Proof of DeMoivre's TheoremDate: 05/01/97 at 20:17:29 From: LIYA Subject: Proof of Demoivre's Theorem Dr. Math, Could you please help me prove DeMoivre's theorem by mathematical induction? I figured out the first part of the proof, but I can't prove that assuming that if it's true for n = k, then it's true for n = k+1. Evelina Date: 05/02/97 at 13:17:50 From: Doctor Wilkinson Subject: Re: Proof of deMoivre's Theorem Evelina, This depends on the addition formulas for the sine and cosine: sin(x + y) = sin(x) cos(y) + sin(y) cos(x) cos(x + y) = cos(x) cos(y) - sin(x) sin(y) DeMoivre's theorem states that: (cos(x) + i sin(x))^n = cos(nx) + i sin(nx) In your case, you are assuming this holds for n = k and are trying to prove that it holds for n = k + 1, so you are looking at: (cos(x) + i sin(x)) ^(k + 1) This is just: (cos(x) + i sin(x))^k * (cos(x) + i sin(x)) Using the induction hypothesis, this is: (cos(kx) + i sin(kx)) ( cos(x) + i sin(x)) Multiplying out and using the fact that i^2 = -1, we get: cos(kx) cos(x) - sin(kx) sin(x) + i (sin(kx) cos(x) + cos(kx) sin(x)) This is just cos((k+1) x) + i sin((k+1) x) by the addition formulas mentioned above. I recommend that you study this proof carefully, as it is very typical of induction proofs. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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