Question

Prove that \( \frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x - y}{2} \).

Hint:

This identity is a useful simplification in trigonometry, often used in solving problems involving sine and cosine differences.

Answer 1

Step 1: Use trigonometric identities.
We will use the following sum and difference trigonometric identities: \[ \sin x - \sin y = 2 \cos \left( \frac{x + y}{2} \right) \sin \left( \frac{x - y}{2} \right) \] \[ \cos x + \cos y = 2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right) \]
Step 2: Simplify the expression.
Substitute these identities into the left-hand side of the equation: \[ \frac{\sin x - \sin y}{\cos x + \cos y} = \frac{2 \cos \left( \frac{x + y}{2} \right) \sin \left( \frac{x - y}{2} \right)}{2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)} \]
Step 3: Cancel common terms.
Canceling \( 2 \cos \left( \frac{x + y}{2} \right) \) from both the numerator and the denominator: \[ = \frac{\sin \left( \frac{x - y}{2} \right)}{\cos \left( \frac{x - y}{2} \right)} = \tan \left( \frac{x - y}{2} \right) \]
Step 4: Conclusion.
Thus, we have proved that: \[ \frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x - y}{2} \]