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Explore the Back-StoryThis is a Test of Mathematics Solution Subjective 127 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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

|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 . Hence we have two possibilities:

- OR
- or

The above situations can happen when when

or or , where k is any integer.

Thus we need to plot the class of lines , and , 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).

**What is this topic:**Graphing**What are some of the associated concept:**Trigonometric Identities**Where can learn these topics:**I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Calculus’ module.**Book Suggestions:**Play With Graphs (Arihant Publication)

This is a Test of Mathematics Solution Subjective 127 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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

|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 . Hence we have two possibilities:

- OR
- or

The above situations can happen when when

or or , where k is any integer.

Thus we need to plot the class of lines , and , 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).

**What is this topic:**Graphing**What are some of the associated concept:**Trigonometric Identities**Where can learn these topics:**I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Calculus’ module.**Book Suggestions:**Play With Graphs (Arihant Publication)

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