cosine approximation?

Not much more help, Mathematica has merely taken an approximation for cos(theta) and substituted (a*x) into it, it'll likely do the same with (a+x).

As I understand it Don has convenient, acurate, values of cos(theta) for only certain values of theta (possibly from a lookup table) and is looking for a rapid way to interpolate.

I bet there's a way to get a good approximation for

cos(theta) given good values for

sin(theta-a) and cos(theta-a) sin(theta+b) and cos(theta+b) where 0

cos(x)=1 - x^2/2! + x^4/4! - x^6/6! etc from handbook of chemistry and physics. But also need sin(x) = x - x^3/3! + x^5/5! - x^7/7! etc. Then use Taylor's series for f(x+h), translating to cos(x+a) = cos(x)

- a*sin(x) - (a^2/2!)*cos(x) + etc if i have done my signs correctly. Finally, expand the terms from above; use 6 terms minimum on all and keep only the first 4 as desired.

I think it is easier to add a and x (=z) first, determine the modulo

2pi from z and then use a lookup table or series expansion to determine cos(z).

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Come again?

Mathematica 5.2 for Mac OS X (64 bit) Copyright 1988-2005 Wolfram Research, Inc. -- Terminal graphics initialized --

In[1]:= Series[Cos[a+x],{x,0,3}]

2 3 Cos[a] x Sin[a] x 4 Out[1]= Cos[a] - Sin[a] x - --------- + --------- + O[x] 2 6

This is the same answer that others have derived by more direct means.

Only Cos[a] and Sin[a] are needed.

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Dave Seaman
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concerning case of Mumia Abu-Jamal.