Exact Value For Cos 36

Consider the diagram below where  
\[\alpha = 36^o\]
.

calculate exact value cos 36

Triangles ABC and CBE are both right angles, have a side in common and a corresponding equal angle  
\[\alpha\]
. They are congruent triangles.
Let  
\[AC=AD=x\]
  then  
\[AE=ED=x/2\]
.
Using trigonometry on triangle  
\[ABE\]
  gives  
\[AB=\frac{x/2}{cos \alpha}\]
  then  
\[CB=\frac{x/2}{cos \alpha}\]
  too.
From triangle  
\[BCD\]
  with  
\[\alpha=36^o\]
,  
\[\angle DBC=180^o-3 \times 36=72^o=2 \alpha\]
.
Then triangle  
\[BCD\]
  is isosceles and  
\[CD=\frac{x/2}{cos \alpha}\]
.
Also,  
\[CD=x-\frac{x/2}{cos \alpha}\]
.

calculate exact value cos 36

Now use the Cosine Rule on triangle  
\[BCD\]
.
\[(x-\frac{x/2}{cos \alpha})^2=(\frac{x/2}{cos \alpha})^2+(\frac{x/2}{cos \alpha})^2-2(\frac{x/2}{cos \alpha})(\frac{x/2}{cos \alpha})^2 cos \alpha\]

Expanding the brackets, Dividing by  
\[x^2\]
  and multiplying by  
\[4cos^2 \alpha\]
  gives
\[4cos^2 \alpha-4cos \alpha +1=2-2cos \alpha \rightarrow 4cos^2 \alpha-2 cos \alpha-1=0\]
The from the quadratic formula  
\[cos \alpha=\frac{--2 \pm \sqrt{20}}{2 \times 4}= \frac{1 + \sqrt{5}}{4}\]
.
We ignore the minus option because this returns a negative value for  
\[cos \alpha\]
  but  
\[cos \alpha \gt 0\]
.