Question

\(\cos\frac{\pi}{12} + \cos\frac{17\pi}{12} + \cos\frac{11\pi}{12}\) is equal to

• A. 1
• B. -1
• C. 0
• D. \(\frac{1-\sqrt{3}}{2\sqrt{2}}\)

Hint:

Convert angles to standard values (like \(75^\circ\)) for easy evaluation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We use the reduction formulas to simplify angles that are in different quadrants to their corresponding first-quadrant values.
Step 2: Detailed Explanation:
Let the expression be \( E = \cos \frac{\pi}{12} + \cos \frac{17\pi}{12} + \cos \frac{11\pi}{12} \).
1. Simplify \( \cos \frac{11\pi}{12} \): \[ \cos \frac{11\pi}{12} = \cos (\pi - \frac{\pi}{12}) = -\cos \frac{\pi}{12} \] Substituting this into E: \[ E = \cos \frac{\pi}{12} + \cos \frac{17\pi}{12} - \cos \frac{\pi}{12} = \cos \frac{17\pi}{12} \] 2. Simplify \( \cos \frac{17\pi}{12} \): \[ \cos \frac{17\pi}{12} = \cos (\frac{3\pi}{2} - \frac{\pi}{12}) \] Using \( \cos(3\pi/2 - \theta) = -\sin \theta \): \[ E = -\sin \frac{\pi}{12} \] 3. Evaluate \( \sin \frac{\pi}{12} \) (which is \( \sin 15^\circ \)): \[ \sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \] \[ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}} \] Therefore, \( E = -\frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}} \).
Step 3: Final Answer:
The result is \( \frac{1 - \sqrt{3}}{2\sqrt{2}} \).