Question Number 80941 by jagoll last updated on 08/Feb/20
$$\mathrm{cos}\:{x}−\mathrm{2cos}\:{y}=−\sqrt{\mathrm{3}} \\ $$$$\mathrm{sin}\:\left({x}−{y}\right)=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${what}\:{is}\:\mathrm{sin}\:{x}−\mathrm{2sin}\:{y}\:? \\ $$
Commented by john santu last updated on 08/Feb/20
$${let}\:\mathrm{sin}\:{x}−\mathrm{2sin}\:{y}\:=\:{t} \\ $$$${squaring}\: \\ $$$$\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{4cos}\:{x}\mathrm{cos}\:{y}+\mathrm{4cos}\:\:^{\mathrm{2}} {y}=\mathrm{3} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{4sin}\:{x}\mathrm{sin}\:{y}+\mathrm{4sin}\:^{\mathrm{2}} {y}\:={t}^{\mathrm{2}} \\ $$$${sum}\::\:\mathrm{5}\:−\mathrm{4cos}\:\left({x}−{y}\right)\:={t}^{\mathrm{2}} +\mathrm{3} \\ $$$$\mathrm{2}−{t}^{\mathrm{2}} \:=\:\mathrm{4cos}\:\left({x}−{y}\right) \\ $$$${from}\:\mathrm{sin}\:\left({x}−{y}\right)=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\:\Rightarrow\mathrm{cos}\:\left({x}−{y}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\therefore\:\mathrm{2}−{t}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\mathrm{2}−\mathrm{3}{t}^{\mathrm{2}} =\mathrm{0} \\ $$$${t}\:=\:\pm\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$
Commented by jagoll last updated on 08/Feb/20
$${thank}\:{you} \\ $$