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) proof

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.


IMPORTANT FACTS ABOUT THE DIAGRAM

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…


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

How to derive the formula for the area of an equilateral triangle

In this post I’ll be showing you how to derive the formula for the area of an equilateral triangle – in easy steps. In order to understand this derivation properly, you need to be familiar with Pythagoras’ theorem and also a few algebraic rules. What you’ll also need is a ruler, pair of compasses, a pencil and a sheet of paper.

Ready? Let me begin…

Step 1: Put a point on a blank sheet of paper and name it A.

Point A

Step 2: Put the needle of your compass on the point A and draw a circle around it.

Circle...

Step 3: Add a point B to this circle, on its edge.

Point B added...

Step 4: Put the needle of your compass on the point B and your pencil on the point A.

Step 4...

Step 5: Draw another circle with a radius the length AB.

Another circle...

Step 6: Now add a few extra points to your drawing. Call these points C and D.

Points C and D added to drawing...

Step 7: Connect the points A, B and C forming a triangle.

Points A, B and C connected.

Step 8: Draw a line going through the points C and D.

Line through points C and D.

Step 9: Where the line going through C and D intersects the triangle, place the point E.

Point E added...

Step 10: Now look at your latest work very carefully… What you will notice is that the lengths AB, AC and BC are all equal to one another. This is because both the circles you drew – are exactly the same size. They each have radiuses equal in proportion. In simple terms, AB=AC=BC.

What you have to do now is name these lengths (r) for radius. Here’s the thing though, because the line going through C and D splits the triangle (equilateral, as each of its sides has the same length) down its middle, the length AE is equal to 1/2 x r, and similarly the length BE is equal to 1/2 x r. Together, the length AE + BE = AB = r.

Step 11: Remember that I said that the line going through C and D splits the triangle down its middle. Also, notice that this exact line is perpendicular to the length AB. Now, because of this, at the point E, you’ve got two right angles. Name these two right angles big R.

[Knowing that these two angles are equal to 90 degrees is vital – because you’ll be able to use Pythagoras’ theorem to find the length CE.]

Step 12: Find the length CE using Pythagoras’ theorem, Adjacent² + Opposite² = Hypotenuse². You will need this length to find the area of the equilateral triangle you’ve produced.

*Algebraic skills will be required from this point…

{ AE }^{ 2 }+{ CE }^{ 2 }={ AC }^{ 2 }\\ \\ \Rightarrow \quad { \left( \frac { 1 }{ 2 } r \right)  }^{ 2 }+{ CE }^{ 2 }={ r }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }={ r }^{ 2 }-{ \left( \frac { 1 }{ 2 } r \right)  }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 4r^{ 2 } }{ 4 } -\frac { { r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 3{ r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad CE=\sqrt { \frac { 3{ r }^{ 2 } }{ 4 }  } \\ \\ \therefore \quad CE=\frac { r\sqrt { 3 }  }{ 2 } 

Length CE found...

Step 13: Derive the formula for the area (A) of the equilateral triangle. Remember that the area of a right angled triangle is L x W x 1/2.

A=\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 }  }{ 2 } \cdot \frac { 1 }{ 2 } +\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 }  }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } +\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =2\cdot \frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =\frac { 1 }{ 4 } { r }^{ 2 }\sqrt { 3 } 

Presto!!! Keep in mind that you can transform the variable (r) into any variable you wish. This variable (r) is the length of each side of the equilateral triangle you were working with. The formula you’ve derived can be used to find the area of any equilateral triangle.

Proof: Thales’ Theorem

In this post I’ll be demonstrating how you can prove that Thales’ Theorem is true. To follow the steps in this post (11 in total), what you will require is a ruler, pair of compasses and a pencil.

Step 1: Draw a random line on a sheet of paper.

Random line....

Step 2: Place your compass needle on this line, and form a circle.

A circle...

Step 3: Add 4 points to your drawing, as shown below…

Add 4 points to your drawing....

Step 4: Name the points A, B, C and D as shown…

Points A, B, C and D

Step 5: Connect the points A, B and D together to form an isosceles triangle

Points A, B and D connected.

Step 6: Name the lines AB and BD the radius (r)…

Label the radiuses...

Step 7: Since the lines AB and BD are equal to one another, it follows that the angles ∠BAD and ∠BDA are equivalent. This is because the angles below the apex of an isosceles triangle are equal. You must name these angles alpha (α).

Label alpha angles...

Step 8: Now connect the points BC and CD together to form another isosceles triangle…

Another isosceles triangle...

Step 9: The line BC is equal to r… Now label the line BC…

Label the line BC...

Step 10: Because the line BC and BD are both equal to r, the triangle BCD is an isosceles triangle. This means that the angles ∠BCD and ∠BDC must both be equivalent. Call these angles beta (β).

Label the angles beta...

Step 11: Prove that the angle at point D is equal to 90 degrees.

Thales’ Theorem is as follows:

Because AC is the diameter of the circle you drew, the angle at the point D (α+β) must be equal to 90 degrees. In more specific and general terms, if you have the points A, C and D lying on a circle – and the line AC is in fact the diameter of this circle – then the angle at point D (α+β) must be a right angle.

Proof (which must be derived using the diagram you’ve created):

Diagram.

All angles within a triangle (in 2 space) must add up to 180 degrees.

Mathematically, this means that:

\alpha +\alpha +\beta +\beta =180\\ \\ \Rightarrow \quad 2\alpha +2\beta =180\\ \\ \Rightarrow \quad 2\left( \alpha +\beta  \right) =180\\ \\ \Rightarrow \quad \frac { 2\left( \alpha +\beta  \right)  }{ 2 } =\frac { 180 }{ 2 } \\ \\ \therefore \quad \alpha +\beta =90

And as a result, Thales’ theorem must be true. The angle α+β is the angle at point D.

How to prove that the two angles below the apex of an isosceles triangle are equivalent

*You will need a pair of compasses, a ruler, pen and pencil to formulate this proof.

How would you go about proving that an isosceles triangle has two angles (below its apex) equal to one another?

Well, first of all – let’s start off by drawing a circle…

A circle.

Now… We can tell that the circle we’ve just drawn has a centre (point at the centre). Next, what we have to do is add a couple of points to the edge of this circle. Like this…

Circle with points

Let’s name all these points A, B and C…

Points a, b and c

Now, let’s connect these points together with a few lines – to create an isosceles triangle ABC…

Isosceles triangle

Ok… So far, so good… What you will need to do now is – place the needle of your compass on the point C and your pencil on the point B, like this…

Points B and C

Now spin your compass – and create an arc…

An arc...

Next, get the needle of your compass and place it on the point B and put your pencil on the point C…

Points B and C...

Draw another arc, like this…

2nd arc

Where the two arcs you’ve just drawn intersect, create a point… Call this point D…

Now, draw a line going through the points A and D. Call this line L. Line L will be perpendicular to the line BC…

Line going through points A and D...

Where the line L intersects the line BC, create a point E…

Point E

Now, it turns out, within the isosceles triangle ABC, we’ve created two right angles… This is because the line L is perpendicular to the line BC. Remember that the line L cuts the isosceles triangle down its centre. Let’s name these right angles big R…

Right angles within isosceles triangle...

If you look at the diagram above carefully, what you will notice is that the radius of the circle is equal in length to the line AB and also the line AC. Let’s name the lines AB and AC… We’ll call them r.

Circle contains radiuses...

Let’s also name the line BC… We’ll call it x. This means that the line BE is equal to half of x, and because of this, the line CE must also be equal to half of x…

Line x...

Finally (I know you must be tired of drawing), let’s call the angle ABC alpha and the angle ACB beta…

alpha and beta angles...

With our diagram complete, we can now prove that alpha and beta are equivalent to each other.

*You will need to know a bit of trigonometry to pass this point. SOH CAH TOA rules to be precise.

It turns out that:

\cos { \left( \alpha  \right)  } =\frac { A }{ H } =\frac { \frac { x }{ 2 }  }{ r } =\frac { x }{ 2r } 

And also:

\cos { \left( \beta  \right)  } =\frac { A }{ H } =\frac { \frac { x }{ 2 }  }{ r } =\frac { x }{ 2r } 

This means that:

\cos { \left( \alpha  \right)  } =\cos { \left( \beta  \right)  } \\ \\ \therefore \quad \alpha =\beta 

Hence, we’ve proven that an isosceles triangle has two angles (below its apex) equal to one another.

Proof: Opposite angles formed when two lines intersect, are equal to one another

How can we prove that opposite angles (when two lines intersect) are in fact equal to one another?

Well, first of all – let’s draw a circle…

A circle

We know that in a full circle, there are 360 degrees. This is an indisputable fact. Now, what happens if we split this circle in two with a straight line (going through its centre)?

Circle split in two with straight line...

Well, each half of the circle (top and bottom) – will now contain 180 degrees. We know this because:

\frac { 360 }{ 2 } =180

Ok, so far so good… Now, let’s draw another line through the circle (going through its centre) which intersects the first line we have drawn…

Two lines intersecting....

As we can see, because we have done this, we now have 4 different angles. Let’s name the two angles which are situated in the top half of the circle α and β

Name two top angles alpha and beta...

Earlier in this demonstration, we remarked that the top half of the circle (when it was split in two) contained 180 degrees. Mathematically and logically speaking, as this is the case, we must say that:

\alpha +\beta =180

Great, now let’s name the angles in the bottom half of the circle x and y

x and y in the bottom half of the circle...

It follows, because the angles in the top half of the circle add up to 180 degrees, we must deduce that:

x+y=180

So, it turns out we now have two useful equations:

\alpha +\beta =180

x+y=180

Do we have enough to form our proof though? Unfortunately, not quite… We have to look at our most recent figure again, but this time from a different perspective…

You see, there are different top halves and bottom halves…

x and y in the bottom half of the circle...

  • There exists top halves α+β and also… β+y
  • There exists bottom halves x+y but also α+x

You may ask, why is this important? Well, here’s what’s crucial:

\beta +y=180

\alpha +x=180

And now we have 4 different equations, 3 of which – will help us finally complete our proof.

\alpha +\beta =180

x+y=180

\beta +y=180

\alpha +x=180

Here’s why we need these equations…

α+β and α+x are equivalent (180 degrees), so we can deduce that:

\alpha +\beta =\alpha +x\\ \\ \therefore \quad \beta =x

*Subtract α from both sides of the equation.

α+β and β+y are equivalent (180 degrees), so we can deduce that:

\alpha +\beta =\beta +y\\ \\ \therefore \quad \alpha =y

*Subtract β from both sides of the equation.

Hence, we’ve proven that: Opposite angles (when two lines intersect) are equal to one another.

 β=x and α=y:

Opposite angles (when two lines intersect) are equal to one another.

How to quickly double the area of a square (simple geometry lesson)

In this post, I’ll be demonstrating how you can quickly double the area of a square using a simple geometrical trick.

Let’s say you have an ordinary square, like the one below…

Firstly, what you have to do is name the area of this square “A”…

Then, what you do next is divide this square (diagonally) into 4 equal parts…

After you have done this, you then name each part of this square “1/4 x A”…

Remember that A is the entire area of the square, therefore 1/4 x A is a quarter of the area of the square.

Notice now, that to double the area of this square, all you have to do, is double the number of the 1/4 x A right angled triangles which currently exist – then configure them – like this…

As you can see, you’ve now got eight of these 1/4 x A right angled triangles neatly configured…

Not only are you left with a new square, double the size of your original square (follow the lines on the outside of the shape), but a handy equation, which proves that you doubled the area of the square you started off with…

8\times \frac { 1 }{ 4 } A=2A

How about that? 🙂

How to derive the formula for a circle from scratch

If you’d like to derive the formula for a circle from absolute scratch, then your best option would be to draw a diagram such as the one below:

formula for a circle
A circle on the x, y plane.

If you look at this diagram carefully, what you will notice is:

  • A circle exists and each point on this circle has the coordinate (x, y).
  • The centre of the circle can be found at (a, b).
  • The circle has a radius ‘r’.
  • The right angled triangles in the diagram each have an adjacent length, opposite length and hypotenuse (r).

Once you’ve prepared a similar diagram, your next aim should be to turn your attention towards the right angled triangles which exist within the circle. You should also think about the many different right angled triangles which could fit within the circle provided they emanate from the centre point (a, b).

The reason I’ve mentioned these right angled triangles is because according to Pythagoras’ theorem, when you have a right angled triangle – its adjacent length squared plus its opposite length squared is equal to the length of its hypotenuse squared:

Adjacent²+Opposite²=Hypotenuse²

Now, in this case – the adjacent lengths of the right angled triangles which can fit within the circle on the diagram can be described using the expression:

\left( x-a \right)  or \left| x-a \right|

The opposite lengths can be described using the expression:

\left( y-b \right)  or \left| y-b \right| 

Also, very interestingly:

  • Each of the right angled triangles you can think of has a hypotenuse ‘r’.
  • { \left( x-a \right)  }^{ 2 }={ \left| x-a \right|  }^{ 2 }
  • { \left( y-b \right)  }^{ 2 }={ \left| y-b \right|  }^{ 2 }

When you combine all the information above, what you get is a neat formula which looks like this:

{ \left( x-a \right)  }^{ 2 }+{ \left( y-b \right)  }^{ 2 }={ r }^{ 2 }

And it turns out… This is the formula for a circle on the x, y plane, whereby, (a, b) is the centre of the circle and ‘r’ is the length of its radius. How spectacular is that? 🙂

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…pythagoras_2

*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…

pythagoras' diagram

*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:part_1

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:

part_2

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:

part_3

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:

part_4

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:

image_2

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