What if answer comes an irrational number

Let the required product be $\alpha$ . Now,

Using The Property $\sin { \theta } =\sin { \left( \pi -\theta \right) }$ ,

We convert The given Problem into,

$\alpha$ = ${ \left( \sin { \frac { \pi }{ 14 } } \cdot \sin { \frac { 3\pi }{ 14 } } \cdot \sin { \frac { 5\pi }{ 14 } } \right) }^{ 2 }$

Then Using The Property $\sin { \left( \frac { \pi }{ 2 } -\theta \right) } =\cos { \theta }$

We Get ,

$\alpha$ = ${ \left( \cos { \frac { 2\pi }{ 14 } } \cdot \cos { \frac { 4\pi }{ 14 } } \cdot \cos { \frac { 8\pi }{ 14 } } \right) }^{ 2 }$

Then We Know That

$\prod_{j=0}^{k-1}\cos(2^j x)=\frac{\sin(2^k x)}{2^k\sin(x)}$

And Here $x = \frac { 2\pi }{ 14 }$ .

Using That We Get

$\alpha$ = ${ \left( \frac { \sin { \left( { 2 }^{ 3 }\cdot \left( \frac { 2\pi }{ 14 } \right) \right) } }{ { 2 }^{ 3 }\cdot \sin { \left( \frac { 2\pi }{ 14 } \right) } } \right) }^{ 2 }$

Also We Know That $\sin { \frac { 2\pi }{ 14 } } =\sin { \frac { 16\pi }{ 14 } }$

The Expression Reduces To

$\alpha$ = ${ \left( \frac { 1 }{ { 2 }^{ 3 } } \right) }^{ 2 }$ .

So We Get $\alpha$ = $\frac { 1 }{ 64 }$ .

Proof of $\prod_{j=0}^{k-1}\cos(2^j x)=\frac{\sin(2^k x)}{2^k\sin(x)}$

$\prod_{j=0}^{k-1}\cos{2^jx}=\dfrac{\sin{2^kx}}{2^k\sin{x}}$

$\begin{aligned}\prod_{j=0}^{k-1}\cos{2^jx}&=\cos{x}\prod_{j=1}^{k-1}\cos{2^jx}\\&=\dfrac{2\sin{x}\cos{x}}{2\sin{x}}\prod_{j=1}^{k-1}\cos{2^jx}\\&=\dfrac{\sin{2x}\cos{2x}}{2\sin{x}}\prod_{j=2}^{k-1}\cos{2^jx}\\&=\dfrac{\sin{4x}\cos{4x}}{2^2\sin{x}}\prod_{j=3}^{k-1}\cos{2^jx}\\& \vdots\\ &\vdots\end{aligned}$

Doing this repeatedly, we get $\prod_{j=0}^{k-1}\cos{2^jx}=\dfrac{\sin{2^kx}}{2^k\sin{x}}$

Let the required product be $P$ . Consider $\alpha=e^{i2\pi/14}$ and $\beta=e^{i2\pi/7}$ .

Now, $\begin{aligned}\alpha^k&=\cos{\dfrac{2k\pi}{14}}+i\sin{\dfrac{2k\pi}{14}}\\ 1-\alpha^k&=1-\cos{\dfrac{2k\pi}{14}}-i\sin{\dfrac{2k\pi}{14}}\\ &=2\sin{\dfrac{k\pi}{14}}\left(\cos {\dfrac{k\pi}{14}}-i\sin{\dfrac{k\pi}{14}}\right)\\ \left|1-\alpha^k\right|&=2\sin{\dfrac{k\pi}{14}}\\ \prod_{k=0}^{6}\left|1-\alpha^{2k+1}\right|&=2^7P \ \ \ \ \ \ \ \ (1)\end{aligned}$

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As we know, $\displaystyle\prod_{k=1}^{13}\left|x-\alpha^k\right|=\left|\displaystyle\sum_{k=0}^{13}x^k\right|$ . Hence $\begin{aligned}\prod_{k=1}^{13}\left|1-\alpha^k\right|&=14\\ \left(\prod_{k=0}^{6}\left|1-\alpha^{2k+1}\right|\right)\left(\prod_{k=1}^{6}\left|1-\alpha^{2k}\right|\right)&=14\end{aligned}$

But, $\begin{aligned}\displaystyle\prod_{k=1}^{6}\left|1-\alpha^{2k}\right|=\displaystyle\prod_{k=1}^{6}\left|1-\beta^{k}\right|&=7\\ \therefore \ \ \prod_{k=0}^{6}\left|1-\alpha^{2k+1}\right|&=\dfrac{14}{7}=2 \ \ \ \ \ \ \ \ (2) \end{aligned}$

And finally, from $(1)$ and $(2)$ , $P=\dfrac{2}{2^7}=\boxed{\dfrac{1}{64}}$