Fina the exactialie of sin((19pi )/(12)) using the sum/differentle identity. sin((19x)/(12))=square (sqrt (6)+sqrt (2))/(4) (-sqrt (6)-sqrt (2))/(4) sqrt (6)-sqrt (2) sqrt (2)-sqrt (6)

To find the value of $\sin(\frac{19\pi}{12})$ using the sum/difference identity, we can break it down into two angles whose sine values we know.<br /><br />We can express $\frac{19\pi}{12}$ as the sum of $\frac{16\pi}{12}$ and $\frac{3\pi}{12}$, which simplifies to $\frac{4\pi}{3}$ and $\frac{\pi}{4}$ respectively.<br /><br />Using the sum identity for sine, we have:<br /><br />$\sin(\frac{19\pi}{12}) = \sin(\frac{4\pi}{3} + \frac{\pi}{4})$<br /><br />Using the sum identity for sine, we have:<br /><br />$\sin(\frac{4\pi}{3} + \frac{\pi}{4}) = \sin(\frac{4\pi}{3})\cos(\frac{\pi}{4}) + \cos(\frac{4\pi}{3})\sin(\frac{\pi}{4})$<br /><br />Now, we can substitute the known values of $\sin(\frac{4\pi}{3})$, $\cos(\frac{\pi}{4})$, $\cos(\frac{4\pi}{3})$, and $\sin(\frac{\pi}{4})$:<br /><br />$\sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2}$, $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, $\cos(\frac{4\pi}{3}) = -\frac{1}{2}$, and $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$<br /><br />Substituting these values, we get:<br /><br />$\sin(\frac{19\pi}{12}) = (-\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2}) + (-\frac{1}{2})(\frac{\sqrt{2}}{2})$<br /><br />Simplifying, we get:<br /><br />$\sin(\frac{19\pi}{12}) = -\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$<br /><br />Therefore, the correct answer is $\frac{-\sqrt{6}-\sqrt{2}}{4}$.