Evaluate $\cos 36^\circ - \cos 72^\circ$ without the aid of a calculator I have a quick question about a difficult trigonometric functions problem that I have been assigned. The problem is as follows: Evaluate $$\cos36° - \cos72°$$ without the aid of a calculator. In terms of my attempts at the problem, I have converted $\cos 36°$ into 2$\cos^{2}18°$ - 1, and I have converted $\cos72°$ into 1 - 2$\sin^{2}36°$. By the property that  $\sin x$ = $\cos(90° - x)$, the latter expression becomes 1 - 2$\cos^{2}54°$. Unfortunately, I've hit a roadblock and don't really know what to do from here. Would, perhaps, adding the two determined expressions yield anything of use? Thanks for all advice. 
EDIT: Or, perhaps, would it be wiser to convert $\cos72°$ into 2$\cos^{2}36°$ - 1 and to convert $\cos36°$ into 1 - 2$\sin^{2}18°$?
 A: Inscribe a regular pentagon inside the unit circle with one vertex at $(1,0)$, the other four vertices are located at 
$$(\cos 72^\circ, \pm \sin 72^\circ)
\quad\text{ and }\quad (\cos 144^\circ, \pm \sin 144^\circ ) = (-\cos 36^\circ, \pm \sin 36^\circ)$$
By symmetry, the center of mass of the pentagon is located at the origin. Looking at the $x$-component alone, one find
$$1 + 2\cos72^\circ - 2\cos 36^\circ = 0 \quad\implies\quad \cos 36^\circ - \cos 72^\circ = \frac12$$
A: First write it as $$\cos  36°-\cos  72°=2\sin { \frac { 72°-36° }{ 2 }  } \sin { \frac { 36°+72° }{ 2 }  } =2\sin { 18°\sin { 54° }  } =2\sin { 18° } \cos { 36° } $$
then let calculate $\sin { 18° } $ 
$$\cos { 36° } =\sin { 54° } \\ \cos { 2\cdot 18°= } \sin { 3\cdot 18° } \\ \theta =18°\\ \cos { 2\theta  } =\sin { 3\theta  } \\ 1-2\sin ^{ 2 }{ \theta  } =3\sin { \theta -4\sin ^{ 3 }{ \theta  }  } \\ 4\sin ^{ 3 }{ \theta -2\sin ^{ 2 }{ \theta  } -3\sin { \theta  } +1 } =0\\ \left( \sin { \theta  } -1 \right) \left( 4\sin ^{ 2 }{ \theta +2\sin { \theta  } -1 }  \right) =0$$ 
from here we get $\sin { 18°=\frac { \sqrt { 5 } -1 }{ 4 }  } $
so

$$\cos  36°-\cos  72°=2\sin { \frac { 72°-36° }{ 2 }  } \sin { \frac { 36°+72° }{ 2 }  } =2\sin { 18°\sin { 54° }  } =2\sin { 18° } \cos { 36° } =\\ =2\sin { 18°\left( 1-2\sin ^{ 2 }{ 18° }  \right)  } =2\sin { 18°-4\sin ^{ 3 }{ 18° }  } =2\left( \frac { \sqrt { 5 } -1 }{ 4 }  \right) -4{ \left( \frac { \sqrt { 5 } -1 }{ 4 }  \right)  }^{ 3 }\\ $$

A: You can compute $\cos72°$ with a bit of geometry or some algebra. Note that $\alpha=72°=360°/5$, so $x=\cos\alpha+i\sin\alpha$ is the fifth root of $1$ with the smaller argument. Since $x^4+x^3+x^2+x+1=0$, we also have
$$
x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0
$$
that can also be rewritten as
$$
\left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-1=0
$$
that means
$$
4\cos\alpha^2+2\cos\alpha-1=0
$$
and so
$$
\cos72°=\cos\alpha=\frac{-1+\sqrt{5}}{4}
$$
Then
$$
\cos36°=\sqrt{\frac{1+\cos72°}{2}}=
\sqrt{\frac{2\sqrt{5}+6}{16}}=\frac{\sqrt{5}+1}{4}
$$
