## Exact Value For Cos 36

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

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,
$CD=x-\frac{x/2}{cos \alpha}$
.

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
$4cos^2 \alpha-4cos \alpha +1=2-2cos \alpha \rightarrow 4cos^2 \alpha-2 cos \alpha-1=0$
$cos \alpha=\frac{--2 \pm \sqrt{20}}{2 \times 4}= \frac{1 + \sqrt{5}}{4}$
$cos \alpha$
$cos \alpha \gt 0$