Question

In \(△ABC\), if \(\angle C = \frac{π}{3}\), then the value of \(cos^2A+cos^2B+cosAcosB\) is...

• A. \(\frac{3}{4}\)
• B. \(\frac{5}{4}\)
• C. \(\frac{-3}{4}\)
• D. \(\frac{-5}{4}\)

Hint:

Use A + B = 120 degrees and the identity for cos(A + B).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a triangle \(A + B + C = \pi\). With \(C = \frac{\pi}{3}\) we get \(A + B = \frac{2\pi}{3}\), so \(\cos(A+B) = -\frac{1}{2}\).

Step 2: Key Formula or Approach:
\(\cos(A+B) = \cos A\cos B - \sin A\sin B = -\frac12\).

Step 3: Detailed Explanation:
Let \(x = \cos A\) and \(y = \cos B\). Then \(\sin A\sin B = xy + \frac12\).
Square \(\sin A \sin B\): \((1 - x^2)(1 - y^2) = \left(xy + \frac12\right)^2\).
Expand: \(1 - x^2 - y^2 + x^2y^2 = x^2y^2 + xy + \frac14\).
So \(x^2 + y^2 + xy = 1 - \frac14 = \frac34\).
\[ \cos^2 A + \cos^2 B + \cos A\cos B = \frac{3}{4} \]

Final Answer:
The value is \(\frac{3}{4}\), option (A). \[ \boxed{\frac{3}{4}} \]