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# distribute/ square/ sinus

Posted 10 years ago
 Good afternoon, I have this function: Sin[Sqrt[b^2 ((Sqrt[k\[Sigma]] m)/(a b) - m^2/a^2)] \[Pi]]  I want to distrabute the b² into the parenthese : Sin[\[Pi] Sqrt[ (b Sqrt[k\[Sigma]] m)/a - (b^2 m^2)/a^2] ]  The Mathematica instruction Distribute allowed to make this, but for one function without the square, and without sinus. I make one Mathematica file, as an attachment. Greetings. Attachments:
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Posted 10 years ago
 Every time pattern matching in Mathematica seems not to work, e.g. one has In[10]:= Expand[b^2 ((Sqrt[k\[Sigma]] m)/(a b) - m^2/a^2)] Out[10]= (b Sqrt[k\[Sigma]] m)/a - (b^2 m^2)/a^2 but gets nevertheless In[9]:= Sin[Sqrt[b^2 ((Sqrt[k\[Sigma]] m)/(a b) - m^2/a^2)] \[Pi]] /. Sqrt[o_] -> Sqrt[Expand[o]] Out[9]= Sin[Sqrt[b^2 ((Sqrt[k\[Sigma]] m)/(a b) - m^2/a^2)] \[Pi]] the internal representation of the expression using FullForm should be inspected: In[17]:= FullForm[Sin[Sqrt[b^2 ((Sqrt[k\[Sigma]] m)/(a b) - m^2/a^2)] \[Pi]]] Out[17]//FullForm= doing so you proceed as follows In[20]:= Sin[Sqrt[b^2 ((Sqrt[k\[Sigma]] m)/(a b) - m^2/a^2)] \[Pi]] /. Power[o_, Rational[1, 2]] :> Power[Expand[o], Rational[1, 2]] Out[20]= Sin[Sqrt[(b Sqrt[k\[Sigma]] m)/a - (b^2 m^2)/a^2] \[Pi]] 
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