# Solution

If $\sin t = 1/4$, and the terminal side of angle $t$ is in quadrant II, find $\cos t$.

Combining the Pythagorean Identity $\cos^2 t + \sin^2 t = 1$ and the given $\sin t = 1/4$, we have $$\cos^2 t + \left( \frac{1}{4} \right)^2 = 1$$ But then if we isolate the $\cos^2 t$ term, we have $$\cos^2 t = 1 - \frac{1}{16} = \frac{15}{16}$$ Hence, $$\cos t = \pm \sqrt{\frac{15}{16}} = \pm \frac{\sqrt{15}}{4}$$
Note that $\cos t$ is negative in quadrant II, so only one of these two values makes any sense $$\cos t = -\frac{\sqrt{15}}{4}$$