# Thread: Prove a trigonometric identity

1. ## Prove a trigonometric identity

Hi !

(I know a proof, so it's like a challenge question)

Prove that $\sin \tfrac \pi m \cdot \sin \tfrac{2 \pi}{m} \cdots \sin \tfrac{(m-1) \pi}{m}=\tfrac{m}{2^{m-1}}$, where $m \in \mathbb{N}$ and $m \geqslant 2$

2. See this blog post of mine from last February for a proof of this, as a corrolary of a geometric proof using complex numbers.

--Kevin C.

3. Originally Posted by Moo
Hi !

(I know a proof, so it's like a challenge question)

Prove that $\sin \tfrac \pi m \cdot \sin \tfrac{2 \pi}{m} \cdots \sin \tfrac{(m-1) \pi}{m}=\tfrac{m}{2^{m-1}}$, where $m \in \mathbb{N}$ and $m \geqslant 2$
I posted a proof a long time ago here.

Another interesting one (from number theory) is if $p\equiv 1(\bmod 4)$ then:
$\cos \frac{1^2\pi}{p} + \cos \frac{2^2\pi}{p} + \cos \frac{3^2\pi}{p} + ... + \cos \frac{(p-1)^2\pi}{p} = \sqrt{p}$

(I do not think this is solvable by ordinary trigonometry techinques it is a special result from number theory).
It generalizes to when $p$ is any positive integer integer.

4. Originally Posted by ThePerfectHacker

Another interesting one (from number theory) is if $p\equiv 1(\bmod 4)$ then:
$\cos \frac{1^2\pi}{p} + \cos \frac{2^2\pi}{p} + \cos \frac{3^2\pi}{p} + ... + \cos \frac{(p-1)^2\pi}{p} = \sqrt{p}$

(I do not think this is solvable by ordinary trigonometry techinques it is a special result from number theory).
It generalizes to when $p$ is any positive integer integer.
are you sure you wrote the sum correctly?! i think if $p$ is any odd integer, then $S_p=\sum_{j=1}^{p-1} \cos \left(\frac{j^2 \pi}{p} \right)=0.$ the reason is that $S_p=\sum_{j=1}^{\frac{p-1}{2}}\left[\cos \left(\frac{j^2 \pi}{p} \right)+\cos \left(\frac{(p-j)^2 \pi}{p} \right) \right],$

and it's easy to see that $\cos \left(\frac{j^2 \pi}{p} \right)+\cos \left(\frac{(p-j)^2 \pi}{p} \right)=0.$

5. Originally Posted by NonCommAlg
are you sure you wrote the sum correctly?! i think if $p$ is any odd integer, then $S_p=\sum_{j=1}^{p-1} \cos \left(\frac{j^2 \pi}{p} \right)=0.$ the reason is that $S_p=\sum_{j=1}^{\frac{p-1}{2}}\left[\cos \left(\frac{j^2 \pi}{p} \right)+\cos \left(\frac{(p-j)^2 \pi}{p} \right) \right],$

and it's easy to see that $\cos \left(\frac{j^2 \pi}{p} \right)+\cos \left(\frac{(p-j)^2 \pi}{p} \right)=0.$
Sorry , $\sum_{t=0}^{p-1} \cos \frac{2\pi t^2}{p} = \sqrt{p}$.

There is a similar one with $\sum_{t=0}^{p-1}\cos \frac{2\pi t^3}{p}$ but this one was only recently solved, $p\equiv 1(\bmod 3)$
(Known as Kummer's problem)