conversion of sin to cos

Hello all,

Here is my problem. I have a very long product of cosines with
arguments of the form k*Pi, k being integers.
I want to convert this expression to a sum of cosines. TrigReduce
works fine, except that it converts cos to sin.
So, I want all the sin (after the TrigReduce) to be converted to cos
and the arguments be of the form p*Pi, p being rational of course.
It has something to do with Hold or HoldAll, but I don't know the
details. Or perhaps the TrigReduce can be overridden
somehow.
Can someone help?
Thank's for your time.

Cos[J Pi] is + or - depending on whether J is respy even/odd . If
there are m odd Pis and n even Pis, then J = m  alone decides the
sign, - for odd m and + for even m. HTH

It might be helpful to give a small actual example of what you want to 
achieve. As you have described it, I would have thought the answer would 
reduce to +1 or -1.

David Bailey
 http://www.**--****.com/ 

Hi Glenn,

Please, could you post an actual (short) example of such a product as 
well as what you get having trigreduced it?

Meanwhile, Simplify/FullSimplify with the option ComplexityFunction 
might be what you are looking for.

Regards,
-- 
Jean-Marc

Cos[k*Pi] == (-1)^k, so your expression is probably a bit more 
complicated than that :-)


Have you tried

expression /. Sin[a_] :> Cos[a - Pi/2]

or

expression /. Sin[a_] :> Cos[Collect[a - Pi/2, Pi]]

or

expression /. Sin[a_] :> Cos[Collect[a - Pi/2, Pi, Together]]

?  It would be easier to answer if you posted a short sample of a 
Sin-expression that you need converted to a Cos-expression.


Hold[] just prevents evaluation of its arguments.  HoldAll is not a 
function, but an attribute.  It is the attribute that gives Hold[] its 
magic.

Hi Glenn,

you don't need a CAS for this. I'll just use Mathematica syntax here:

 For integer k: Cos[k*Pi]==(-1)^k.

If your product is Product[Cos[k[i]*Pi],{i,1,n}], then it can be written as
(-1)^Mod[Sum[k[i],{i,1,n}],2] or if you insist on cosines:
Cos[Mod[Sum[k[i],{i,1,n}],2]*Pi].


Peter

Hi,

you may simply replace all Sin[x] -> Cos[x-Pi/2]

hope this helps, Daniel