**Problem: **Find all (x, y) such that sin x + sin y = sin (x+y) and |x| + |y| = 1

**Discussion: **

|x| + |y| =1 is easier to plot. We have to treat the cases separately.

- First quadrant: x +y = 1
- Second quadrant: -x + y = 1 (since |x| = -x when x is negative)
- Third Quadrant: -x-y =1
- Fourth Quadrant: x – y =1

Now we work on sin x + sin y = sin (x + y).

This implies \(\displaystyle{2 \sin \left ( \frac{x+y} {2} \right ) \cos \left ( \frac{x-y} {2} \right ) = 2 \sin \left ( \frac{x+y} {2} \right ) \cos \left ( \frac{x+y} {2} \right ) }\). Hence we have two possibilities:

- \(\displaystyle{\sin \left ( \frac{x+y} {2} \right ) = 0 }\) OR
- \(\displaystyle{\cos \left ( \frac{x+y} {2} \right ) – \cos \left ( \frac{x-y} {2} \right ) = 0 }\) or \(\displaystyle { \sin \frac{x}{2} \sin \frac{y}{2} } = 0 \)

The above situations can happen when when

\(\displaystyle{ \frac{x +y}{2} = k \pi } \) or \(\displaystyle{\frac {x}{2} = k \pi }\) or \(\displaystyle{ \frac{y}{2} = k \pi }\), where k is any integer.

Thus we need to plot the class of lines \(\displaystyle{ x + y = 2 k \pi } \), \(\displaystyle{ x = 2k\pi } \) and \(\displaystyle{ y = 2k\pi } \), and consider the intersection points of these lines with the graph of |x| + |y| = 1.

Clearly only for k=0, such intersection points can be found.

Hence required points are (0,1), (0,-1), (1,0), (-1,0), (1/2, -1/2), (-1/2, 1/2).

## Chatuspathi:

**What is this topic:**Graphing**What are some of the associated concept:**Trigonometric Identities**Where can learn these topics:**Cheenta**Book Suggestions:**Play With Graphs (Arihant Publication)

this is great… easy to understand & the solutions given in details….