Find the exact value of sin345^circ using the sum/difference identity. sin345^circ =square (sqrt (6)+sqrt (2))/(4) (-sqrt (6)-sqrt (2))/(4) (sqrt (6)-sqrt (2))/(4) (sqrt (2)-sqrt (6))/(4) 2+sqrt (3) -2-sqrt (3) 2-sqrt (3) sqrt (3)-2

To find the exact value of $\sin 345^{\circ}$ using the sum/difference identity, we can use the identity for the sine of a difference of angles:<br /><br />$\sin (A - B) = \sin A \cos B - \cos A \sin B$<br /><br />Let $A = 360^{\circ}$ and $B = 15^{\circ}$. Then $A - B = 345^{\circ}$.<br /><br />Using the known values of $\sin 360^{\circ} = 0$, $\cos 360^{\circ} = 1$, $\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$, and $\cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$, we have:<br /><br />$\sin 345^{\circ} = \sin (360^{\circ} - 15^{\circ}) = \sin 360^{\circ} \cos 15^{\circ} - \cos 360^{\circ} \sin 15^{\circ} = 0 \cdot \frac{\sqrt{6} + \sqrt{2}}{4} - 1 \cdot \frac{\sqrt{6} - \sqrt{2}}{4} = -\frac{\sqrt{6} - \sqrt{2}}{4}$<br /><br />Therefore, the exact value of $\sin 345^{\circ}$ is $\boxed{\frac{-\sqrt{6} - \sqrt{2}}{4}}$.