Producto de los cosenos: $ \prod_{r=1}^{7} \cos \frac{r\pi}{15} $

Question

Evaluar

$$ \prod_{r=1}^{7} \cos {\dfrac{r\pi}{15}} $$

Traté de identidades trigonométricas de producto de los cosenos, yo.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$

pero no pude encontrar el producto.

Cualquier ayuda será apreciada.
Gracias.

Answer 1

Deje $\displaystyle\text{C}=\prod_{r=1}^{7}\cos{\left(\dfrac{r\pi}{15}\right)}$

y

$\displaystyle\text{S}=\prod_{r=1}^{7}\sin{\left(\dfrac{r\pi}{15}\right)}$

Ahora,

$\text{C}\cdot\text{S}=\left(\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\right)\cdot \ldots \cdot\left(\sin{\dfrac{7\pi}{15}} \cdot \cos{\dfrac{7\pi}{15}}\right) $

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \left(2\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(2\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\right)\cdot \ldots \cdot\left(2\sin{\dfrac{7\pi}{15}} \cdot \cos{\dfrac{7\pi}{15}}\right) $

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \ \sin{\dfrac{2\pi}{15}}\cdot\sin{\dfrac{4\pi}{15}}\cdot \ldots \cdot\sin{\dfrac{14\pi}{15}} $

$\{\because \sin(2x) = 2\sin (x) \cos (x) \}$

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \ \sin{\dfrac{\pi}{15}}\cdot\sin{\dfrac{2\pi}{15}} \cdot \ldots \cdot \sin{\dfrac{7\pi}{15}} \\\\ \{\because \sin(\pi-x)=\sin(x)\} $

$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \cdot \text{S}$

desde $\text{S} \neq 0$,

$\therefore \boxed{\text{C}=\dfrac{1}{2^7}}$

Answer 2

Creo que vale la pena señalar que el producto es también evaluables recordando, además de la conocida $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2},$ poco agradable $$\displaystyle\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}=\frac{\phi}{2}$$ (where $\phi$ is the golden section) and iterating the sum/difference formula for the cosine and the product formula you mention. Your product is, once we rearrange factors and simplify fractions, equal to $${\color\red{\cos\frac{\pi}{3}\cdot\cos\frac{\pi}{5}}}\cdot\color\orange{\cos\frac{2\pi}{5}} \cdot\color\navy{\cos\frac{\pi}{15}\cdot\cos\frac{4\pi}{15}}\cdot\color\green{\cos\frac{2\pi}{15}\cdot\cos\frac{7\pi}{15}} \\ ={\color\red{\frac{\phi}{4}}}\color\orange{\left(2\cos^2\frac{\pi}{5}-1\right)}\color\navy{\frac{1}{2}\left(\cos\frac{\pi}{3}+\cos\left(-\frac{\pi}{5}\right)\right)}\color\green{\frac{1}{2}\left(\cos\left(-\frac{\pi}{3}\right)+\cos\frac{3\pi}{5}\right)} \\ = \frac{\phi}{16}\left(\frac{\phi^2}{2}-1\right)\frac{\phi+1}{2}\left(\frac{1}{2}+\cos\frac{\pi}{5}\cdot\left(2\cos^2\frac{\pi}{5}-1\right)-2\cos\frac{\pi}{5}\left(1-\cos^2\frac{\pi}{5}\right)\right)\\=\frac{\phi(\phi^2-1)}{64}\left(\frac{1}{2}+\frac{\phi(\phi^2-2)}{4}-\frac{\phi(4-\phi^2)}{4}\right)\\=\frac{\phi^2}{64}\left(\frac{1}{2}+\frac{\phi(\phi-1)-\phi(3-\phi)}{4}\right)\\=\frac{\phi+1}{64}\left(\frac{1}{2}+\frac{2-2\phi}{4}\right)=\frac{\phi+1}{128}-\frac{\phi^2-1}{128}=\frac{\phi+1}{128}-\frac{\phi}{128}=\frac{1}{128}.$$

Answer 3

Desde una elegante solución ya ha sido proporcionada, voy a ir a por una exageración.

Desde el coseno de Fourier de la serie de $\log\cos x$ tenemos: $$ \log\cos x = -\log 2-\sum_{n\geq 1}^{+\infty}\frac{(-1)^n\cos(2n x)}{n}\tag{1} $$ pero para cualquier $n\geq 1$ tenemos: $$ 15\nmid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = -\frac{1}{2},\quad 15\mid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = 7\tag{2}$$ así: $$ \sum_{k=1}^{7}\log\cos\frac{k\pi}{15} = -7\log 2+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n}-\frac{15}{2}\sum_{n\geq 1}\frac{(-1)^n}{15n}=-7\log 2\tag{3}$$ y por exponentiating la línea anterior:

$$ \prod_{k=1}^{7}\cos\left(\frac{\pi k}{15}\right) = \color{red}{\frac{1}{2^7}}.\tag{4}$$

Answer 4

Nota: Aquí es otra variación inspirado por una respuesta a esta pregunta.

Consideramos que las raíces de la unidad $e^{\frac{2\pi i k}{15}}, 0\leq k < 15$ del polinomio

$$p(z)=z^{15}-1=\prod_{k=0}^{14}(z-e^{\frac{2\pi i k}{15}})$$

Obtenemos

\begin{align*} -p(-z)=z^{15}+1&=\prod_{k=0}^{14}(z+e^{\frac{2\pi i k}{15}})\\ &=(z+1)\prod_{k=1}^{7}\left[(z+e^{\frac{2\pi i k}{15}})(z+e^{-\frac{2\pi i k}{15}})\right]\\ \end{align*}

Evaluar el polinomio $-p(-z)$ $z=1$ da

\begin{align*} 1&=\prod_{k=1}^{7}\left[(1+e^{\frac{2\pi i k}{15}})(1+e^{-\frac{2\pi i k}{15}})\right]\\ &=\prod_{k=1}^{7}\left[(e^{-\frac{\pi i k}{15}}+e^{\frac{\pi i k}{15}})e^{\frac{\pi i k}{15}}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})e^{-\frac{\pi i k}{15}}\right]\\ &=\prod_{k=1}^{7}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})^2\tag{1}\\ &=\prod_{k=1}^{7}\left(2\cos\left(\frac{k \pi}{15}\right)\right)^2\tag{2}\\ \end{align*}

En (1) utilizamos la fórmula $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.

Llegamos a la conclusión de (2) \begin{align*} \prod_{k=1}^{7}\cos\left(\frac{k\pi}{15}\right)=\frac{1}{2^7} \end{align*}

Nota: Escrito $-p(-z)$

\begin{align*} -p(-z)=\prod_{k=1}^{7}\left[z^2+\left(e^{\frac{2\pi i k}{15}}+e^{-\frac{2\pi i k}{15}}\right)z+1\right]\\ \end{align*}

y evaluar el polinomio $-p(-z)$ $z=i$ obtenemos la fórmula basada en la

\begin{align*} \prod_{k=1}^{7}\cos\left(\frac{2k\pi}{15}\right)=\frac{1}{2^7} \end{align*}

Duplicar el argumento no cambia el valor del producto.

Answer 5

El uso de la fórmula de Moivre,

$$\cos(2n+1)x+i\sin(2n+1)x=(\cos x+i\sin x)^{2n+1}=\sum_{r=0}\binom{2n+1}r(\cos x)^{2n+1-r}(i\sin x)^r$$

Igualando las partes reales, $$\cos(2n+1)x$$

$$=(\cos x)^{2n+1}-\binom{2n+1}2(\cos x)^{2n+1-2}(\sin x)^2+\binom{2n+1}r(\cos x)^{2n+1-4}(\sin x)^4+\cdots+\binom{2n+1}{2n-2}(-1)^{n-1}(\cos x)^3(\sin x)^{2n-2}+(-1)^n\cos x\sin^{2n}x$$

$$=(\cos x)^{2n+1}-\binom{2n+1}2(\cos x)^{2n+1-2}(1-\cos^2x)+\binom{2n+1}r(\cos x)^{2n+1-4}(1-\cos^2x)^2+\cdots+\binom{2n+1}{2n-2}(-1)^{n-1}(\cos x)^3(1-\cos^2x)^{n-1}+(-1)^n\cos x(1-\cos^2x)^nx$$

$$\implies\cos(2n+1)x=(\cos x)^{2n+1}\left[1+\binom{2n+1}2+\binom{2n+1}4+\cdots+\binom{2n+1}{2n}\right]+\cdots+\binom{2n+1}{2n}(-1)^n\cos x$$

Ahora $\displaystyle1+\binom{2n+1}2+\binom{2n+1}4+\cdots+\binom{2n+1}{2n}=\sum_{r=0}^n\binom{2n+1}{2r}=\sum_{r=0}^n\binom{2n+1}{2n+1-2r}=\dfrac{(1+1)^{2n+1}}2=2^{2n}$

$$\implies\cos(2n+1)x=2^{2n}(\cos x)^{2n+1}+\cdots+(-1)^n(2n+1)\cos x$$

Ahora si $\cos(2n+1)x=\cos A,(2n+1)x=2m\pi\pm A$ donde $m$ es cualquier entero

y $x=\dfrac{2m\pi\pm A}{2n+1}$ donde $m\equiv0,\pm1,\pm2\cdots,\pm n\pmod{2n+1}$

Así, las raíces de $$2^{2n}(\cos x)^{2n+1}+\cdots+(-1)^n(2n+1)\cos x-\cos A=0\ \ \ \ (1)$$ $\cos\dfrac{2m\pi\pm A}{2n+1}$ donde $m\equiv0,\pm1,\pm2\cdots,\pm n\pmod{2n+1}$

El uso de Vieta de la fórmula, $$\cos\dfrac A{2n+1}\prod_{m=1}^n\cos\dfrac{2m\pi\pm A}{2n+1}=\dfrac{\cos A}{2^{2n}}$$

Si $A=0,$ $$\prod_{m=1}^n\cos^2\dfrac{2m\pi}{2n+1}=\dfrac1{2^{2n}}$$

Ahora $\cos\dfrac{2m\pi}{2n+1}=-\cos\left(\pi-\dfrac{2m\pi}{2n+1}\right)=-\cos\dfrac{(2n+1-2m)\pi}{2n+1}$

$$\implies(-1)^n\prod_{m=0}^{2n}\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^{2n}}$$

Como $\displaystyle\cos\dfrac{(2n+1-m)\pi}{2n+1}=\cdots=-\cos\dfrac{m\pi}{2m+1}$

$\displaystyle\implies(-1)^n\prod_{m=1}^n\cos^2\dfrac{m\pi}{2n+1}(-1)^n=\dfrac1{2^{2n}}\implies\prod_{m=1}^n\cos^2\dfrac{m\pi}{2n+1}=\dfrac1{2^{2n}}$

De nuevo $0\le\dfrac{m\pi}{2n+1}\le\dfrac\pi2\iff0\le2m\le2n+1\iff0\le m\le n$

$\displaystyle\implies\prod_{m=1}^n\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^n}$ $\cos\dfrac{m\pi}{2n+1}>0$ $0<m\le n$