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Question Number 197583 by cortano12 last updated on 23/Sep/23

   { ((((sin x)/(cos (x+y))) = −((√2)/2))),((((cos y)/(cos (x+y))) = ((√2)/2))) :}     find the solution

$$\:\:\begin{cases}{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)}\:=\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\frac{\mathrm{cos}\:\mathrm{y}}{\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{cases} \\ $$$$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\: \\ $$

Answered by Sutrisno last updated on 23/Sep/23

(((sinx)/(cos(x+y)))/((cosy)/(cos(x+y))))=((−((√2)/2))/((√2)/2))→((sinx)/(cosy))=−1→sinx=sin(270^o +y)→x=270^o +y  ((cos(270^o +y))/(cos(270^o +2y)))=((√2)/2)  ((siny)/(sin2y))=((√2)/2)  ((siny)/(2sinycosy))=((√2)/2)  y=45^o  and x=315^o

$$\frac{\frac{{sinx}}{{cos}\left({x}+{y}\right)}}{\frac{{cosy}}{{cos}\left({x}+{y}\right)}}=\frac{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\rightarrow\frac{{sinx}}{{cosy}}=−\mathrm{1}\rightarrow{sinx}={sin}\left(\mathrm{270}^{{o}} +{y}\right)\rightarrow{x}=\mathrm{270}^{{o}} +{y} \\ $$$$\frac{{cos}\left(\mathrm{270}^{{o}} +{y}\right)}{{cos}\left(\mathrm{270}^{{o}} +\mathrm{2}{y}\right)}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{{siny}}{{sin}\mathrm{2}{y}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\frac{{siny}}{\mathrm{2}{sinycosy}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${y}=\mathrm{45}^{{o}} \:{and}\:{x}=\mathrm{315}^{{o}} \\ $$

Commented by mr W last updated on 24/Sep/23

generally:  x=−45°+k×180°  y=45°+m×360°  k,m∈Z

$${generally}: \\ $$$${x}=−\mathrm{45}°+{k}×\mathrm{180}° \\ $$$${y}=\mathrm{45}°+{m}×\mathrm{360}° \\ $$$${k},{m}\in{Z} \\ $$

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