Number of solutions of $\sin x+\sin y=\sin (x+y)$ and $|x|+|y|=1$ is

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Professor PC Thomas Classes · Answer

$ \begin{aligned} & \sin x+\sin y=\sin (x+y) \ & 2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)=2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x+y}{2}ight) \ & \sin \left(\frac{x+y}{2}ight)\left(\cos \frac{x-y}{2}-\cos \frac{x+y}{2}ight)=0 \ & \frac{x+y}{2}=0 \quad 	ext { (or) } \ & -2 \sin \frac{x}{2} \cdot \sin \left(\frac{-y}{2}ight)=0 \ & y=-x \quad x=0 \quad 	ext { (or) } y=0 \quad 	ext { Number of solutions }: 6 \end{aligned} $