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

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(D) Given equations are $\sin x + \sin y = \sin(x + y)$ and $|x| + |y| = 1$. Using the sum-to-product formula: $2 \sin \frac{x+y}{2} \cos \frac{x-y}{2

Vedclass · Accepted Answer

$6$

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(D) Given equations are $\sin x + \sin y = \sin(x + y)$ and $|x| + |y| = 1$. Using the sum-to-product formula: $2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} = 2 \sin \frac{x+y}{2} \cos \frac{x+y}{2}$. This implies $\sin \frac{x+y}{2} [\cos \frac{x-y}{2} - \cos \frac{x+y}{2}] = 0$. Using $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$,we get $\sin \frac{x+y}{2} [2 \sin \frac{x}{2} \sin \frac{y}{2}] = 0$. Thus,$\sin \frac{x+y}{2} = 0$ or $\sin \frac{x}{2} = 0$ or $\sin \frac{y}{2} = 0$. This leads to $x + y = 0$ or $x = 0$ or $y = 0$. Case $1$: If $x + y = 0$,then $|x| + |-x| = 1$ $\Rightarrow 2|x| = 1$ $\Rightarrow x = \pm \frac{1}{2}$. Pairs: $(\frac{1}{2}, -\frac{1}{2}), (-\frac{1}{2}, \frac{1}{2})$. Case $2$: If $x = 0$,then $|0| + |y| = 1$ $\Rightarrow |y| = 1$ $\Rightarrow y = \pm 1$. Pairs: $(0, 1), (0, -1)$. Case $3$: If $y = 0$,then $|x| + |0| = 1$ $\Rightarrow |x| = 1$ $\Rightarrow x = \pm 1$. Pairs: $(1, 0), (-1, 0)$. Total distinct ordered pairs are $2 + 2 + 2 = 6$.

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

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