Prove Tan π/(2n + 1) Tan 2π/(2n + 1) Tan 3π/(2n + 1) .... Tan nπ/(2n + 1) = √(2n + 1)

Question

Tan π/(2n + 1) Tan 2π/(2n + 1) Tan 3π /(2n + 1) .... Tan nπ /(2n + 1) = √(2n + 1

Dividing by √ (2n + 1)

(Tanπ/(2n + 1) Tan2π/(2n + 1) Tan 3π /(2n + 1) .... Tan nπ/(2n + 1))/(√(2n + 1)) = 1

To be equal to 1, the top term must be 1 or equal to (√ (2n + 1)

Putting

1 = tg (pi / 4) = tg45th (Tanπ/(2n + 1) Tan2π/(2n + 1) Tan 3π / (2n + 1) .... Tannπ/(2n + 1)) = opposite collared (√(2n + 1) = adjacent collet

Note that in a rectangular triangle if it is an angle is 45º, the other is also, characterizing the triangle as isosceles, both sides are equal, so (Tan π/(2n + 1) Tan2π/(2n + 1) Tan 3π/(2n + 1) .... Tan nπ/(2n + 1)) = (√ (2n + 1)

Can anyone show me anything better than this? (other than complex) or Who can prove using De Moivre's formula?

Answer 1

Use the identity that $$\left(\frac{x^{2n+1}-1}{x-1}\right)=\left(x^2-2x\cos\frac{2\pi}{2n+1}+1\right).\left(x^2-2x \cos\frac {4\pi}{2n+1}+1\right).... \left(x^2-2x\cos\frac{2n\pi}{2n+1}+1\right). $$ Let us take limit $x\rightarrow 1$, we get $$2^{2n}\sin^2\frac{\pi}{2n+1} \sin^2 \frac{2\pi}{2n+1}...\sin^2 \frac{n\pi}{2n+1}=(2n+1)~~~~~~(1).$$ Next take limit $x \rightarrow -1$, we get $$2^{2n}\cos^2\frac{\pi}{2n+1} \cos^2 \frac{2\pi}{2n+1}...\cos^2 \frac{n\pi}{2n+1}=1~~~~~~(2).$$ By dividing (1) by (2) and taking square root, we get $$\tan \frac{\pi}{2n+1} \tan \frac{2\pi}{2n+1}...\tan \frac{n\pi}{n+1}=\sqrt{(2n+1)}~~~~~~(3).$$