Question

If $ \cos A + \cos B + \cos C = 0 $ and $ \sin A + \sin B + \sin C = 0 $, then $ \cos(A - B) = $

• A. 0
• B. $ \frac{1}{2} $
• C. $ -\frac{2}{3} $
• D. $ -\frac{1}{2} $

Hint:

Squaring and adding equations involving sums of sines and cosines can simplify the problem using $ \sin^2 \theta + \cos^2 \theta = 1 $ and the cosine difference formula.

Answer 1

The Correct Option is D

Step 1: Use the given equations.
$ \cos A + \cos B = -\cos C $
$ \sin A + \sin B = -\sin C $
Step 2: Square and add the equations.
$ (\cos A + \cos B)^2 + (\sin A + \sin B)^2 = (-\cos C)^2 + (-\sin C)^2 $ $ (\cos^2 A + 2 \cos A \cos B + \cos^2 B) + (\sin^2 A + 2 \sin A \sin B + \sin^2 B) = \cos^2 C + \sin^2 C $
Step 3: Simplify using trigonometric identities.
$ (\cos^2 A + \sin^2 A) + (\cos^2 B + \sin^2 B) + 2 (\cos A \cos B + \sin A \sin B) = 1 $
$ 1 + 1 + 2 \cos(A - B) = 1 $ 
Step 4: Solve for $ \cos(A - B) $.
$ 2 + 2 \cos(A - B) = 1 $
$ 2 \cos(A - B) = -1 $
$ \cos(A - B) = -\frac{1}{2} $
Step 5: Conclusion.
The value of $ \cos(A - B) $ is $ -\frac{1}{2} $.