Law of TangentsDate: 01/22/99 at 01:23:14 From: Matthew Veatch Subject: Law of tangents For my pre-calc class, I am trying to prove and derive the law of tangents. Please help me. Thank you. Date: 01/22/99 at 12:44:51 From: Doctor Floor Subject: Re: Law of tangents Hi Matthew, Thank you for sending your question to Dr. Math! Let a and b be two sides of a triangle. Let A be the angle opposite a and B be the angle opposite b. The law of tangents says: a+b tan(0.5(A+B)) --- = ------------- a-b tan(0.5(A-B)) To derive this formula, first consider the sum and subtraction formulas for sines: sin(t+u) = sin(t)cos(u) + cos(t)sin(u) sin(t-u) = sin(t)cos(u) - cos(t)sin(u) Adding and subtracting these two gives: sin(t+u) + sin(t-u) = 2sin(t)cos(u) sin(t+u) - sin(t-u) = 2cos(t)sin(u) Now let t = 0.5(A+B) and u = 0.5(A-B), then t+u = A and t-u = B. This gives: sin(A) + sin(B) = 2sin(0.5(A+B))cos(0.5(A-B)) sin(A) - sin(B) = 2sin(0.5(A-B))cos(0.5(A+B)) So we can derive the following: tan(0.5(A+B)) sin(0.5(A+B))cos(0.5(A-B)) ------------- = -------------------------- tan(0.5(A-B)) sin(0.5(A-B))cos(0.5(A+B)) 2sin(0.5(A+B))cos(0.5(A-B)) = --------------------------- 2sin(0.5(A-B))cos(0.5(A+B)) sin(A) + sin(B) = --------------- sin(A) - sin(B) And using the law of sines this equals: a + b = ----- a - b as desired! If you have a math question again, please send it to Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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