Determining $\sin(2x)$

Question

Given that

$$\sin (y-x)\cos(x+y)=\dfrac 1 2$$

$$\sin (x+y)\cos (x-y) = \dfrac 1 3 $$

Determine $\sin (2x)$.

As stated in my perspective, the question does not make any sense. We know that the double angle identity for $\sin(2x)$ is given by

$$\sin(2x) = 2\sin\cos$$

Let us try simpiflying the second equation

$$\sin(x+y)-\cos(x-y)=\sin x \cos y+\cos x \sin y-\cos x \cos y-\sin x\sin y= \cos x(\sin y-\cos y)+\sin x(\cos y-\sin y)=\color{blue}{(\cos x-\sin x)(\sin y-\cos y})$$

However, there seems to be nothing useful.

Answer 1

Recall that by Product to sum identity

$$2\sin \theta \cos \varphi = {{\sin(\theta + \varphi) + \sin(\theta - \varphi)} }$$

that is

$$\begin{cases}\sin (y-x)\cos(x+y)=\frac12\sin (2y)+\frac12 \sin (-2x)=\dfrac 1 2\\\sin (x+y)\cos (x-y) = \frac12\sin (2x)+\frac12 \sin (2y)=\dfrac 1 3\end{cases} $$

$$\begin{cases} \sin (2y)-\sin (2x)=1\\ \sin (2x)+\sin (2y)=\dfrac 2 3 \end{cases} $$

and subtracting the first equation from the second we obtain

$$2\sin (2x)=-\frac13 \implies \sin (2x)=-\frac16$$

Answer 2

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

Answer 3

$$ 2x = ( x+y) + (x-y)$$

$$ \sin ( \alpha + \beta ) = \sin (\alpha) \cos (\beta) + \cos (\alpha) \sin (\beta) $$

$$ \sin ( 2x ) = \sin (x+y) \cos (x-y) + \cos (x+y) \sin (x-y)=1/3-1/2 =-1/6$$

Answer 4

You were on the right track but the equations contain the product, not the difference.

\begin{align} \frac12 &= \sin (y-x)\cos(x+y)\\ &=(\sin y\cos x - \sin x \cos y)(\cos x\cos y - \sin x\sin y) \\ &= \sin y\cos y - \cos x \sin x \end{align}

\begin{align} \frac13 &= \sin (x+y)\cos(x-y)\\ &=(\sin x\cos y + \sin y \cos x)(\cos x\cos y + \sin x\sin y) \\ &= \sin y\cos y + \cos x \sin x \end{align}

So subtracting them gives

$$\sin 2x = 2\cos x\sin x = ( \sin y\cos y + \cos x \sin x) - ( \sin y\cos y - \cos x \sin x) = \frac13 - \frac12 = -\frac16$$