## Trigonometry (11th Edition) Clone

$$\sin\theta=-\frac{\sqrt{15}}{5}$$
$$\tan\theta=-\frac{\sqrt6}{2}\hspace{1cm}\cos\theta\gt0$$ From a Pythagorean Identity $$\tan^2\theta+1=\sec^2\theta$$ combined with a Reciprocal Identity $$\sec\theta=\frac{1}{\cos\theta}$$ we have $$\tan^2\theta+1=\frac{1}{\cos^2\theta}$$ $$\frac{1}{\cos^2\theta}=(-\frac{\sqrt 6}{2})^2+1=\frac{6}{4}+1=\frac{10}{4}$$ $$\cos^2\theta=\frac{4}{10}$$ $$\cos\theta=\frac{2}{\sqrt{10}}=\frac{2\sqrt{10}}{10}=\frac{\sqrt{10}}{5}$$ Also, from Quotient Identities, $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ which means $$\sin\theta=\tan\theta\times\cos\theta$$ $$\sin\theta=(-\frac{\sqrt6}{2})\times(\frac{\sqrt{10}}{5})=-\frac{\sqrt{60}}{10}=-\frac{2\sqrt{15}}{10}=-\frac{\sqrt{15}}{5}$$