Proving Laws of Sines, Cosines

Date: 10/01/2003 at 04:19:53
From: Michael 
Subject: Trigonometry

Hi,

I am having a little trouble following the logic of the proofs of the
sine and cosine laws, 

  a/SinA = b/SinB = c/SinC 

and

  a^2 = b^2 + c^2 - 2bcosA

Can you help? 

Date: 10/09/2003 at 07:28:29
From: Doctor Jerry
Subject: Re: Trigonometry

Hi Michael,

There are several ways of proving these laws.  I'll give you the
proofs that I remember.

For the law of cosines, I'll use a coordinate-system approach.  A
general triangle can be plotted as follows:  Put vertex C at the
origin, put vertex B on the positive x-axis at (a,0), and put vertex A 
anywhere above the x-axis. 

I prefer to put it in the second quadrant, above and to the left of C.
The angle C is in "standard position" and so the point A will have
coordinates (b*cos(C),b*sin(C)). 

Now calculate the square of the distance from A to B. Because it is
opposite angle C we call it c^2; from the distance formula we have

  c^2 = (b*cos(C)-a)^2 + (b*sin(C)-0)^2

Expanding,

  c^2 = b^2*cos^2(C) - 2*a*b*cos(C) + a^2 + b^2*sin^2(C).

The terms with b^2 can be combined:

  c^2 = b^2(cos^2(C) + sin^2(C)) - 2*a*b*cos(C) + a^2 

  c^2 = a^2 + b^2 - 2*a*b*cos(C).

This is the law of cosines.

For the law of sines, draw a triangle with all angles less than 90
deg.  Other triangles can be handled using the same ideas, but it's a
little harder to describe in words.

Label the vertices A, B, and C.  Label the sides opposite these
vertices a, b, and c.

Choose a vertex, say A.  Draw an altitude from A to the opposite side.
 Let D be the point of intersection.  Let h be the length of AD. 
Then 

  sin(C) = h/b   and   sin(B) = h/c.

Solve these two equations for h and set the results equal:

  b*sin(C) = c*sin(B).

By symmetry, we can also show that

  a*sin(B) = b*sin(A)

and

  c*sin(A) = a*sin(C).

From these three equations we can see that

  sin(A)/a = sin(B)/b = sin(C)/c.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/