has emerged… As a pattern, we get tidy little sums. The kind of sums that Carl Friedrich Gauss was able to add up, thanks to diagrams such as the one below…

Diagram Explanation:

To add up the sum 1+2+3+4, you simply have to multiply 5 (which is the variable ‘s’ in this case) by (5-1) which is 4, then divide their product ( 5 x (5-1) ) by 2.

(5×4)/2 = 10 = 1+2+3+4.

Notice that:

• When there was one student in the cafeteria, there were 0 handshakes. (0) is 1 less than the number 1.
• When there were two students in the cafeteria, there was 1 handshake. (1) is 1 less than 2.
• When there were 3 students in the cafeteria, there were 3 handshakes. 3 =1+(2). 2 is 1 less than 3.
• When there were 4 students in the cafeteria, there were 6 handshakes. 6=1+2+(3). 3 is 1 less than 4.
• When there were 5 students in the cafeteria, there were 10 handshakes. 10=1+2+3+(4). 4 is 1 less than 5.

Also notice that:

*To understand the pattern below and how it was intuitively discovered, see the diagram which helped Carl Friedrich Gauss neatly add up sums such as 1+2+3+4.

• [ 1 x (1-1) ] / 2 = 0 which is the same as : [ 1 x 0 ] / 2 = 0
• [ 2 x (2-1) ] / 2 = 1 which is the same as : [ 2 x 1 ] / 2 = 1
• [ 3 x (3-1) ] / 2 = 3 which is the same as : [ 3 x 2 ] / 2 = 3
• [ 4 x (4-1) ] / 2 = 6 which is the same as : [ 4 x 3 ] / 2 = 6
• [ 5 x (5-1) ] / 2 = 10  which is the same as : [ 5 x 4 ] / 2 = 10

With this information, we can conclude that:

s = number of students

h = handshakes

[ s x (s-1) ] / 2 = h

And this is the formula we can use to solve all student handshake problems such as the one mentioned at the top of this post. If we plug the value 66 into this formula, we will discover how many students there were in the cafeteria whereby 66 handshakes took place. At the beginning of this post, I said that I would reveal the answer to the main question. To reveal it though, I will have to solve a quadratic equation by completing the square. I will also have to turn the variable ‘h’ into 66. Let’s do this…

$\frac { s\left( s-1 \right) }{ 2 } =66\\ \\ s\left( s-1 \right) =132\\ \\ { s }^{ 2 }-s=132\\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=132\\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=132+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }$

${ \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=132+\frac { 1 }{ 4 } \\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 4\left( 100+30+2 \right) }{ 4 } +\frac { 1 }{ 4 } \\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 400+120+8 }{ 4 } +\frac { 1 }{ 4 } \\ \\ { \left( s-\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 529 }{ 4 } \\ \\ s-\frac { 1 }{ 2 } =\sqrt { \frac { 529 }{ 4 } } \\ \\ s-\frac { 1 }{ 2 } =\frac { 23 }{ 2 } \\ \\ s=\frac { 23 }{ 2 } +\frac { 1 }{ 2 } \\ \\ s=\frac { 24 }{ 2 } \\ \\ \therefore \quad s=12$

We now know that there were 12 students in the cafeteria. Obviously, this problem could have been solved when we knew that s x (s-1) = 132, because 12 x 11 = 132. However, if you get a larger problem, you will need to produce a quadratic formula and complete the square to get an answer.

I hope that this post has shed light on how to solve handshake / people problems. If you have any questions or feedback, please leave a comment below. 🙂

# The quickest Sine Rule proof

In this post I’ll be demonstrating how to prove that the Sine Rule is true in the quickest manner possible.

First of all, let’s begin with writing down the 3 formulas which can be used to find the area of a triangle:

$A=\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 }$

$A=\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 }$

$A=\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 }$

Now, let’s make the first two formulas above equivalent to one another…

$\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 } =\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 }$

Alright, now watch what happens when we multiply both sides of the equation by a handy expression…

$\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 } \cdot \frac { 2 }{ c\cdot \sin { \left( A \right) } \cdot \sin { \left( B \right) } } =\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } \cdot \frac { 2 }{ c\cdot \sin { \left( A \right) \cdot \sin { \left( B \right) } } }$

If we do this, what we’re going to be left with is…

$\frac { b }{ \sin { \left( B \right) } } =\frac { a }{ \sin { \left( A \right) } }$

So far so good! Let’s now make these two area formulas equivalent to one another…

$\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } =\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 }$

And now, let’s multiply both sides of the equation we’ve just created by a handy expression…

$\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } \cdot \frac { 2 }{ a\cdot \sin { \left( B \right) } \cdot \sin { \left( C \right) } } =\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 } \cdot \frac { 2 }{ a\cdot \sin { \left( B \right) } \cdot \sin { \left( C \right) } }$

If we do this, what we’re going to be left with is…

$\frac { c }{ \sin { \left( C \right) } } =\frac { b }{ \sin { \left( B \right) } }$

And it turns out, because:

$\frac { b }{ \sin { \left( B \right) } } =\frac { a }{ \sin { \left( A \right) } }$

$\frac { c }{ \sin { \left( C \right) } } =\frac { b }{ \sin { \left( B \right) } }$

We can say that:

$\frac { a }{ \sin { \left( A \right) } } =\frac { b }{ \sin { \left( B \right) } } =\frac { c }{ \sin { \left( C \right) } }$

I’ve made a video related to this Sine Rule proof. You can watch it below if you wish.

Hope you enjoyed reading this post! 🙂

# Rolling 3 dice… What is most likely to happen?

Understanding probabilities / chance can sometimes be difficult. This is why I’ve written (in detail) an article about what kind of sums you should expect to get when you roll 3 dice and why you should expect to get them. This article reveals some of the mechanics behind our probabilistic theories and should help maths students gain a deeper understanding of the nature of randomness and chance.

Recently, I’ve been doing a bit of coding in HTML, CSS, PHP and MySQL. I’ll be looking to improve mathsvideos.net and also provide students with more mathematics resources. I’m also providing businesses with automation solutions in order to fund my maths project.

Thanks for stopping by!! Tiago. 🙂

For more mathematics proofs, visit https://plus.google.com/b/100450538547176385655/communities/106007058741903558109.

Like probabilities? Why not check out the “Hannah sweets” problem?

Hannah Sweets Maths Problem – Edexcel (June 2015)

# How to derive Euler’s Identity using the Maclaurin Series

In this post I’ll be showing you how to derive Euler’s identity using the Maclaurin Series. It turns out that the Maclaurin series looks like this:

And expanded, it looks like this:

[*A larger version of this image can be found here.]

Now, since we want to derive Euler’s identity, we first have to find out what the formula for e^x looks like. In order to get this formula we must use the table below:

Derivatives of e^x When x=0
${ f }^{ \left( 0 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 0 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 1 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 1 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 2 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 2 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 3 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 3 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 4 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 4 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 5 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 5 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 6 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 6 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 7 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 7 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 8 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 8 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 9 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 9 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 10 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 10 \right) }\left( 0 \right) ={ e }^{ 0 }=1$

Ok. So we’ve got a useful table just above. Let’s write out the function of e^x in its Maclaurin Series form:

$f\left( x \right) ={ e }^{ x }\\ \\ =\frac { 1 }{ 0! } { f }^{ \left( 0 \right) }\left( 0 \right) { x }^{ 0 }+\frac { 1 }{ 1! } { f }^{ \left( 1 \right) }\left( 0 \right) { x }^{ 1 }+\frac { 1 }{ 2! } { f }^{ \left( 2 \right) }\left( 0 \right) { x }^{ 2 }\\ \\ +\frac { 1 }{ 3! } { f }^{ \left( 3 \right) }\left( 0 \right) { x }^{ 3 }+\frac { 1 }{ 4! } { f }^{ \left( 4 \right) }\left( 0 \right) { x }^{ 4 }+\frac { 1 }{ 5! } { f }^{ \left( 5 \right) }\left( 0 \right) { x }^{ 5 }\\ \\ +\frac { 1 }{ 6! } { f }^{ \left( 6 \right) }\left( 0 \right) { x }^{ 6 }+\frac { 1 }{ 7! } f^{ \left( 7 \right) }\left( 0 \right) { x }^{ 7 }+\frac { 1 }{ 8! } { f }^{ \left( 8 \right) }\left( 0 \right) { x }^{ 8 }\\ \\ +\frac { 1 }{ 9! } { f }^{ \left( 9 \right) }\left( 0 \right) { x }^{ 9 }+\frac { 1 }{ 10! } { f }^{ \left( 10 \right) }\left( 0 \right) { x }^{ 10 }+...$

Now, let’s replace

with the values from the table. If we do this, the formula for e^x will become:

Alright, so far, so good. We are certainly on the right track. Our next goal will be to discover what e^(i*x) is. This is because to produce Euler’s identity, we need to come up with:

${ e }^{ ix }=\cos { \left( x \right) } +i\sin { \left( x \right) }$

To come up with the formula above, we will need the table below, because our latest e^(x) formula will have to be transformed. x will be turned into i*x.

Imaginary Numbers Exponentiated
$\sqrt { -1 } =i$
$i\cdot i={ i }^{ 2 }=-1$
${ i }^{ 3 }={ i }^{ 2 }\cdot i=-i$
${ i }^{ 4 }={ i }^{ 3 }\cdot i=-i\cdot i=-{ i }^{ 2 }=1$
${ i }^{ 5 }={ i }^{ 4 }\cdot i=i$
${ i }^{ 6 }={ i }^{ 5 }\cdot i=i\cdot i={ i }^{ 2 }=-1$
${ i }^{ 7 }={ i }^{ 6 }\cdot i=-i$
${ i }^{ 8 }={ i }^{ 7 }\cdot i=-i\cdot i=-{ i }^{ 2 }=1$
${ i }^{ 9 }={ i }^{ 8 }\cdot i=i$
${ i }^{ 10 }={ i }^{ 9 }\cdot i={ i }^{ 2 }=-1$
${ i }^{ 11 }={ i }^{ 10 }\cdot i=-i$

As we’ve got the table above, we can figure out what the formula e^(i*x) would look like:

Since:

This means that:

${ e }^{ ix }=\cos { \left( x \right) } +i\sin { \left( x \right) }$

And finally, when x=π:

${ e }^{ i\pi }=\cos { \left( \pi \right) } +i\sin { \left( \pi \right) } \\ \\ \therefore \quad { e }^{ i\pi }=-1\\ \\ \therefore \quad { e }^{ i\pi }+1=0$

This is because:

$\cos { \left( \pi \right) } =-1\\ \\ \sin { \left( \pi \right) } =0$

You have produced Euler’s identity from almost absolute scratch. Give yourself a pat on the back! 🙂

Related:

Deriving the Taylor Series from scratch

# Deriving the Taylor Series from scratch

[Please note: In order to derive the Taylor Series, you will need to understand how to differentiate. If you know how to differentiate, finding the Taylor Series won’t be much of a problem. You also need to know that 0!=1, 1!=1, 2!=2, 3!=6, x^0=1, x^1=x.]

In this post I will be demonstrating how one can produce the Taylor Series from absolute scratch.

First of all, let’s look at the diagram above. Now, let’s suppose that the equation of the function above is:

$f\left( x+a \right) ={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }+...$

Ok, so we have the equation for the function, however, it isn’t complete. C_0, C_1, C_2, C_3 etc are hidden constants. This means that our second task will be to discover these constants. We need to discover these constants to find the complete equation of the function so that we can arrive at the Taylor Series. Fortunately, this task won’t be too difficult. Let me show you how C_0, C_1, C_2, C_3 etc can be found fairly easily…

When x=0:

${ f }^{ \left( 0 \right) }\left( a \right) ={ C }_{ 0 }\quad \therefore \quad \frac { 1 }{ 0! } { f }^{ \left( 0 \right) }\left( a \right) ={ C }_{ 0 }$

Now:

${ f }^{ \left( 1 \right) }\left( x+a \right) ={ C }_{ 1 }+2{ C }_{ 2 }x+3{ C }_{ 3 }{ x }^{ 2 }+...$

When x=0:

${ f }^{ \left( 1 \right) }\left( a \right) ={ C }_{ 1 }\quad \therefore \quad \frac { 1 }{ 1! } { f }^{ \left( 1 \right) }\left( a \right) ={ C }_{ 1 }$

Also:

${ f }^{ \left( 2 \right) }\left( x+a \right) =2{ C }_{ 2 }+6{ C }_{ 3 }x+...$

When x=0:

${ f }^{ \left( 2 \right) }\left( a \right) =2{ C }_{ 2 }\quad \therefore \quad \frac { 1 }{ 2! } { f }^{ \left( 2 \right) }\left( a \right) ={ C }_{ 2 }$

And, finally:

${ f }^{ \left( 3 \right) }\left( x+a \right) =6{ C }_{ 3 }+...$

When x=0:

${ f }^{ \left( 3 \right) }\left( a \right) =6{ C }_{ 3 }\quad \therefore \quad \frac { 1 }{ 3! } { f }^{ \left( 3 \right) }\left( a \right) ={ C }_{ 3 }$

Alright, so now that we have discovered the hidden constants C_0, C_1, C_2 and C_3, our third task is to write down the complete equation of the function f(x+a). Thanks to the information we have above, the fact that x^0=1 and x^1=x, plus our ability to spot patterns, we will be able to do this quite quickly…

[*Image can be seen here if it appears to be too small on this page.]

And it turns out that the equation we have just above is the Taylor Series function. It can be simplified to look like this…

What is also interesting is that if we transform a=0, we get the Maclaurin Series function which can be used to discover formulas for things such as e^x.

If you have any questions regarding this post, please leave your comments below. Once again, thanks for stopping by! 🙂

Related:

How to derive Euler’s Identity using the Maclaurin Series