4
votes
2answers
78 views

Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$

I would like to show that $$ \sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64} $$ I've been working on this for a few ...
3
votes
2answers
37 views

Product of $1-\operatorname{cis}(2k\pi/n)$

I'm in a question about polygonals and got stuck at a part. I have to prove that $$\prod_{k=1}^{n-1} \left(1 - \operatorname{cis}(\frac{2k\pi}{n})\right) = n$$ I've tried to multiply it to make ...
0
votes
1answer
45 views

Trigonometry and complex numbers

Suppose $z_0=e^{i\theta_0}$ a complexe number as $\theta_0\in ]-\pi,\pi[ \setminus\{0\}$. For $n\in \mathbb{N}$, we pose $z_{n+1}=\dfrac{|z_n|+z_n}{2}$ and $z_n=r_ne^{i\theta_n}$ with ...
1
vote
1answer
45 views

For what $k$ is $P(k):=\prod_{j=1}^{13}\cos\frac{\pi kj}{13}$ negative?

Let $k>0$ be an integer. For what $k$ is $P(k):=\prod_{j=1}^{13}\cos\frac{\pi kj}{13}$ negative? Since $13$ is prime, and for $\gcd(m,13)=1$, $P(2m)=P(2)=2^{-12}$ (can be shown by considering the ...
4
votes
2answers
325 views

General expression for $\sin(2^n x)$

Are there general expressions for $\sin(2^n x)$ and $\cos(2^n x)$ that only involve $\sin x$ and $\cos x$, and that moreover involve only polynomial (in $n$) number of terms? Edit: $2^n$ is not ...
3
votes
1answer
59 views

Show that the following product equals 1 (involves trig)

How can I show that: $$\prod_{k=1}^{n}\left ( 1+2\cos\frac{2\pi .3^{k}}{3^{n}+1} \right )=1$$ Could you please explain to me how to approach this problem? Thank you.
19
votes
3answers
394 views

Finding $ \prod_{n=1}^{999}\sin\left(\frac{n \pi}{1999}\right)$

I would appreciate if somebody could help me with the following problem. How can we find the product $$ \prod_{n=1}^{999}\sin\left(\frac{n \pi}{1999}\right)$$
12
votes
1answer
494 views

How does one calculate the product of $\tan 1^{\circ} … \tan 45^{\circ}?$

I have seen a question asked on yahoo asking to find the value of $\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \dots \cdot \tan 45^{\circ}$ (in degrees) I have seen various results concerning ...
8
votes
2answers
217 views

How to find finite trigonometric products

I wonder how to prove ? $$\prod_{k=1}^{n}\left(1+2\cos\frac{2\pi 3^k}{3^n+1} \right)=1$$ give me a tip
3
votes
0answers
208 views

Product of sines

I am looking to evaluate $$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n}$$ without using complex numbers. I can show the result if $n$ is a power of $2$, but if $n$ is anything else I reach a point where I ...
9
votes
4answers
1k views

Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$

Recently, I ran across a product that seems interesting. Does anyone know how to get to the closed form: $$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$ I ...
7
votes
1answer
395 views

Evaluation of a product of sines [duplicate]

Possible Duplicate: Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ I am looking for a closed form for this product of sines: \begin{equation} \sin ...
6
votes
2answers
668 views

How was Euler able to create an infinite product for sinc by using its roots?

In the Wikipedia page for the Basel problem, it says that Euler, in his proof, found that $$\begin{align*} \frac{\sin(x)}{x} &= \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 ...
16
votes
3answers
821 views

A “fast” way for computing $ \prod \limits_{i=1}^{45}(1+\tan i^\circ) $?

Which is the fastest paper-pencil approach to compute the product $$ \prod \limits_{i=1}^{45}(1+\tan i^\circ) $$
3
votes
2answers
558 views

Proving: $\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A … \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } $

$$\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } $$ I am very much inquisitive to see how this trigonometrical identity can be ...
15
votes
2answers
2k views

Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

Using n_th root of unity $$\Large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$ Prove that $$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$