4
$\begingroup$

The question is simple as that. I have:

$Assumptions[v \[Element] Reals]
x[u_, v_] := {f[u], g[u] Cos[v], g[u] Sin[v]}
pil[{x_, y_, z_}, {a_, b_, c_}] := x*a + y*b - z*c
pil[D[x[u, v], u], D[x[u, v], v]]
-2 Cos[v] g[u] Sin[v] Derivative[1][g][u]

and for some reason, Simplify won't turn the last expression into -Sin[2v] Derivative[1][g][u]. I looked a bit around before asking, so I also tried FullSimplify and $Assumptions[v \[Element] Reals], to no avail. Why is all of this going wrong? Can someone explain to me, please? (I find a bit hard to trust a program that can't use the simple fact that $\sin(2v) = 2 \sin v \cos v$.)

$\endgroup$
4
  • 2
    $\begingroup$ Try using TrigReduce $\endgroup$
    – RunnyKine
    Oct 12, 2014 at 22:19
  • $\begingroup$ It worked. Thank you very much! Anyway, I'll leave the question here, in case someone shows up to give a more technical explanation for the problem. $\endgroup$
    – Ivo Terek
    Oct 12, 2014 at 22:21
  • 1
    $\begingroup$ The reason is Mathematica is such a large system with a lot of rewrite rules and special cases of those rules. So sometimes you'll have to use available special functions to force it to do certain simplifications as in this case. $\endgroup$
    – RunnyKine
    Oct 12, 2014 at 22:26
  • $\begingroup$ That makes it clearer! $\endgroup$
    – Ivo Terek
    Oct 12, 2014 at 22:28

1 Answer 1

6
$\begingroup$

You can use TrigReduce to do what you want:

TrigReduce[-2 Cos[v] g[u] Sin[v] Derivative[1][g][u]]
-g[u] Sin[2 v] Derivative[1][g][u]
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.