Approximation of $\pi$
$\sin\frac{60}{2^x}(3(2^x))=\pi$
Polygons inscribed in a circle of a diameter of 1 unit:Let there be an equilateral triangle inscribed in a circle and the measure of the sides of the triangle is a measure of an angle of sin of 60°, as the measures of the angles of the triangles decrease by a multiple of 2 the sides of the triangles increase by a multiple of 2. From 3 sides to 6 sides to 12 sides to 24 etc... The sides of the polygons are a measure of angles with isosceles triangles in each polygons and in Isosceles triangle you only need one angle to find the lenght of the sides of the triangles since the sides correspond to the measure of an angle.
The following statement hold true for isosceles triangles:
$\cos C=1-((\cos A)^2 + (\cos B)^2)$
1)The apex of an equilateral triangle which is called also Isosceles triangle inscribed with in the circle of a diameter of 1 unit is 60° and the base facing the apex is sin of 60°. 180° divided by 60° is equal to 3 as well, which are the 3 sides of the triangle.2)For an hexagon with 6 equal angles of 120°(the internal angles of a hexagon) and one of the angle being sin of 120° which is the apex facing the base and the side of the equilateral triangle with the hexagon vertex having angles of 30° and 30° and apex of 120°. 120°+30°+30°=180° and 180÷6=30° representing 6 equal sides.3)next is a polygon with 12 equal sides and 180°÷12=15°. 15°+15°+150°=180°.In this case the apex is 150° and sin of 150° is the base of the triangle which is the side of the hexagon equal to 0.5 unit of lenght.4)next level is a polygon with 24 equal sides. 180°÷24=7.5°7.5°+7.5°+165°=180° and sin of 165° is the apex with sin of 15° being the base and the side of the polygon with 12 sides.
$\sin B\cos C+\sin C\cos B=\sin A$
$\sin B\sin C-\cos C\cos B=\cos A$
$\sin A\cos C+\sin C\cos A=\sin B$
$\sin A\sin C-\cos C\cos A=\cos B$
$\sin A\cos B+\sin B\cos A=\sin C$
$\sin A\sin B-\cos A\cos B=\cos C$
Length of the sides of the triangles:
$b\times \cos C+c\times \cos B=a$
$a\times \cos C+c\times \cos A=b$
$a\times \cos B+b\times \cos A=c$
Step by step starting with:
$\sin (60)=\sqrt {0.75}$
$\sqrt {(1-\sin (60)^2)}= \cos (60)$
$\sqrt {\frac {1-\cos (60)}{2}}=\sin (30)$
$\sqrt {(1-\sin (30)^2)}= \cos (30)$
$\sqrt {\frac {1-\cos (30)}{2}}=\sin (15)$
$\sqrt {(1-\sin (15)^2)}= \cos (15)$
$\sqrt {\frac {1-\cos (15)}{2}}=\sin (7.5)$
$\sqrt {(1-\sin (7.5)^2)}= \cos (7.5)$
$\sqrt {\frac {1-\cos (7.5)}{2}}=\sin (3.75)$
$\sqrt {(1-\sin (3.75)^2)}= \cos (3.75)$
$\sqrt {\frac {1-\cos (3.75)}{2}}=\sin (1.875)$
$\sqrt {(1-\sin (1.875)^2)}= \cos (1.875)$
$\sqrt {\frac {1-\cos (1.875)}{2}}=\sin (0.9375)$
$\sqrt {(1-\sin (0.9375)^2)}= \cos (0.9375)$
$\frac {180}{60}=3$
$\frac {180}{30}=6$
$\frac {180}{15}=12$
$\frac {180}{7.5}=24$
$\frac {180}{3.75}=48$
$\frac {180}{1.875}=96$
$\frac {180}{0.9375}=192$
$\frac {180}{0.468750}=384$
