## Trigonometry (11th Edition) Clone

$$\cot\theta=\frac{\sqrt5}{5}$$
$$\tan\theta=-\frac{\sqrt5}{2}\hspace{1.5cm}90^\circ\lt\theta\lt180^\circ\hspace{1.5cm}\cot\frac{\theta}{2}=?$$ To find $\cot\frac{\theta}{2}$, first we need to find $\tan\frac{\theta}{2}$, which involves using 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 $90^\circ\lt\theta\lt180^\circ$, meaning the angle $\theta$ lies in quadrant II, where sines are positive but cosines are negative. Thus, $\sin\theta\gt0$ and $\cos\theta\lt0$. - Pythagorean Identities: $$\sec^2\theta=1+\tan^2\theta=1+\Big(-\frac{\sqrt5}{2}\Big)^2=1+\frac{5}{4}=\frac{9}{4}$$ $$\sec\theta=\pm\frac{3}{2}$$ - Reciprocal Identities: $$\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\pm\frac{3}{2}}=\pm\frac{2}{3}$$ But $\cos\theta\lt0$, so $$\cos\theta=-\frac{2}{3}$$ - Quotient Identities: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ $$\sin\theta=\tan\theta\times\cos\theta=-\frac{\sqrt5}{2}\times\Big(-\frac{2}{3}\Big)=\frac{\sqrt5}{3}$$ Therefore, $$\sin\theta=\frac{\sqrt5}{3}\hspace{2cm}\cos\theta=-\frac{2}{3}$$ 2) Find $\tan\frac{\theta}{2}$ Now we can apply the identity to find $\tan\frac{\theta}{2}$ $$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$$ $$\tan\frac{\theta}{2}=\frac{\frac{\sqrt5}{3}}{1-\frac{2}{3}}$$ $$\tan\frac{\theta}{2}=\frac{\frac{\sqrt5}{3}}{\frac{1}{3}}$$ $$\tan\frac{\theta}{2}=\frac{\sqrt5}{1}=\sqrt5$$ 3) Find $\cot\frac{\theta}{2}$ - Reciprocal Identities: $$\cot\frac{\theta}{2}=\frac{1}{\tan\frac{\theta}{2}}=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}$$