Draw the Triangle

To find $\cos\left(\arcsin\left(\frac{3}{5}\right)\right)$, give the inner angle a name: θ = arcsin(3/5). Then $\sin\left(\theta\right) = \frac{3}{5}$, and since 3/5 is positive, θ is between 0° and 90°: it is an angle of a right triangle with opposite side 3 and hypotenuse 5.

Pythagoras gives the third side: ${3}^{2}+{b}^{2} = {5}^{2}$, so b = 4. Now read the outer function off the triangle: $\cos\left(\theta\right) = \frac{4}{5}$, so $\cos\left(\arcsin\left(\frac{3}{5}\right)\right) = \frac{4}{5}$. No calculator, and the angle itself is never needed.