## Trigonometry (11th Edition) Clone

$$\cos\frac{\theta}{2}=-\frac{\sqrt5}{5}$$
$$\sin\theta=-\frac{4}{5}\hspace{1.5cm}180^\circ\lt \theta\lt270^\circ\hspace{1.5cm}\cos\frac{\theta}{2}=?$$ Apply the half-angle identity for cosine $$\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos \theta}{2}}$$ As $180^\circ\lt\theta\lt270^\circ$, we deduce that $90^\circ\lt\frac{\theta}{2}\lt135^\circ$, the area of quadrant II. In quadrant II, cosine is negative, thus $\cos\frac{\theta}{2}\lt0$. So we need to select the negative square root. $$\cos\frac{\theta}{2}=-\sqrt{\frac{1+\cos \theta}{2}}$$ However, we don't have the value of $\cos\theta$ now. So we need to find $\cos\theta$. 1) Find $\cos\theta$ - Pythagorean Identities: $$\cos^2\theta=1-\sin^2\theta=1-\Big(-\frac{4}{5}\Big)^2=1-\frac{16}{25}=\frac{9}{25}$$ $$\cos\theta=\pm\frac{3}{5}$$ $180^\circ\lt\theta\lt270^\circ$ denotes that the position of $\theta$ is in quadrant III, where cosines are negative. Thus, $\cos\theta\lt0$. $$\cos\theta=-\frac{3}{5}$$ 2) Find $\cos\frac{\theta}{2}$ Now we can find $\cos\frac{\theta}{2}$ according to the mentioned identity. $$\cos\frac{\theta}{2}=-\sqrt{\frac{1+\cos\theta}{2}}$$ $$\cos\frac{\theta}{2}=-\sqrt{\frac{1-\frac{3}{5}}{2}}$$ $$\cos\frac{\theta}{2}=-\sqrt{\frac{\frac{2}{5}}{2}}$$ $$\cos\frac{\theta}{2}=-\sqrt{\frac{2}{5\times2}}$$ $$\cos\frac{\theta}{2}=-\sqrt{\frac{1}{5}}$$ $$\cos\frac{\theta}{2}=-\frac{\sqrt5}{5}$$