Ptolemy's Theorem Shows Compound Angle Formula

Date: 7/7/96 at 0:25:0
From: Alan Yu
Subject: Ptolemy Theorem Shows Compound Angle Formula

Hi Dr.Math,

From an article about history of mathematics, I saw the Ptolemy 
Theorem (the sum of the products of the opposite sides of a concyclic 
quadrilateral equals the product of its diagonals).

	The article suggested that the theorem can be extended to show the 
Sin(A+B) and the Cos(A+B) formula if the concyclic quadrilateral is 
constructed so that one of its sides is the diameter. (The proof was 
not shown there.)

	I have been trying to show the formula with this idea, but I am 
not able do it.  Would you please show me the formula with the Ptolemy 
theorem?

Best Regards,
Alan

Date: 7/7/96 at 15:29:31
From: Doctor Anthony
Subject: Re: Ptolemy Theorem Shows Compound Angle Formula

In fact you need to make one of the diagonals a diameter of the 
circle. Draw the diagram I describe and refer to it as I derive the 
sin(X+Y) formula.

Let the cyclic quadrilateral be ABCD (lettered clockwise or anti-
clockwise) but keep the letters in cyclic order.  I have the diagonal 
BD as a diameter of the circle.  Since BD is a diameter the angles at 
A and C will be right angles. Let angle ABD = X and angle CBD = Y. (so 
angle B = X+Y) We can now write down the lengths of various sides in 
terms of sines and cosines of X and Y. Note that BD = 2r, where 
r = radius of the circle.

AD = 2r sin(X)  AB = 2r cos(X)

BC = 2r cos(Y)  CD = 2r sin(Y)

Now, using the sine formula on triangle DAC we have AC/sin(D) = 2r 
(remember that in the formula: a/sin(A) = b/sin(B) = c/sin(C) for 
triangle ABC, each of these ratios is equal to the diameter of the 
circumcircle of triangle ABC) and so AC = 2r sin(D)  

Now B and D are supplementary angles since they are opposite angles of 
a cyclic quadrilateral, and so Sin(D) = sin(B) = sin(X+Y)

We can say therefore  AC = 2r sin(X+Y)

We now write down Ptolemy's formula for the cyclic quadrilateral.

  AC*BD = AD*BC + AB*CD   and putting in the values found above:

 2r sin(X+Y) * 2r = 2r sin(X)*2r cos(Y) + 2r cos(X) * 2r sin(Y)

Divide through by 4r^2 and we have:

    sin(X+Y) = sin(X)*cos(Y) + cos(X)*sin(Y) 


-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/