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Ptolemy's Theorem and Double-angle Formulas

Date: 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,


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
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Associated Topics:
High School Calculus

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