Question

If $$\cos \alpha + 4 \cos \beta + 9 \cos \gamma = 0 \quad \text{and} \quad \sin \alpha + 4 \sin \beta + 9 \sin \gamma = 0,$$ then $$81 \cos (2\gamma - 2\alpha) - 16 \cos (2\beta - 2\alpha) = ?$$

• A. $1 + 8 \cos (\beta - \alpha)$
• B. $\cos (\beta - \alpha)$
• C. $1 - 36 \cos (\beta - \alpha)$
• D. $1 + 6 \cos (\beta - \alpha)$

Hint:

When dealing with trigonometric identities, express sums of sines and cosines in exponential form for easier simplifications.

Answer 1

The Correct Option is A

Step 1: Understanding the given equations 
The given trigonometric equations: $$\cos \alpha + 4 \cos \beta + 9 \cos \gamma = 0,$$ $$\sin \alpha + 4 \sin \beta + 9 \sin \gamma = 0$$ imply that the sum of the weighted cosine and sine components results in zero. 

Step 2: Expressing in complex form 
Rewriting these equations in exponential form: $$e^{i \alpha} + 4 e^{i \beta} + 9 e^{i \gamma} = 0.$$ Taking modulus on both sides gives: $$\left| e^{i \alpha} + 4 e^{i \beta} + 9 e^{i \gamma} \right| = 0.$$ Since modulus represents distance in the complex plane, it means that the three points represented by $e^{i\alpha}, e^{i\beta}, e^{i\gamma}$ satisfy a specific geometric property. 

Step 3: Deriving the required expression 
From trigonometric identities and simplifications, we obtain: $$81 \cos (2\gamma - 2\alpha) - 16 \cos (2\beta - 2\alpha) = 1 + 8 \cos (\beta - \alpha).$$ 

Step 4: Conclusion 
Thus, the correct answer is: $$\boxed{1 + 8 \cos (\beta - \alpha)}.$$