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Question Number 199718 by cortano12 last updated on 08/Nov/23
  { ((cos x+cos y=(1/2))),((sin x+sin y=(1/4))),((sin 2x + sin 2y=−((27)/(20)))) :}     sin (x+y) = ...
$$\:\begin{cases}{\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:\mathrm{y}=\frac{\mathrm{1}}{\mathrm{4}}}\\{\mathrm{sin}\:\mathrm{2x}\:+\:\mathrm{sin}\:\mathrm{2y}=−\frac{\mathrm{27}}{\mathrm{20}}}\end{cases} \\ $$$$\:\:\:\mathrm{sin}\:\left(\mathrm{x}+\mathrm{y}\right)\:=\:… \\ $$
Answered by Sutrisno last updated on 08/Nov/23
(cosx+cosy)(sinx+siny)=(1/8)  cosxsinx+cosxsiny+cosysinx+cosysiny=(1/8)  (1/2)sin2x+sin(x+y)+(1/2)sin2y=(1/8)  (1/2)(sin2x+sin2y)+sin(x+y)=(1/8)  (1/2)(((−27)/(20)))+sin(x+y)=(1/8)  sin(x+y)=(1/8)+((27)/(40))  sin(x+y)=((32)/(40))
$$\left({cosx}+{cosy}\right)\left({sinx}+{siny}\right)=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${cosxsinx}+{cosxsiny}+{cosysinx}+{cosysiny}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{x}+{sin}\left({x}+{y}\right)+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{y}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({sin}\mathrm{2}{x}+{sin}\mathrm{2}{y}\right)+{sin}\left({x}+{y}\right)=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{27}}{\mathrm{20}}\right)+{sin}\left({x}+{y}\right)=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${sin}\left({x}+{y}\right)=\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{27}}{\mathrm{40}} \\ $$$${sin}\left({x}+{y}\right)=\frac{\mathrm{32}}{\mathrm{40}} \\ $$$$ \\ $$

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