Ptolemy's Theorem and Double-angle FormulasDate: 7/11/96 at 7:27:42 From: Alan Yu Subject: Ptolemy Theorem and Double-angle Formulas Dr. Maths, Thanks for your previous instructions on how to show the compound angle formula Sin(A+B)= Sin(A)Cos(B) + Cos(A)Sin(B). I followed your steps and found out the proof. Thank you so much for your instructions. Is it possible to show the formula Sin(A-B), Cos(A+B) or Cos(A-B) with Ptolemy's Theorem? Look forward to receiving your reply. Best Regards, Alan Date: 7/11/96 at 12:46:57 From: Doctor Anthony Subject: Re: Ptolemy Theorem and Double-angle Formulas Once you have established the formula for sin(A+B), the others can be derived directly from that. You need to know that sin(-B) = -sin(B), cos(-B)=cos(B), sin(pi/2 - A) = cos(A), cos(pi/2 - A) = sin(A). Then: sin(A+(-B)) = sin(A)cos(-B) + cos(A)sin(-B) sin(A-B) = sin(A)cos(B) - cos(A)sin(B) sin(pi/2 -(A+B)) = sin((pi/2-A) -B) cos(A+B) = sin(pi/2-A)cos(-B) + cos(pi/2-A)sin(-B) = cos(A)cos(B) - sin(A)sin(B) sin(pi/2 -(A-B)) = sin(pi/2-A + B) cos(A-B) = sin(pi/2-A)cos(B) + cos(pi/2-A)sin(B) = cos(A)cos(B) + sin(A)sin(B) -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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