Question

With usual notations, in a triangle $ABC$, if $\theta$ is any real number, then $a \cos(B - \theta) + b \cos(A + \theta)$ is

• A. $a \cos \theta$
• B. $b \cos \theta$
• C. $\cos \theta$
• D. $c \cos \theta$

Hint:

The projection rule $c = a \cos B + b \cos A$ is very useful whenever you see sides and cosines mixed in a linear sum.

Answer 1

The Correct Option is D

Step 1: Concept
Use the expansion of $\cos(X \pm Y)$: $\cos(X \pm Y) = \cos X \cos Y \mp \sin X \sin Y$.

Step 2: Meaning
Expand both terms: $a(\cos B \cos \theta + \sin B \sin \theta) + b(\cos A \cos \theta - \sin A \sin \theta)$.

Step 3: Analysis
Group $\cos \theta$ and $\sin \theta$ terms: $\cos \theta(a \cos B + b \cos A) + \sin \theta(a \sin B - b \sin A)$. By the projection rule, $a \cos B + b \cos A = c$. By the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} \implies a \sin B = b \sin A$, so $a \sin B - b \sin A = 0$.

Step 4: Conclusion
The expression simplifies to $c \cos \theta + 0 = c \cos \theta$. Final Answer: (D)