## Trigonometry (11th Edition) Clone

$$\sin\theta=-\frac{\sqrt{105}}{11}$$
$$\sec\theta=\frac{11}{4}$$ *Step 1: Find $\cos\theta$ according to the identity $$\sec\theta=\frac{1}{\cos\theta}$$ $$\cos\theta=\frac{1}{\sec\theta}$$ Now apply $\sec\theta=\frac{11}{4}$, we have $$\cos\theta=\frac{1}{\frac{11}{4}}=\frac{4}{11}$$ *Step 2: Find $\sin\theta$ according to the identity $$\sin^2\theta=1-\cos^2\theta$$ Apply $\cos\theta=\frac{4}{11}$ $$\sin^2\theta=1-(\frac{4}{11})^2=1-\frac{16}{121}=\frac{105}{121}$$ $$\sin\theta=\pm\frac{\sqrt{105}}{11}$$ As $\cot\theta\lt0$, that means $$\frac{\cos\theta}{\sin\theta}\lt0$$ But we already see that $\cos\theta=\frac{4}{11}\gt0$, so $\sin\theta$ must be $\lt0$ Therefore, $$\sin\theta=-\frac{\sqrt{105}}{11}$$