9. 0/1 points details my notes sprecalc7 7.3.027. use an appropriate half - angle formula to find the exact…

Explanation:

Step1: Rewrite the angle

We know that $\frac{13\pi}{12}=\frac{\frac{13\pi}{6}}{2}$. Also, $\frac{13\pi}{6}=2\pi+\frac{\pi}{6}$, so $\sin(\frac{13\pi}{12})=\sin(\frac{\frac{13\pi}{6}}{2})$, and $\cos(\frac{13\pi}{6})=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$ since cosine has a period of $2\pi$.

Step2: Use the half - angle formula for sine

The half - angle formula for sine is $\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{2}}$. Since $\frac{13\pi}{12}$ is in the second quadrant where sine is positive, and $\alpha=\frac{13\pi}{6}$, we have $\sin(\frac{13\pi}{12})=\sqrt{\frac{1-\cos(\frac{13\pi}{6})}{2}}$.

Step3: Substitute the value of cosine

Substitute $\cos(\frac{13\pi}{6})=\frac{\sqrt{3}}{2}$ into the formula: $\sin(\frac{13\pi}{12})=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2 - \sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}$.

Answer:

$\frac{\sqrt{2 - \sqrt{3}}}{2}$