How to calculate the difference cos 30 -sin 60, using Pythagorean theorem?

Let ABC be a right angle triangle with A = 90  degrees and
B =30 and  C= 60 degrees . This is possible as angles A+B+C = 90+30+60 =
180.

Let  D be the mod point of BC. Then we can draw a
semicircle with D as centre and radius as CD = DB as radius and the semi circle has to
pass through A, as in a right a right angled triangle, we can always draw a circle with
the mid point on the side opposite to right angle a circle which circumscribe the
triangle.Therefore DB=DA = DC = R.  Now C = 60 degree DC = DB . SO ADC is isosceles.
Therefore, Angle C = angle DAC = 60. So in triangle ADC all angles are 60 . Therefore
the isosceles triangle ABD is an equilateral triangle . Therefore AB =AD = DC = R
,

Also in the right angled triangle ABC, AC = R , BC = 2R .
Therefore by Pythagorus theorem,  AB = sqrt(BC^2-BC^2) = sqrt[(2R)^2-R^2 ] = sqrt(3R^2)
= (sqrt3)R

Therefore in triangle
ABC,

AB  = (sqrt3)R side opposite angle C =60
degree.

AC = R  side opposite to angle B = 30
degree.

BC , the hypotenuse =
2R,

Threfore cos30 = AC/BC = R/(2R) =
1/2

sin60 = AC/BC = R/(2R) =
1/2.

Therefore cos 30 - sin60 = 1/2 - 1/2 =  (1-1)/2 =
0.