Call the short side length 1, and the long side length x.

Make an infinite series of triangles which subdivide the main triangle in such a way so that the long side length, x, can be equated to an infinite series of x, specifically x = 1 + 1/x^2 + 1/x^4 + 1/x^6 + ...

This leads to x^3 - x^2 - x = 0 with roots {0, (1+√5)/2, (1-√5)/2}

Since this is geometry and we know from our diagram that x > 1, x=(1+√5)/2

Now go back to the original triangle, bisect the smaller angle to get a right triangle with short side 1/2 and long side (1+√5)/2

We now know that sin(18) = (1/2)/((1+√5)/2) = 1/(1+√5) = (√5-1)/4

Let's double every side of that triangle to get rid of the denominators.

The short side is 1 and the hypoteneuse is (1+√5)

The other leg of this triangle is sqrt( (1+√5)^2 - 1^2 ) = sqrt(5+2√5)

We need cos(18) for the sine half angle formula.

And cos (18) = sqrt(5+2√5) / (1+√5); [ = 0.951056516295153 checked with spreadsheet ]

Sin(9) = sqrt( (1 - cos(18))/2  ); [ = 0.156434465040231 checked with spreadsheet ]

So, sin(9) = sqrt( (1 - (sqrt(5+2√5) / (1+√5)) )/2  )

Or, sin(9) = sqrt( (1 - (sqrt(5+2√5)(√5-1)/4) )/2  )

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Method of subdividing the triangle:

Call vertex A the one which is 36 degrees.

Step 1:  Bisect vertex B, one of the 72 degree angles.  This splits the main triangle into a smaller 36-72-72 one with sides 1,1,1/x and a 36-36-108 triangle.  

Step 2:  Split the 108 degree angle into 72 and 36 so that the 36 degree angle shares a triangle with vertex A.  This creates a 36-72-72 triangle which includes vertex B and a 36-36-108 triangle which includes vertex A.

Step n: Continue splitting 108 degree angles the same way.