# Would someone mind explaining to me why I got this question wrong? Thank you so much for your help!

Jun 23, 2018

$- \frac{\sqrt{5}}{5}$

#### Explanation:

$\text{using the "color(blue)"half angle identity for cos}$

•color(white)(x)cos(theta/2)=+-sqrt((1+costheta)/2)

$\text{given "pi< theta <(3pi)/2" then}$

$\frac{\pi}{2} < \frac{\theta}{2} < \frac{3 \pi}{4} \leftarrow \textcolor{b l u e}{\text{second quadrant}}$

$\text{where } \cos \left(\frac{\theta}{2}\right) < 0$

$\cos \left(\frac{\theta}{2}\right) = - \sqrt{\frac{1 - \frac{3}{5}}{2}}$

$\textcolor{w h i t e}{\times \times \times} = - \sqrt{\frac{1}{5}}$

$\textcolor{w h i t e}{\times \times \times} = - \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = - \frac{\sqrt{5}}{5}$

Jun 23, 2018

See explanation.

#### Explanation:

Think about it logically. $\pi < \theta < \frac{3 \pi}{2}$, so $\theta$ lies in the 3rd quadrant. To get to the 3rd quadrant, you take the basic acute angle and add $\pi$ or ${180}^{\circ}$ to it. If you halve an angle between $\pi$ and $\frac{3 \pi}{2}$, it will always be greater than $\frac{\pi}{2}$ because the original angle was greater than $\pi$ to begin with. Now, realize that since $\frac{\theta}{2}$ is in the 2nd quadrant, $\cos \left(\frac{\theta}{2}\right)$ will always be negative because $\cos \theta$ is negative in the 2nd and 3rd quadrant. This is why the correct answer should be $- \frac{\sqrt{5}}{5}$.