Answer
$$\sin\theta=\frac{\sqrt{15}}{4}$$
$$\sec\theta=-4$$
$$\csc\theta=\frac{4\sqrt{15}}{15}$$
$$\tan\theta=-\sqrt{15}$$
$$\cot\theta=-\frac{\sqrt{15}}{15}$$
Work Step by Step
$$\cos\theta=-\frac{1}{4}\hspace{1.5cm}\sin\theta\gt0$$
1) Find $\sin\theta$
- Pythagorean Identities:
$$\sin^2\theta=1-\cos^2\theta=1-(-\frac{1}{4})^2=1-\frac{1}{16}=\frac{15}{16}$$
$$\sin\theta=\pm\frac{\sqrt{15}}{4}$$
As $\sin\theta\gt0$, $$\sin\theta=\frac{\sqrt{15}}{4}$$
2) Find $\sec\theta$ and $\csc\theta$
- Reciprocal Identities:
$$\sec\theta=\frac{1}{\cos\theta}=\frac{1}{-\frac{1}{4}}=-4$$
$$\csc\theta=\frac{1}{\sin\theta}=\frac{1}{\frac{\sqrt{15}}{4}}=\frac{4}{\sqrt{15}}=\frac{4\sqrt{15}}{15}$$
3) Find $\tan\theta$ and $\cot\theta$
- Quotient Identities:
$$\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{-\frac{1}{4}}{\frac{\sqrt{15}}{4}}=-\frac{1}{\sqrt{15}}=-\frac{\sqrt{15}}{15}$$
- Reciprocal Identities:
$$\tan\theta=\frac{1}{\cot\theta}=\frac{1}{-\frac{\sqrt{15}}{15}}=-\frac{15}{\sqrt{15}}=-\sqrt{15}$$
