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Question Number 137811 by bramlexs22 last updated on 07/Apr/21

Given { ((cos (x−y)=−1+(1/2)cos x)),((cos (x+y)= 1+(1/3)cos x)) :} where 270° < y< 360° . find sin 2y .

$${Given}\:\begin{cases}{\mathrm{cos}\:\left({x}−{y}\right)=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{x}}\\{\mathrm{cos}\:\left({x}+{y}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:{x}}\end{cases} \\ $$ $${where}\:\mathrm{270}°\:<\:{y}<\:\mathrm{360}°\:.\:{find} \\ $$ $$\mathrm{sin}\:\mathrm{2}{y}\:. \\ $$

Answered by EDWIN88 last updated on 07/Apr/21

⇔ cos (x−y)+cos (x+y)=(5/6)cos x ⇔ 2cos x cos y = (5/6)cos x ⇔ cos x [2cos y−(5/6) ]= 0 { ((cos y=(5/(12)))),((sin y=−(√(1−((25)/(144)))) =−((√(119))/(12)))) :} then sin 2y = 2sin y cos y = −2[((5(√(119)))/(144)) ] = −((5(√(119)))/(72)) .

$$\Leftrightarrow\:\mathrm{cos}\:\left({x}−{y}\right)+\mathrm{cos}\:\left({x}+{y}\right)=\frac{\mathrm{5}}{\mathrm{6}}\mathrm{cos}\:{x} \\ $$ $$\Leftrightarrow\:\mathrm{2cos}\:{x}\:\mathrm{cos}\:{y}\:=\:\frac{\mathrm{5}}{\mathrm{6}}\mathrm{cos}\:{x} \\ $$ $$\Leftrightarrow\:\mathrm{cos}\:{x}\:\left[\mathrm{2cos}\:{y}−\frac{\mathrm{5}}{\mathrm{6}}\:\right]=\:\mathrm{0} \\ $$ $$ \:\begin{cases}{\mathrm{cos}\:{y}=\frac{\mathrm{5}}{\mathrm{12}}}\\{\mathrm{sin}\:{y}=−\sqrt{\mathrm{1}−\frac{\mathrm{25}}{\mathrm{144}}}\:=−\frac{\sqrt{\mathrm{119}}}{\mathrm{12}}}\end{cases} \\ $$ $${then}\:\mathrm{sin}\:\mathrm{2}{y}\:=\:\mathrm{2sin}\:{y}\:\mathrm{cos}\:{y}\:=\:−\mathrm{2}\left[\frac{\mathrm{5}\sqrt{\mathrm{119}}}{\mathrm{144}}\:\right] \\ $$ $$=\:−\frac{\mathrm{5}\sqrt{\mathrm{119}}}{\mathrm{72}}\:. \\ $$

Answered by mr W last updated on 07/Apr/21

cos (x−y)+cos (x+y)=(5/6) cos x cos x cos y=(5/(12)) cos x cos x (cos y−(5/(12)))=0 ⇒cos x=0 or ⇒cos y=(5/(12)) with cos x=0: sin x=±1 cos (x−y)=−1+(1/2)cos x sin x sin y=−1 ⇒sin y=−(1/(sin x))=±1 ⇒no solution for y∈(270°,360°) with cos y=(5/(12)): sin y=−((√(12^2 −5^2 ))/(12))=−((√(119))/(12)) sin 2y=−2×(5/(12))×((√(119))/(12))=−((5(√(119)))/(72))

$$\mathrm{cos}\:\left({x}−{y}\right)+\mathrm{cos}\:\left({x}+{y}\right)=\frac{\mathrm{5}}{\mathrm{6}}\:\mathrm{cos}\:{x} \\ $$ $$\mathrm{cos}\:{x}\:\mathrm{cos}\:{y}=\frac{\mathrm{5}}{\mathrm{12}}\:\mathrm{cos}\:{x} \\ $$ $$\mathrm{cos}\:{x}\:\left(\mathrm{cos}\:{y}−\frac{\mathrm{5}}{\mathrm{12}}\right)=\mathrm{0} \\ $$ $$\Rightarrow\mathrm{cos}\:{x}=\mathrm{0}\:{or} \\ $$ $$\Rightarrow\mathrm{cos}\:{y}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$ $$ \\ $$ $${with}\:\mathrm{cos}\:{x}=\mathrm{0}: \\ $$ $$\mathrm{sin}\:{x}=\pm\mathrm{1} \\ $$ $$\mathrm{cos}\:\left({x}−{y}\right)=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:{x} \\ $$ $$\mathrm{sin}\:{x}\:\mathrm{sin}\:{y}=−\mathrm{1} \\ $$ $$\Rightarrow\mathrm{sin}\:{y}=−\frac{\mathrm{1}}{\mathrm{sin}\:{x}}=\pm\mathrm{1} \\ $$ $$\Rightarrow{no}\:{solution}\:{for}\:{y}\in\left(\mathrm{270}°,\mathrm{360}°\right) \\ $$ $$ \\ $$ $${with}\:\mathrm{cos}\:{y}=\frac{\mathrm{5}}{\mathrm{12}}: \\ $$ $$\mathrm{sin}\:{y}=−\frac{\sqrt{\mathrm{12}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }}{\mathrm{12}}=−\frac{\sqrt{\mathrm{119}}}{\mathrm{12}} \\ $$ $$\mathrm{sin}\:\mathrm{2}{y}=−\mathrm{2}×\frac{\mathrm{5}}{\mathrm{12}}×\frac{\sqrt{\mathrm{119}}}{\mathrm{12}}=−\frac{\mathrm{5}\sqrt{\mathrm{119}}}{\mathrm{72}} \\ $$

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