Evaluate $ \cos\frac\pi{2n+1} \cos\frac{3\pi}{2n+1}\cdots \cos\frac {(2n-1)\pi}{2n+1} $

Question

Question: How to evaluate the trigonometric product $$ P = \cos\left(\frac\pi{2n + 1}\right) \cos\left(\frac{3\pi}{2n + 1}\right)\cdots \cos\left(\frac{\left[2n - 1\right]\pi}{2n+1}\right) $$ where $n$ is a positive integer ?.

• I tried to write the cosine components of the product in their exponential form as $$ \cos\left(\frac{k\pi}{2n + 1}\right) = {1 \over 2}\left(% {\rm e}^{{\rm i}k\pi/\left[2n + 1\right]} - {\rm e}^{-{\rm i}k\pi/\left[2n + 1\right]} \right)$$ and then rewrite the product $P$ as a sum of exponentials.
• However, it is hard to keep track of the orders of all the exponential terms.
• Besides, the summation does not necessarily make the evaluation easier.

I would like to know any other better methods.

Answer 1

Per the following Chebyshev factorization

$$ \frac{\sin[(2n+1)x]}{\sin x}= U_{2n} (\cos x)=2^n\prod_{k=1}^n \bigg(\cos2x+\cos\frac{(2k-1)\pi}{2n+1} \bigg) $$ set $x=\frac\pi4$ to obtain the product $$P_n = \prod_{k=1}^n \cos\frac{(2k-1)\pi}{2n+1} = \frac1{2^{n}}\bigg( \sin\frac{n\pi}2+ \cos\frac{n\pi}2\bigg)=\frac1{2^n}(-1)^{\frac{n(n-1)}2} $$

Answer 2

Let's define $$Q=\prod_{k=1}^n\sin\left(\frac{2k-1}{2n+1} \pi \right)$$ Then $$\begin{align} 2^nPQ &= 2^n\cdot\prod_{k=1}^n\cos\left(\frac{2k-1}{2n+1} \pi \right)\cdot \prod_{k=1}^n\sin\left(\frac{2k-1}{2n+1} \pi \right)\\ &=\prod_{k=1}^n\left(2 \cos\left(\frac{2k-1}{2n+1} \pi \right) \sin\left(\frac{2k-1}{2n+1} \pi \right)\right)\\ &=\prod_{k=1}^n\sin\left(\frac{2(2k-1)}{2n+1} \pi \right) \\ &=\prod_{k=1}^{\lfloor\frac{n+1}{2} \rfloor}\sin\left(\frac{2(2k-1)}{2n+1} \pi \right) \cdot \prod_{k=\lfloor\frac{n+1}{2} \rfloor+1}^{n}\sin\left(\frac{2(2k-1)}{2n+1} \pi \right)\tag{1}\\ &=\prod_{k=1}^{\lfloor\frac{n+1}{2}\rfloor}\sin\left(\color{red}{\pi-}\frac{2(2k-1)}{2n+1} \pi \right) \cdot \prod_{k=\lfloor\frac{n+1}{2} \rfloor+1}^{n}\color{red}{(-1)}\cdot\sin\left(\frac{2(2k-1)}{2n+1} \pi \color{red}{ - \pi} \right)\tag{2}\\ &=(-1)^{n-\lfloor\frac{n+1}{2}\rfloor}\cdot\prod_{k=1}^{\lfloor\frac{n+1}{2} \rfloor}\sin\left(\frac{2(n+2-2k)-1}{2n+1} \pi \right) \cdot \prod_{k=\lfloor\frac{n+1}{2} \rfloor+1}^{n}\sin\left(\frac{2(2k-1-n)-1}{2n+1} \pi \right)\tag{3}\\ &= (-1)^{n-\lfloor\frac{n+1}{2}\rfloor}\cdot\prod_{k=1}^{n}\sin\left(\frac{2k-1}{2n+1} \pi \right) \tag{4}\\ 2^nPQ&= (-1)^{n-\lfloor\frac{n+1}{2}\rfloor}\cdot Q \tag{5} \end{align}$$

Remark:

$(1)$ : The first product contains $\lfloor\frac{n+1}{2} \rfloor$ terms that satisfy $\frac{2(2k-1)}{2n+1}\pi <\pi \iff k<\frac{n+1.5}{2} \iff k \le \lfloor\frac{n+1}{2} \rfloor$, the second product contains $n-\lfloor\frac{n+1}{2} \rfloor$ that satisfy $\frac{2(2k-1)}{2n+1}\pi >\pi$.

$(2)$: apply the formula $ \sin(\pi - x)=\sin(x)$ for the terms of the first product and the formula $\sin(x-\pi) = -\sin(x)$ for the terms of the second product

$(3)\implies (4)$: all $n$ expressions inside the sin function of the two products take values in the set $\left\{\frac{1}{2n+1}\pi,\frac{3}{2n+1}\pi,...,\frac{2n-1}{2n+1}\pi \right\}$ and are different from each other ( suppose by contracdiction that there are two indices $p,q\in \mathbb{N}$ such that $\frac{2(n+2-2p)-1}{2n+1} \pi =\frac{2(2q-1-n)-1}{2n+1} \pi $, we must have then $2n = 2(p+q)-3$: can't occurs as $p,q,n \in \mathbb{N}$). As a consequence, the $n$ terms are different from each other and take values in the set of $n$ elements $\left\{\sin\left( \frac{1}{2n+1}\pi\right),\sin\left( \frac{3}{2n+1}\pi\right),...,\sin\left( \frac{2n-1}{2n+1}\pi \right) \right\}$

From $(5)$, we conclude that

$$\color{red}{P = \frac{(-1)^{n-\lfloor\frac{n+1}{2}\rfloor}}{2^n}} \tag{6}$$

Remark: $$n-\lfloor\frac{n+1}{2}\rfloor \equiv \frac{n(n-1)}{2} \pmod 2$$ Then, the result $(6)$ is equivalent to the result suggested by @Semiclassical, i.e

$$\color{red}{P = \frac{(-1)^{\frac{n(n-1)}{2}}}{2^n}} $$

Answer 3

I start with an elementary claim.

Claim: $\frac{(-1)^n}{2^{2n}} = \prod_{r=1}^{2n} \cos\left(\frac{r}{2n+1}\pi\right)$

Proof: Consider the $2n+1$-th roots of unity other than $1$ ie $$\begin{align} \frac{x^{2n+1} - 1}{x-1} &= \prod_{r=1}^{2n} \left(x-e^{2i\pi\frac{r}{2n+1}}\right) \overset{x=-1}{\implies} \\ 1 &= (-1)^{2n} \prod_{r=1}^{2n} \left(1+e^{2i\pi\frac{r}{2n+1}}\right) \implies\\ 1 &= \prod_{r=1}^{2n}\left|1 + \cos\left(\frac{2r\pi}{2n+1}\right) + i \sin\left(\frac{2r\pi}{2n+1}\right)\right| \\ 1 &= \prod_{r=1}^{2n} \sqrt{\left(1 + \cos\left(\frac{2r\pi}{2n+1}\right)\right)^2 + \sin^2\left(\frac{2r\pi}{2n+1}\right)} \\ 1 &= \prod_{r=1}^{2n} 2 \sqrt{\cos^2\left(\frac{r\pi}{2n+1}\right)} \\ 1 &= 2^{2n}(-1)^n \prod_{r=1}^{2n} \cos\left(\frac{r\pi}{2n+1}\right) \tag{0.1} \\ \frac{(-1)^n}{2^{2n}} &= \prod_{r=1}^{2n} \cos\left(\frac{r}{2n+1}\pi\right) \end{align}$$ $(0.1)$ is true since $\sqrt{x^2} = |x|$ and there are exactly $n$ negative cosines dealt in the product.

---

We are interested in $P := \prod_{k=1}^{n} \cos\left(\frac{2k-1}{2n+1}\pi\right)$. But note that $$\begin{align} \prod_{k=1}^{2n} \cos\left(\frac{k}{2n+1}\pi\right) &= P \prod_{k=1}^{n}\cos\left(\frac{2k}{2n+1}\pi\right) \tag{1.1} \\ &= P \prod_{k=1}^n -\cos\left(\pi - \frac{2k}{2n+1}\pi\right)\tag{1.2}\\ &= (-1)^nP\prod_{k=1}^n \cos\left(\frac{2n+1-2k}{2n+1}\pi\right) \\ &= (-1)^nP\prod_{k=1}^n \cos\left(\frac{2k-1}{2n+1}\pi\right) \tag{1.3}\\ &= (-1)^n P^2 \\ P^2 &= \frac1{2^{2n}} \\ \left|P\right| &= \frac1{2^n}\\ P &= \frac{(-1)^{n+1 - \lceil \frac{2n+3}4 \rceil }}{2^n} \tag{1.4}\\ P &= \frac{(-1)^{\frac{n(n-1)}2}}{2^n} \tag{1.5} \end{align}$$

$(1.1)$ follows by splitting the product into two based on the parity of $k$.

$(1.2)$ uses $\cos(\pi - x) = -\cos(x)$.

$(1.3)$ re-writes the product in the previous line from the end to start, ie re-indexing $k\to n-k+1$.

$(1.4)$ similar to $(0.1)$ we count the number of negative cosines appearing in our product; there are exactly $n + 1 - \lceil \frac{2n+3}{4}\rceil$ such cosines.

$(1.5)$ uses that $\bmod 2: n+1- \lceil \frac{2n+3}{4}\rceil \equiv \lfloor \frac{n}2\rfloor \equiv \frac{n(n-1)}2$.

Answer 4

Use the fact(Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$ or $\prod_{k=1}^{n-1}\cos\left(\theta+\frac{k\pi}{n}\right)$): $$\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\frac{\pi n}{2})}{2^{n-1}},\quad n=1,2,\cdots,$$ make change $n\to2n+1$, we get $$\prod_{k=1}^{2n}\cos\left(\frac{k\pi}{2n+1}\right)=\frac{\sin\left(\frac{\pi(2n+1)}{2}\right)}{2^{2n}}=\frac{(-1)^n}{2^{2n}},\quad n=1,2,\cdots.$$ By the formular $\cos(\pi-x)=-\cos(x)$, we can get $$(-1)^n\prod_{k=1}^{n}\cos^2\left(\frac{k\pi}{2n+1}\right)=\frac{(-1)^n}{2^{2n}},\quad n=1,2,\cdots.$$ So $$\prod_{k=1}^{n}\cos^2\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2^{2n}},\quad n=1,2,\cdots,$$ i.e., $$\left|\prod_{k=1}^{n}\cos\left(\frac{k\pi}{2n+1}\right)\right|=\frac{1}{2^n},\quad n=1,2,\cdots,$$ In the following we determine the sign of the product: $$P_n=\prod_{k=1}^{n}\cos\left(\frac{k\pi}{2n+1}\right).$$ Divide into two cases:

(1) if $n=2m-1$ is odd, then $2n+1=4m-1$ and there are $2m-1$ terms in the product $P_n$ and $(m-1)$ terms are negative: $$\cos\frac{2m-1}{4m-1}\pi,\cos\frac{2m+1}{4m-1}\pi,\cdots,\cos\frac{4m-3}{4m-1}\pi.$$ So the sign of the product $P_n$ in this case is $$(-1)^{m-1}=\left[(-1)^{2m-1}\right]^{m-1}=(-1)^{\frac{(2m-2)(2m-1)}2}=(-1)^{\frac{(n-1)n}2}.$$ (2) if $n=2m$ is even, then $2n+1=4m+1$ and there are $2m$ terms in the product $P_n$ and $m$ terms are negative: $$\cos\frac{2m+1}{4m+1}\pi,\cos\frac{2m+3}{4m+1}\pi,\cdots,\cos\frac{4m-1}{4m+1}\pi.$$ So the sign of the product $P_n$ in this case is $$(-1)^{m}=\left[(-1)^{2m-1}\right]^{m}=(-1)^{\frac{(2m-1)(2m)}2}=(-1)^{\frac{(n-1)n}2}.$$

All in all, we have $$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{2n+1}\right)= \frac{(-1)^{\frac{(n-1)n}{2}}}{2^n}.$$ (Dute to @ Semiclassical pointing out this result by Mathematica)

Answer 5

The equation $\cos (2n+1) \theta =-1$ has roots $\pi, \pm \dfrac{\pi}{2n+1}, \pm \dfrac{3\pi}{2n+1},\ldots, \pm\dfrac{(2n-1)\pi}{2n+1}$

Using results from Chebyshev polynomials we have that the roots of

$2^{2n}t^{2n+1}-\ldots +1=0$ has roots $-1, \cos \dfrac{(2k-1)\pi}{2n+1}, 1\le k \le 2n-1$ where all roots except $t=-1$ are double roots.

Thus if the required product is denoted by $P$ we have $|P| = \dfrac{1}{2^n}$

We further note that if $n$ is even then the negative terms in $P$ are $\cos \dfrac{(n+1)\pi}{2n+1},\ldots \cos \dfrac{(2n-1)\pi}{2n+1} $ i.e. $\dfrac{n}{2}$ of them. Similarly if $n$ is odd, we have $\dfrac{n-1}{2}$ negative terms

Its now clear that $P = \dfrac{(-1)^{\frac{n(n-1)}{2}}}{2^n}$

Answer 6

Another variation. In order to show \begin{align*} \color{blue}{\prod_{q=1}^{n}\cos\left(\frac{2q-1}{2n+1}\pi\right)=(-1)^{\frac{n(n-1)}{2}}\frac{1}{2^n}}\tag{1} \end{align*} we consider the roots of unity $e^{\frac{2\pi i q}{2n+1}}, 0\leq q \leq 2n$ of the polynomial

\begin{align*} p(z)=z^{2n+1}-1=\prod_{q=0}^{2n}\left(z-e^{\frac{2\pi i q}{2n+1}}\right) \end{align*}

We obtain \begin{align*} -p(-z)&=z^{2n+1}+1=\prod_{q=0}^{2n}\left(z+e^{\frac{2\pi i q}{2n+1}}\right)\\ &=(z+1)\prod_{q=1}^{n}\left[\left(z+e^{\frac{2\pi i q}{2n+1}}\right)\left(z+e^{-\frac{2\pi i q}{2n+1}}\right)\right]\\ &=(z+1)\prod_{q=1}^n\left[z^2+\left(e^{\frac{2\pi i q}{2n+1}}+e^{-\frac{2\pi i q}{2n+1}}\right)z+1\right]\tag{2} \end{align*}

Evaluating the polynomial $-p(-z)$ at $z=i$ in (2) gives \begin{align*} {\color{blue}{\frac{i^{2n+1}+1}{i+1}}} &=\prod_{q=1}^n\left[2i\cos\left(\frac{2q}{2n+1}\,\pi\right)\right]\tag{3}\\ &\color{blue}{=2^ni^n\prod_{q=1}^n\cos\left(\frac{2q}{2n+1}\,\pi\right)}\tag{4}\\ &=2^ni^n\prod_{q=1}^n\left[-\cos\left(\pi-\frac{2q}{2n+1}\,\pi\right)\right]\\ &=2^n\left(-i\right)^n\prod_{q=1}^n\cos\left(\frac{2n+1-2q}{2n+1}\,\pi\right)\\ &\color{blue}{=2^n(-i)^n\prod_{q=1}^n\cos\left(\frac{2q-1}{2n+1}\,\pi\right)}\tag{5}\\ \end{align*}

In (3) we use the identity $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$. In (5) we change the order of multiplication: $q\to n-q+1$.

We conclude from (3) and (5) \begin{align*} \color{blue}{\prod_{q=1}^{n}\cos\left(\frac{2q-1}{2n+1}\,\pi\right)} &=\frac{1+(-1)^ni}{1+i}\,\frac{1}{(-i)^n}\,\frac{1}{2^n}\\ &=\frac{1}{2}\,\left(1+(-1)^ni\right)(1-i)i^n\,\frac{1}{2^n}\\ &=\left(\frac{1+(-1)^n}{2}\,i^n-\frac{1-(-1)^n}{2}\,i^{n+1}\right)\frac{1}{2^n}\\ &\,\,\color{blue}{=(-1)^{\frac{n(n-1)}{2}}\,\frac{1}{2^n}} \end{align*} and the claim (1) follows.

Note:

The comparison of (4) with (5) also results in \begin{align*} {\color{blue}{\prod_{q=1}^{n}\cos\left(\frac{2q}{2n+1}\,\pi\right)}} &=(-1)^{\frac{n(n-1)}{2}+n}\,\frac{1}{2^n}\\ &\,\,\color{blue}{=(-1)^{\frac{n(n+1)}{2}}\,\frac{1}{2^n}} \end{align*} and multiplication of both identities gives \begin{align*} {\color{blue}{\prod_{q=1}^{2n}\cos\left(\frac{q}{2n+1}\,\pi\right)}} &=\left[\prod_{q=1}^{n}\cos\left(\frac{2q}{2n+1}\,\pi\right)\right] \left[\prod_{q=1}^{n}\cos\left(\frac{2q-1}{2n+1}\,\pi\right)\right]\\ &=\left[(-1)^{\frac{n(n+1)}{2}}\,\frac{1}{2^n}\right]\left[(-1)^{\frac{n(n-1)}{2}}\,\frac{1}{2^n}\right]\\ &=(-1)^{n^2}\,\frac{1}{2^{2n}}\\ &\,\,\color{blue}{=(-1)^n\,\,\frac{1}{2^{2n}}} \end{align*}