## Trigonometry (11th Edition) Clone

$$\sin\theta=-\frac{\sqrt{33}}{6}$$
$$\cos(-\theta)=\frac{\sqrt 3}{6}\hspace{1cm}\cot(\theta)\lt0$$ First, $\cos\theta=\cos(-\theta)=\frac{\sqrt 3}{6}$ Using this Pythagorean Identity, $$\cos^2\theta+\sin^2\theta=1$$ we have, $$\sin^2\theta=1-\cos^2\theta$$ $$\sin^2\theta=1-(\frac{\sqrt 3}{6})^2=1-\frac{3}{36}=\frac{33}{36}$$ Therefore, $$\sin\theta=\pm\frac{\sqrt{33}}{6}$$ We also know that $$\cot\theta\lt0$$ $$\frac{\cos\theta}{\sin\theta}\lt0\hspace{1cm}\text{(Quotient Identities)}$$ That means the signs of $\cos\theta$ and $\sin\theta$ must be opposite from each other. $\cos\theta=\frac{\sqrt 3}{6}\gt0$. So, $\sin\theta\lt0$. That means, $$\sin\theta=-\frac{\sqrt{33}}{6}$$