OK, I have a version that this works for now. Something I noticed is
that there could be simplifications applied inside some of the cos and
sin functions - because of the factor of I. How do you simplify inside
a cos or sin? Alternatively, how do I get the expression with common
factors of I in the exponentials?

all the best
--cjc

On 31 Mar, 16:30, "Aaron S. Meurer" <[email protected]> wrote:
> Use expr.rewrite(cos), as in
>
> In [22]: print a.rewrite(cos)
> 3**(1/2)*(I*sin(k_1/4) + cos(k_1/4))/45 + 3**(1/2)*(-I*sin(k_1/4) + 
> cos(k_1/4))/45 + 3**(1/2)*(-I*sin(I*(I*k_1/4 - I*k_2*3**(1/2)/4)) + 
> cos(I*(I*k_1/4 - I*k_2*3**(1/2)/4)))/45 + 3**(1/2)*(-I*sin(I*(-I*k_1/4 + 
> I*k_2*3**(1/2)/4)) + cos(I*(-I*k_1/4 + I*k_2*3**(1/2)/4)))/45 + 
> 3**(1/2)*(-I*sin(I*(I*k_1/8 + I*k_2*3**(1/2)/8)) + cos(I*(I*k_1/8 + 
> I*k_2*3**(1/2)/8)))/45 + 3**(1/2)*(I*sin(-I*(-I*k_1/8 - I*k_2*3**(1/2)/8)) + 
> cos(-I*(-I*k_1/8 - I*k_2*3**(1/2)/8)))/45
>
> You could then use expand() to simplify things.
>
> Aaron Meurer
> On Mar 31, 2010, at 9:18 AM, Colin wrote:
>
>
>
> > Dear list,
> >   I have expressions coming from my calculation of the type:
>
> > (1/45)*3**(1/2)*exp(-1/4*I*k_1 + (1/4)*I*k_2*3**(1/2)) +
> > (1/45)*3**(1/2)*exp(-1/8*I*k_1 - 1/8*I*k_2*3**(1/2)) +
> > (1/45)*3**(1/2)*exp((1/4)*I*k_1 - 1/4*I*k_2*3**(1/2)) +
> > (1/45)*3**(1/2)*exp((1/8)*I*k_1 + (1/8)*I*k_2*3**(1/2)) +
> > (1/45)*3**(1/2)*exp(-1/4*I*k_1) + (1/45)*3**(1/2)*exp((1/4)*I*k_1)
>
> > where k_1,k_2 are real numbers. What is the best way to automatically
> > rewrite this as a weighted sum of cosines?
>
> > all the best
> > --Colin
>
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