So I have this identity or result from studying polygons and complex numbers which states:

$$\prod_{j=0}^{n-1}\sqrt{\bigg(\cos\frac{\theta+2j\pi}{n}-\cos\frac{\theta}{n}\bigg)^2 +\bigg(\sin\frac{\theta+2j\pi}{n}-\sin\frac{\theta}{n}\bigg)^2}=n $$

I know how to prove this directly using trigonometric identities, and roots of unity of complex numbers, which is quite a lengthy process in my opinion. But my question is, how do I prove this using induction?

For instance going to the induction step straight way give us:

Step I: Let $n = k,$ if $P(k)$ holds true then for $P(k+1)$ we have: $$P(k)=\prod_{j=0}^{k-1}\sqrt{\bigg(\cos\frac{\theta+2j\pi}{k}-\cos\frac{\theta}{k}\bigg)^2 +\bigg(\sin\frac{\theta+2j\pi}{k}-\sin\frac{\theta}{k}\bigg)^2}=k $$ $$P(k+1)=\prod_{j=0}^{k}\sqrt{\bigg(\cos\frac{\theta+2j\pi}{k+1}-\cos\frac{\theta}{k+1}\bigg)^2 +\bigg(\sin\frac{\theta+2j\pi}{k+1}-\sin\frac{\theta}{k+1}\bigg)^2} $$

I am not sure if I am doing this correctly. Furthermore I do not know what to do next after the induction step like how it is carried out in normal induction. I would appreciate some guidance or a possible root to carry out the induction on this result.

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    $\begingroup$ Is $P(k)$ a proposition or a number? Anyway induction does not strike me as a suitable method here. $\endgroup$ – Lord Shark the Unknown May 14 at 4:52
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    $\begingroup$ Why would knowing $\frac{\theta+2j\pi}{k}$, $j=0,1,\dots,k-1$ tell you anything about $\frac{\theta+2j\pi}{k+1}$, $j=0,1,\dots,k$? $\endgroup$ – user10354138 May 14 at 4:52
  • $\begingroup$ Wait let me edit my post $\endgroup$ – Aurora Borealis May 14 at 4:52
  • $\begingroup$ So is the above product not possible to show using induction, in such a way that the LHS gives us $k+1$ with some algebra like standard high school induction? $\endgroup$ – Aurora Borealis May 14 at 4:54
  • $\begingroup$ math.stackexchange.com/questions/8385/… $\endgroup$ – lab bhattacharjee May 14 at 4:57

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