## Trigonometry (10th Edition)

$\sin\theta=\frac{-3}{5}$
1.We get this information from the sine value $x=4$ $r=5$ 2. We can use x and r to find y (we need to remember that sign of y depends on quadrant, since it ends in quadrant IV, it must be negative) $r=\sqrt {y^{2}+x^{2}}$ $y=-\sqrt {r^{2}-x^{2}}=-\sqrt {5^{2}-4^{2}}=-\sqrt {25-16}=-\sqrt {9}=-3$ 4. Then insert values, $\sin\theta=\frac{-3}{5}$