A Question about a trigonometric identity

Martin,

You have to do it yourself.  Here is a partial derivation, reverse the steps and you can finish it yourself.

(out23) a*sin(x-b)
(%i24) trigexpand(%);

(out24) a*(cos(b)*sin(x)-sin(b)*cos(x))   /* law of sum and difference between 2 angles is applied by trigexpand */
(%i25) %/cos(b);

(out25) a*(cos(b)*sin(x)-sin(b)*cos(x))/cos(b)
(%i26) expand(%);

(out26) a*sin(x)-a*sin(b)*cos(x)/cos(b)
(%i27) ratsubst(tan(b),sin(b)/cos(b),%);

(out27) a*sin(x)-a*tan(b)*cos(x)
(%i28) ratsubst(B,a*tan(b),%);

(out28) a*sin(x)-cos(x)*B


Etc...

Now go figure out A.  B=a*tan(b)

This is the basic idea

Rich


  ----- Original Message ----- 
  From: Stavros Macrakis 
  To: Richard Hennessy 
  Cc: Martin Mallinson ; maxima at math.utexas.edu 
  Sent: Monday, November 16, 2009 7:04 PM
  Subject: Re: [Maxima] A Question about a trigonometric identity


  Yes, but as far as I know there is no easy way in Maxima to convert in this direction:

          a*sin(x)+b*cos(x) => q*sin(x+r)

  The other direction is of course trivial -- expand(trigexpand(...))

               -s



  On Mon, Nov 16, 2009 at 6:17 PM, Richard Hennessy <rich.hennessy at verizon.net> wrote:

    a*sin(b)+c*cos(b) is always a phase shift with a possible amplitude change
    as well.

    Rich



    ----- Original Message -----
    From: "Martin Mallinson" <Martin at mallinson.homeip.net>
    To: <maxima at math.utexas.edu>
    Sent: Monday, November 16, 2009 1:35 PM
    Subject: A Question about a trigonometric identity


    >I am new to Maxima, never having used this mailing list before, and I
    > wonder
    > if anyone can help clarify this small problem I have?.
    >
    > eq1: R = (3^(3/2)*sin(z)-3*cos(z))/4;
    > trigrat(%);
    > print("Maxima fails to simplify the expression. But...");
    > subst(z1+%pi*2/3,z,eq1);
    > trigrat(%);
    > subst(z-%pi*2/3,z1,%);
    > print("Maxima now shows us that eq1 was in fact just a phase shift");
    > print("How can I get Maxima to show this without the two subst
    > operations?");
    >
    > My problem of course, is that I had to guess that the expression was a
    > simple
    > phase shift, had I not guessed this, I would never have got there...
    >
    > Thanks for any help
    > Martin M
    >
    > By the way - thank you all for supporting Maxima. I had the pleasure of
    > working with Macsyma on the Symbolic XL1200 -- I have the original
    > Symbolics Macsyma documentation someplace. It is wonderful to have
    > access to Maxima as an open source project. Thank you.
    >
    >
    > _______________________________________________
    > Maxima mailing list
    > Maxima at math.utexas.edu
    > http://www.math.utexas.edu/mailman/listinfo/maxima
    >


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