My idea for q-Trigonometry was simply to substitute QFactorial for Factorial in the identity z! (-z)! == π z/Sin[π z]. (z = -1/2 is where we get (-1/2)! = √π.) This determines the "q-extension" of π to be q^(1/4) QFactorial[-1/2,q^2]^2== π*(1+(q-1)+1/6 (q-1)^2-1/180 (q-1)^4+1/180 (q-1)^5-(37 (q-1)^6)/7560+4/945 (q-1)^7-(119 (q-1)^8)/32400+(9 (q-1)^9)/2800+O[q-1]^10) (Arguably, q^(1/8) QFactorial[-1/2,q]^2). This expansion (which should be in powers of 1-q) dates from an ancient Macsyma. I was unable to drag it out of Mathematica, which is woefully ignorant of expanding *q*-things near the all-important q➝1 limit. Likewise, other points on the unit circle. But Mathematica should be able to average π_q for us over the unit disk. For every triplet of distinct positive integers a,b,c, there is a polynomial identity relating π_q^a, π_q^b, and π_q^c, because π_q==(1-q^2) 𝜂(q^2)^4)/𝜂(q)^2, where 𝜂(q) := q^(1/24) Product[1-q^n,{n,∞}] = DedekindEta[Log@q/2/I/π] and 𝜂(q) satisfies an equivalent infinitude of three term identities. The simplest π_q identity is Out[95]= 4==-(1+q^4)^2 (π_q^2)^2/(π_q^4)^2+(1+q^2)^2 (1+q^4) (π_q)^2/π_q^2/π_q^4 (Converted to 𝜂 functions, this is Jacobi's Æquatio Identica Satis Abstrusa. But the *least* abstruse, it turns out!). Testing with our π_q expansion near 1, π_q:>π*(q+1/6 (-1+q)^2-1/180 (-1+q)^4+1/180 (-1+q)^5-(37 (-1+q)^6)/7560+4/945 (-1+q)^7-(119 (-1+q)^8)/32400+(9 (-1+q)^9)/2800) % /. q -> q^2; % /. q -> q^2; (My need to special case q^2 and q^4 affirms that the design of Series is seriously deficient.) %95 /. {%, %%, %%%} <Big mess> In[117]:= Series[Activate[%],{q,1,9}] Out[117]= 4==4+O[q-1]^10 Quod erat orandum. Why idolize a mere number? —rwg