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

$$Given\{\begin{array}{l}\cos (x-y)=-1+\frac{1}{2}\cos x\\ \cos (x+y)=1+\frac{1}{3}\cos x\end{array}$$ $$where270°<y<360°.find$$ $$\sin 2y.$$

Answered by EDWIN88 last updated on 07/Apr/21

$$\iff \cos (x-y)+\cos (x+y)=\frac{5}{6}\cos x$$ $$\iff 2\cos x\cos y=\frac{5}{6}\cos x$$ $$\iff \cos x[2\cos y-\frac{5}{6}]=0$$ $$\{\begin{array}{l}\cos y=\frac{5}{12}\\ \sin y=-\sqrt{1-\frac{25}{144}}=-\frac{\sqrt{119}}{12}\end{array}$$ $$then\sin 2y=2\sin y\cos y=-2\left[\frac{5\sqrt{119}}{144}\right]$$ $$=-\frac{5\sqrt{119}}{72}.$$

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

$$\cos (x-y)+\cos (x+y)=\frac{5}{6}\cos x$$ $$\cos x\cos y=\frac{5}{12}\cos x$$ $$\cos x(\cos y-\frac{5}{12})=0$$ $$\Rightarrow \cos x=0or$$ $$\Rightarrow \cos y=\frac{5}{12}$$ $$$$ $$with\cos x=0:$$ $$\sin x=\pm 1$$ $$\cos (x-y)=-1+\frac{1}{2}\cos x$$ $$\sin x\sin y=-1$$ $$\Rightarrow \sin y=-\frac{1}{\sin x}=\pm 1$$ $$\Rightarrow nosolutionfory\in (270°,360°)$$ $$$$ $$with\cos y=\frac{5}{12}:$$ $$\sin y=-\frac{\sqrt{{12}^2-5^2}}{12}=-\frac{\sqrt{119}}{12}$$ $$\sin 2y=-2\times \frac{5}{12}\times \frac{\sqrt{119}}{12}=-\frac{5\sqrt{119}}{72}$$

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