## Trigonometry (11th Edition) Clone

$$\sin\theta=-\frac{2\sqrt5}{5}$$
$$\cos(-\theta)=\frac{\sqrt 5}{5}$$ Remember that $\cos(-x)=\cos x$. So, $$\cos(-\theta)=\cos\theta=\frac{\sqrt 5}{5}$$ We would apply Pythagorean identities: $$\sin^2\theta+\cos^2\theta=1$$ $$\cos^2\theta=1-\sin^2\theta=1-\Big(\frac{\sqrt 5}{5}\Big)^2=1-\frac{5}{25}=1-\frac{1}{5}=\frac{4}{5}$$ $$\sin\theta=\pm\frac{2}{\sqrt5}=\pm\frac{2\sqrt 5}{5}$$ Another information tells us that $\tan\theta\lt0$. Since $\tan \theta=\frac{\sin\theta}{\cos\theta}$, this means that the signs of $\sin\theta$ and $\cos\theta$ must be opposite with each other. So, as $\cos\theta\gt0$, $\sin\theta\lt0$. Therefore, $$\sin\theta=-\frac{2\sqrt5}{5}$$