## Trigonometry (11th Edition) Clone

$$\sin\theta=-\frac{3\sqrt 5}{7}$$
$$\sec\theta=\frac{7}{2}\hspace{1cm}\tan\theta\lt0$$ From a Reciprocal Identity, $$\sec\theta=\frac{1}{\cos\theta}$$ So, $$\cos\theta=\frac{1}{\sec\theta}$$ $$\cos\theta=\frac{1}{\frac{7}{2}}=\frac{2}{7}$$ From a Pythagorean Identity, we also have $$\cos^2\theta+\sin^2\theta=1$$ $$\sin^2\theta=1-\cos^2\theta$$ $$\sin^2\theta=1-(\frac{2}{7})^2=1-\frac{4}{49}=\frac{45}{49}$$ $$\sin\theta=\pm\frac{\sqrt{45}}{7}=\pm\frac{3\sqrt5}{7}$$ From a Quotient Identity: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ As $\tan\theta\lt0$, this identity means that the signs of $\sin\theta$ and $\cos\theta$ must be opposite from each other. Therefore, since $\cos\theta=\frac{2}{7}\gt0$, $\sin\theta\lt0$. $$\sin\theta=-\frac{3\sqrt 5}{7}$$