Evaluate Expression Involving Product of Cosine Functions

To find the value of $5 - \alpha^2$, we first simplify the expression given for $\alpha$:

$\alpha = \left(1 - 2 \cos\left(\frac{\pi}{11}\right)\right)\left(1 - 2 \cos\left(\frac{3\pi}{11}\right)\right)\left(1 - 2 \cos\left(\frac{9\pi}{11}\right)\right)\left(1 - 2 \cos\left(\frac{27\pi}{11}\right)\right)\left(1 - 2 \cos\left(\frac{81\pi}{11}\right)\right)$

We use the trigonometric identity: $\cos(3y) = 4\cos^3(y) - 3\cos(y) = \cos(y)(4\cos^2(y) - 3) = \cos(y)(2(1 + \cos(2y)) - 3) = \cos(y)(2\cos(2y) - 1)$

Dividing both sides by $\cos(y)$, we get: $2\cos(2y) - 1 = \frac{\cos(3y)}{\cos(y)}$

Multiplying by $-1$ and substituting $2y = x \implies y = \frac{x}{2}$, we obtain: $1 - 2\cos(x) = -\frac{\cos\left(\frac{3x}{2}\right)}{\cos\left(\frac{x}{2}\right)}$

Now, applying this identity to each term in the product for $\alpha$:

1. For $x = \frac{\pi}{11}$: $1 - 2\cos\left(\frac{\pi}{11}\right) = -\frac{\cos\left(\frac{3\pi}{22}\right)}{\cos\left(\frac{\pi}{22}\right)}$

2. For $x = \frac{3\pi}{11}$: $1 - 2\cos\left(\frac{3\pi}{11}\right) = -\frac{\cos\left(\frac{9\pi}{22}\right)}{\cos\left(\frac{3\pi}{22}\right)}$

3. For $x = \frac{9\pi}{11}$: $1 - 2\cos\left(\frac{9\pi}{11}\right) = -\frac{\cos\left(\frac{27\pi}{22}\right)}{\cos\left(\frac{9\pi}{22}\right)}$

4. For $x = \frac{27\pi}{11}$: $1 - 2\cos\left(\frac{27\pi}{11}\right) = -\frac{\cos\left(\frac{81\pi}{22}\right)}{\cos\left(\frac{27\pi}{22}\right)}$

5. For $x = \frac{81\pi}{11}$: $1 - 2\cos\left(\frac{81\pi}{11}\right) = -\frac{\cos\left(\frac{243\pi}{22}\right)}{\cos\left(\frac{81\pi}{22}\right)}$

Multiplying all 5 factors together yields a telescoping product: $\alpha = (-1)^5 \cdot \frac{\cos\left(\frac{3\pi}{22}\right)}{\cos\left(\frac{\pi}{22}\right)} \cdot \frac{\cos\left(\frac{9\pi}{22}\right)}{\cos\left(\frac{3\pi}{22}\right)} \cdot \frac{\cos\left(\frac{27\pi}{22}\right)}{\cos\left(\frac{9\pi}{22}\right)} \cdot \frac{\cos\left(\frac{81\pi}{22}\right)}{\cos\left(\frac{27\pi}{22}\right)} \cdot \frac{\cos\left(\frac{243\pi}{22}\right)}{\cos\left(\frac{81\pi}{22}\right)}$

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

Note that $\frac{243\pi}{22} = 11\pi + \frac{\pi}{22}$. Therefore: $\cos\left(\frac{243\pi}{22}\right) = \cos\left(11\pi + \frac{\pi}{22}\right) = -\cos\left(\frac{\pi}{22}\right)$

Substituting this back into the expression for $\alpha$: $\alpha = -\frac{-\cos\left(\frac{\pi}{22}\right)}{\cos\left(\frac{\pi}{22}\right)} = 1$

Finally, we calculate the required value: $5 - \alpha^2 = 5 - (1)^2 = 4$