# The quickest Sine Rule proof

In this post I’ll be demonstrating how to prove that the Sine Rule is true in the quickest manner possible.

First of all, let’s begin with writing down the 3 formulas which can be used to find the area of a triangle:

$A=\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 }$

$A=\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 }$

$A=\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 }$

Now, let’s make the first two formulas above equivalent to one another…

$\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 } =\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 }$

Alright, now watch what happens when we multiply both sides of the equation by a handy expression…

$\frac { b\cdot c\cdot \sin { \left( A \right) } }{ 2 } \cdot \frac { 2 }{ c\cdot \sin { \left( A \right) } \cdot \sin { \left( B \right) } } =\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } \cdot \frac { 2 }{ c\cdot \sin { \left( A \right) \cdot \sin { \left( B \right) } } }$

If we do this, what we’re going to be left with is…

$\frac { b }{ \sin { \left( B \right) } } =\frac { a }{ \sin { \left( A \right) } }$

So far so good! Let’s now make these two area formulas equivalent to one another…

$\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } =\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 }$

And now, let’s multiply both sides of the equation we’ve just created by a handy expression…

$\frac { a\cdot c\cdot \sin { \left( B \right) } }{ 2 } \cdot \frac { 2 }{ a\cdot \sin { \left( B \right) } \cdot \sin { \left( C \right) } } =\frac { a\cdot b\cdot \sin { \left( C \right) } }{ 2 } \cdot \frac { 2 }{ a\cdot \sin { \left( B \right) } \cdot \sin { \left( C \right) } }$

If we do this, what we’re going to be left with is…

$\frac { c }{ \sin { \left( C \right) } } =\frac { b }{ \sin { \left( B \right) } }$

And it turns out, because:

$\frac { b }{ \sin { \left( B \right) } } =\frac { a }{ \sin { \left( A \right) } }$

$\frac { c }{ \sin { \left( C \right) } } =\frac { b }{ \sin { \left( B \right) } }$

We can say that:

$\frac { a }{ \sin { \left( A \right) } } =\frac { b }{ \sin { \left( B \right) } } =\frac { c }{ \sin { \left( C \right) } }$

I’ve made a video related to this Sine Rule proof. You can watch it below if you wish.

Hope you enjoyed reading this post! 🙂

You can now find out how to derive the 3 main cosine rule formulas through a new document that I’ve created called “Cosine Rule Mastery“.

This document can be downloaded free of charge along with “Sine Rule Mastery” which is another document that explains in detail how to come up with the sine rule formula.

I’d also like to talk about a new video I’ve created (posted below). It’s related to a 4 dimensional hypercube and learning how to train your mind to see things from different mathematical perspectives.

Prior to posting up the video above, I did create another similar video. In the video below, you will see me split a prism into 3 equal parts. This video will interest those who’d like to find the volume of a square based pyramid.

# Sine Rule Derivation In Record Time…

$\frac { 1 }{ 2 } bc\sin { A } =\frac { 1 }{ 2 } ac\sin { B } \\ \\ \frac { 2 }{ c\sin { A\sin { B } } } \cdot \frac { 1 }{ 2 } bc\sin { A } =\frac { 2 }{ c\sin { A\sin { B } } } \cdot \frac { 1 }{ 2 } ac\sin { B } \\ \\ \frac { 2bc\sin { A } }{ 2c\sin { A\sin { B } } } =\frac { 2ac\sin { B } }{ 2c\sin { A\sin { B } } } \\ \\ \frac { 2 }{ 2 } \cdot \frac { c }{ c } \cdot \frac { \sin { A } }{ \sin { A } } \cdot \frac { b }{ \sin { B } } =\frac { 2 }{ 2 } \cdot \frac { c }{ c } \cdot \frac { \sin { B } }{ \sin { B } } \cdot \frac { a }{ \sin { A } } \\ \\ \frac { b }{ \sin { B } } =\frac { a }{ \sin { A } }$

Now:

$\sin { C } =\sin { \left( 90-B+\left( 90-A \right) \right) } \\ \\ \sin { C } =\sin { \left( 180-\left( A+B \right) \right) } \\ \\ \sin { C } =\sin { 180 } \cos { \left( A+B \right) -\cos { 180 } } \sin { \left( A+B \right) } \\ \\ \sin { C } =-\left( -1 \right) \sin { \left( A+B \right) } \\ \\ \sin { C= } \sin { \left( A+B \right) } \\ \\ \sin { C } =\sin { A } \cos { B } +\cos { A } \sin { B } \\ \\ \sin { C= } \frac { 2A }{ bc } \cdot \frac { x }{ a } +\frac { \left( c-x \right) }{ b } \cdot \frac { 2A }{ ac } \\ \\ \sin { C } =\frac { 2Ax }{ acb } +\frac { 2A\left( c-x \right) }{ acb }$

$\\ \\ \sin { C } =\frac { 2Ax+2A\left( c-x \right) }{ acb } \\ \\ \sin { C } =\frac { 2A\left\{ x+\left( c-x \right) \right\} }{ acb } \\ \\ \sin { C= } \frac { 2Ac }{ acb } \\ \\ \frac { ab }{ 2 } \cdot \sin { C } =\frac { 2A }{ ab } \cdot \frac { ab }{ 2 } \\ \\ \frac { 1 }{ 2 } ab\sin { C } =A\\ \\ \therefore \quad \frac { 1 }{ 2 } ab\sin { C } =\frac { 1 }{ 2 } bc\sin { A } \\ \\ \frac { 2 }{ b\sin { A\sin { C } } } \cdot \frac { 1 }{ 2 } ab\sin { C } =\frac { 2 }{ b\sin { A\sin { C } } } \cdot \frac { 1 }{ 2 } bc\sin { A } \\ \\ \frac { a }{ \sin { A } } =\frac { c }{ \sin { C } } \\ \\ \therefore \quad \frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } }$

# Sine Rule Derivation

Use the formulas you’d use to calculate the area of a triangle. See the magic emerge.

$\frac { 1 }{ 2 } bcsinA=\frac { 1 }{ 2 } acsinB\\ \\ bcsinA=acsinB\\ \\ bsinA=asinB\\ \\ \frac { bsinA }{ b } =\frac { asinB }{ b } \\ \\ \frac { sinA }{ 1 } =\frac { asinB }{ b } \\ \\ \frac { sinA }{ 1 } \cdot \frac { 1 }{ a } =\frac { asinB }{ b } \cdot \frac { 1 }{ a } \\ \\ \frac { sinA }{ a } =\frac { sinB }{ b } \\ \\ \\ OR:\\ \\ bsinA=asinB\\ \\ \frac { bsinA }{ sinB } =\frac { asinB }{ sinB } \\ \\ \frac { bsinA }{ sinB } =\frac { a }{ 1 } \\ \\ \frac { bsinA }{ sinB } \cdot \frac { 1 }{ sinA } =\frac { a }{ 1 } \cdot \frac { 1 }{ sinA } \\ \\ \frac { b }{ sinB } =\frac { a }{ sinA }$

# Areas Of Triangles & The Sine Rule

Learn how to come up with the formula for areas of triangles, then use this formula to derive the sine rule formula.

Related:

The quickest Sine Rule proof