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Question Number 127781 by shaker last updated on 02/Jan/21

Answered by bemath last updated on 02/Jan/21

$$\{\begin{array}{l}2\sin \mathrm{xcos}\mathrm{y}=-1\\ 2\cos \mathrm{xsin}\mathrm{y}=1\end{array}$$$$\Rightarrow \sin (\mathrm{x}+\mathrm{y})+\sin (\mathrm{x}-\mathrm{y})=-1$$$$\Rightarrow \sin (\mathrm{x}+\mathrm{y})-\sin (\mathrm{x}-\mathrm{y})=1$$$$\iff 2\sin (\mathrm{x}+\mathrm{y})=0\Rightarrow \sin (\mathrm{x}+\mathrm{y})=0$$$$\Rightarrow \mathrm{x}+\mathrm{y}=\mathrm{n}\pi ;\mathrm{n}\in \mathbb{Z}$$$$\mathrm{and}2\sin (\mathrm{x}-\mathrm{y})=-2;\sin (\mathrm{x}-\mathrm{y})=-1$$$$\mathrm{x}-\mathrm{y}=\frac{3\pi }{2}+\mathrm{n}\pi ,\mathrm{n}\in \mathbb{Z}$$

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