Step 1 : Understanding the Question:
The problem involves a triangle $ABC$ where one angle $C$ is given as $120^\circ$ ($2\pi/3$ radians). We are required to find the numerical value of a trigonometric expression involving the other two angles $A$ and $B$. In any triangle, the sum of internal angles is always $180^\circ$ ($\pi$ radians). This relationship allows us to link $A$ and $B$ to the known value of $C$, making it possible to simplify the expression using standard trigonometric identities.
Step 2 : Key Formulas and approach:
We use the angle sum property: $A + B + C = \pi \Rightarrow A + B = \pi - \frac{2\pi}{3} = \frac{\pi}{3} (60^\circ)$.
Key identities:
1. $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$
2. $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$
3. $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$
Our approach is to expand the squares and products in the given expression and substitute the known value of $(A+B)$.
Step 3 : Detailed Explanation:
Since $A + B = \frac{\pi}{3}$, we have $\cos(A+B) = \cos(60^\circ) = \frac{1}{2}$.
Consider the first part: $\cos^2 A + \cos^2 B = \frac{1 + \cos 2A}{2} + \frac{1 + \cos 2B}{2} = 1 + \frac{1}{2} (\cos 2A + \cos 2B)$.
Applying sum-to-product formula: $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$.
Substitute $\cos(A+B) = 1/2$: $2(1/2) \cos(A-B) = \cos(A-B)$.
So, $\cos^2 A + \cos^2 B = 1 + \frac{1}{2} \cos(A-B)$.
Consider the second part: $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] = \frac{1}{2} [\frac{1}{2} + \cos(A-B)] = \frac{1}{4} + \frac{1}{2} \cos(A-B)$.
Combining both parts into the expression: $[1 + \frac{1}{2} \cos(A-B)] - [\frac{1}{4} + \frac{1}{2} \cos(A-B)]$.
The terms containing $\cos(A-B)$ cancel out exactly.
The final value is $1 - 1/4 = 3/4$.
Step 4 : Final Answer:
The value of the expression is $3/4$.