Category Archives: Mathematics Proofs

Another way to express the golden ratio mathematically

In this post I’m going to be proving that…

\varphi =\frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }

So, here I go…

x=\frac { 1+\sqrt { 5 } }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 1+2\sqrt { 5 } +5 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 6+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 6+2\sqrt { 5 } }{ 4 } -\frac { 4 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2 }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 }

Wait for it…

\Rightarrow \quad { x }^{ 2 }-1=1\cdot x\\ \\ \Rightarrow \quad { x }^{ 2 }-1=x\\ \\ \Rightarrow \quad { x }^{ 2 }=x+1\\ \\ \Rightarrow \quad \frac { { x }^{ 2 } }{ x } =\frac { x }{ x } +\frac { 1 }{ x } \\ \\ \Rightarrow \quad x=1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ x } } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } } \\ \\ \therefore \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }

This expression for the golden ratio is quite common, however, before I produced this post – I think it would’ve been very hard to figure out how to derive it from scratch. There aren’t many quirky proofs like this one on the internet – I am quite certain. I hope you liked reading this post! 😀

Deriving the formula for an ellipse

In this post, I’ll be demonstrating how one can derive the formula for an ellipse from absolute scratch.

To derive the formula for an ellipse, what we must first do is create a diagram like the one below.

Ellipse diagram

 

 

 

 

 

 

** Click on the image above to see it in full size.

Now, the first thing we’ve got to acknowledge here is that:

{ D }_{ 1 }+{ D }_{ 2 }=2a

What we’re basically saying is that D_1 + D_2 is equal to the length from -a to a in the diagram above.

This formula can be understood by watching the video below…

These photographs can also help the formula sink into your mind…

Ellipse Image 1:

Ellipse Image 1

 

 

 

 

 

 

Ellipse Image 2:

Ellipse Image 2

 

 

 

 

 

 

Now, look at the diagram at the top of this page once again…

What you will notice is that:

{ \left( c+x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 1 }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }^{ 2 }={ c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }=\sqrt { { c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 } } \\ \\ { \left( c-x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 2 }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }^{ 2 }={ c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }=\sqrt { { c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 } }

If this is the case, we can say that:

part of ellipse workings

 

 

 

 

 

 

 

** Click on the image of the workings to see it in full size.

Alright, so far so good… Now, it turns out – if you look at the diagram at the top of this page carefully, you will discover that:

{ b }^{ 2 }+{ c }^{ 2 }={ a }^{ 2 }\\ \\ \therefore \quad { c }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }

And this ultimately means that:

{ a }^{ 4 }+\left( { a }^{ 2 }-{ b }^{ 2 } \right) { x }^{ 2 }={ a }^{ 2 }\left( { a }^{ 2 }-{ b }^{ 2 } \right) +{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ \therefore \quad { a }^{ 4 }+{ a }^{ 2 }{ x }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 }={ a }^{ 4 }-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ -{ b }^{ 2 }{ x }^{ 2 }=-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ { b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }\\ \\ \frac { { b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } +\frac { { a }^{ 2 }{ y }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } \\ \\ \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\\ \\ { \left( \frac { x }{ a } \right) }^{ 2 }+{ \left( \frac { y }{ b } \right) }^{ 2 }=1

The formula you see just above is the formula for an ellipse. You’ve derived it from scratch!!

Finding the formulas for areas of triangles

In this post I’ll be demonstrating how one can derive the three formulas which can be used to find the areas of triangles.

These formulas are in fact:

A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }

To begin with, let’s start by looking at the diagram below:

Triangle Diagram

Now, if you look at the diagram carefully – you will notice that the area of the triangle is:

A=\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 }

This can be simplified into:

\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { x\cdot CN+\left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { CN\left\{ x+\left( c-x \right) \right\} }{ 2 } \\ \\ =\frac { CN\cdot c }{ 2 }

Because of SOH CAH TOA, what we can also say is that:

\sin { \left( A \right) } =\frac { O }{ H } =\frac { CN }{ b } \\ \\ \therefore \quad b\cdot \sin { \left( A \right) } =CN\\ \\ \sin { \left( B \right) =\frac { O }{ H } } =\frac { CN }{ a } \\ \\ \therefore \quad a\cdot \sin { \left( B \right) } =CN

Now because:

A=\frac { CN\cdot c }{ 2 }

This ultimately means that:

A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } \\ \\ A=\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) } \\ \\ \therefore \quad A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) =\frac { 1 }{ 2 } ac } \cdot \sin { \left( B \right) }

Alright, so far so good… Now we must put the icing on the cake and attach the final piece of the jigsaw puzzle to the formula above. In order to find the three equations which can be used to find the areas of triangles, we must now discover the expression for sin(C). We can discover its expression by first saying that:

C=\left( 90-A \right) +\left( 90-B \right) \\ \\ =90-A+90-B\\ \\ =180-A-B\\ \\ =180-\left( A+B \right) \\ \\ \therefore \quad \sin { \left( C \right) } =\sin { \left( 180-\left( A+B \right) \right) }

And if we use the trigonometric identity below:

\sin { \left( \alpha -\beta \right) } =\sin { \left( \alpha \right) \cdot \cos { { \left( \beta \right) } } -\cos { \left( \alpha \right) \cdot \sin { \left( \beta \right) } } }

We will reach the conclusion:

\sin { \left( 180-\left( A+B \right) \right) } =\sin { \left( 180 \right) \cdot \cos { \left( A+B \right) -\cos { \left( 180 \right) \cdot \sin { \left( A+B \right) } } } } 

But because:

\sin { \left( 180 \right) =0 } ,\quad \cos { \left( 180 \right) =-1 } \\ \\ \sin { \left( 180-\left( A+B \right) \right) =-\left( -1 \right) \cdot \sin { \left( A+B \right) } } \\ \\ \therefore \quad \sin { \left( C \right) =\sin { \left( A+B \right) } }

Now, sin(A+B) as a trigonometric identity, is:

\sin { \left( A+B \right) =\sin { \left( A \right) \cdot \cos { \left( B \right) +\cos { \left( A \right) \cdot \sin { \left( B \right) } } } } }

And, thanks to SOH CAH TOA…

\sin { \left( A+B \right) =\sin { \left( C \right) } } \\ \\ \sin { \left( A \right) =\frac { CN }{ b } } \\ \\ \cos { \left( B \right) =\frac { A }{ H } } =\frac { \left( c-x \right) }{ a } \\ \\ \cos { \left( A \right) =\frac { A }{ H } =\frac { x }{ b } } \\ \\ \sin { \left( B \right) =\frac { CN }{ a } }

Which means that…

\sin { \left( C \right) =\frac { CN }{ b } \cdot \frac { \left( c-x \right) }{ a } +\frac { x }{ b } \cdot \frac { CN }{ a } } \\ \\ =\frac { CN\left( c-x \right) }{ ab } +\frac { CN\cdot x }{ ab } \\ \\ =\frac { CN\left( c-x \right) +CN\cdot x }{ ab } \\ \\ =\frac { CN\left\{ \left( c-x \right) +x \right\} }{ ab } \\ \\ =\frac { CN\cdot c }{ ab } \\ \\ \therefore \quad ab\cdot \sin { \left( C \right) =CN\cdot c } \\ \\ \therefore \quad \frac { 1 }{ 2 } ab\cdot \sin { \left( C \right) =\frac { CN\cdot c }{ 2 } =A }

As this is the case, we can conclude that:

A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }

Latest Mathematics Proofs: August 2016

Recently I discovered a few more proofs, some related to A Level mathematics.  You can access these proofs by clicking on the links below.

Derive the formula of an ellipsehttps://plus.google.com/+mathsvideosforweb/posts/ZF8D3ghSNRD

Discover the distance between point A and B on the edge of a circle: https://plus.google.com/+mathsvideosforweb/posts/9fmuEJ4qeXY

Find the area of a sector within a circle: https://plus.google.com/+mathsvideosforweb/posts/cHt9t9PeSfg

Formula for a torus, derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/TS84TL44BYd

Create the mathematical singularity shown in movie documentaries: https://plus.google.com/+mathsvideosforweb/posts/7p2onSzNYte

Parameterised formula for a torus derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/AZrVMiPVdbv

Derive the formula for an ellipsoid (normal and parameterised) from scratch: https://plus.google.com/+mathsvideosforweb/posts/Wijf94ztn3v

Other news:

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Free A Level Mathematics Proofs

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As always, I strive to produce the clearest A Level maths proofs on the internet, so that you can pass your C1, C2, C3 and C4 exams with flying colours. I’m not the kind of tutor who simply tells students to plug numbers into formulas. I make sure my students understand the problems they’re dealing with – using proofs and logical models.

Thankyou for your attention. 🙂