## Trigonometry (11th Edition) Clone

$$\tan\frac{\theta}{2}=-\sqrt7$$
$$\tan\theta=\frac{\sqrt7}{3}\hspace{1.5cm}180^\circ\lt\theta\lt270^\circ\hspace{1.5cm}\tan\frac{\theta}{2}=?$$ To find $\tan\frac{\theta}{2}$, we need to apply half-angle identity for tangents: $$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$$ However, both $\sin\theta$ and $\cos\theta$ are not known now, meaning we need to find them. 1) Find $\sin\theta$ and $\cos\theta$ The question states that $180^\circ\lt\theta\lt270^\circ$, meaning the angle $\theta$ lies in quadrant III, where both sines and cosines are negative. Thus, $\sin\theta\lt0$ and $\cos\theta\lt0$. - Pythagorean Identities: $$\sec^2\theta=1+\tan^2\theta=1+\Big(\frac{\sqrt7}{3}\Big)^2=1+\frac{7}{9}=\frac{16}{9}$$ $$\sec\theta=\pm\frac{4}{3}$$ - Reciprocal Identities: $$\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\pm\frac{4}{3}}=\pm\frac{3}{4}$$ But $\cos\theta\lt0$, so $$\cos\theta=-\frac{3}{4}$$ - Quotient Identities: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ $$\sin\theta=\tan\theta\times\cos\theta=\frac{\sqrt7}{3}\times\Big(-\frac{3}{4}\Big)=-\frac{\sqrt7}{4}$$ Therefore, $$\sin\theta=-\frac{\sqrt7}{4}\hspace{2cm}\cos\theta=-\frac{3}{4}$$ 2) Find $\tan\frac{\theta}{2}$ Now we can apply the identity to find $\tan\frac{\theta}{2}$ $$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$$ $$\tan\frac{\theta}{2}=\frac{-\frac{\sqrt7}{4}}{1-\frac{3}{4}}$$ $$\tan\frac{\theta}{2}=\frac{-\frac{\sqrt7}{4}}{\frac{1}{4}}$$ $$\tan\frac{\theta}{2}=-\frac{\sqrt7}{1}=-\sqrt7$$