I've been looking around and I see no formulas given in any of the sources I've been able to find for the infinite product representing $\tan\left(x\right)$. Is it simply the ratio of the infinite products for $\sin\left(x\right)$ and $\cos\left(x\right)$? In which case, it should look like $$ \frac{x\cdot\prod\limits_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)}{\prod\limits_{n=1}^{\infty}\left(1-\frac{4x^2}{\pi^2\left(2n-1\right)^2}\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}=\tan\left(x\right)? $$ Forgive me if there is an obvious solution to this problem that I haven't seen as of yet. I've taken up to differential equations and multivariable calc so far in my major.


  • $\begingroup$ It's fine, I remember the numerator and I imagine the denominator works. Be aware that the evident factoring(s) lead to products that do not converge. As long as you keep the squares in place all is good. $\endgroup$ – Will Jagy Sep 28 '14 at 4:08
  • $\begingroup$ A bunch of $\pi$'s got lost, of course. $\endgroup$ – user138530 Sep 28 '14 at 4:55
  • $\begingroup$ Woops, sorry! I'll add them... $\endgroup$ – jm324354 Sep 28 '14 at 5:58
  • $\begingroup$ Now it is okay. $\endgroup$ – Lucian Sep 28 '14 at 15:30
  • The upper product should start at $n=1$, otherwise you have division by $0$.
  • The x before the infinite product in the numerator should be changed to $\pi x$.
  • The arguments of the sine, cosine, and tangent functions should be $\pi x$ rather than x. Unless you want to imply that $~\sin(0)=\sin(\pm1)=\sin(\pm2)=\sin(\pm3)=\ldots=0$, as opposed to $\sin(0)=\sin(\pm\pi)=\sin(\pm2\pi)=\sin(\pm3\pi)=\ldots=0$.
  • $\begingroup$ Yeah someone commented above and said that the pi's were missing. I just copied it wrong from the formula sheet and haven't fixed it yet. $\endgroup$ – jm324354 Sep 28 '14 at 15:11
  • $\begingroup$ Fixed it now, at least according to Wolfram. $\endgroup$ – jm324354 Sep 28 '14 at 15:15
  • $\begingroup$ I didn't know you could do step three...neat, thanks. $\endgroup$ – jm324354 Sep 28 '14 at 22:36

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