Question

Solve for \( a \) and \( b \) given the equations: \[ \sin x + \sin y = a, \quad \cos x + \cos y = b, \quad x + y = \frac{2\pi}{3} \]

• A. \( a = \frac{1}{2}, \quad b = \frac{1}{2} \)
• B. \( a = \frac{1}{2}, \quad b = \frac{1}{\sqrt{2}} \)
• C. \( a = 0, \quad b = 1 \)
• D. \( a = 1, \quad b = 1 \)

Hint:

For trigonometric equations involving sums of sines and cosines, use sum identities to simplify the expressions and solve for the variables.

Answer 1

The Correct Option is B

We are given two equations involving trigonometric functions: \[ \sin x + \sin y = a \quad \text{and} \quad \cos x + \cos y = b \] with the additional condition that \( x + y = \frac{2\pi}{3} \). We will use the sum identities for sine and cosine to express the given equations in terms of \( x + y \) and \( x - y \).
1. Use the sum identity for sine: \[ \sin x + \sin y = 2 \sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right) \] Substituting \( x + y = \frac{2\pi}{3} \) into the equation, we get: \[ a = 2 \sin \left( \frac{\pi}{3} \right) \cos \left( \frac{x - y}{2} \right) \] Since \( \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \), we have: \[ a = \sqrt{3} \cos \left( \frac{x - y}{2} \right) \]
2. Similarly, use the sum identity for cosine: \[ \cos x + \cos y = 2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right) \] Substituting \( x + y = \frac{2\pi}{3} \) again: \[ b = 2 \cos \left( \frac{\pi}{3} \right) \cos \left( \frac{x - y}{2} \right) \] Since \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \), we get: \[ b = \cos \left( \frac{x - y}{2} \right) \] From the equations for \( a \) and \( b \), we can solve for \( a \) and \( b \). Given that \( \cos \left( \frac{x - y}{2} \right) = \frac{1}{\sqrt{2}} \), we find that: \[ a = \frac{1}{2}, \quad b = \frac{1}{\sqrt{2}} \]