Step 1: Understanding the Concept:
We are given an expression involving sides and angles of a triangle with an arbitrary angle $\theta$.
The approach is to expand the compound trigonometric terms using addition/subtraction formulas and then simplify using standard triangle properties like the Sine Rule and Projection Formula.
Step 2: Key Formula or Approach:
Compound angle formulas:
$\cos(x - y) = \cos x \cos y + \sin x \sin y$
$\cos(x + y) = \cos x \cos y - \sin x \sin y$
Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow a \sin B = b \sin A$.
Projection Formula: $c = a \cos B + b \cos A$.
Step 3: Detailed Explanation:
Let the given expression be $E = a \cos(B - \theta) + b \cos(A + \theta)$.
Expand the cosine terms using the compound angle formulas:
\[ E = a (\cos B \cos \theta + \sin B \sin \theta) + b (\cos A \cos \theta - \sin A \sin \theta) \]
Distribute the sides $a$ and $b$:
\[ E = a \cos B \cos \theta + a \sin B \sin \theta + b \cos A \cos \theta - b \sin A \sin \theta \]
Group the terms by $\cos \theta$ and $\sin \theta$:
\[ E = (a \cos B + b \cos A) \cos \theta + (a \sin B - b \sin A) \sin \theta \]
Now apply properties of triangles.
From the Sine Rule, we know that $\frac{a}{\sin A} = \frac{b}{\sin B}$, which cross-multiplies to $a \sin B = b \sin A$.
Therefore, the coefficient of $\sin \theta$ becomes zero:
$a \sin B - b \sin A = 0$.
From the Projection Formula, we know that $a \cos B + b \cos A$ is equal to side $c$.
Therefore, the coefficient of $\cos \theta$ becomes $c$:
$a \cos B + b \cos A = c$.
Substitute these findings back into the expression for $E$:
\[ E = (c) \cos \theta + (0) \sin \theta \]
\[ E = c \cos \theta \]
Step 4: Final Answer:
The expression is equal to $c \cos \theta$.