Answer
$cos(arcsin\frac{4}{5})=\frac{3}{5}$
Work Step by Step
Let $u=arcsin\frac{4}{5}$. Then:
$sin~u=\frac{4}{5}$
The range $arcsin~x$ is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. So, since $arcsin\frac{4}{5}\gt0$, then $0\lt u\lt\frac{\pi}{2}$ (First Quadrant)
$cos^2u+sin^2u=1$
$cos^2u=1-sin^2u$
$cos^2u=1-(\frac{4}{5})^2$
$cos^2u=1-\frac{16}{25}=\frac{9}{25}$
$cos~u=\frac{3}{5}$ (First Quadrant)
$cos(arcsin\frac{4}{5})=\frac{3}{5}$
