I would appreciate if somebody could help me with the following problem:
Q: proof? (by mathematical induction)
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n}=\frac{n}{2^{n-1}}~(n\geq 2)$$
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Is this homework? – Katie Dobbs Mar 08 '13 at 11:25
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Do you really only want a solution if it uses induction? It seems very difficult to apply here... – Ragib Zaman Mar 08 '13 at 11:29
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I wonder very much it uses induction – Young Mar 08 '13 at 11:33
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For a non inductive proof use the fact that $\sin x=(e^{ix}-e^{-ix})/2i$ in the LHS. – Mar 08 '13 at 11:40
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Oh, I see that you have already seen http://math.stackexchange.com/questions/8385/prove-that-prod-k-1n-1-sin-frack-pin-fracn2n-1 – Mar 08 '13 at 13:45
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1Possible duplicate of Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ – mlc Feb 19 '18 at 06:42
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Let $$S_n=\prod_{k=1}^{n-1}\sin \frac{k\pi}{n}.$$ We solve the equation $(z+1)^n=1$ for $z\in\mathbb{C}$ we find $$z=e^{i2k\pi/n}-1=2ie^{ik\pi/n}\sin\frac{k\pi}{n}=z_k,\quad 0,\ldots,n-1.$$ Moreover $(x+1)^n-1=x\left((x+1)^{n-1}+(x+1)^{n-2}+\cdots+(x+1)+1\right)=xP(x).$
The roots of $P$ are $z_k, k=1,\ldots ,n-1$. By the relation between polynomial's cofficients of $P$ and its roots we have $$\sigma_{n-1}=(-1)^{n-1}n=\prod_{k=1}^{n-1}z_k.$$
In another way, we have $$\prod_{k=1}^{n-1}z_k=2^{n-1}i^{n-1}\left(\prod_{k=1}^{n-1}e^{ik\pi/n}\right)\left(\prod_{k=1}^{n-1}\sin \frac{k\pi}{n}\right)=2^{n-1}i^{n-1}e^{i\pi(1+2+\cdots+(n-1)/n)}S_n=2^{n-1}(-1)^{n-1}S_n.$$ We conclude.
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If you want a proof without induction:
Let $\displaystyle P(X)=\sum\limits_{i=0}^{n-1} X^i$. Notice that the roots of $P$ are the $\displaystyle e^{i2k\pi/n}$ for $1 \leq k \leq n-1$, hence $\displaystyle P(X)= \prod\limits_{i=1}^{n-1} \left(X-e^{i 2k\pi/n} \right)$. Therefore, $$n=P(1)= \prod\limits_{i=1}^n \left( 1- e^{i2k\pi/n} \right)= \prod\limits_{i=1}^n e^{ik\pi/n} \left( e^{-ik\pi/n}-e^{ik\pi/n} \right) =2^{n-1} \prod\limits_{i=1}^{n-1} \sin \left( \frac{k\pi}{n} \right)$$
Seirios
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Here is a non inductive proof:
$LHS=\prod(\frac{e^{ik\pi/n}-e^{ik\pi/n}}{2i})$
$=(\frac{1}{2i})^{n-1}\prod e^{ik\pi/n}\prod(1-e^{i2k\pi/n})$
$=(\frac{1}{2i})^{n-1}e^{\frac{i\pi}{n}(\frac{n(n-1)}{2})}\prod(1-e^{i2k\pi/n})$
$=\frac{1}{2^{n-1}}\prod(1-e^{i2k\pi/n})$.
Now $\prod(1-e^{i2k\pi/n})=n$ can be proved as follows: Note that each $r_k:=e^{i2k\pi/n}$ is a $n$th root of unity which is not 1. Also the roots of $x^{n-1}+\cdots x+1$ are precisely these numbers $r_k$. So $(x-r_1)\cdots(x-r_{n-1})\equiv x^{n-1}+\cdots x+1$. Letting $x=1$ we get $\prod(1-e^{i2k\pi/n})=n$ which proves the result.