We are given the system of equations: \[ \sin x + \sin y = \sin(x + y) \quad \text{and} \quad |x| + |y| = 1. \]
Step 1: Solve the trigonometric equation. We start with the equation \( \sin x + \sin y = \sin(x + y) \). Using the trigonometric identity for \( \sin(x + y) \), we have: \[ \sin(x + y) = \sin x \cos y + \cos x \sin y. \] Substituting this into the equation, we get: \[ \sin x + \sin y = \sin x \cos y + \cos x \sin y. \] Rearranging the terms: \[ \sin x + \sin y - \sin x \cos y - \cos x \sin y = 0. \] Factorizing: \[ \sin x(1 - \cos y) = \sin y(\cos x - 1). \] This is a complicated trigonometric equation, but by testing special values for \( x \) and \( y \), we can find solutions.
Step 2: Analyze the second equation. Next, we are given that \( |x| + |y| = 1 \).
This equation represents a geometric constraint where \( (x, y) \) lies within the square with vertices at \( (1, 0) \), \( (-1, 0) \), \( (0, 1) \), and \( (0, -1) \).
Step 3: Consider possible values of \( x \) and \( y \).
We test various values of \( x \) and \( y \) within the constraint \( |x| + |y| = 1 \).
The points on the boundary of the square where this condition is satisfied are: - \( (1, 0) \), - \( (0, 1) \), - \( (-1, 0) \), - \( (0, -1) \), - \( \left( \frac{1}{2}, \frac{1}{2} \right) \), - \( \left( -\frac{1}{2}, \frac{1}{2} \right) \).
Step 4: Conclusion. There are 6 distinct ordered pairs that satisfy both equations.
Therefore, the number of solutions is: \[ \boxed{6}. \]
The direction cosines of two lines are connected by the relations \( 1 + m - n = 0 \) and \( lm - 2mn + nl = 0 \). If \( \theta \) is the acute angle between those lines, then \( \cos \theta = \) ?