How to calculate the difference cos 30 -sin 60, using Pythagorean theorem?
The theorem of 30 degrees angle states that in a right angle triangle, where the lengthes of the cathetus are b and c and the hypothenuse is a, the length of the opposite cathetus, to the angle of 30, is half from hypothenuse.
If b is the opposite cathetus to 30 angle,namely B angle, that means that it's length is b=a/2. In this way, we can find the other cathetus' length, using Pythagorean theorem.
a^2=b^2 + c^2
a^2 = a^2/4 + c^2
a^2 - a^2/4 = c^2
3*a^2/4 = c^2
sin 60 = opposite cathetus/hypotenuse
sin 60= c/a
After simplifying, we'll get:
sin 60= sqrt3/2
cos 30 = adjacent cathetus/hypotenuse
cos 30 = c/a
cos 30 = sqrt3/2
cos 30 -sin 60=sqrt3/2 -sqrt3/2=0
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.