Date: 12/29/97 at 16:11:56 From: phil Subject: Cosine of 40 degrees You derived a cubic equation for cosine of 40 degrees. Did you know the following approximation? cos (2x/3) = approx 0.5*(1 + cos x) so if x = 60 degrees, cos 40 = approx 0.5*(1+0.5) = 0.75 versus actual cos 40 = 0.76604, for 2 percent accuracy. Phil F.
Date: 01/03/98 at 04:19:19 From: Doctor Pete Subject: Re: Cosine of 40 degrees Hi, This approximation can be derived as well: Let k = x/3. Then Cos[3k] = Cos[k + 2k] = Cos[k]Cos[2k] - Sin[k]Sin[2k] = Sqrt[(1+Cos[2k])/2]Cos[2k] - Sqrt[(1-Cos[2k])/2]Sqrt[1-Cos[2k]^2] = Sqrt[(1+m)/2]m - Sqrt[(1-m)/2]Sqrt[1-m^2] where m = Cos[2k]. Simplifying, we obtain Cos[3k] = (Sqrt[1+m]m - Sqrt[1+m](1-m))/Sqrt = Sqrt[1+m](m - (1-m))/Sqrt = Cos[k](2*Cos[2k]-1). Now, suppose that k is a small positive quantity. Then Cos[k] ~ 1, and so Cos[3k] ~ 2*Cos[2k]-1, or (1+Cos[x])/2 ~ Cos[2x/3]. In particular, when x = Pi/3 = 60 degrees, then we obtain the approximation you mentioned. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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