Trig Functions E7E9Be

1. **State the problem:** We are given $$\sin \theta = -\frac{3}{4}$$ and $$\cos \theta < 0$$. We need to find $$\cos \theta$$, $$\tan \theta$$, $$\csc \theta$$, $$\sec \theta$$, and $$\cot \theta$$. 2. **Identify the quadrant:** Since $$\sin \theta$$ is negative and $$\cos \theta$$ is negative, $$\theta$$ lies in the third quadrant. 3. **Use the Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$. 4. **Calculate $$\cos \theta$$:** $$\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(-\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16}$$ Since $$\cos \theta < 0$$ in the third quadrant, $$\cos \theta = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4}$$. 5. **Calculate $$\tan \theta$$:** $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{3}{4}}{-\frac{\sqrt{7}}{4}} = \frac{-3}{4} \times \frac{4}{-\sqrt{7}} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$$ (rationalized). 6. **Calculate reciprocal functions:** - $$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{3}{4}} = -\frac{4}{3}$$ - $$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{\sqrt{7}}{4}} = -\frac{4}{\sqrt{7}} = -\frac{4\sqrt{7}}{7}$$ (rationalized) - $$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{3\sqrt{7}}{7}} = \frac{7}{3\sqrt{7}} = \frac{\sqrt{7}}{3}$$ (rationalized). **Final answers:** $$\cos \theta = -\frac{\sqrt{7}}{4}, \quad \tan \theta = \frac{3\sqrt{7}}{7}, \quad \csc \theta = -\frac{4}{3}, \quad \sec \theta = -\frac{4\sqrt{7}}{7}, \quad \cot \theta = \frac{\sqrt{7}}{3}.$$