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Sine of 36 Degrees

Date: 11/18/2001 at 21:07:38
From: Steven Kilby
Subject: Trig


   Methods of Computing Trig Functions   

Dr. Rick wrote that Ptolemy geometrically calculated the sine of 36 
degrees using the construction of a regular pentagon. How would this 

Date: 11/19/2001 at 16:11:22
From: Doctor Rick
Subject: Re: Trig

Hi, Steven.

Draw a regular pentagon. Its vertex angles are 180*3/5 = 108. Draw the 
diagonals. The triangle formed by one diagonal and two sides of the 
pentagon has angles of 108, 36 and 36 degrees. The triangle formed by 
one side and parts of the two closest diagonals has the same angles, 
so it is similar to the other triangle. The larger triangle is also 
divided into two of the smaller triangles and another isosceles 
triangle. Call the sides of the pentagon x, and the equal sides of the 
smaller isosceles triangle y. Thus:

                 x   /  /  \  \   x
                  /    /y  y\    \
              /       /      \       \
          A     y   D          E   y      C

You can show that triangle BCD is isosceles, so that DC = x. Then the 
similarity of ABC and ADB gives this proportion:

  x/y = (x+y)/x

Rearrange to get a quadratic equation in y:

  x^2 = y(x+y)
  x^2 = xy + y^2
  y^2 + xy - x^2 = 0

Using the quadratic formula,

  y = (-x +or- sqrt(x^2 + 4*1*x^2))/2
    = x(-1 +or- sqrt(5))/2

The negative solution is spurious since y is a length, so

  y/x = (sqrt(5)-1)/2

The base of the big isosceles triangle is x+y:

  x+y = x(1+y/x)
      = x(sqrt(5)+1)/2

Now, draw a perpendicular from B to AC. It bisects AC, so we have a 
right triangle with hypotenuse x and leg (x+y)/2. Therefore the cosine 
of angle BAC is 

  cos(BAC) = (x+y)/(2x)
           = (sqrt(5)+1)/4

Angle BAC is 36 degrees. The sine of 36 degrees can now be found using 
the Pythagorean theorem. 

There is probably a neater way to derive this, but it's what I come up 
with on the fly. Our Dr. Math Archives can tell you more about 
pentagons, their construction, and the ratio (x+y)/x, which happens to 
be the Golden Ratio.

- Doctor Rick, The Math Forum   
Associated Topics:
High School Geometry
High School History/Biography
High School Triangles and Other Polygons
High School Trigonometry

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