Let
$$a = \left(1-2\cos\dfrac{\pi}{11}\right)\!\left(1-2\cos\dfrac{3\pi}{11}\right)\!\left(1-2\cos\dfrac{5\pi}{11}\right)\!\left(1-2\cos\dfrac{7\pi}{11}\right)\!\left(1-2\cos\dfrac{9\pi}{11}\right).$$
Then the value of $5 - a^2$ is __________.
Step-by-Step Solution
Key Concept: Products of the form $\prod(1-2\cos\theta_k)$ for $\theta_k = k\pi/p$ (odd $k$, prime $p$) equal $\pm1$ by the theory of cyclotomic polynomials evaluated at $z=1$.
**Step 1: Link the product to cyclotomic polynomials**
The angles $k\pi/11$ for odd $k = 1,3,5,7,9$ arise from the 22nd cyclotomic polynomial. The identity $\prod_{k=1,3,5,7,9}(x - 2\cos\frac{k\pi}{11})$ evaluated at $x=1$ gives the product $a$.
**Step 2: Evaluate using the known result**
By properties of cyclotomic polynomials, $\prod_{k=1}^{5}\left(1-2\cos\frac{(2k-1)\pi}{11}\right) = \pm1$, so $a^2 = 1$.
**Step 3: Compute the answer**
$5 - a^2 = 5 - 1 = \mathbf{4}$.
Correct Answer: