Date: 09/13/2004 at 03:57:39 From: Ippy Subject: Trigonometric Proofs How do I prove that (sinA - sinB)/(cos A + cosB) = tan(A/2 - B/2)? How can I obtain: 1) sinA - sinB = 2sin(A/2 - B/2)cos(A/2 + B/2) and 2) cosA + cosB = 2cos(A/2 + B/2)cos(A/2 - B/2) using standard trigonometric proofs?
Date: 09/13/2004 at 04:33:36 From: Doctor Luis Subject: Re: Trigonometric Proofs Hi Ippy, You can come up with (1) and (2) from the sum and difference identities by using a clever substitution: A = u + v, B = u - v Then, sin(A) - sin(B) = sin(u + v) - sin(u - v) = [sin(u)cos(v) + cos(u)sin(v)] - [sin(u)cos(v) - cos(u)sin(v)] = sin(u)cos(v) + cos(u)sin(v) - sin(u)cos(v) + cos(u)sin(v) = 2 cos(u)sin(v) So now we need to know u and v in terms of A and B. Fortunately, that's easy A + B = (u + v) + (u - v) = 2u ==> u = (A + B)/2 A - B = (u + v) - (u - v) = 2v ==> v = (A - B)/2 To summarize, sin(A) - sin(B) = 2 cos((A + B)/2) sin((A - B)/2) which is exactly (1). Using this method, you should be able to prove (2). Once you have these two results, you can prove your original identity. Please write back if you have further questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/
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