How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
 A: Note that: $\cos{2x} = \cos^{2}{x} - \sin^{2}{x} = 2\:\cos^{2}{x} - 1$. Therefore you have $\cos \frac{2\pi}{5} = 2\:\cos^{2}\frac{\pi}{5} - 1$
Now,
\begin{align*}
\cos\frac{\pi}{5} - \cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - 2\: \cos^{2}\frac{\pi}{5}+1
\end{align*}
This is a quadratic equation of the form $2 x^{2} - x -1 =0$ and solving this will give you the value of $\cos\frac{\pi}{5}$ from which you can find the above value which you need.
A: For the first equality.
\begin{split} \cos(\pi/5) - \cos(2\pi/5) &=\cos(3\pi/10-\pi/10) - \cos(3\pi/10+\pi/10)\\
&=2\sin(\pi/10)\sin(3\pi/10)=2\sin(\pi/10)\cos(\pi/5)\\
&=\frac{2\sin(\pi/10)\cos(\pi/10)\cos(\pi/5)}{\cos(\pi/10)}\\
&=\frac{\sin(\pi/5)\cos(\pi/5)}{\cos(\pi/10)}\\
&=\frac{\sin(2\pi/5)}{2\cos(\pi/10)}=\frac 12. \end{split}
Here 's the latter equality:
\begin{split}
\cos(\pi/5)\cos(2\pi/5) &= \frac{\sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)}{\sin(\pi/5)}\\ 
&=\frac{\sin(2\pi/5)\cos(2\pi/5)}{2\sin(\pi/5)}\\
&=\frac{ \sin(4\pi/5)}{4\sin(\pi/5)}\\
&=\frac 14.\end{split}
A: I was not able (yet) to follow Chandrasekar's solution, but noticed this while trying to understand the argument (how it could possibly lead to the solution, or how exactly he arrives at $2x^2-x-1$ for $x=\cos\frac{\pi}{5}$, which to me seems non-obvious and even a fallacious deduction from his equations and prose -- apologies if I am just being dense)...perhaps it is what Chandrasekar meant all along, but in any case, it does seem to be the most elementary solution available.
Apply the double angle formula
$\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$
to $\theta=\frac{\pi}{5}$ and $\frac{2\pi}{5}$,
with $a=\cos\frac{\pi}{5}$ and $b=\frac{2\pi}{5}$
for convenience, recalling also that
$\cos(\pi\pm\theta)=-\cos\theta$:
$$
b=\cos\frac{2\pi}{5}=2\,\cos^2\frac{\pi}{5}-1=2a^2-1
$$
$$
-a=\cos\frac{4\pi}{5}=2\,\cos^2\frac{2\pi}{5}-1=2b^2-1
$$
Next, subtracting the equations
$$
\matrix{
2a^2=1+b\\
2b^2=1-a}
$$
we get
$$
\eqalign{
2\left(a^2-b^2\right)&=b+a\\
2\left(a+b\right)\left(a-b\right)&=b+a\\
2\left(a-b\right)&=1\\
a-b&=\frac12\,.
}
$$
Furthermore, multiplying, we get
$$
4(ab)^2=(1+b)(1-a)=1+(b-a)-ab=1+\left(-\tfrac12\right)-ab
$$
giving us the quadratic equation
$$4(ab)^2+(ab)-\tfrac12=0$$
$$8(ab)^2+2(ab)-1=0$$
$$\left(4ab-1\right)\left(2ab+1\right)=0$$
so that $ab=\frac14$ or $-\frac12$,
from which we can choose the former
since we know that $0&lta&ltb$.
A: $$\cos(\pi/5)\cos(2\pi/5)=A$$
$$\Longrightarrow\quad A = \frac{\sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)}{\sin(\pi/5)}=\frac{\sin(2\pi/5)\cos(2\pi/5)}{2\sin(\pi/5)}$$
$$A = \frac{\sin(4\pi/5)} {2\cdot 2\cdot\sin(\pi/5)}=\frac{1}{4}$$ 
since $\sin(4\pi/5)=\sin(\pi/5)$.
A: Here is a geometric way to show that from one of my friends. I don't know if it counts as a proof but it looks beautiful.

A: The complex roots of $x^5-1$ are:
$$
\begin{align}
x_1&=1\\
x_2&=\cos\frac{2\pi}5+i\sin\frac{2\pi}5\\
x_3&=-\cos\frac{\pi}5+i\sin\frac{\pi}5\\
x_4&=-\cos\frac{\pi}5-i\sin\frac{\pi}5\\
x_5&=\cos\frac{2\pi}5-i\sin\frac{2\pi}5
\end{align}
$$
using Vieta's formulas you get
$$0=x_1+x_2+\dots+x_5=1+2\left(\cos\frac{2\pi}5-\cos\frac\pi5\right)=0,$$
which yields your first equation. 

From now on, let $\varphi=\frac{\pi}5$ (for brevity).
We know that
$\cos2\varphi-\cos\varphi+\frac12=(2\cos^2\varphi-1)-\cos\varphi+\frac12=2\cos^2\varphi-\cos\varphi-\frac12=0$, i.e.
$$2\cos^2\varphi-\cos\varphi=\frac12.$$
Now from $\cos2\varphi=\cos\varphi-\frac12$ you get
$$\cos\varphi\cos2\varphi=\cos^2\varphi-\frac{\cos\varphi}2=\frac{2\cos^2\varphi-\cos\varphi}2=\frac14.$$
(Or, as suggested in Chandrasekhar's answer, from $2\cos^2\varphi-\cos\varphi=\frac12$ you can find the value of $\cos\varphi$ by solving the quadratic equation and taking the positive root. Once you know $\cos\varphi$, you can compute $\cos2\varphi$ and many other things. If you try it this way, you can check your result e.g. here.)
A: If $a=\cos\frac{\pi}5$ and $b=\cos\frac{2\pi}5=2a^2-1$ (by the double-angle identity for cosine), we want to show that
$$
0=(x-a)(x+b)=x^2-(a-b)x-ab=x^2-\tfrac12x-\tfrac14,
$$
i.e., that $a$ and $-b$ are roots of $4x^2-2x-1$.
Note also that $4x^2+2x-1=4(x+a)(x-b)$ has roots $-a$ and $b$.
What is special about the numbers $\{a,b,-b,-a\}$?
They are the $x$-coordinates (real parts) of the nonreal $10$th roots of unity,
$$
x+iy=e^{\pm\pi i\cdot\frac{k}5}
\qquad
\text{for}
\qquad
k\not\equiv0\pmod5.
$$
But these satisfy the equation $(x+iy)^5=(-1)^k=\pm1$.
Taking the imaginary part, we have
$$
\eqalign{
0 &= \Im\left[(x+iy)^5\right] \\
  &= \Im\left[x^5+5x^4(iy)+10x^3(iy)^2+10x^2(iy)^3+5x(iy)^4+(iy)^5\right] \\
  &= \Im\left[iy\left(5x^4+10x^2(iy)^2+(iy)^4\right)\right] \\
  &= y\left[5x^4-10x^2y^2+y^4\right] \\
\implies
0 &= \left[5x^4-10x^2(1-x^2)+(1-x^2)^2\right] \\
  &= 16x^4-12x^2+1 \\
  &= 16x^4-8x^2+1 ~-~4x^2 \\
  &= \left( 4x^2-1 \right) - \left( 2x \right)^2 \\
  &= \left( 4x^2+2x-1\right)\left( 4x^2-2x-1\right) \,.
}
$$
So far, we have shown this has roots $\{\pm a,\pm b\}$.
It only remains to show that
$\{-a,b\}$ are the roots of the first factor
and that the desired pair $\{a,-b\}$
splits the second quadratic factor into linear terms
(as we wrote at the outset).
Perhaps there is a clever way to infer the correct linear order
of this set of roots by noticing that $a>\cos\frac\pi4>b$ so that $a^2>\frac12>b^2$ and combining this with our identity $b=2a^2-1$ above.
I propose in stead to use that $a>b$ but then to notice
that our quadratic factors are in fact parabolas,
with roots that are very easy to order on the $x$-axis.
If we let $t=2x$ (which preserves order), then we have
$$
t^2\pm t-1 = t\,(t\pm1)-1
$$
which both pass through $(0,-1)$
and alternately pass through $(\mp1,-1)$,
shown respectively in red and blue below;
the result follows.

It's only a small extra step to note that the roots of our factor,
$t^2-t-1$, are $t=\frac{1\pm\sqrt5}{2}$, or $x=\frac{1\pm\sqrt5}{4}$.
A: $$\cos(\pi/5) - \cos(2\pi/5)= {\rm Re}(e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ \overline{e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ e^{-i\frac{\pi}{5}}-e^{-i\frac{2\pi}{5}})=\frac{e^{\frac{-2i\pi}{5}}}{2}( -e^{i\frac{4\pi}{5}}+e^{i\frac{3\pi}{5}}+ e^{i\frac{\pi}{5}}-1)$$
To simplify the computations, let $\omega=e^{\frac{i \pi}{5}}$. Note that $\omega^5=-1$.
Then
$$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\omega^4-\omega^3-\omega+1)= \frac{-1}{2\omega^2}(\omega^4-\omega^3+\omega^2-\omega+1-\omega^2)$$
$$\cos(\pi/5) - \cos(2\pi/5)=  \frac{-1}{2\omega^2}(\frac{\omega^5+1}{\omega+1} -\omega^2)=\frac{-1}{2\omega^2}(0 -\omega^2)=\frac{1}{2}$$
A: Here is a solution using only elementary trigonometric relationships.  We do not need complex variables or advance knowledge of the cosine values.
Let $x = \cos (\pi/5) - \cos (2\pi/5)$.  Compute $x^2$:
$$x^2 = \cos^2 (\pi/5) - 2 \cos (\pi/5)\cos (2\pi/5) + \cos^2 (2\pi/5)$$
Use the standard trigonometric identities:
\begin{split} \cos^2 (\pi/5) + \cos^2 (2\pi/5) &= \frac{1 + \cos (2\pi/5)}{2}
    + \frac{1 + \cos (4\pi/5)}{2} \\
&= 1 - \frac 12(\cos (\pi/5) - \cos (2\pi/5))\\
&= 1 - \frac x2\end{split}
and 
\begin{split} \cos (\pi/5) \cos (2\pi/5) &= \frac 12(\cos (3\pi/5) + \cos(-\pi/5))\\
&= \frac 12 (\cos (\pi/5) - \cos (2\pi/5))\\
&= \frac x2 
\end{split}
Then $x^2 = (1 - \frac x2 ) - 2(\frac x2)$, hence
$$2x^2 - 3x -2 = 0.$$
Thus $x = \cos (\pi/5) - \cos (2\pi/5) =1/2$, the only positive root of the above quadratic equation (clearly $x$ is positive).  Also the product of the cosines is $x/2 = 1/4$.
A: There is a neat trick:
\begin{align}
\cos 36^\circ - \cos72 ^\circ &=
(\cos^2 36^\circ - \cos^2 72 ^\circ) / (\cos 36^\circ + \cos72 ^\circ) \\
&= (\cos^2 36^\circ - \cos^2 72 ^\circ) / ((1 - 2\sin^2 18^\circ) + (2\cos^2 36^\circ - 1)) \\
&= (\cos^2 36^\circ - \cos^2 72 ^\circ) / 2(\cos^2 36^\circ - \sin^2 18^\circ) \\
&= (\cos^2 36^\circ - \cos^2 72 ^\circ) / 2(\cos^2 36^\circ - \cos^2 72^\circ) \\
&= 1/2
\end{align}
A: MUCH SIMPLER: 
Just construct a regular pentagon (it might help to construct it within a circle in order to appreciate better the angles and symmetry). So let $ABCDE$ be a regular pentagon. Draw the line $BE$. Then let $M$ and $N$ be the perpendicular projection of $C$ and $D$ on $BE$, and $P$ the perpendicular projection of $A$ on $BE$. 
Because of symmetry one has: $NP= \frac{CD}2 = \frac{AB}2$. Now $BP = AB \cos\left(\frac{\pi}5\right) = BN+NP$. And $BN = BC \cos\left(\frac{2 \pi}5\right) = AB \cos \left( \frac{2\pi}5\right)$. Hence: $\cos\left(\frac{\pi}5\right)= \cos\left(\frac{2 \pi}5\right) + \frac12$
A: If $\displaystyle x=\frac\pi5,5x=\pi\implies3x=\pi-2x\implies\cos3x=\cos(\pi-2x)\  \   \  \ (1)$
But as $\displaystyle\cos(\pi-2x)=-\cos2x$, (1)$\displaystyle\implies\cos3x+\cos2x=0\ \ \ \ (2)$
If $\displaystyle\cos x=c(2)$ becomes $4c^3+2c^2-3c-1=0\  \ \ \ (3)$
Again  from $(1)\implies3x=2n\pi\pm(\pi-2x)$ 
$\displaystyle'-'\implies x=(2n-1)\pi\implies\cos x=-1$
$\displaystyle'+'\implies x=\frac{(2n+1)\pi}5$ where $n=0,1,2,3,4$
Observe that $\displaystyle\cos\frac{(2\cdot2+1)\pi}5=\cos\pi=-1$(which converges with $'-'$ case )
and $\displaystyle\cos\frac{(2\cdot0+1)\pi}5=\cos\frac{(2\cdot4+1)\pi}5$ as $\displaystyle\cos(2\pi-y)=\cos y$
Similarly,  $\displaystyle\cos\frac{(2\cdot1+1)\pi}5=\cos\frac{(2\cdot3+1)\pi}5$
So, the quadratic equation whose roots are $\displaystyle\cos\frac{\pi}5=\cos\frac{9\pi}5,\cos\frac{3\pi}5=\cos\frac{7\pi}5$ will be $\displaystyle\frac{4c^3+2c^2-3c-1}{c+1}=0\iff4c^2-2c-1=0\  \ \ \ (4)$
Now use the fact $\displaystyle c^2=\cos^2x=\frac{1+\cos x}2$
A: We can factorize $z^5+1$, using the roots of $-1$,
 $$z^5+1=(z+1)(z-e^{i\pi/5})(z-e^{i2\pi/5})(z-e^{i3\pi/5})(z-e^{i4\pi/5}).$$
The coefficient of degree $4$ is the sum of the roots, so that
$$1-e^{i\pi/5}-e^{i2\pi/5}-e^{i3\pi/5}-e^{i4\pi/5}=0.$$
Taking the real part and noticing that there are complementary angles
$$2\ (\cos\frac\pi5-\cos\frac{2\pi}5)=1.$$
A: Just like my other answer,
If $\displaystyle x=\frac\pi5,5x=\pi\implies3x=\pi-2x\implies\sin3x=\sin(\pi-2x)\  \   \  \ (1)$
But as $\displaystyle\sin(\pi-y)=\sin y, (1)\displaystyle\implies\sin3x=\sin2x\ \ \ \ (2)$
which $\displaystyle\implies3x=n\pi+(-1)^n2x$
For even $n(=2m)$(say),  $\displaystyle x=2m\pi\implies\sin x=0$
For odd $n(=2m+1)$(say),  $\displaystyle x=\frac{(2m+1)\pi}5$ where $m=0,1,2,3,4$
Observe that for $\displaystyle m=2, \sin x=\sin\pi=0$
Again from $(2),2\sin x\cos x=3\sin x-4\sin^3x$ 
If  $\displaystyle\sin x\ne0, 2\cos x=3-4\sin^2x=3-4(1-\sin^2x)\iff4\cos^2x-2\cos x-1=0$
whose roots are $\displaystyle\cos\frac{(2\cdot0+1)\pi}5=\cos\frac{(2\cdot4+1)\pi}5$ and $\displaystyle\cos\frac{(2\cdot1+1)\pi}5=\cos\frac{(2\cdot3+1)\pi}5$ (Reason for equality has been explained in my other answer)
Hope you can take it home from here.
A: Put $A=18^\circ,36^\circ$ in $$\sin2A=2\sin A\cos A$$ and use the fact $$\sin72^\circ=\cos18^\circ$$ to find  $$1=4\sin18^\circ\cos36^\circ$$
Now $2\sin18^\circ\cos36^\circ=\sin(18+36)^\circ-\sin(36-18)^\circ=\cos ?-\cos ?$
A: The imaginary parts of $\left(\cos\left(\frac{k\pi}5\right)+i\sin\left(\frac{k\pi}5\right)\right)^5=(-1)^k$ give
$$
\sin^5\left(\frac{k\pi}5\right)-10\sin^3\left(\frac{k\pi}5\right)\cos^2\left(\frac{k\pi}5\right)+5\sin\left(\frac{k\pi}5\right)\cos^4\left(\frac{k\pi}5\right)=0
$$
Dividing by $\sin\left(\frac{k\pi}5\right)$ and substituting $\sin^2\left(\frac{k\pi}5\right)=1-\cos^2\left(\frac{k\pi}5\right)$, we get
$$
16\cos^4\left(\frac{k\pi}5\right)-12\cos^2\left(\frac{k\pi}5\right)+1=0
$$
which, using the quadratic formula, has the solutions
$$
\pm\frac{1\pm\sqrt5}4
$$
Thus, since $\cos(x)$ is decreasing from $1$ to $-1$ for $x\in\left[0,\pi\right]$,
$$
\begin{align}
\cos\left(\frac\pi5\right)&=\frac{1+\sqrt5}4\\
\cos\left(\frac{2\pi}5\right)&=\frac{-1+\sqrt5}4\\
\cos\left(\frac{3\pi}5\right)&=\frac{1-\sqrt5}4\\
\cos\left(\frac{4\pi}5\right)&=\frac{-1-\sqrt5}4\\
\end{align}
$$
From which we get
$$
\cos\left(\frac\pi5\right)-\cos\left(\frac{2\pi}5\right)=\frac12
$$
A: Note that $\cos(\frac{\pi}5)=\frac{1+\sqrt5}4=\frac{\phi}2$ where $\phi$ is Golden ratio . also note that $\phi^2=\phi+1$, we use this in the following calculations
$$\cos(\frac{\pi}5)-\cos(\frac{2\pi}5)=\frac{\phi}2-(2(\frac{\phi}2)^2-1)=\frac{\phi-(\phi+1)+2}{2}=\frac12$$
$$\cos(\frac{\pi}5)\;\cos(\frac{2\pi}5)=\frac{\phi}2\times(\frac{\phi^2}2-1)=\frac{\phi}2\times\frac{\phi-1}2=\frac{\phi^2-\phi}4=\frac14$$
A: $$***\textrm{  Proof without words by Lai}*** $$

Wish you like the solution!
