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If I do

FullSimplify[Reduce[Sin[p1] == 0 && Cos[p1] == 0, Reals]]

I get

 Cos[p1] == 0 && Sin[p1] == 0

while I would expect False. why? is there a way to have mathematica compute that both sin and cos cannot be 0?

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1  
Reduce[Sin[p1] == 0 && Cos[p1] == 0, p1] returns the expected result. –  J. M. Nov 10 '12 at 17:03
    
Domain specification Reals is unnecessary. –  Artes Nov 10 '12 at 17:12
    
Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2)Read the FAQs! 3) When you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign` –  chris Nov 10 '12 at 17:21
    
Thank you @J.M., If you write it as an answer I will mark it as definitive –  Fabio Dalla Libera Nov 10 '12 at 17:33
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2 Answers

up vote 5 down vote accepted

As noted, specifying the variable of interest within Reduce[] yields the expected result:

Reduce[Sin[p1] == 0 && Cos[p1] == 0, p1]
   False
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Because sin(0)=0 and cos(0)=1 (and cos(π/2)=0 while sin(π/2)=1), it is clear that the result is false, as expected.

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Yes, exactly: as the result is obvious, why does Mathematica not state it explicitly? J.M. has given the answer, but here you simply restate the premise of the question. Sorry that I have to downvote your first post here, but please don't take it personally--thanks for contributing and welcome to the site! –  Oleksandr R. Nov 10 '12 at 18:23
    
You are right, thank you. –  Bujanca Mihai Nov 12 '12 at 7:03
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