Trig Proof - Newton's FormulaDate: 04/09/98 at 23:37:14 From: Jay Subject: Newton's Formula The problem I've been having trouble with is: (a - b) sin [(1/2)(A - B)] ------- = ------------------- c cos [(1/2)(C)] I've tried desperately to solve this problem analytically but it just turned into a mess. Is it possible that you (Dr. Math) could tell me an easier way to prove this formula (i.e. graphically)? If so, please let me know. Jay Date: 04/10/98 at 13:15:23 From: Doctor Anthony Subject: Re: Newton's Formula Start with the Sine formula: sin(A) sin(B) sin(C) ------ = ------ = ------ a b c From the note below, we find that: sin(A)-sin(B) sin(C) ------------- = ------- a-b c Thus: 2cos((A+B)/2)*sin((A-B)/2) sin(C) -------------------------- = ------- (Equation 1) a - b c Since (A+B+C)/2 = 90: (A+B)/2 = 90 - C/2 So: cos((A+B)/2) = cos(90 - C/2) = sin(C/2) (Equation 2) Substituting Equation B into Equation A, we find that: 2sin(C/2)*sin((A-B)/2) 2sin(C/2)cos(C/2) ---------------------- = ---------------- a - b c sin((A-B)/2) cos(C/2) ------------ = -------- a - b c sin((A-B)/2) a - b ------------ = ----- cos(C/2) c as required. NOTE ---- If: a c --- = ---- = k b d then: a = bk and c = dk So: a-c bk - dk k(b-d) a c ----- = --------- = -------- = k = --- = --- b-d b-d b-d b d -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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