Product of tangents when angles are in AP

Asked May 27 '21 at 02:52

Active May 27 '21 at 02:52

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$$\displaystyle \tan\bigg(\frac{\pi}{25}\bigg)\cdot \tan\bigg(\frac{2\pi}{25}\bigg)\cdot \tan\bigg(\frac{3\pi}{25}\bigg)\cdots\cdots \tan\bigg(\frac{12\pi}{25}\bigg)$$ I know how to evaluate this using Complex numbers and then using Viete's theorem. Is there a way that this can be found out only using Trigonometry?

asked May 27 '21 at 02:52

satyam sharma

Duplicate – Sathvik May 27 '21 at 02:54
@SathvikAcharya Are you suggesting me to just post my comment on the original question? – satyam sharma May 27 '21 at 02:56
No, it adds value to the question by providing solutions already presented by users on the site. – Sathvik May 27 '21 at 03:00
In theory, you could repeatedly apply the tan addition formula and hope it simplifies nicely. – Eric May 27 '21 at 03:14
Not sure if this helps, but: Using $\cos(x)=\sin\left(\frac\pi2-x\right),$ $$\tan \frac{n\pi}{25}=\frac{\sin\frac{n\pi}{25}}{\sin\frac{(25-2n)\pi}{50}}$$
So the product is $$\prod_{n=1}^{24} \left(\sin\frac{n\pi}{50}\right)^{(-1)^n}$$
– Thomas Andrews May 27 '21 at 03:29
It seems like you might be able to use Chebyshev polynomials. That might be what the Vieta argument is hiding. But it won’t be any easier than Vieta. – Thomas Andrews May 27 '21 at 03:41
There is also a similar question here, here and here. – May 27 '21 at 10:15
See https://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series – lab bhattacharjee May 29 '21 at 07:35

Product of tangents when angles are in AP

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