Advanced Search

Question: trouble with "applyrule"

Question:trouble with "applyrule"

jeilerts 50
January 27 2013
1

I have a question about the "applyrule" function. I have an expression that looks like:

 

exprsn1:=sin(z)(Acos(z) + Bcos(x)cos(2z) + Ccos(x));

 

I would like to express cos(2z)sin(z) in the form 1/2(sin(3z)-sin(z)). However, when I use the "applyrule" as:

exprsn2 := applyrule(sin(z)cos(2z)=1/2(sin(3z)-sin(z), exprsn1);

it returns the same expression. Is there something else I should be doing before I use applyrule?

 

 

 
Please Wait...