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why does the following instruction ok in mathematica, and it returns a boolean value

In[398]:= catenary[x, 1, 0, 0] == catenary[-x, 1, 0, 0]
Out[398]:= True

and not this one?

In[406]:= catenary[-x, 1, 0, 0] == -catenary[x, 1, 0, 0]
Out[406]:= Cosh[x] == -Cosh[x]

I wanted to have a boolean value too.. to prove the symmetry.. Thanks for your help!

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    $\begingroup$ Wrap the entire thing in TrueQ[] if need be. Less facetiously: your last equation is true for x an odd multiple of $\dfrac{\pi i}{2}$ and false otherwise. $\endgroup$ – J. M.'s discontentment Jun 19 '12 at 15:52
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I think you should use three "="s, i.e.,

Cosh[x]===Cosh[-x]

(* True *)

If you need the Boolean value, then

Cosh[x] === Cosh[-x] // Boole

(* 1 *)

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    $\begingroup$ Cosh[x] == Cosh[-x] works fine; Mathematica is not too dumb to not know that the hyperbolic cosine is even. $\endgroup$ – J. M.'s discontentment Jun 19 '12 at 16:30
  • $\begingroup$ @J.M. But Sinh[x]==Sinh[-x] return itself, which is not the purpose of the programmer, whereas, Sinh[x]===Sinh[-x] works fine. $\endgroup$ – yulinlinyu Jun 21 '12 at 1:28
  • $\begingroup$ Again: Sinh[] is not the function in the OP. $\endgroup$ – J. M.'s discontentment Jun 21 '12 at 1:29
  • $\begingroup$ This is an example to illustrate the way to cope with the problem. I don't think the author need Cosh ONLY but no other functions. $\endgroup$ – yulinlinyu Jun 21 '12 at 1:31
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    $\begingroup$ You'd think that, until you see OP's other questions... $\endgroup$ – J. M.'s discontentment Jun 21 '12 at 1:32

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