Berasiniz1

2022-01-28

Given that
$\mathrm{sin}\left(y-x\right)\mathrm{cos}\left(x+y\right)=\frac{1}{2}$
$\mathrm{sin}\left(x+y\right)\mathrm{cos}\left(x-y\right)=\frac{1}{3}$
Determining $\mathrm{sin}\left(2x\right)$
As stated in my perspective, the question does not make any sense. We know that the double angle identity for $\mathrm{sin}2x$ is given by

Let us try simpiflying the second equation
$\mathrm{sin}\left(x+y\right)-\mathrm{cos}\left(x-y\right)=\mathrm{sin}x\mathrm{cos}y+\mathrm{cos}x\mathrm{sin}y-\mathrm{cos}x\mathrm{cos}y-\mathrm{sin}x\mathrm{sin}y=\mathrm{cos}x\left(\mathrm{sin}y-\mathrm{cos}y\right)+\mathrm{sin}x\left(\mathrm{cos}y-\mathrm{sin}y\right)=\left(\mathrm{cos}x-\mathrm{sin}x\right)\left(\mathrm{sin}y-\mathrm{cos}y\right)$

Gordon Stephens

Expert

Hint
$\mathrm{sin}2x=\mathrm{sin}\left(\left(x+y\right)+\left(x-y\right)\right)=\mathrm{sin}\left(x+y\right)\mathrm{cos}\left(x-y\right)+\mathrm{cos}\left(x+y\right)\mathrm{sin}\left(x-y\right)$
We know the value of the first summand. For the second use the fact that sine is odd.
$\left(\mathrm{sin}-x\right)=-\mathrm{sin}\left(x\right)$
so that
$\mathrm{sin}\left(y-x\right)=-\mathrm{sin}\left(x-y\right)$

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