Using the half-angle formula for cosine:
cos(theta/2) = +-sqrt((1+costheta)/2)
Therefore...with
theta = arcsin(3/5)
cos((1/2)arcsin(3/5)) = +-sqrt(((1+cos(arcsin(3/5)))/2)
Now, set up a right triangle having theta as one of the angles that is not the right angle. Since theta = arcsin(3/5), we realize that theta is an angle whose sine is 3/5. Since "sine equals opposite over hypotenuse," mark the hypotenuse of the triangle with a 5, and the side opposite angle theta with a 3.
This is a 3-4-5 triangle. The length of the remaining side is 4.
The cosine of the angle is adjacent over hypotenuse:
cos(arcsin(3/5)) = 4/5.
Now 1 + 4/5 = 9/5. Dividing 9/5 by 2 gives 9/10.
This is the radicand. Back to the half-angle formula,
cos((1/2)arcsin(3/5)) = +-sqrt(((1+cos(arcsin(3/5)))/2)) = +-sqrt(9/10).
If we wish to rationalize the radical, sqrt(9/10) = 3/sqrt10 = (3sqrt10)/10.
So...
cos((1/2)arcsin(3/5)) = (3sqrt10)/10.
Since arcsinx is in Quadrant 1 when x > 0, we need not worry about the =- symbol. We know that the angle is in the first quadrant where cosine is positive. Observe that I removed the =- symbol for that reason.
