A geometry problem by Priyanshu Mishra

$\begin{aligned} P & = \prod_{k=1}^{11} \cos \frac {k\pi}{11} \\ & = \prod_{k=1}^5 \cos \frac {k\pi}{11} \cdot {\color{#3D99F6} \prod_{k=6}^{10} \cos \frac {k\pi}{11}} \cdot {\color{#D61F06} \cos \frac{11\pi}{11}} & \small \color{#3D99F6} \text{Note that } \cos (\pi - x) = - \cos x \\ & = \prod_{k=1}^5 \cos \frac {k\pi}{11} \cdot {\color{#3D99F6} \prod_{k=1}^5 \left( -\cos \frac {k\pi}{11}\right)} \cdot {\color{#D61F06} \cos \pi} \\ & = \left(\prod_{k=1}^5 \cos \frac {k\pi}{11}\right)^2 \\ & = \left(\cos \frac \pi{11} \cos \frac {2 \pi}{11} {\color{#3D99F6}\cos \frac {3 \pi}{11}} \cos \frac {4 \pi}{11} {\color{#D61F06} \cos \frac {5 \pi}{11}} \right)^2 \\ & = \left(\cos \frac \pi{11} \cos \frac {2 \pi}{11} \cos \frac {4 \pi}{11} {\color{#3D99F6}\left( - \cos \frac {8 \pi}{11} \right)} {\color{#D61F06} \left( - \cos \frac {16 \pi}{11}\right)} \right)^2 \\ & = \left(\frac {{\color{#3D99F6}\sin \frac \pi{11}}\cos \frac \pi{11} \cos \frac {2 \pi}{11} \cos \frac {4 \pi}{11} \cos \frac {8 \pi}{11} \cos \frac {16 \pi}{11}}{\color{#3D99F6}\sin \frac \pi{11}} \right)^2 \\ & = \left(\frac {\sin \frac {2\pi}{11} \cos \frac {2 \pi}{11} \cos \frac {4 \pi}{11} \cos \frac {8 \pi}{11} \cos \frac {16 \pi}{11}}{2 \sin \frac \pi{11}} \right)^2 \\ & = \left(\frac {\sin \frac {4\pi}{11} \cos \frac {4 \pi}{11} \cos \frac {8 \pi}{11} \cos \frac {16 \pi}{11}}{4 \sin \frac \pi{11}} \right)^2 \\ & = \left(\frac {\sin \frac {8\pi}{11} \cos \frac {8 \pi}{11} \cos \frac {16 \pi}{11}}{8 \sin \frac \pi{11}} \right)^2 \\ & = \left(\frac {\sin \frac {16\pi}{11} \cos \frac {16 \pi}{11}}{16 \sin \frac \pi{11}} \right)^2 \\ & = \left(\frac {\color{#3D99F6}\sin \frac {32\pi}{11}}{32 \sin \frac \pi{11}} \right)^2 & \small \color{#3D99F6} \text{Note that } \sin \frac {32\pi}{11} = \sin \frac \pi{11} \\ & = \left(\frac 1{32} \right)^2 = \frac 1{1024}\end{aligned}$

$\implies a+b = 1+1024 = \boxed{1025}$

$\begin{aligned} \prod _{ k=1 }^{ 11 }{ \cos { \frac { k\pi }{ 11 } } } & = \prod _{ k=1 }^{ 5 }{ { \cos ^{ 2 }{ \frac { k\pi }{ 11 } } } } \\ & ={ \left( \prod _{ k=1 }^{ 5 }{ \cos { \frac { k\pi }{ 11 } } } \right) }^{ 2 } \end{aligned}$ .

Applying the formula

$\prod _{ r=0 }^{ n-1 }{ \cos { { 2 }^{ r }A } } =\frac { \sin { { 2 }^{ n }A } }{ { 2 }^{ n }\sin { A } }$ ,

where $A, n$ is first angle and number of terms respectively, we get

$\large\ { \left( \frac { \sin { \frac { 10\pi }{ 11 } } }{ 32\sin { \frac { \pi }{ 11 } } } \right) }^{ 2 } = \frac 1{1024}$

In general, $\begin{aligned}\prod_{k=1}^{n-1}\cos{\dfrac{k\pi}{n}}=\left(\frac{\iota}{2}\right)^{n-1}\end{aligned}$