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I would like to simplify a fraction of 2 trigonometric functions

EqAng1={Cos[Subscript[β, 0][t]] Sin[Subscript[α, 0][t]] == 
  Cos[Subscript[β, i][t]] Sin[Subscript[α, i][t]], 
 Cos[Subscript[α, 0][t]] Cos[Subscript[β, 0][t]] == 
  Cos[Subscript[α, i][t]] Cos[Subscript[β, i][t]], 
 Sin[Subscript[β, 0][t]] == Sin[Subscript[β, i][t]]}

Simplify[EqAng1[[1]]/EqAng1[[2]]]

The result obtained is not simplified.

Is there a function enabling a direct simplification of this fraction with trigonometric functions ?

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  • $\begingroup$ Most likely, you want to avoid using subscripts. They are pretty tricky if you do not exactly know what you are doing (that's a general remark). Besides that, Simplify and FullSimplify work better if you provide Assumptions. Also, what do you want to happen to the Equals in your list? Do you want the method to give the condition under which Equal returns true? $\endgroup$ – Lukas Feb 20 '16 at 19:13
  • $\begingroup$ How I can remove "the subscripts" when I cite a piece of code? $\endgroup$ – Bendesarts Feb 20 '16 at 20:10
  • $\begingroup$ @Lukas If I have eq1: A=B and eq2: C=D with A,B,C,D trigonometric expressions, i wanted to know if there is a function which can do eq1/eq2 which can give A/B=C/D $\endgroup$ – Bendesarts Feb 20 '16 at 20:12
  • $\begingroup$ It was not just about getting rid of the subscripts in the question, I meant to not use them in your code. They are not used like symbols, see e.g. here (mathematica.stackexchange.com/q/1004/21606). So you just want to rewrite your equations? Would rewrite[eq1_,eq2_]:=eq1[[1]]/eq1[[2]]==eq2[[1]]/eq2[[2]]be what you want? $\endgroup$ – Lukas Feb 20 '16 at 21:03
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Format[α[i_]] := Subscript[α, i];
Format[β[i_]] := Subscript[β, i];

EqAng1 = {
   Cos[β[0][t]] Sin[α[0][t]] == 
    Cos[β[i][t]] Sin[α[i][t]],
   Cos[α[0][t]] Cos[β[0][t]] == 
    Cos[α[i][t]] Cos[β[i][t]],
   Sin[β[0][t]] == Sin[β[i][t]]};

You cannot divide one equation by another. Define a function to do what you want.

divideEqn[eqn1_Equal, eqn2_Equal] :=
  eqn1[[1]]/eqn2[[1]] == eqn1[[-1]]/eqn2[[-1]];

divideEqn[EqAng1[[1]], EqAng1[[2]]]

enter image description here

divideEqn[EqAng1[[1]], EqAng1[[3]]]

enter image description here

divideEqn[EqAng1[[2]], EqAng1[[3]]]

enter image description here

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