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RE: Re[a + I b] = a
A member of the group wanted to do the following: 
> for (n) an integer number Sin[n x] results 0 >

Dave Wothoff (of Wolfram Research) replied with: 
Algorithmically, this is a much harder (and largely unsolved) problem.
If your needs are not too elaborate, though, you may be able to
program it yourself, as in

In[4]:= Unprotect[Sin] ;

In[5]:= Sin[p_ Pi] := 0 /; IntegerQ[p] 
In[6]:= Protect[Sin] ;

In[7]:= IntegerQ[n] ^= True ;

In[8]:= Sin[n Pi]

Out[8]= 0


Dave, Please be patient and read on.
The lines below are a small part of a package I am working on. Why
doesn't Mathematica have the Rules below and many more built in? I
doubt Mathematica will ever have all known transformations stored in a
list of Rules, but I would think Wolfram Research could easily provide
lists with many of the important Rules.
You say "Algorithmically, this is a much harder (and largely unsolved)
problem".
I don't understand. Can you give me an example of why this is very
difficult to achieve.
Ted Ersek
(Mathematica Input, Output with remarks below)

RealNumericQ[expr], and VariableQ[expr] are useful in and of themselves.
The function names pretty much explain what they do.
In[1]:=
RealNumericQ[expr_?NumericQ]:=
FreeQ[ComplexExpand[expr,TargetFunctions>{Re,Im}],Complex[_,_]]/;
FreeQ[expr,_?VaribleQ,{1},Heads>False]
VaribleQ[symb_]:=(Head@symb===Symbol)&&Not@NumericQ[symb]
In the next Cell ArcSin[7] doesn't simplify, but RealNumericQ knows
it's Complex.
However it is not known if (Sqrt[x]+Sin[7]) is Complex.
In[3]:=
Clear[x];
{RealNumericQ[Sqrt[2]+ArcSin[1/7]],
RealNumericQ[Sqrt[2]+ArcSin[7]],
RealNumericQ[Sqrt[x]+Sin[7]]}
Out[3]=
{True, False, RealNumericQ[Sqrt[x] + Sin[7]]}
Note: My command Declare[.....]
is an extension of a package in mathsource.
In[4]:=
AssumedInterval[x_]:=x
SetAttributes[Declare, HoldAll]
Declare[x_Symbol,IntegerQ]/;(NumericQ[x]===False):=
(AssumedInterval[x]^=Interval[{Infinity,Infinity}];
IntegerQ[x]^=True;)
Declare[x_Symbol,Positive]/;(NumericQ[x]===False):=
(AssumedInterval[x]^=Interval[{0,Infinity}];
Sign[x]^=1;)
Declare[LessEqual[a_?RealNumericQ,x_?VaribleQ,b_?RealNumericQ]]/;
(a<b):= (AssumedInterval[x]^=Interval[{a,b}])
In[9]:=
RuleList={
Sin[_?IntegerQ*Pi]:>0,
Cos[_?EvenQ*Pi]:>1,
Cos[_?OddQ*Pi]:>1,
Sign[x_?VaribleQ]/;(Min[AssumedInterval[x]]>0):>1,
Sign[x_?VaribleQ]/;(Max[AssumedInterval[x]]<0):>1,
Arg[x_?VaribleQ]/;((Sign[x]/.RuleList)===1):>0,
Arg[x_?VaribleQ]/;((Sign[x]/.RuleList)===1):>1,
Positive[x_?VaribleQ]/;((Sign[x]/.RuleList)===1):>True,
Positive[x_?VaribleQ]/;((Sign[x]/.RuleList)!=1):>False,
Sqrt[x_^n_]/;(n*(Arg[x]/.RuleList)<=Pi):>x^(n/2)
};
Note: The Rules for simplifying expressions don't slow down normal
evaluation. They are only used when called on.
In[10]:=
Clear[n];
Declare[n,IntegerQ];
{Sin[n Pi], Sin[n Pi]/.RuleList}
Out[10]=
{Sin[n*Pi], 0}
Note: The UpValue for Sign[x] also effects the result of Positive[x].
In[11]:=
Clear[x];
Declare[x,Positive];
{Sign[x], Positive[x]}
Out[11]=
{1,True}
After I Declare that (a) is in a Positive Interval, the RuleList knows
how to handle Arg[a], Sign[a], Positive[a].
In[12]:=
Clear[a];
Declare[Sqrt[3]<=a<=Sqrt[6]];
{Arg[a]/.RuleList,
Sign[a]/.RuleList,
Positive[a]/.RuleList}
Out[12]=
{0,1,True}
RuleList also knows when Sqrt[x^n] can be simplified.
In[13]:=
Sqrt[a^2]/.RuleList
Out[13]=
a
In[14]:=
Clear[z];
Arg[z]^=Pi/6;
Sqrt[z^4]/.RuleList
Out[14]=
z^2
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