Solving the Trigonometric Equation \(\sin x + \sin y = \cos y - \cos x\)

The problem asks us to find the value of \( \tan \left( \frac {x - y} 2 \right) \) given the equation \( \sin x + \sin y = \cos y - \cos x \) and the conditions \( 0 < y < x < \rm \frac \pi 2 \).

Using Sum-to-Product Formulas

We can use the sum-to-product trigonometric identities to simplify the given equation. The relevant identities are:

• \( \sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) \)
• \( \cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) \)

Let's apply these identities to the terms in the given equation \( \sin x + \sin y = \cos y - \cos x \).

The left side, \( \sin x + \sin y \), becomes: \(\sin x + \sin y = 2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)\)

The right side, \( \cos y - \cos x \), becomes. Note the order of the terms in the difference formula: \( \cos y - \cos x \) corresponds to \( A=y \) and \( B=x \). \(\cos y - \cos x = -2 \sin \left( \frac{y+x}{2} \right) \sin \left( \frac{y-x}{2} \right)\) Since \( \sin(-\theta) = -\sin(\theta) \), we can write \( \sin \left( \frac{y-x}{2} \right) = -\sin \left( \frac{x-y}{2} \right) \). Also \( \frac{y+x}{2} = \frac{x+y}{2} \). So, the right side simplifies to: \(\cos y - \cos x = -2 \sin \left( \frac{x+y}{2} \right) \left( -\sin \left( \frac{x-y}{2} \right) \right) = 2 \sin \left( \frac{x+y}{2} \right) \sin \left( \frac{x-y}{2} \right)\)

Equating the Simplified Expressions

Now, we set the simplified left side equal to the simplified right side: \(2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right) = 2 \sin \left( \frac{x+y}{2} \right) \sin \left( \frac{x-y}{2} \right)\)

We can rearrange the equation to group all terms on one side: \(2 \sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right) - 2 \sin \left( \frac{x+y}{2} \right) \sin \left( \frac{x-y}{2} \right) = 0\) Factor out the common term \( 2 \sin \left( \frac{x+y}{2} \right) \): \(2 \sin \left( \frac{x+y}{2} \right) \left[ \cos \left( \frac{x-y}{2} \right) - \sin \left( \frac{x-y}{2} \right) \right] = 0\)

Analyzing the Conditions

This equation implies that either \( \sin \left( \frac{x+y}{2} \right) = 0 \) or \( \cos \left( \frac{x-y}{2} \right) - \sin \left( \frac{x-y}{2} \right) = 0 \). Let's consider the given conditions: \( 0 < y < x < \rm \frac \pi 2 \).

From \( 0 < y < x < \frac{\pi}{2} \), we can deduce the range for \( \frac{x+y}{2} \) and \( \frac{x-y}{2} \). For \( \frac{x+y}{2} \): \( 0 + 0 < y + x < \frac{\pi}{2} + \frac{\pi}{2} \) \( 0 < x+y < \pi \) \( 0 < \frac{x+y}{2} < \frac{\pi}{2} \) In the interval \( \left( 0, \frac{\pi}{2} \right) \), the value of \( \sin \left( \frac{x+y}{2} \right) \) is strictly positive. Therefore, \( \sin \left( \frac{x+y}{2} \right) \neq 0 \).

For \( \frac{x-y}{2} \): \( y < x \implies x-y > 0 \) \( x < \frac{\pi}{2} \) and \( y > 0 \) Consider the maximum value of \( x-y \). As \( x \) approaches \( \frac{\pi}{2} \) and \( y \) approaches \( 0 \), \( x-y \) approaches \( \frac{\pi}{2} \). So, \( 0 < x-y < \frac{\pi}{2} \). Therefore, \( 0 < \frac{x-y}{2} < \frac{\pi}{4} \).

Solving for \( \tan \left( \frac{x-y}{2} \right) \)

Since \( \sin \left( \frac{x+y}{2} \right) \neq 0 \), the equation \( 2 \sin \left( \frac{x+y}{2} \right) \left[ \cos \left( \frac{x-y}{2} \right) - \sin \left( \frac{x-y}{2} \right) \right] = 0 \) implies: \(\cos \left( \frac{x-y}{2} \right) - \sin \left( \frac{x-y}{2} \right) = 0\) \(\cos \left( \frac{x-y}{2} \right) = \sin \left( \frac{x-y}{2} \right)\)

We want to find \( \tan \left( \frac{x-y}{2} \right) \). Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). We can divide both sides of the equation \( \cos \left( \frac{x-y}{2} \right) = \sin \left( \frac{x-y}{2} \right) \) by \( \cos \left( \frac{x-y}{2} \right) \), provided that \( \cos \left( \frac{x-y}{2} \right) \neq 0 \). From the range we found for \( \frac{x-y}{2} \), which is \( 0 < \frac{x-y}{2} < \frac{\pi}{4} \), we know that the cosine value is strictly positive in this interval. Thus, \( \cos \left( \frac{x-y}{2} \right) \neq 0 \).

Dividing by \( \cos \left( \frac{x-y}{2} \right) \): \(\frac{\sin \left( \frac{x-y}{2} \right)}{\cos \left( \frac{x-y}{2} \right)} = 1\) \(\tan \left( \frac{x-y}{2} \right) = 1\)

Thus, the value of \( \tan \left( \frac{x - y} 2 \right) \) is 1.

Revision Table: Key Concepts

Additional Information: Angle Ranges and Trigonometric Values

Understanding the range of angles and the sign of trigonometric functions within those ranges is crucial for solving such problems.

• The condition \( 0 < y < x < \frac{\pi}{2} \) means both \( x \) and \( y \) are in the first quadrant.
• For \( 0 < \theta < \frac{\pi}{2} \), \( \sin \theta > 0 \) and \( \cos \theta > 0 \).
• The average angle \( \frac{x+y}{2} \) lies between \( 0 \) and \( \frac{\pi}{2} \), so \( \sin \left( \frac{x+y}{2} \right) > 0 \).
• The half difference angle \( \frac{x-y}{2} \) lies between \( 0 \) and \( \frac{\pi}{4} \). In this specific range \( \left( 0, \frac{\pi}{4} \right) \), both \( \sin \left( \frac{x-y}{2} \right) \) and \( \cos \left( \frac{x-y}{2} \right) \) are positive, and \( \cos \left( \frac{x-y}{2} \right) > \sin \left( \frac{x-y}{2} \right) \) unless they are equal. At \( \frac{\pi}{4} \), they are equal.
• Our result \( \tan \left( \frac{x-y}{2} \right) = 1 \) implies \( \frac{x-y}{2} = \frac{\pi}{4} \) (since the angle is acute), which means \( x-y = \frac{\pi}{2} \). This is consistent with the derived equation \( \cos \left( \frac{x-y}{2} \right) = \sin \left( \frac{x-y}{2} \right) \).