Ptolemy's Theorem Shows Compound Angle FormulaDate: 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/ |
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