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Graphing relations (Tomato Subjective 127)

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)
November 20, 2015

1 comment

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