Question 12

Using the identity $1 - 2 \cos \theta = - \frac{\cos(3\theta/2)}{\cos(\theta/2)}$, we have:

$\alpha = \prod_{k=0}^4 (1 - 2 \cos(3^k \frac{\pi}{11})) = \prod_{k=0}^4 \left( -\frac{\cos(3^{k+1} \frac{\pi}{22})}{\cos(3^k \frac{\pi}{22})} \right)$

$\alpha = (-1)^5 \frac{\cos(3 \frac{\pi}{22})}{\cos(\frac{\pi}{22})} \frac{\cos(9 \frac{\pi}{22})}{\cos(3 \frac{\pi}{22})} \frac{\cos(27 \frac{\pi}{22})}{\cos(9 \frac{\pi}{22})} \frac{\cos(81 \frac{\pi}{22})}{\cos(27 \frac{\pi}{22})} \frac{\cos(243 \frac{\pi}{22})}{\cos(81 \frac{\pi}{22})}$

$\alpha = - \frac{\cos(\frac{243\pi}{22})}{\cos(\frac{\pi}{22})}$.

Since $\frac{243\pi}{22} = 11\pi + \frac{\pi}{22}$, we have $\cos(11\pi + \frac{\pi}{22}) = -\cos(\frac{\pi}{22})$.

Thus, $\alpha = - \frac{-\cos(\frac{\pi}{22})}{\cos(\frac{\pi}{22})} = 1$.

The required value is $5 - \alpha^2 = 5 - 1^2 = 4$.