sines law proof
The goal is to prove the sine law:
and where is the radius of the circumcircle that encloses our triangle.
Let’s add a couple of lines and define more variables.
So, we now know that
and, therefore, we need to prove
From geometry, we can see that
So the proof is reduced to proving that
This is easily seen as true after examining the top half of the unit circle. So, putting all of our results together, we get
The same logic may be followed to show that each of these fractions is also equal to .
For the final step of the proof, we must show that
We begin by defining our coordinate system. For this, it is convenient to find one side that is not shorter than the others and label it with length . (The concept of a “longest” side is not well defined in equilateral and some isoceles triangles, but there is always at least one side that is not shorter than the others.) We then define our coordinate system such that the corners of the triangle that mark the ends of side are at the coordinates and . Our third corner (with sides labelled alphbetically clockwise) is at the point . Let the center of our circumcircle be at . We now have
Combining equations (3) and (2), we find
Substituting this into equation (2) we find that
Combining equations (4) and (5) leaves us with
where we have applied the cosines law in the second to last step.
|Title||sines law proof|
|Date of creation||2013-03-22 11:57:41|
|Last modified on||2013-03-22 11:57:41|
|Last modified by||drini (3)|