Category Archives: Trigonometry

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. 😀

Useful trigonometric formulas for finding areas of circles

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. 🙂

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) }

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…


Related Videos:


Related posts:

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

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