arcsin(sqrt(3)/2) = pi/3arcsin(√32)=π3 radians = 60^o=60o
So cos(2 arcsin(sqrt(3)/2)) = cos((2 pi)/3) = -1/2cos(2arcsin(√32))=cos(2π3)=−12
Picture an equilateral triangle with sides of length 1. Cut the triangle in two to get two right-angled triangles.
Each of these right-angled triangles has internal angles of 30^o30o, 60^o60o and 90^o90o (or pi/6π6, pi/3π3 and pi/2π2 radians if you prefer).
The shortest side (which is opposite the 30^o30o (pi/6) angle) of one of these triangles has length 1/212 and the hypotenuse is of length 11.
So sin 30^o = sin (pi/6) = cos 60^o = cos (pi/3) = 1/2sin30o=sin(π6)=cos60o=cos(π3)=12.
The remaining side of the right angled triangle can be found using Pythagoras theorem to be
sqrt(1^2 - (1/2)^2) = sqrt(3/4) = sqrt(3)/2√12−(12)2=√34=√32.
Hence sin 60^o = sin (pi/3) = cos 30^o = cos (pi/6) = sqrt(3)/2sin60o=sin(π3)=cos30o=cos(π6)=√32.
So when you see values of sin thetasinθ or cos thetacosθ of the form +-1/2±12 or +-sqrt(3)/2±√32 then you know that thetaθ is a multiple of 30^o30o but not a multiple of 90^o90o. Then look at a rough graph of sin thetasinθ or cos thetacosθ to spot which multiples you need.
