Question

If \( \sin A + \sin B = a \) and \( \cos A + \cos B = b \), then \( \cos(A + B) \) equals?

• A. \( \frac{a^2 + b^2}{b^2 - a^2} \)
• B. \( \frac{2ab}{a^2 + b^2} \)
• C. \( \frac{b^2 - a^2}{a^2 + b^2} \)
• D. \( \frac{a^2 - b^2}{a^2 + b^2} \)

Hint:

To simplify expressions involving \( \sin A + \sin B \) and \( \cos A + \cos B \), use the sum-to-product identities. Also, remember the identity: \[ \cos(A + B) = \cos^2\left(\frac{A + B}{2}\right) - \sin^2\left(\frac{A + B}{2}\right) \]

Answer 1

The Correct Option is C

Given: \( \sin A + \sin B = a \), and \( \cos A + \cos B = b \)
Step 1: Use trigonometric identities for sum
\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) = a \] \[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) = b \] Step 2: Eliminate \( \cos\left(\frac{A - B}{2}\right) \)
Let us square and add both expressions: \[ (\sin A + \sin B)^2 + (\cos A + \cos B)^2 = a^2 + b^2 \] \[ = 4 \left[\sin^2\left(\frac{A + B}{2}\right) + \cos^2\left(\frac{A + B}{2}\right)\right] \cdot \cos^2\left(\frac{A - B}{2}\right) \] \[ = 4 \cdot 1 \cdot \cos^2\left(\frac{A - B}{2}\right) \Rightarrow \cos^2\left(\frac{A - B}{2}\right) = \frac{a^2 + b^2}{4} \] Step 3: Use expression for \( \cos(A + B) \)
\[ \cos(A + B) = \cos^2\left(\frac{A + B}{2}\right) - \sin^2\left(\frac{A + B}{2}\right) \] Now divide both the original expressions: \[ \frac{\cos A + \cos B}{\sin A + \sin B} = \frac{2 \cos(\frac{A + B}{2})}{2 \sin(\frac{A + B}{2})} = \cot\left(\frac{A + B}{2}\right) = \frac{b}{a} \Rightarrow \tan\left(\frac{A + B}{2}\right) = \frac{a}{b} \] Use: \[ \cos(A + B) = \frac{1 - \tan^2(\frac{A + B}{2})}{1 + \tan^2(\frac{A + B}{2})} = \frac{1 - \left(\frac{a}{b}\right)^2}{1 + \left(\frac{a}{b}\right)^2} = \frac{b^2 - a^2}{b^2 + a^2} \]