Answer
$ \dfrac{\sqrt{{2-\sqrt2}}}{2}$
Work Step by Step
The half-angle Identity for cosine can be expressed as:
$\cos {(\dfrac{\theta}{2})}=\pm\sqrt{\dfrac{1+\cos{\theta}}{2}}$
Since, $\cos (-\dfrac{3 \pi}{4} )=- \dfrac{\sqrt 2}{2}$
Therefore, $\cos (-\dfrac{3\pi}{8})=\sqrt{\dfrac{1 +\cos(\dfrac{-3 \pi}{4})}{2}} \\= \sqrt{\dfrac{1-\dfrac{\sqrt2}{2}}{2}}\\= \sqrt{\dfrac{2-\sqrt2}{4}}\\ = \dfrac{\sqrt{{2-\sqrt2}}}{2}$
