Proving $\tan^2\frac{\pi}{4n}\cdot\tan^2\frac{3\pi}{4n}\cdot\,\cdots\,\cdot\tan^2\frac{(2n-1)\pi}{4n}=1$

Question

$$\tan^2\frac{\pi}{4n}\cdot\tan^2\frac{3\pi}{4n}\cdot\cdots\cdot\tan^2\frac{(2n-1)\pi}{4n}=1, \quad \forall n \in \Bbb{N}^*$$

original problem image

I tried putting $z=\cos(\pi/4)+i\sin(\pi/4)$, and expressing $\tan$ of $\pi/4$, $3\pi/4$ etc. The problem is that the expression is really hard to compute.

The official solution starts from the equation $$(2n+iz)^{2n}+(2n-iz)^{2n}$$ which has solutions that you can multiply using Vieta’s formulas and get the desired result. However, HOW am I supposed to GUESS an equation that has those solutions?

If you have a logical answer please help me! Thanks in advance.

score 4 · Answer 1 · answered Oct 09 '19 at 08:18

4

Hint:

I thought of a really simple argument that may lead to the proof. Notice the inputs to the first and last terms being multiplied. They add up to $\pi/2$ as shown: $\pi/4n+(2n-1)\pi/4n=\pi/2$. This means that their product cancels to $1$ as follows: $\tan(\pi/4n)\cdot\tan((2n-1)\pi/4n)=1$. Similarly you may see that each of such pair of terms give $1$.

answered Oct 09 '19 at 08:18

Paras Khosla

6,481

1

The interesting case is when $n$ is odd, so there are an odd number of terms. – Toby Mak Oct 09 '19 at 08:23
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That's what I was thinking about.Then the middle term would have to be $\tan(\pi/4)$ and the pattern confirms that. – Paras Khosla Oct 09 '19 at 08:24

Proving $\tan^2\frac{\pi}{4n}\cdot\tan^2\frac{3\pi}{4n}\cdot\,\cdots\,\cdot\tan^2\frac{(2n-1)\pi}{4n}=1$

1 Answers1