Using the half-angle formula for cosine:
cos(theta/2) = +-sqrt((1+costheta)/2)cos(θ2)=±√1+cosθ2
Therefore...with
theta = arcsin(3/5)θ=arcsin(35)
cos((1/2)arcsin(3/5)) = +-sqrt(((1+cos(arcsin(3/5)))/2)cos((12)arcsin(35))=±
⎷⎛⎜⎝1+cos(arcsin(35))2⎞⎟⎠
Now, set up a right triangle having thetaθ as one of the angles that is not the right angle. Since theta = arcsin(3/5)θ=arcsin(35), we realize that thetaθ is an angle whose sine is 3/535. Since "sine equals opposite over hypotenuse," mark the hypotenuse of the triangle with a 5, and the side opposite angle thetaθ 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)) = 4/5cos(arcsin(35))=45.
Now 1 + 4/5 = 9/51+45=95. Dividing 9/595 by 2 gives 9/10910.
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)cos((12)arcsin(35))=±
⎷⎛⎜⎝1+cos(arcsin(35))2⎞⎟⎠=±√910.
If we wish to rationalize the radical, sqrt(9/10)√910 = 3/sqrt103√10 = (3sqrt10)/103√1010.
So...
cos((1/2)arcsin(3/5)) = (3sqrt10)/10cos((12)arcsin(35))=3√1010.
Since arcsinxarcsinx is in Quadrant 1 when x > 0x>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.
