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.


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.


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…




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)

2 thoughts on “How to prove that sin(A-B)=sin(A)cos(B)-cos(A)sin(B) geometrically”

  1. This tutorial is very straight forward and helpful. I’m really thankful for this. I’ve been searching around for ways to do this. I actually got how to prove cos(A+B) and sin(A+B).

    However, cos(A-B) and sin(A-B) have been giving me problems. Thanks to this tutorial, I finally understand it. I also watched your video and subscribed to your channel. Hope to see more soon.

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