Question

If \( \left( 1 + \cos \frac{\pi}{8} \right) \left( 1 + \cos \frac{7\pi}{8} \right) \) =

• A. \( \frac{1}{\sqrt{2}} (1 + \sqrt{2}) \)
• B. \( \frac{1}{2\sqrt{2}} (1 + \sqrt{2}) \)
• C. \( \frac{1}{2\sqrt{2}} (\sqrt{2} - 1) \)
• D. \( \frac{1}{\sqrt{2}} (\sqrt{2} - 1) \)
• E. \( \frac{1}{2} (\sqrt{2} + 1) \)

Hint:

Use the identity \( \cos(\pi - x) = -\cos(x) \) to simplify trigonometric expressions involving angles that sum to \( \pi \). Also, recognize the difference of squares to easily simplify products of trigonometric terms.

Answer 1

The Correct Option is C

Step 1: We are given the expression: \[ \left( 1 + \cos \frac{\pi}{8} \right) \left( 1 + \cos \frac{7\pi}{8} \right). \] We can simplify the second cosine term using the identity \( \cos \left( \pi - x \right) = -\cos x \): \[ \cos \frac{7\pi}{8} = \cos \left( \pi - \frac{\pi}{8} \right) = - \cos \frac{\pi}{8}. \] Thus, the expression becomes: \[ \left( 1 + \cos \frac{\pi}{8} \right) \left( 1 - \cos \frac{\pi}{8} \right). \] Step 2: Recognizing this as a difference of squares, we can simplify it as: \[ 1^2 - \left( \cos \frac{\pi}{8} \right)^2 = 1 - \cos^2 \frac{\pi}{8}. \] Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can write: \[ 1 - \cos^2 \frac{\pi}{8} = \sin^2 \frac{\pi}{8}. \] Step 3: Thus, the given expression simplifies to: \[ \sin^2 \frac{\pi}{8}. \] Step 4: Now, we use the known value for \( \sin \frac{\pi}{8} \), which is \( \frac{\sqrt{2 - \sqrt{2}}}{2} \). 
Therefore, we can write: \[ \sin^2 \frac{\pi}{8} = \left( \frac{\sqrt{2 - \sqrt{2}}}{2} \right)^2 = \frac{2 - \sqrt{2}}{4}. \] Step 5: Simplifying this, we get: \[ \frac{1}{2\sqrt{2}} (\sqrt{2} - 1). \] Thus, the correct answer is option (C).