Trigonometric exercise proved by using Geometric Approach

Exercise:

Show that, $\sin \left( {{18}^{\circ }} \right)=\frac{\sqrt{5}-1}{4}$ using Geometric Construction

Solution: Construct an isosceles triangle $ABC$of vertex $\hat{A}=36$

Hence $AB=BC$ and $\hat{B}=\hat{C}=72$

Now Construct the bisector $\left[ BD \right)$ of angle $A\hat{B}C$

Thus the triangle $ABD$ is an isosceles of vertex $D$

Hence $AD=BD$ thus $D$ equidistant from $A\,\And \,B$

So the height issued from $D$ to $AB$ turned to be the perpendicular bisector of $\left[ AB \right]$

And suppose that $BC=BD=1\,\,\And \,\,AB=x$Hence $BE=EA=\frac{x}{2}$ and $BC=BD=AD=1$

Similarly the height will turned to the perpendicular bisector of $BC$ as $ABC$ is an isosceles

Thus $BF=CF=\frac{1}{2}$ also by congruent triangles( $\vartriangle AFB\,=\vartriangle AFC$ by $S.S.S$) we get

$\left[ AF \right)$ is a bisector of the angle $B\hat{A}C$

Observe that in triangle $EDB$ we have $\cos 36=\frac{x}{2}$ and

in triangle $AFB$ we have $\sin 18=\frac{\frac{1}{2}}{x}=\frac{1}{2x}$

Now by using $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-{{\sin }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta $

Hence $\cos 36=1-2{{\sin }^{2}}18\Leftrightarrow \frac{x}{2}=1-2\frac{1}{4{{x}^{2}}}=1-\frac{1}{2{{x}^{2}}}=\frac{2{{x}^{2}}-1}{2{{x}^{2}}}\Leftrightarrow 2{{x}^{3}}=4{{x}^{2}}-2$

$\Rightarrow 2{{x}^{3}}-4{{x}^{2}}+2=0\Rightarrow 2\left( {{x}^{3}}-2{{x}^{2}}+1 \right)=0\Leftrightarrow {{x}^{3}}-2{{x}^{2}}+1=0$

Now by using rational root theorem that states ${{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+....+{{a}_{1}}x+{{a}_{0}}=0$

 if ${{a}_{n\,}}\,,\,\,{{a}_{0}}\ne 0$ then each rational solution $x$can be written as $x=\frac{p}{q}$ such that $\gcd (p,q)=1$

where $p$ is an integer factors ${{a}_{0}}$ and $q$ is an integer factors of ${{a}_{n}}$

so in our case we have ${{a}_{n}}=1\,\,\,\And \,\,{{a}_{0}}=1$ thus $x=1$ hence $x-1$ is a factor

Write $P\left( x \right)=\left( x-1 \right)\left( a{{x}^{2}}+bx+c \right)={{x}^{3}}-2{{x}^{2}}+1$

$\Rightarrow a{{x}^{3}}+b{{x}^{2}}+cx-a{{x}^{2}}-bx-c=a{{x}^{3}}+{{x}^{2}}\left( b-a \right)+x\left( c-b \right)-c={{x}^{3}}-2{{x}^{2}}+1$

So $a=1\,\,,\,\,b-a=-2\,\,,\,c-b=0\,\,\And \,\,\,-c=1$ hence $\left\{ a,b,c \right\}=\left\{ 1,-1,-1 \right\}$

Thus $p\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}-x-1 \right)=0$

So the roots are $x=1\,\,or\,\,{{x}^{2}}-x+1=0\Leftrightarrow {{x}^{2}}-x-1+\frac{1}{4}-\frac{1}{4}=0$

$\Leftrightarrow {{\left( x-\frac{1}{2} \right)}^{2}}=\frac{5}{4}\Leftrightarrow x-\frac{1}{2}=\frac{\sqrt{5}}{2}\Leftrightarrow x=\frac{1+\sqrt{5}}{2}$

But $\sin 18=\frac{1}{2x}=\frac{1}{2}\left( \frac{2}{1+\sqrt{5}} \right)=\frac{\sqrt{5}-1}{4}$