# Solution

Find the value of each of the following:

1. $\sin \frac{5\pi}{4}$

2. $\sin \frac{17\pi}{6}$

3. $\cos \, (-\frac{\pi}{3})$

4. $\sin \, ( -\frac{5\pi}{2})$

1. In standard position, the terminal side of $\displaystyle{\frac{5\pi}{4}}$ is in quadrant III with reference angle $\displaystyle{\frac{\pi}{4}}$, and $\displaystyle{\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}}$

In quadrant III, sine values (i.e., $y$-coordinates) are negative.

Thus, $\displaystyle{\sin \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}}$.

2. In standard position, the terminal side of $\displaystyle{\frac{17\pi}{6}}$ is in quadrant II with reference angle $\displaystyle{\frac{\pi}{6}}$, and $\displaystyle{\sin \frac{\pi}{6} = \frac{1}{2}}$

In quadrant II, sine values (i.e., $y$-coordinates) are positive.

Thus, $\displaystyle{\sin \frac{17\pi}{6} = \frac{1}{2}}$.

3. In standard position, the terminal side of $\displaystyle{-\frac{\pi}{3}}$ is in quadrant IV with reference angle $\displaystyle{\frac{\pi}{3}}$, and $\displaystyle{\cos \frac{\pi}{3} = \frac{1}{2}}$

In quadrant III, cosine values (i.e., $y$-coordinates) are positive.

Thus, $\displaystyle{\cos \left( -\frac{\pi}{3} \right) = \frac{1}{2}}$.

4. In standard position, the terminal side of $\displaystyle{-\frac{5\pi}{2}}$ is on the negative $y$-axis, giving a sine value (i.e., $y$-coordinate) of negative one.

Thus, $\displaystyle{\sin \left( -\frac{5\pi}{2} \right) = -1}$.