Due to the fact that $$\Gamma(x)\Gamma(1-x)\sin \pi x=\pi$$


is this necessarily true:


$$\Gamma(x)\Gamma(1-x)\sin(\Gamma(1+x)\Gamma(1-x)\sin(\cdots))=\pi$$


Is this at all used as a tactic to create different identities in general?
(The Reflection Property example was just to illustrate my point)



When are/ if recursive identities used?
As follows is what I assume you are asking. Taking


$$A = \Gamma(z)\Gamma(1-z)\sin(z\pi) = \pi$$


then rather trivially substituting $A$ for $\pi$


$$\Gamma(z)\Gamma(1-z)\sin(z\Gamma(z)\Gamma(1-z)\sin(\pi z)) = \pi$$


Continually making the substitution for $\pi$ on the inside of the $\sin$ with $A$ produces the exact same effect. Wherein you have a rather trivial limit. 


written more concisely using two variables we have


$$\phi(z,s) = \Gamma(1+z)\Gamma(1-z)\sin(s)$$


that


$$\sin(\phi(z,\phi(z,\phi(z,...(n\,times)...\phi(z,\pi z)) = \pi \sin(\pi z)$$


namely that your limit is true, but rather more interestingly that


$$\sin(\Gamma(1+z)\Gamma(1-z)\sin(\Gamma(1+z)\Gamma(1-z)\sin(... = \pi \sin\pi z$$


which seems rather bewildering to be honest.


Edit:


As to if this technique is used in functional identities: On infinite processes similar techniques are used (infinite products, infinite sums), however modified from the above (respectfully if multiplication or addition, instead of composition, is used). On infinite compositions (compositions of a function with itself infinite times as in how I wrote my answer) yes this technique is used occasionally. However it's usually much more diverse in the requirements of the proof.


