Question

In \(\triangle ABC\), if \(\angle C = \frac{2\pi}{3}\), then the value of \(\cos^2 A + \cos^2 B - \cos A \cos B\) is:

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{4} \)
• C. \( 1 \)
• D. \( \frac{3}{2} \)

Hint:

For triangle-based trig expressions, try using \(A+B = \pi - C\) first. It often reduces everything to a single known angle and cancels complicated terms quickly.

Answer 1

The Correct Option is B

Concept: In any triangle, \[ A + B + C = \pi \] We use trigonometric identities:
• \( \cos^2 x = \frac{1 + \cos 2x}{2} \)
• \( \cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)] \)

Step 1: Using angle sum property.
\[ A + B = \pi - C = \pi - \frac{2\pi}{3} = \frac{\pi}{3} \] So, \[ \cos(A+B) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Step 2: Simplify \( \cos^2 A + \cos^2 B \).
\[ \cos^2 A + \cos^2 B = \frac{1 + \cos 2A}{2} + \frac{1 + \cos 2B}{2} \] \[ = 1 + \frac{1}{2}(\cos 2A + \cos 2B) \] Using sum-to-product: \[ \cos 2A + \cos 2B = 2\cos(A+B)\cos(A-B) \] So, \[ \cos^2 A + \cos^2 B = 1 + \cos(A+B)\cos(A-B) \] Substitute \( \cos(A+B) = \frac{1}{2} \): \[ \cos^2 A + \cos^2 B = 1 + \frac{1}{2}\cos(A-B) \quad \cdots (1) \]

Step 3: Simplify \( \cos A \cos B \).
\[ \cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)] \] \[ = \frac{1}{2}\left[\frac{1}{2} + \cos(A-B)\right] \] \[ = \frac{1}{4} + \frac{1}{2}\cos(A-B) \quad \cdots (2) \]

Step 4: Compute required expression.
Let \[ E = \cos^2 A + \cos^2 B - \cos A \cos B \] Substitute (1) and (2): \[ E = \left(1 + \frac{1}{2}\cos(A-B)\right) - \left(\frac{1}{4} + \frac{1}{2}\cos(A-B)\right) \] \[ E = 1 - \frac{1}{4} = \frac{3}{4} \]