Question

If \(\cos x + \cos y = \tfrac{2}{3}\) and \(\sin x - \sin y = \tfrac{3}{4}\), then \[ \sin(x - y) \;+\; \cos(x - y) \;=\;? \]

• A. \(\tfrac{161}{145}\)
• B. \(\tfrac{127}{145}\)
• C. \(\tfrac{1}{2}\)
• D. \(\tfrac{8}{9}\)

Hint:

Sum/difference of sines or cosines can be turned into products.
- Then revert to \(\sin(\alpha\pm\beta)\), \(\cos(\alpha\pm\beta)\) forms to find the final expression in \((x-y)\).

Answer 1

The Correct Option is B

Step 1: Known sum-to-product identities.
\[ \cos x + \cos y = 2\cos\!\Bigl(\tfrac{x+y}{2}\Bigr)\,\cos\!\Bigl(\tfrac{x-y}{2}\Bigr), \] \[ \sin x - \sin y = 2\cos\!\Bigl(\tfrac{x+y}{2}\Bigr)\,\sin\!\Bigl(\tfrac{x-y}{2}\Bigr). \] From these, we can solve for \(\cos\!\bigl(\tfrac{x-y}{2}\bigr)\) and \(\sin\!\bigl(\tfrac{x-y}{2}\bigr)\), and also find \(\cos\!\bigl(\tfrac{x+y}{2}\bigr)\). Step 2: Express \(\sin(x-y) + \cos(x-y)\).
Recall: \[ \sin(x-y) = 2\sin\!\Bigl(\tfrac{x-y}{2}\Bigr)\,\cos\!\Bigl(\tfrac{x-y}{2}\Bigr), \quad \cos(x-y) = \cos^2\!\Bigl(\tfrac{x-y}{2}\Bigr) - \sin^2\!\Bigl(\tfrac{x-y}{2}\Bigr). \] We can then combine them. Step 3: Numerics (outline).
A direct approach is to treat \(\cos x+\cos y=\frac{2}{3}\) and \(\sin x - \sin y=\frac{3}{4}\) as a system in terms of \(\tfrac{x+y}{2}\) and \(\tfrac{x-y}{2}\). After solving carefully, one arrives at \[ \sin(x - y) + \cos(x - y) = \frac{127}{145}. \]