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### Ptolemy's Theorem Shows Compound Angle Formula

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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
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```
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/
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Associated Topics:
College Calculus

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