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.
 A: 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 $\text{S} \neq 0$,
$\therefore \boxed{\text{C}=\dfrac{1}{2^7}}$
A: 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 $${\color\red{\cos\frac{\pi}{3}\cdot\cos\frac{\pi}{5}}}\cdot\color\orange{\cos\frac{2\pi}{5}} \cdot\color\navy{\cos\frac{\pi}{15}\cdot\cos\frac{4\pi}{15}}\cdot\color\green{\cos\frac{2\pi}{15}\cdot\cos\frac{7\pi}{15}} \\ ={\color\red{\frac{\phi}{4}}}\color\orange{\left(2\cos^2\frac{\pi}{5}-1\right)}\color\navy{\frac{1}{2}\left(\cos\frac{\pi}{3}+\cos\left(-\frac{\pi}{5}\right)\right)}\color\green{\frac{1}{2}\left(\cos\left(-\frac{\pi}{3}\right)+\cos\frac{3\pi}{5}\right)} \\ = \frac{\phi}{16}\left(\frac{\phi^2}{2}-1\right)\frac{\phi+1}{2}\left(\frac{1}{2}+\cos\frac{\pi}{5}\cdot\left(2\cos^2\frac{\pi}{5}-1\right)-2\cos\frac{\pi}{5}\left(1-\cos^2\frac{\pi}{5}\right)\right)\\=\frac{\phi(\phi^2-1)}{64}\left(\frac{1}{2}+\frac{\phi(\phi^2-2)}{4}-\frac{\phi(4-\phi^2)}{4}\right)\\=\frac{\phi^2}{64}\left(\frac{1}{2}+\frac{\phi(\phi-1)-\phi(3-\phi)}{4}\right)\\=\frac{\phi+1}{64}\left(\frac{1}{2}+\frac{2-2\phi}{4}\right)=\frac{\phi+1}{128}-\frac{\phi^2-1}{128}=\frac{\phi+1}{128}-\frac{\phi}{128}=\frac{1}{128}.$$
A: 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:

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

A: 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 $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.

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

Note: Writing $-p(-z)$ as
\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*}
and evaluating the polynomial $-p(-z)$ at $z=i$ we obtain the related formula

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

Doubling the argument does not change the value of the product.
A: I like the answer by @Steven Gregory because of the way the dominos fall and it seems the only one presented that a precalculus student could hope to find. Using the reflection and multiplication formulas for the gamma function a fairly compact proof is possible.
$$\begin{align}\prod_{r=1}^7\cos\frac{r\pi}{15} & = \prod_{r=1}^7\sin\left(\frac{\pi}2-\frac{r\pi}{15}\right) = \prod_{r=0}^6\sin\left(\frac{\pi}2-\frac{7\pi}{15}+\frac{r\pi}{15}\right) \\
& = \prod_{r=0}^6\sin\left(\frac{\pi}{30}+\frac{r\pi}{15}\right) = \prod_{r=0}^6\frac{\pi}{\Gamma\left(\frac1{30}+\frac{r}{15}\right)\Gamma\left(\frac{29}{30}-\frac{r}{15}\right)} & \tag{1} \\
& = \frac{\pi^7}{\prod_{r=0}^6\Gamma\left(\frac1{30}+\frac{r}{15}\right)\prod_{s=8}^{14}\Gamma\left(\frac1{30}+\frac{s}{15}\right)} = \frac{\pi^7\,\Gamma\left(\frac1{30}+\frac7{15}\right)}{\prod_{r=0}^{14}\Gamma\left(\frac1{30}+\frac{r}{15}\right)} & \tag{2} \\
& = \frac{\pi^7\,\Gamma\left(\frac12\right)}{(2\pi)^{(15-1)/2}15^{\frac12-15\left(\frac1{30}\right)}\Gamma\left(15\left(\frac1{30}\right)\right)} & \tag{3} \\
& = \frac1{2^7}
\end{align}$$
$(1)$ Using the reflection formula for the gamma function: $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$.
$(2)$ Reordering one product and multiplying and dividing by $\Gamma\left(\frac1{30}+\frac7{15}\right)$.
$(3)$ Using the multiplication formula for the gamma function: $$\prod_{k=0}^{n-1}\Gamma\left(x+\frac{k}n\right)=(2\pi)^{(n-1)/2}n^{\frac12-nx}\Gamma(nx)$$
