Question

Let \[ \alpha= \left(1-2\cos\frac{\pi}{11}\right) \left(1-2\cos\frac{3\pi}{11}\right) \left(1-2\cos\frac{9\pi}{11}\right) \left(1-2\cos\frac{27\pi}{11}\right) \left(1-2\cos\frac{81\pi}{11}\right) \] Then the value of \[ 5-\alpha^2 \] is ________.

Hint:

Products involving: \[ 1-2\cos\theta \] are often simplified using: \[ z=e^{i\theta} \] and roots of unity identities.

Answer 1

Correct Answer: 4

Step 1: Reduce the angles modulo \(2\pi\).
Since: \[ 27\equiv5\pmod{22} \] and \[ 81\equiv15\pmod{22} \] we get: \[ \cos\frac{27\pi}{11}=\cos\frac{5\pi}{11} \] \[ \cos\frac{81\pi}{11}=\cos\frac{15\pi}{11} \] Also: \[ \cos\frac{15\pi}{11} = -\cos\frac{4\pi}{11} \] Thus: \[ \alpha= \prod_{k=1}^{5}\left(1-2\cos\frac{m_k\pi}{11}\right) \] where: \[ m_k=1,3,5,9,15 \]

Step 2: Use roots of unity identity.
Using the standard identity: \[ \prod_{r=1}^{5}\left(1-2\cos\frac{(2r-1)\pi}{11}\right)=-1 \] Hence: \[ \alpha=-1 \] Therefore: \[ \alpha^2=1 \]

Step 3: Compute the required value.
\[ 5-\alpha^2 = 5-1 \] \[ =4 \]

Step 4: Identify the final answer.
Therefore: \[ \boxed{4} \]