Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $

Evaluate

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

I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$

but I couldn't find the product.

Any help will be appreciated.
Thanks.

Solution 1:

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

and

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

Now,

$\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)\} $

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

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

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

Solution 2:

I think it's worth noting the product is also evaluable just remembering, besides the well known $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2},$ the somewhat nice $$\displaystyle\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}=\frac{\phi}{2}$$ (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 $${\color\red{\cos\frac{\pi}{3}\cdot\cos\frac{\pi}{5}}}\cdot\color\orange{\cos\frac{2\pi}{5}} \cdot\color\navy{\cos\frac{\pi}{15}\cdot\cos\frac{4\pi}{15}}\cdot\color\green{\cos\frac{2\pi}{15}\cdot\cos\frac{7\pi}{15}} \\ ={\color\red{\frac{\phi}{4}}}\color\orange{\left(2\cos^2\frac{\pi}{5}-1\right)}\color\navy{\frac{1}{2}\left(\cos\frac{\pi}{3}+\cos\left(-\frac{\pi}{5}\right)\right)}\color\green{\frac{1}{2}\left(\cos\left(-\frac{\pi}{3}\right)+\cos\frac{3\pi}{5}\right)} \\ = \frac{\phi}{16}\left(\frac{\phi^2}{2}-1\right)\frac{\phi+1}{2}\left(\frac{1}{2}+\cos\frac{\pi}{5}\cdot\left(2\cos^2\frac{\pi}{5}-1\right)-2\cos\frac{\pi}{5}\left(1-\cos^2\frac{\pi}{5}\right)\right)\\=\frac{\phi(\phi^2-1)}{64}\left(\frac{1}{2}+\frac{\phi(\phi^2-2)}{4}-\frac{\phi(4-\phi^2)}{4}\right)\\=\frac{\phi^2}{64}\left(\frac{1}{2}+\frac{\phi(\phi-1)-\phi(3-\phi)}{4}\right)\\=\frac{\phi+1}{64}\left(\frac{1}{2}+\frac{2-2\phi}{4}\right)=\frac{\phi+1}{128}-\frac{\phi^2-1}{128}=\frac{\phi+1}{128}-\frac{\phi}{128}=\frac{1}{128}.$$

Solution 3:

Since an elegant solution has already been provided, I will go for an overkill.

From the Fourier cosine series of $\log\cos x$ we have: $$ \log\cos x = -\log 2-\sum_{n\geq 1}^{+\infty}\frac{(-1)^n\cos(2n x)}{n}\tag{1} $$ but for any $n\geq 1$ we have: $$ 15\nmid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = -\frac{1}{2},\quad 15\mid n\rightarrow\sum_{k=1}^{7}\cos\left(\frac{2n k \pi}{15}\right) = 7\tag{2}$$ so: $$ \sum_{k=1}^{7}\log\cos\frac{k\pi}{15} = -7\log 2+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n}-\frac{15}{2}\sum_{n\geq 1}\frac{(-1)^n}{15n}=-7\log 2\tag{3}$$ and by exponentiating the previous line:

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

Solution 4:

Note: Here's another variation inspired by an answer to this question.

We consider the roots of unity $e^{\frac{2\pi i k}{15}}, 0\leq k < 15$ of the polynomial

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

We obtain

\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*}

Evaluating the polynomial $-p(-z)$ at $z=1$ gives

\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*}

In (1) we use the formula $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.