Answer
$$\cos\frac{\theta}{2}=-\frac{\sqrt5}{5}$$
Work Step by Step
$$\sin\theta=-\frac{4}{5}\hspace{1.5cm}180^\circ\lt \theta\lt270^\circ\hspace{1.5cm}\cos\frac{\theta}{2}=?$$
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.
$$\cos\frac{\theta}{2}=-\sqrt{\frac{1+\cos\theta}{2}}$$
$$\cos\frac{\theta}{2}=-\sqrt{\frac{1-\frac{3}{5}}{2}}$$
$$\cos\frac{\theta}{2}=-\sqrt{\frac{\frac{2}{5}}{2}}$$
$$\cos\frac{\theta}{2}=-\sqrt{\frac{2}{5\times2}}$$
$$\cos\frac{\theta}{2}=-\sqrt{\frac{1}{5}}$$
$$\cos\frac{\theta}{2}=-\frac{\sqrt5}{5}$$
