Prove that $\cos 72=\frac{\phi}{2}$

Question

anonymous · Answer

where $\phi$ is the $$\color{brown}{Golden  \space Ratio}$$

ganeshie8 · Answer

$$\large   x = e^{i \dfrac{360}{5}t }   , ~~0\leq   t\leq 4$$
are the five roots of polynomial $\large x^5-1$

ganeshie8 · Answer

we can work out the roots by factoring $\large x^5-1 $ and comparing it with the five roots of unity complete work is here http://www.integraltec.com/math/math.php?f=cosine72.html

anonymous · Answer

YES YES ACTUALLY THE REAL ORIGINAL PROBLEM IS FINDING THE ROOTS OF UNITY
$$z_k=\cos \frac{k\theta }{n}+i \sin \frac{k\theta}{n} $$

$$0\le n\le k-1$$
 for k=5

anonymous · Answer

i was wondering if i can compute cos 72 without a calculator,somehow

ganeshie8 · Answer

Ahh I see :) 
the algebraic method shown in that site is pretty neat -
they are dividing x-1 and find finding the remainign four roots by factoring

anonymous · Answer

very NEAT solution,thank you @ganeshie8 apperiated

ganeshie8 · Answer

here is another way to work it using simple trig http://mathforum.org/library/drmath/view/53992.html this is also looking good to me if we just want the value of cos(72) :)

ganeshie8 · Answer

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