The number of pairs (x, y) satisfying the equ | vedclass

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(C) The equation $\sin x + \sin y = \sin (x + y)$ can be written as $2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} = 2 \sin \frac{x+y}{2} \cos \frac{x+y}{2}

Vedclass · Accepted Answer

$6$

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(C) The equation $\sin x + \sin y = \sin (x + y)$ can be written as $2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} = 2 \sin \frac{x+y}{2} \cos \frac{x+y}{2}$.This implies $\sin \frac{x+y}{2} = 0$ or $\cos \frac{x-y}{2} = \cos \frac{x+y}{2}$.Case $1$: $\sin \frac{x+y}{2} = 0 \implies x+y = 0$. Given $|x| + |y| = 1$,if $x+y=0$,then $y=-x$,so $|x| + |-x| = 2|x| = 1 \implies |x| = 1/2$. Thus,$(1/2, -1/2)$ and $(-1/2, 1/2)$ are solutions.Case $2$: $\cos \frac{x-y}{2} = \cos \frac{x+y}{2} \implies \frac{x-y}{2} = \pm \frac{x+y}{2} + 2n\pi$. For $n=0$,this gives $x-y = x+y \implies y=0$ or $x-y = -(x+y) \implies x=0$.If $y=0$,$|x| + |0| = 1 \implies x = \pm 1$. Thus,$(1, 0)$ and $(-1, 0)$ are solutions.If $x=0$,$|0| + |y| = 1 \implies y = \pm 1$. Thus,$(0, 1)$ and $(0, -1)$ are solutions.Combining these,we have $6$ pairs: $(1/2, -1/2), (-1/2, 1/2), (1, 0), (-1, 0), (0, 1), (0, -1)$.

The number of pairs $(x, y)$ satisfying the equations $\sin x + \sin y = \sin (x + y)$ and $|x| + |y| = 1$ is

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