Cosines And Pies

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 $\text{S} \neq 0$ ,

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

---

In general,

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

$\prod_{k=1}^{n}2|\cos(\frac{k\pi}{n})|=\prod_{k=1}^{n}|e^{ik\pi/n}+e^{-ik\pi/n}|=\prod_{k=1}^{n}|-1-e^{2ik\pi/n}|=2$ for odd $n$ , since

$\prod_{k=1}^{n}(x-e^{2ik\pi/n})=x^n-1$ . Now $\prod_{k=1}^{(n-1)/2}\cos(\frac{k\pi}{n})=\frac{1}{2^{(n-1)/2}}$ . For $n=15$ this is $2^{\boxed{-7}}$ .

$\begin{aligned} P & = \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \color{#3D99F6}{\cos \frac{5\pi}{15}} \cos \frac{6\pi}{15} \color{#D61F06}{\cos \frac{7\pi}{15}} \\ & = \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \color{#3D99F6}{\left(\frac{1}{2}\right)} \cos \frac{6\pi}{15} \color{#D61F06}{\left(-\cos \frac{8\pi}{15}\right)} \\ & = -\frac{1}{2} \left(\cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15} \right) \left(\cos \frac{3\pi}{15} \cos \frac{6\pi}{15} \right) \\ & = -\frac{1}{2} \left(\frac{\sin \frac{\pi}{15} \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15}}{\sin \frac{\pi}{15}} \right) \left(\frac{\sin \frac{\pi}{5} \cos \frac{\pi}{5} \cos \frac{2\pi}{5}}{\sin \frac{\pi}{5}} \right) \\ & = -\frac{1}{2} \left(\frac{\sin \frac{2\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15}}{2 \sin \frac{\pi}{15}} \right) \left(\frac{\sin \frac{2\pi}{5} \cos \frac{2\pi}{5}}{2 \sin \frac{\pi}{5}} \right) \\ & = -\frac{1}{2} \left(\frac{\sin \frac{4\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15}}{2^2 \sin \frac{\pi}{15}} \right) \left(\frac{\sin \frac{4\pi}{5}}{2^2 \sin \frac{\pi}{5}} \right) \\ & = -\frac{1}{2} \left(\frac{\sin \frac{8\pi}{15} \cos \frac{8\pi}{15}}{2^3 \sin \frac{\pi}{15}} \right) \left(\frac{\sin \frac{\pi}{5}}{2^2 \sin \frac{\pi}{5}} \right) \\ & = -\frac{1}{2} \left(\frac{\sin \frac{16\pi}{15}}{2^4 \sin \frac{\pi}{15}} \right) \left(\frac{1}{2^2} \right) \\ & = -\frac{1}{2} \left(\frac{-\sin \frac{\pi}{15}}{2^4 \sin \frac{\pi}{15}} \right) \left(\frac{1}{2^2} \right) \\ & = -\frac{1}{2} \left(\frac{-1}{2^4} \right) \left(\frac{1}{2^2} \right) \\ & = \frac{1}{2^7} \end{aligned}$

$\implies n = \boxed{-7}$

All the angles are in degrees for the sake of convenience. Consider

( $cos 60^\circ-\theta$ )( $cos \theta$ )( $cos 60^\circ+\theta$ ) = $\dfrac{1}{4}$ $cos 3\theta$ (The above is easy to prove by expanding.)

The given problem is of the form

P= $cos 12^\circ$ $cos 24^\circ$ $cos 36^\circ$ $cos 48^\circ$ $cos 60^\circ$ $cos 72^\circ$ $cos 84^\circ$

Regrouping,

P=( $cos 36^\circ$ $cos 24^\circ$ $cos 84^\circ$ )( $cos 48^\circ$ $cos 12^\circ$ $cos 72^\circ$ ) $cos 60^\circ$

P=( $cos 60^\circ-24^\circ$ )( $cos 24^\circ$ )( $cos 60^\circ+24^\circ$ )( $cos 60^\circ-12^\circ$ )( $cos 12^\circ$ $cos 60^\circ+12^\circ$ )( $cos 60^\circ$ )

P= $\dfrac{1}{4}$ $cos 72^\circ$ $\dfrac{1}{4}$ $cos 36^\circ$ $\dfrac{1}{2}$

P= $\dfrac{1}{2^{5}}$ ( $\dfrac{\sqrt{5}-1}{4}$ )( $\dfrac{\sqrt{5}+1}{4}$ )

P= $\dfrac{1}{2^{7}}$

Hence the answer is -7.

$\Large \displaystyle \Pi_{k=1}^{n-1}Cos\frac{\frac{k*\pi} 2} n=\dfrac{Sin\frac{n*\pi} 2}{2^{n-1}}\\ But~for ~ odd~ n, ~~ \Large \displaystyle \Pi_{k=1}^{\frac {n-1} 2}Cos\dfrac{\frac{k*\pi}2} n=\color{#3D99F6}{ - } \Pi_{k= \frac{ n+1} 2 }^{n-1 }Cos\dfrac{\frac{k*\pi} 2} n \\ \therefore \Large \displaystyle \Pi_{k=1}^{14}Cos\dfrac{Sin(\frac{k*\pi} 2)}{14} =\sqrt {\dfrac1 {2^{14}}}=2^{-7}$

Generalization: $\displaystyle\prod_{k=1}^{n-1}\cos\dfrac{k\pi}{n}=\left(\dfrac{i}{2}\right)^{n-1}$ where n is odd natural number