Linked Questions

13
votes
1answer
3k 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 \left(\frac{\pi}{n}\...
8
votes
2answers
4k views

Prove that: $\sin{\frac{\pi}{n}} \sin{\frac{2\pi}{n}} …\sin{\frac{(n-1)\pi}{n}} =\frac{n}{2^{n-1}}$ [duplicate]

Using that: $$ x^{n - 1} + x^{n - 2} + \cdots + x + 1 = \left(x - w\right)\left(x - w^{2}\right)\ldots\left(x - w^{n - 1}\right) $$ Prove that: $$ \sin\left(\pi \over n\right)\sin\left(2\pi \over n\...
4
votes
3answers
288 views

Find the value of $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$ [duplicate]

Find the value of $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$ $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\...
5
votes
2answers
262 views

Evaluate $\prod_{i=1}^{89} \sin (i)$ [duplicate]

Evaluate: $$\prod_{i=1}^{89} \sin (i)$$ My attempt:We know that $\sin \alpha \cdot \sin(90- \alpha)=\sin \alpha \cos \alpha=\frac{1}{2}\sin 2\alpha$ Then we have: $$ (\sin1\cdot\sin89)(\sin2\cdot\...
5
votes
2answers
104 views

$\sin(\frac{\pi}{n})\sin(\frac{2\pi}{n})…\sin(\frac{(n-1)\pi}{n})=\frac{n}{2^{n-1}}$ [duplicate]

Prove that $$\sin\left(\frac{\pi}{n}\right)\sin\left(\frac{2\pi}{n}\right)\sin\left(\frac{3\pi}{n}\right).....\sin\left(\frac{(n-1)\pi}{n}\right)=\frac{n}{2^{n-1}}$$ Is there a proof without using ...
4
votes
1answer
308 views

how to prove $\prod_{k=1}^{p-1} \sin(\frac{\pi k}{p}) = \frac{p}{2^{p-1}}$? [duplicate]

i found this relation whilst trying to evaluate the norm (over $\mathbb{Q}$) of $1-\zeta$ for $\zeta$ a primitive $p$-th root of unity ($p$ supposed prime) $$ \prod_{k=1}^{p-1} \sin(\frac{\pi k}{p}) = ...
3
votes
0answers
162 views

Prove that $\prod^{n}_{k=1}\sin\left(\frac{k\pi}{2n+1}\right) = \frac{\sqrt{2n+1}}{2^{n}}$ [duplicate]

Prove that $$\prod^{n}_{k=1}\sin\left(\frac{k\pi}{2n+1}\right) = \frac{\sqrt{2n+1}}{2^{n}}$$ $\bf{My\; Try::}$ Let $\displaystyle \cos \left(\frac{k\pi}{2n+1}\right)+i\sin \left(\frac{k\pi}{2n+1}\...
-1
votes
1answer
40 views

Evaluating a product of sines [duplicate]

I saw this product from a question, but got deleted. $$\prod_{k=1}^{n-1}2\sin\frac{k\pi}{n}$$ Naturally, I was curious, and evaluated this in mathematica, which suprisingly turns out to be: $$\prod_{k=...
3
votes
0answers
46 views

$2^{199}\sin(π/199)\cdots\sin(198π/199)$ Sine Product Series [duplicate]

What will be the value of $2^{199}\sin(\pi/199)\cdots\sin(198\pi/199)$ ? I could have found in case the functions were cosine but what should i do in case of sine?
262
votes
63answers
20k views

Funny identities [closed]

Here is a funny exercise $$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$ (If you prove it don't publish it here please). Do you have similar examples?
25
votes
4answers
2k views

How to prove those “curious identities”?

How to prove $$ \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$ and $$ \prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$
31
votes
5answers
24k views

Probability that n points on a circle are in one semicircle

Choose n points randomly from a circle, how to calculate the probability that all the points are in one semicircle? Any hint is appreciated.
20
votes
3answers
780 views

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

I would appreciate if somebody could help me with the following problem. How can we find the product $$ \prod_{n=1}^{999}\sin\left(\frac{n \pi}{1999}\right)$$
9
votes
4answers
379 views

How to prove that $ \sum_{k=1}^{n-1} \frac{1}{1-e^{2 \pi i k/n}} = \frac{n-1}{2}$?

I came across the fact that $$ \sum_{k=1}^{n-1} \frac{1}{1-e^{2 \pi i k/n}} = \frac{n-1}{2}.$$ How can we prove this identity?
10
votes
3answers
2k views

Ahlfors “Prove the formula of Gauss”

He says: Prove the formula of Gauss: $$ (2\pi)^\frac{n-1}{2} \Gamma(z) = n^{z - \frac{1}{2}}\Gamma(z/n)\Gamma(\frac{z+1}{n})\cdots\Gamma(\frac{z+n-1}{n}) $$ This is an exercise out of Ahlfors. ...

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