Congratulations to lfdahl, MarkFL and kaliprasad for their correct solutions which you may find below.
lfdahl's solution:
[sp] I will make use of the following rules:
\[\sin \alpha + \sin \beta = 2\sin \left ( \frac{\alpha +\beta }{2} \right )\cos \left ( \frac{\alpha - \beta }{2} \right )\;\;\;(1).\]
\[\cos (\alpha \pm \beta ) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \;\;\; (2).\]
\[\left.\begin{matrix} \cos (\pi - \alpha ) = - \cos \alpha \\ \sin (\pi - \alpha ) = \sin \alpha \end{matrix}\right\} \;\;\;(3).\]
Furthermore, since $x+y+z = \pi$, we have from $(3).$: $\sin z = \sin (\pi – (x+y)) = \sin (x+y)$. In the same manner, we get: $\cos z = -\cos (x+y).$
First, I will rewrite the sum $\sin 2x + \sin 2y$:
\[\sin 2x + \sin 2y = 2\sin (x+y)\cos (x-y). \;\;\; (by \;\;1.) \\\\= 2\sin (x+y)\left ( \cos x \cos y + \sin x \sin y \right ) \;\;\;(by \;\;2.) \\\\= 2\sin z\left ( \cos x \cos y + \sin x \sin y \right ) \;\;\;(by \;\;3.) \\\\= 2 \sin x\sin y\sin z +2\sin z\cos x\cos y\]
Next, I rewrite the expression $\sin 2z$:
\[\sin 2z = 2 \sin z \cos z = -2\sin z \cos (x+y)\;\;\;(by \;\; 3.) \\\\ = -2\sin z\left ( \cos x \cos y - \sin x\sin y \right ) (by \;\; 2.) \\\\ = 2 \sin x\sin y\sin z-2\sin z \cos x \cos y\]
Finally, adding the expressions gives the desired result:
\[ \sin 2x + \sin 2y + \sin 2z = \\\\4 \sin x\sin y \sin z + 2\sin z\cos x\cos y - 2\sin z\cos x\cos y \\\\= 4 \sin x\sin y \sin z.\][/sp]
MarkFL's solution:
[sp] Consider that:
$$2z=2\pi-2(x+y)$$
And:
$$\sin(2\pi-\theta)=-\sin(\theta)$$
Which allows us to write:
$$\sin(2x)+\sin(2y)+\sin(2z)=\sin(2x)+\sin(2y)-\sin(2(x+y))$$
Now consider the sum to product identity:
$$\sin(\alpha)+\sin(\beta)=2\sin\left(\frac{\alpha+beta}{2}\right)\cos\left(\frac{\alpha-beta}{2}\right)$$
Which allows us to write:
$$\sin(2x)+\sin(2y)+\sin(2z)=2\sin(x+y)\cos(x-y)-\sin(2(x+y))$$
Now consider the double angle identity for sine:
$$\sin(2\theta)=2\sin(\theta)\cos(\theta)$$
Which allows us to write:
$$\sin(2x)+\sin(2y)+\sin(2z)=2\sin(x+y)\cos(x-y)-2\sin(x+y)\cos(x+y)$$
Factor:
$$\sin(2x)+\sin(2y)+\sin(2z)=2\sin(x+y)\left(\cos(x-y)-\cos(x+y)\right)$$
Now consider the sum to product identity:
$$\cos(\alpha)-\cos(\beta)=-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$$
Which allows us to write:
$$\sin(2x)+\sin(2y)+\sin(2z)=2\sin(x+y)\left(-2\sin(x)\sin(-y)\right)$$
$$\sin(2x)+\sin(2y)+\sin(2z)=4\sin(x+y)\sin(x)\sin(y)$$
Consider:
$$x+y=\pi-z$$
And:
$$\sin(\pi-\theta)=\sin(\theta)$$
Which allows us to write:
$$\sin(2x)+\sin(2y)+\sin(2z)=4\sin(z)\sin(x)\sin(y)$$
$$\sin(2x)+\sin(2y)+\sin(2z)=4\sin(x)\sin(y)\sin(z)\quad\checkmark$$
Shown as desired.[/sp]
kaliprasad's solution:
[sp] We derive certain identities under above constraints to use below
$\sin (x+y) = \sin(180-z)=\sin\,z\cdots(1)$
$\cos(z) = -\cos(180-z) = -\cos(x+y)\cdots(2)$
using $\sin\, A + \sin\, B$ for 1st 2 terms and $\sin\ 2c$ for $3^{rd}$ term
$\sin 2x + \sin 2y + \sin 2z = 2\ sin (x+y) \cos (x-y) + 2 \sin\, z\ cos\, z$
$= 2\sin\, z \cos (x-y) - 2 \sin\, z \cos(x+y)$ (using (1) and (2))
$= 2\sin\, z (\cos (x-y) - \cos (x+ y)$
$= 2\sin\, z ( 2 \sin\, x \sin\, y)$ using (cos A - cos B) identity
$= 4 \sin\, x \sin\, y \sin\, z $,[/sp]
