Question

If \(\sin \mathrm{A}+\sin \mathrm{B}=x\) and \(\cos \mathrm{A}+\cos \mathrm{B}=y\), then \(\sin (A+B)=\)

• A. \(\frac{2xy}{x^{2}+y^{2}}\)
• B. \(\frac{xy}{x^{2}+y^{2}}\)
• C. \(\frac{2xy}{y^{2}-x^{2}}\)
• D. \(\frac{xy}{y^{2}-x^{2}}\)

Hint:

Convert sums to products, divide to get tan of (A+B)/2, then use the tangent half-angle form of sine.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(x=\sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\) and \(y=\cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\).

Step 2: Divide
\(\dfrac xy=\tan\dfrac{A+B}{2}=t\).

Step 3: Use the half-angle formula
\(\sin(A+B)=\dfrac{2t}{1+t^2}=\dfrac{2x/y}{1+x^2/y^2}\).

Step 4: Simplify
\(\sin(A+B)=\dfrac{2xy}{x^2+y^2}\).

Final Answer:
\(\sin(A+B)=\dfrac{2xy}{x^2+y^2}\). \[ \boxed{\text{(A) }\dfrac{2xy}{x^{2}+y^{2}}} \]