1. **Problem:** Calculate $\sin \alpha$ given $\cos \alpha = \frac{12}{37}$ and $270^\circ \leq \alpha \leq 360^\circ$. 2. **Formula and rules:** - Use the Pythagorean identity: $$\sin^2 \alpha + \cos^2 \alpha = 1$$ - Since $\cos \alpha$ is positive and $\alpha$ is in the fourth quadrant ($270^\circ$ to $360^\circ$), $\sin \alpha$ is negative. 3. **Calculate $\sin \alpha$:** $$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{12}{37}\right)^2 = 1 - \frac{144}{1369} = \frac{1369 - 144}{1369} = \frac{1225}{1369}$$ 4. **Simplify:** $$\sin \alpha = -\sqrt{\frac{1225}{1369}} = -\frac{35}{37}$$ 5. **Answer:** $$\sin \alpha = -\frac{35}{37}$$ This negative sign is because $\alpha$ is in the fourth quadrant where sine is negative.