Tag Archives: algebra

Algebra, Complex Numbers, Further Pure Maths, Imaginary Numbers, Magnitudes, Modulus, Proofs, Roots, Surds

|z_1|/|z_2|=|z_1/z_2| proof (algebraic)

November 4, 2018 Tiago Hands

In this post I’ll be proving to you that:

$\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right|$

Firstly, I’ll say that:

${ z }_{ 1 }=x+iy,\quad \therefore \quad \left| { z }_{ 1 } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

And also that:

${ z }_{ 2 }=p+iq,\quad \therefore \quad \left| { z }_{ 2 } \right| =\sqrt { { p }^{ 2 }+{ q }^{ 2 } }$

If this is the case, then:

$\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { x+iy }{ p+iq } \\ \\ =\frac { \left( x+iy \right) }{ \left( p+iq \right) } \cdot \frac { \left( p-iq \right) }{ \left( p-iq \right) } \\ \\ =\frac { px-iqx+ipy-{ i }^{ 2 }qy }{ { p }^{ 2 }-ipq+ipq-{ i }^{ 2 }{ q }^{ 2 } } \\ \\ =\frac { \left( px+qy \right) +i\left( py-qx \right) }{ { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left( \frac { px+qy }{ { p }^{ 2 }+{ q }^{ 2 } } \right) +i\left( \frac { py-qx }{ { p }^{ 2 }+{ q }^{ 2 } } \right)$

And as this is in the form:

$z=a+ib$

I would have to conclude that:

$RHS=\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| \\ \\ =\sqrt { { \left( \frac { px+qy }{ { p }^{ 2 }+{ q }^{ 2 } } \right) }^{ 2 }+{ \left( \frac { py-qx }{ { p }^{ 2 }+{ q }^{ 2 } } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { { \left( px+qy \right) }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } +\frac { { \left( py-qx \right) }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { { \left( px+qy \right) }^{ 2 }+{ \left( py-qx \right) }^{ 2 } }{ \left( { p }^{ 2 }+{ q }^{ 2 } \right) ^{ 2 } } } \\ \\ =\sqrt { \frac { { p }^{ 2 }{ x }^{ 2 }+2pqxy+{ q }^{ 2 }{ y }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }-2pqxy+{ q }^{ 2 }{ x }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ \left( { p }^{ 2 }+{ q }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } } \\ \\ =\sqrt { \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { p }^{ 2 }+{ q }^{ 2 } } } \\ \\ =\frac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ \sqrt { { p }^{ 2 }+{ q }^{ 2 } } } \\ \\ =\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } =LHS$

Hence, I have my proof.

Algebra, Complex Numbers, Further Pure Maths, Imaginary Numbers, Mathematics Proofs, Modulus, Pythagoras

How to prove that |z_1||z_2|=|z_1z_2|, Complex Numbers

October 14, 2018 Tiago Hands

In this post I’ll be showing you how to prove that:

$\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| =\left| { z }_{ 1 }{ z }_{ 2 } \right|$

Firstly, let’s say that:

${ z }_{ 1 }=x+iy$

${ z }_{ 2 }=p+iq$

If this is the case, then according to the rules of complex numbers:

$\left| { z }_{ 1 } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

$\left| { z }_{ 2 } \right| =\sqrt { { p }^{ 2 }+{ q }^{ 2 } }$

Secondly, let’s determine what ${ z }_{ 1 }{ z }_{ 2 }$ is…

${ z }_{ 1 }{ z }_{ 2 }\\ \\ =\left( x+iy \right) \left( p+iq \right) \\ \\ =px+iqx+ipy+{ i }^{ 2 }qy\\ \\ =px+iqx+ipy-qy\\ \\ =\left( px-qy \right) +i\left( qx+py \right)$

As you can see, we get the result above – which is another complex number.

This means that:

$RHS\\ \\ =\left| { z }_{ 1 }{ z }_{ 2 } \right| \\ \\ =\sqrt { { \left( px-qy \right) }^{ 2 }+{ \left( qx+py \right) }^{ 2 } } \\ \\ =\sqrt { \left( px-qy \right) \left( px-qy \right) +\left( qx+py \right) \left( qx+py \right) } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }-2pqxy+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+2pqxy+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } \\ \\ =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \cdot \sqrt { { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left| { z }_{ 1 } \right| { \left| { z }_{ 2 } \right| }\\ \\ =LHS$

Therefore we’ve proven that:

$\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| =\left| { z }_{ 1 }{ z }_{ 2 } \right|$

You can watch a video related to this proof below…

Algebra, SOH CAH TOA, Trigonometric Identities, Trigonometry

sin(0°) to sin(90°), cos(0°) to cos(90°) and tan(0°) to tan(90°) derivations. 15° steps.

September 23, 2017 Tiago Hands

Hello. In this post I’ll be showing you how to derive sin(0°), sin(15°), sin(30°), sin(45°), sin(60°), sin(75°), sin(90°), cos(0°), cos(15°), cos(30°), cos(45°), cos(60°), cos(75°), cos(90°), tan(0°), tan(15°), tan(30°), tan(45°), tan(60°), tan(75°) and tan(90°) from absolute scratch.

sin(30°), sin(60°), cos(30°) and cos(60°):

Now, I’ll first start off by showing you how to derive sin(30°), sin(60°), cos(30°) and cos(60°) with the use of an equilateral triangle (image above). This equilateral triangle has lengths equal to 2. If you look at the diagram above and its properties carefully, you should conclude that:

$\sin { \left( { 30 } \right) } =\frac { O }{ H } =\frac { 1 }{ 2 } \\ \\ \sin { \left( { 60 } \right) } =\frac { O }{ H } =\frac { \sqrt { 3 } }{ 2 } \\ \\ \cos { \left( { 30 } \right) } =\frac { A }{ H } =\frac { \sqrt { 3 } }{ 2 } \\ \\ \cos { \left( { 60 } \right) } =\frac { A }{ H } =\frac { 1 }{ 2 }$

sin(45°) and cos(45°):

Alright, so far so good. Next, have a look at this isosceles triangle (image above). If you take its properties into consideration – you’ll discover that:

$\sin { \left( 45 \right) =\frac { O }{ H } } =\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \cdot \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { \sqrt { 2 } }{ 2 } \\ \\ \cos { \left( 45 \right) =\frac { A }{ H } } =\frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \cdot \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { \sqrt { 2 } }{ 2 }$

sin(15°), sin(75°), cos(15°) and cos(75°):

Ok, so I’ve already shown you how to derive sin(30°), sin(45°), sin(60°), cos(30°), cos(45°) and cos(60°) using simple diagrams. It turns out that with the information above and also some trigonometric identities – we can derive sin(15°), sin(75°), cos(15°) and cos(75°). Let me show you what I mean…

$\sin { \left( 15 \right) } \\ \\ =\sin { \left( 45-30 \right) } \\ \\ =\sin { \left( 45 \right) \cos { \left( 30 \right) -\cos { \left( 45 \right) \sin { \left( 30 \right) } } } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } -\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { \sqrt { 6 } }{ 4 } -\frac { \sqrt { 2 } }{ 4 } \\ \\ =\frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 }$

$\sin { \left( 75 \right) } \\ \\ =\sin { \left( 45+30 \right) } \\ \\ =\sin { \left( 45 \right) \cos { \left( 30 \right) +\cos { \left( 45 \right) \sin { \left( 30 \right) } } } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } +\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { \sqrt { 6 } }{ 4 } +\frac { \sqrt { 2 } }{ 4 } \\ \\ =\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 }$

$\cos { \left( 15 \right) } \\ \\ =\cos { \left( 45-30 \right) } \\ \\ =\cos { \left( 45 \right) \cos { \left( 30 \right) +\sin { \left( 45 \right) \sin { \left( 30 \right) } } } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } +\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { \sqrt { 6 } }{ 4 } +\frac { \sqrt { 2 } }{ 4 } \\ \\ =\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 }$

$\cos { \left( 75 \right) } \\ \\ =\cos { \left( 45+30 \right) } \\ \\ =\cos { \left( 45 \right) \cos { \left( 30 \right) -\sin { \left( 45 \right) \sin { \left( 30 \right) } } } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } -\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { \sqrt { 6 } }{ 4 } -\frac { \sqrt { 2 } }{ 4 } \\ \\ =\frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 }$

sin(0°), sin(90°), cos(0°) and cos(90°):

sin(0°), sin(90°), cos(0°) and cos(90°) are values you should already know, so I won’t be demonstrating how to derive them. If you have studied the unit circle – you’ll know that:

$\sin { \left( 0 \right) } =0\\ \\ \sin { \left( 90 \right) } =1\\ \\ \cos { \left( 0 \right) =1 } \\ \\ \cos { \left( 90 \right) } =0$

These values are fairly easy to find.

tan(0°), tan(15°), tan(30°), tan(45°), tan(60°), tan(75°) and tan(90°):

So, this is the moment you’ve been waiting for… The complete set of derivations I said I’d give you. Although it may seem hard to derive tan(0°), tan(15°), tan(30°), tan(45°), tan(60°), tan(75°) and tan(90°) from absolute scratch, or like a tedious task – we have already done most of the hard work. All these tangent values can be derived using the information we’ve already accumulated, because:

$\tan { \left( \theta \right) } =\frac { \sin { \left( \theta \right) } }{ \cos { \left( \theta \right) } }$

Therefore:

$\tan { \left( 0 \right) } =\frac { \sin { \left( 0 \right) } }{ \cos { \left( 0 \right) } } =\frac { 0 }{ 1 } =0$

$\tan { \left( 15 \right) } \\ \\ =\frac { \sin { \left( 15 \right) } }{ \cos { \left( 15 \right) } } \\ \\ =\frac { \frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 } }{ \frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } } \\ \\ =\frac { \left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 4 } \cdot \frac { 4 }{ \left( \sqrt { 6 } +\sqrt { 2 } \right) } \\ \\ =\frac { \left( \sqrt { 6 } -\sqrt { 2 } \right) }{ \left( \sqrt { 6 } +\sqrt { 2 } \right) } \cdot \frac { \left( \sqrt { 6 } -\sqrt { 2 } \right) }{ \left( \sqrt { 6 } -\sqrt { 2 } \right) } \\ \\ =\frac { 6-\sqrt { 12 } -\sqrt { 12 } +2 }{ 6-\sqrt { 12 } +\sqrt { 12 } -2 } \\ \\ =\frac { 8-2\sqrt { 12 } }{ 4 } \\ \\ =\frac { 8-2\sqrt { 4 } \sqrt { 3 } }{ 4 } \\ \\ =\frac { 8-4\sqrt { 3 } }{ 4 } \\ \\ =\frac { 4\left( 2-\sqrt { 3 } \right) }{ 4 } \\ \\ =2-\sqrt { 3 }$

$\tan { \left( 30 \right) } \\ \\ =\frac { \sin { \left( 30 \right) } }{ \cos { \left( 30 \right) } } \\ \\ =\frac { \frac { 1 }{ 2 } }{ \frac { \sqrt { 3 } }{ 2 } } \\ \\ =\frac { 1 }{ 2 } \cdot \frac { 2 }{ \sqrt { 3 } } \\ \\ =\frac { 1 }{ \sqrt { 3 } } \\ \\ =\frac { 1 }{ \sqrt { 3 } } \cdot \frac { \sqrt { 3 } }{ \sqrt { 3 } } \\ \\ =\frac { \sqrt { 3 } }{ 3 }$

$\tan { \left( 45 \right) } \\ \\ =\frac { \sin { \left( 45 \right) } }{ \cos { \left( 45 \right) } } \\ \\ =\frac { \frac { \sqrt { 2 } }{ 2 } }{ \frac { \sqrt { 2 } }{ 2 } } \\ \\ =\frac { \sqrt { 2 } }{ 2 } \cdot \frac { 2 }{ \sqrt { 2 } } \\ \\ =1$

$\tan { \left( 60 \right) } \\ \\ =\frac { \sin { \left( 60 \right) } }{ \cos { \left( 60 \right) } } \\ \\ =\frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { 1 }{ 2 } } \\ \\ =\frac { \sqrt { 3 } }{ 2 } \cdot \frac { 2 }{ 1 } \\ \\ =\sqrt { 3 }$

$\tan { \left( 75 \right) } \\ \\ =\frac { \sin { \left( 75 \right) } }{ \cos { \left( 75 \right) } } \\ \\ =\frac { \frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } }{ \frac { \sqrt { 6 } -\sqrt { 2 } }{ 4 } } \\ \\ =\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } \cdot \frac { 4 }{ \sqrt { 6 } -\sqrt { 2 } } \\ \\ =\frac { \left( \sqrt { 6 } +\sqrt { 2 } \right) }{ \left( \sqrt { 6 } -\sqrt { 2 } \right) } \cdot \frac { \left( \sqrt { 6 } +\sqrt { 2 } \right) }{ \left( \sqrt { 6 } +\sqrt { 2 } \right) } \\ \\ =\frac { 6+\sqrt { 12 } +\sqrt { 12 } +2 }{ 6+\sqrt { 12 } -\sqrt { 12 } -2 } \\ \\ =\frac { 8+2\sqrt { 12 } }{ 4 } \\ \\ =\frac { 8+2\sqrt { 4 } \sqrt { 3 } }{ 4 } \\ \\ =\frac { 8+4\sqrt { 3 } }{ 4 } \\ \\ =\frac { 4\left( 2+\sqrt { 3 } \right) }{ 4 } \\ \\ =2+\sqrt { 3 }$

$\tan { \left( 90 \right) } \\ \\ =\frac { \sin { \left( 90 \right) } }{ \cos { \left( 90 \right) } } \\ \\ =\frac { 1 }{ 0 } \\ \\ =undefined$

And now, the set of derivations is complete. 😀

Algebra, Completing the Square, Geometry

Completing the Square: Two Geometrical Proofs

September 20, 2017 Tiago Hands

On this website I previously showed you why the formulas used to complete the square work – and how they can be used to derive formulas such as the quadratic equation. Now, I’ll be doing something different, but related… On this post, I’ll be showing you how to come up with the formulas (2 in total) used to complete the square – geometrically.

Completing the Square Formula (Derivation 1):

In the diagram above, what we can see is that:

${ x }^{ 2 }+2\cdot \frac { b }{ 2 } x+{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }$

This means that:

${ x }^{ 2 }+bx+{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }\\ \\ \therefore \quad { x }^{ 2 }+bx={ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }$

Completing the Square Formula (Derivation 2):

In the diagram above, what we can see is that:

${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }+2\cdot \frac { b }{ 2 } \left( x-\frac { b }{ 2 } \right) +{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }$

This means that:

${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }+b\left( x-\frac { b }{ 2 } \right) +{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { { b }^{ 2 } }{ 2 } +\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { 2{ b }^{ 2 } }{ 4 } +\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-\frac { { b }^{ 2 } }{ 4 } ={ x }^{ 2 }\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }+bx-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }\\ \\ \therefore \quad { x }^{ 2 }-bx={ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }$

I hope these geometrical proofs have helped you better understand why the formulas we use to complete the square are in existence. Thanks for reading! 😀

Algebra, Exponentials, Golden Ratio, Irrational Numbers, Roots, Surds, Tutorials

More ways in which to express the golden ratio

September 19, 2017 Tiago Hands

In this blog post I’ll be revealing more ways (4 in fact) in which to express or come up with the value of the golden ratio…

Number One:

$\varphi =\frac { a+b }{ a } \\ \\ =\frac { a }{ a } +\frac { b }{ a } \\ \\ =1+{ \left( \varphi \right) }^{ -1 }\\ \\ =1+\frac { 1 }{ \varphi }$

Number Two:

$\varphi =1+\frac { 1 }{ \varphi } \\ \\ \Rightarrow \quad { \varphi }^{ 2 }=\varphi +1\\ \\ \Rightarrow \quad { \varphi }^{ 2 }-\varphi =1\\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=1\\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 4 }{ 4 } +\frac { 1 }{ 4 } \\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 5 }{ 4 } \\ \\ \Rightarrow \quad \varphi -\frac { 1 }{ 2 } =\frac { \sqrt { 5 } }{ 2 } \\ \\ \Rightarrow \quad \varphi =\frac { 1 }{ 2 } +\frac { \sqrt { 5 } }{ 2 } \\ \\ \therefore \quad \varphi =\frac { 1+\sqrt { 5 } }{ 2 }$

Number Three:

$\varphi =1+\frac { 1 }{ \varphi } \\ \\ \Rightarrow \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ \varphi } } \\ \\ \Rightarrow \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \varphi } } } \\ \\ \therefore \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

Number Four:

$\varphi =1+\frac { 1 }{ \varphi } \\ \\ { \Rightarrow \quad \varphi }^{ 2 }=\varphi +1\\ \\ \Rightarrow \quad \varphi =\sqrt { \varphi +1 } \\ \\ \Rightarrow \quad \varphi =\sqrt { \sqrt { \varphi +1 } +1 } \\ \\ \Rightarrow \quad \varphi =\sqrt { \sqrt { \sqrt { \varphi +1 } +1 } +1 } \\ \\ \therefore \quad \varphi =\sqrt { \sqrt { \sqrt { \frac { 1+\sqrt { 5 } }{ 2 } +1 } +1 } +1 } \\ \\$

And check out this calculator trick…

If you’re not satisfied with what I’ve already produced, then you can have a go at proving that…

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

Without using the phi (φ) symbol.

Enjoy!!! 😀

Algebra, Continued Fractions, Exponentials, Fractions, Golden Ratio, Irrational Numbers, Mathematics Proofs, Proofs, Roots

Another way to express the golden ratio mathematically

September 17, 2017 Tiago Hands 1 Comment

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! 😀

Algebra, Areas, Ellipses, Mathematics Proofs, Proofs

How to find a neat version of y when you have the equation of an ellipse

August 30, 2017 Tiago Hands

So, you have the equation of the ellipse but you need to completely isolate y. How would you go about doing this? Well, here is a fantastic example…

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

This will come in handy if you’re trying to derive the area of an ellipse from absolute scratch.

Thanks for reading! 🙂

Algebra, Areas, Calculus, Circles, Co-ordinate Geometry, Integration, Pythagoras, Trigonometry

Useful trigonometric formulas for finding areas of circles

August 30, 2017 Tiago Hands

If you’re trying to find the area of a circle using integration methods, then these trigonometric formulas are going to be very useful:

First formulas:

$\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\cos ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\cos ^{ 2 }{ \theta } \right) }$

Second formulas:

$\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\sin ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) }$

These formulas are to be used when you have to transform the expression:

${ y }=\sqrt { { r }^{ 2 }-{ x }^{ 2 } }$

You can either make:

$x=r\sin { \theta }$

Or…

$x=r\cos { \theta }$

The choice is yours. 🙂

Addition, Algebra, Tutorials

If a/b=c/d, then a/b=(a+c)/(b+d) mathematical proof

August 5, 2017 Tiago Hands

In this post I’ll be proving to you that if:

a/b=c/d

Then:

a/b=(a+c)/(b+d)

To prove this, the first thing you need to know is that:

$\frac { a }{ b } =\frac { c }{ d } \\ \\ \therefore \quad ad=bc$

The rest is just mathematical / algebraic trickery… Let me show you…

Proof 1:

$LHS\\ \\ =\frac { a }{ b } \\ \\ =\frac { a }{ b } \cdot 1\\ \\ =\frac { a }{ b } \cdot \frac { \left( b+d \right) }{ \left( b+d \right) } \\ \\ =\frac { ab+ad }{ b\left( b+d \right) } \\ \\ =\frac { ab }{ b\left( b+d \right) } +\frac { ad }{ b\left( b+d \right) } \\ \\ =\frac { a }{ b+d } +\frac { bc }{ b\left( b+d \right) } \\ \\ =\frac { a }{ b+d } +\frac { c }{ b+d } \\ \\ =\frac { a+c }{ b+d } \\ \\ =RHS$

Proof 2:

$LHS\\ \\ =\frac { a }{ b } \\ \\ =\frac { a }{ b } \cdot 1\\ \\ =\frac { a }{ b } \cdot \frac { \left( b+d \right) }{ \left( b+d \right) } \\ \\ =\frac { ab+ad }{ b\left( b+d \right) } \\ \\ =\frac { ab+bc }{ b\left( b+d \right) } \\ \\ =\frac { b\left( a+c \right) }{ b\left( b+d \right) } \\ \\ =\frac { a+c }{ b+d } \\ \\ =RHS$

Proof 1 Video:

Proof 2 Video:

Algebra, Ellipses, Geometry, Mathematics Proofs, Proofs, Pythagoras, Visualising Mathematics

Deriving the formula for an ellipse

July 30, 2017 Tiago Hands

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.

** 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 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:

** 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!!

Maths Proofs, Videos & Tutorials

Tag Archives: algebra

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sin(30°), sin(60°), cos(30°) and cos(60°):

sin(45°) and cos(45°):

sin(15°), sin(75°), cos(15°) and cos(75°):

sin(0°), sin(90°), cos(0°) and cos(90°):

tan(0°), tan(15°), tan(30°), tan(45°), tan(60°), tan(75°) and tan(90°):

Completing the Square: Two Geometrical Proofs

Completing the Square Formula (Derivation 1):

Completing the Square Formula (Derivation 2):

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Another way to express the golden ratio mathematically

How to find a neat version of y when you have the equation of an ellipse

Useful trigonometric formulas for finding areas of circles

If a/b=c/d, then a/b=(a+c)/(b+d) mathematical proof

Deriving the formula for an ellipse

GCSE + A Level Mathematics Proofs, Videos and Tutorials.