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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)
• 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 I.S.I. & C.M.I. course, discusses these topics in the ‘Calculus’ module.
• Book Suggestions: Play With Graphs (Arihant Publication)