Trigonometric sum

[man111]

prove that [tex]\sum_{n=1}^{k}\tan \left(\frac{n\pi}{2k+1}\right) = \sqrt{2k+1}[/tex]

[kerry]

prove that [tex]\sum_{n=1}^{k}\tan \left(\frac{n\pi}{2k+1}\right) = \sqrt{2k+1}[/tex]

maybe you mean [tex]\prod_{n=1}^{k}\tan \left(\frac{n\pi}{2k+1}\right) = \sqrt{2k+1}[/tex]

[man111]

Yes kerry

[man111]

prove that [tex]\prod_{n=1}^{k}\tan \left(\frac{n\pi}{2k+1}\right) = \sqrt{2k+1}[/tex]

after editing

[martin123456]

Prove the following facts:
(1) [tex]\prod_{k=1}^{n-1}\sin{(\frac{k\pi}{n})}=\frac{n}{2^{n-1}}[/tex]
(2) [tex]\prod_{k=1}^{n-1}\cos{(\frac{k\pi}{n})}=\frac{\sin{(\frac{\pi n}{2})}}{2^{n-1}}[/tex]
Divide them to get
(*) [tex]\prod_{k=1}^{n-1}tg{(\frac{k\pi}{n})}=\frac{n}{\sin{(\frac{\pi n}{2})}}[/tex]
Now use the symmetry of tg to conclude the fact stated in the problem.