Start with A an abelian category and form the derived category D(A). Take a triangle (not necessarily distinguished) and take it's cohomology. We obtain a long sequence (not necessarily exact). If the triangle is distinguished it is exact. How about the converse: if the long sequence in cohomology is exact does it follow that the triangle is distinguished? (My guess is no, but I can't find a counterexample).

An important property of the derived category is that distinguished triangles don't just produce long exact sequences in cohomology. If A > B > C > A[1] is an exact triangle and E is another object in the derived category, then you get a long exact sequence ... Hom(A[1],E) > Hom(C,E) > Hom(B,E) > Hom(A,E) > Hom(C[1],E) > ... where these Homsets are sets of maps in the derived category. A particular counterexample is as follows. We can view any abelian group as a chain complex concentrated in degree zero. There is a distinguished triangle as follows: ℤ > ℤ > ℤ/2 > ℤ[1] However, we can take the last map ℤ/2 > ℤ[1] in the sequence (which is not zero in the derived category) and replace it with the zero map. This still gives us a long exact sequence on (co)homology groups. However, if we let E = ℤ/2, then applying maps in the derived category from our new nondistinguished triangle gives us the sequence ... 0 > Hom(ℤ/2,ℤ) > Hom(ℤ,ℤ) > Hom(ℤ,ℤ) > Ext(ℤ/2,ℤ) > 0 ... where the last map is induced by the zero map from ℤ/2[1] to ℤ, and so it must be zero. This sequence can't possibly be exact, and so the new triangle is not distinguished. EDIT: A more subtle question this suggests is: "Suppose I have a triangle and, for any E, applying Map(,E) or Map(E,) gives a long exact sequence. Is this a distinguished triangle?" The answer to this is actually still no. Still considering chain complexes of abelian groups, take the distinguished triangle ℤ > ℤ > ℤ/3 > ℤ[1] where I'll call the last map β (for Bockstein). You can take this distinguished triangle and replace β with its negative β. The new triangle still induces long exact sequences on maps in or maps out (because the maps in the triangle have the same kernel and image). However, as an exercise show that this can't be a disntinguished triangle because it's not isomorphic to the original distinguished triangle. 


I just wanted to point out that this failure is quite standard rather than pathological. As a starting point it can go wrong more generally than Tyler points out. For instance there exist triangles which give long exact sequences under any product preserving homological functor to an [AB4*] abelian category which are not distinguished. In general for a noncontractible distinguished triangle the morphism X > Y > Z > X[1] 0 0  0 v v v v Y > Z >X[1]>Y[1] where the nonlabeled vertical map is produced via [TR3] will have a mapping cone which is such a triangle. More concretely (well a bit) along the lines of Tyler's examples one can consider K^b(Z)^{} the bounded homotopy category of abelian groups but without its class of distinguished triangles and declare that a triangle (u,v,w) (these are the morphisms occurring in it) in K^b(Z)^{1} is distinguished if and only if (u,v,w) is a triangle in K^b(Z) with its usual triangulation. This makes K^b(Z)^{1} a triangulated category (in fact the collection of triangles which good homological functors send to long exact sequences is pretty well behaved and one can prove a bunch of the standard facts in such generality), but these two categories are not triangle equivalent. 


You might be interested in the paper: Vaknin A., Virtual Triangles// KTheory, 22 (2001), no. 2, 161197. 

