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Science & Mathematics by Anonymous 2018-06-09 01:33:25
Social Science
Can anyone prove that sin^2 (A+B) - sin^2 (A-B) = sin 2A sin2B?
4 answers
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Anonymous
no
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Anonymous
Let (s , c) denote (sin , cos). RTP s^2(A+B) -s^2 (A-B) = s(2A)*s(2B)...[1]. LS [1] = [s(A+B)+s(A-B)][s(A+B)- s(A-B)] = [(sAcB + cAsB) + (sAcB - cAsB)][(sAcB + cAsB) - (sAcB - cAsB)] = [2sAcB][2cAsB] = 4[sAcAsBcB] = [2sAcA][2sBcB] = sin(2A)*sin(2B).
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Anonymous
sin^2(A+B)-sin^2(A-B) = [sin(A+B)+sin(A-B)][sin(A+B)-sin(A-B)] = [2sin[2A/2]cos[2B/2]] [2cos[2A/2]sin[2B/2]] = [2sinAcosB][2cosAsinB] = [2sinAcosA][2sinBcosB] = sin(2A)cos(2B)
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Anonymous
p = sinAcosB q = cosAsinB u = sin(A+B) = p+q v = sin(A-B) = p-q u+v = 2p u-v = 2q sin^2 (A+B) - sin^2 (A-B) = u^2 - v^2 = (u+v)(u-v) = (2p)(2q) = 2 sinAcosB 2 cosAsinB = 2 sinAcosA 2 sinBcosB = sin(2A) sin(2B)