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

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°):

equilateral_triangle

 

 

 

 

 

 

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°):

isosceles_triangle

 

 

 

 

 

 

 

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. 😀

Completing the Square: Two Geometrical Proofs

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):

Completing the Square - Geometric 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):

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! 😀

More ways in which to express the golden ratio

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!!! 😀

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! 😀