Diagonals of a regular Pentagon

Here is a method to find out exact value of cos72 degrees.My idea is come from C.F.Gauss.

let cos(a) =2pi/5=x

sin(a)=-sin(2kPi-a) if k=1

sin(a)=-sin(2pi-a)=-sin(5a-a)=-sin4a=-2sin(2a)cos(2a)=-4sin（a)cos(a)cos(2a)

because -4sin(a)  equal 0,so divide the equation by -4sin(a)

simplify we have -1/4=cos(a)cos(2a)

                        -1/4=1/2*[cos(a+2a)+cos(a-2a)]

                        -1/2=cos3a+cos(-a）

because cos(3a)=cos(-3a)=cos(2kpi-3a),when k=1,we have cos(3a)=cos(5a-3a)=cos(2a),and cos(-a)=cos(a)

                    so -1/2=cos3a+cos(-a) $$\Rightarrow$$ -1/2=cos(2a)+cos(a)

remember,I setted up cos(a)=x at the start 

so cos(2a)=cos^2(a)-sin^2(a)=cos^2(a)-[1-cos^2(a)]=2cos^2(a)-1=2x^2-1

change the equation -1/2=cos(2a)+cos(a) to -1/2=2x^2-1+x$$\Rightarrow$$-1=4x^2+2x-2$$\Rightarrow$$0=4x^2+2x-1

ax^2+bx+c=0 a=4 b=2 c=-1 let x1 and x2 are the two soloutionof the equation

x1=[-b+(b^2-4ac)^1/2]/2a=[-2+(2^2-4*4(-1)^1/2]/(4*2)=[-2+20^(1/2)]/8=[-1+5^(1/2)]/4

x2=[-b-(b^2-4ac)^1/2]/2a=[-2-(2^2-4*4(-1)^1/2]/(4*2)=[-2-20^(1/2)]/8=[-1-5^(1/2)]/4

Beacause x=cos(a)=cos 72degrees>0,so cos 72 degrees=-1/4+5^(1/2)/4

What Is the length of the side of this pentagon?

Your answer is in the wording of question 1... But that's because I used the word "size" instead of "lenght".

The relationship between a side length (s) and a diagonal (d) for a regular pentagon is: sin(18°) = s/(2d)

Rearrange and plug in the numbers as required.

.

Yes, but there's another way to find the length of the diagonal. An easier way...

That's the hard way to do it.

The easy way: call d the length of the diagonal, and s the length of the side.

The ratio $${\frac{{\mathtt{d}}}{{\mathtt{c}}}}$$ is equal to φ for any regular Pentagon.

Yes，φ is the simple ratio between the size of the diagonal and the the size of the sides of any regular pentagon.

So,what?Do you know the background information of φ?Do you know understand why the φ is ratio betweeenthe diagonal and the sides?Can you prove it?DO you know how to use φ in a math problem or ideal life? Might be not.

And I want to say there is no always a shortcut to solve a problem.Every formula and shortcut is come from 

ancient peoples through thinking and hard working to figure it out,and reserve it for us.They might figured out the ratio between the diagonal and of any regular pentagon just like I did,or may be in a easier way.

That is it.

Now, I want contiune my work.

In my last post ,we got cos 72 degrees =(-1+$${\sqrt{}}$$5)/4

Any interior angle of regular pentagon=(5-2)*180/5 degrees=108 degrees

cos 108 degrees =-cos（$${\frac{{\mathtt{\pi}}}{{\mathtt{2}}}}$$-72 degrees) =-cos72 degrees=(1-$${\sqrt{}}$$5)/4

In any pentagon ,every sides are equal and every diagonals are equal.

Lets say we are fing the ratio between AB and AC

According to law of cosine 

In triangle ABC,we have AB^2=AC^2+BC^2-2*AC*BC*cosC, rearrange we have

                     2*AC*BC*cosC=AC^2+BC^2-AB^2

                    and because AC and BC are the sides of regular pentagon,so

                    2*AC^2*cosC=2*AC^2-AB^2

Both side divide by 2*AC^2,then we get

                               cosC=1-AB^2/2*AC^2

                AB^2/2*AC^2=1-cosC

                AB^2/2*AC^2=1-(-cos72 degrees)

                AB^2/2*AC^2=1+cos72 degrees

               AB^2/2*AC^2=1+(-1+5)/4

$${\frac{\left({\mathtt{sqrt5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}$$ 

Now,I am going to rearrange the Alan's equation and check

sin18 degrees=s/2d

2*sin 18 degrees=s/d

1/(2*sin18 degrees)=d/s

d/s=1/(2*sin18 degrees)=1.6180339887496196

s = the size of the side of the Pentagon

d = the size of the diagonal

the central angle of the pentagon is $$\small{\text{$\dfrac{360\ensurement{^{\circ}}}{5}=72 \ensurement{^{\circ}}
$}}$$

the incribed angle is half of the central angle is $$\small{\text{$\dfrac{72\ensurement{^{\circ}}}{2}=36 \ensurement{^{\circ}}
$}}$$

so

$$\cos{(36\ensurement{^{\circ}} ) = \dfrac{\frac{d}{2} } {s }
\qquad \rm{or} \qquad \dfrac{ d } {s } = 2\cdot \cos{(36\ensurement{^{\circ}} ) \qquad | \qquad \cos{(36\ensurement{^{\circ}} ) =\dfrac{1}{4}\cdot(1+\sqrt{5})$$

$$\\\dfrac{ d } {s } = 2\cdot \dfrac{1}{4}\cdot(1+\sqrt{5})\\\\\\
\dfrac{ d } {s } = \dfrac{1}{2}\cdot(1+\sqrt{5})=\varphi$$