# 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 related question: How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$?. Perhaps we have some other questions here, where $\cos\frac{\pi}5$ or $\cos\frac{2\pi}5$ is calculated, but I've only found this one. Apr 12, 2012 at 11:44
• @MartinSleziak: I have an elementary answer for both questions (both parts of this question, and the related question as well, since my solution uses the polynomial $4x^2-2x-1=t^2-t-1$ for $t=2x$, which has roots $t=\frac{1\pm\sqrt5}{2}$. Apr 12, 2012 at 16:18
• I am not sure if this is what Chandrasekar had in mind, because I don't follow his argument, but I have another post which is either the simplest solution yet, or else is a long-winded restatement of Chandrasekar's more elegant but potentially hard to follow solution (if correct, which I cannot judge). Apr 12, 2012 at 23:14
• Does anyone know if there's a geometric proof of this? Given that this can be expressed in terms of the vertices of a dodecagon it seems like there ought to be one somewhere... Apr 12, 2012 at 23:32
• I gave a proof of a more general statement than your "related question" here: mathoverflow.net/questions/16583/a-trigonometry-problem
– user641
Apr 13, 2012 at 10:46

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.)

• Dear Martin, There seems to be a typo in the real part of $x_4$. Regards, Apr 12, 2012 at 10:27
• Thanks @MattE, it should be corrected now. Apr 12, 2012 at 10:34

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.

• Perhaps it should be mentioned that the quadratic equation has two solutions; but only one of them is positive, so there is no doubt which of the two solutions is $\cos\frac\pi5$. Apr 12, 2012 at 10:38
• Chandrasekhar: I think the quadratic equation should be $2x^{2}-x-\frac12=0$ or $4x^2-2x-1=0$. (If I understand correctly, you get the equation from $\cos\frac\pi5-\cos\frac{2\pi}5=\frac12$.) Apr 12, 2012 at 10:51
• @Martin: I have just calculated value of $\cos\frac{\pi}{5}$ using that equation. Why should it be $\frac{1}{2}$. That is what i have to prove.
– user9413
Apr 12, 2012 at 11:04
• How is this a quadratic equation of the form $2x^2-x-1=0$ exactly? It seems to me that your main equation is merely what you get when you put the double angle formula relating $\cos\frac\pi5$ and $\cos\frac{2\pi}5$ into the LHS of what we want to be $\frac12$. What is $x$ in this quadratic equation then? $\left(\cos\frac\pi5-\cos\frac{2\pi}5\right)$? You should make this more explicit so others can follow. Apr 12, 2012 at 19:30
• yes... I don't follow this argument either and I don't think its correct. Seems like you mixed up cos(x) and cos(2x) to be the same value?
– user18862
Apr 13, 2012 at 1:47

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}

• the most important one is \frac{a}{b}. I also like to use \cdot or no multiplication symbol at all.
– user18862
Apr 13, 2012 at 1:37

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. • It very much is a proof. Compare this description with considering a regular pentagon as a series of vectors placed end to end forming a ring. The resultant is zero, from which the components parallel to any side give ... . Feb 1, 2022 at 12:30

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 $0<a<b$.

• This is a very nice solution. Apr 13, 2012 at 5:47

$$\cos(\pi/5)\cos(2\pi/5)=A$$

$$\Longrightarrow\quad A = \frac{\sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)}{\sin(\pi/5)}=\frac{\sin(2\pi/5)\cos(2\pi/5)}{2\sin(\pi/5)}$$

$$A = \frac{\sin(4\pi/5)} {2\cdot 2\cdot\sin(\pi/5)}=\frac{1}{4}$$ since $\sin(4\pi/5)=\sin(\pi/5)$.

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 $x=\frac{1\pm\sqrt5}{4}$.

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}

$$\cos(\pi/5) - \cos(2\pi/5)= {\rm Re}(e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ \overline{e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ e^{-i\frac{\pi}{5}}-e^{-i\frac{2\pi}{5}})=\frac{e^{\frac{-2i\pi}{5}}}{2}( -e^{i\frac{4\pi}{5}}+e^{i\frac{3\pi}{5}}+ e^{i\frac{\pi}{5}}-1)$$

To simplify the computations, let $\omega=e^{\frac{i \pi}{5}}$. Note that $\omega^5=-1$.

Then

$$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\omega^4-\omega^3-\omega+1)= \frac{-1}{2\omega^2}(\omega^4-\omega^3+\omega^2-\omega+1-\omega^2)$$

$$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\frac{\omega^5+1}{\omega+1} -\omega^2)=\frac{-1}{2\omega^2}(0 -\omega^2)=\frac{1}{2}$$

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 $x/2 = 1/4$.

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: $\cos\left(\frac{\pi}5\right)= \cos\left(\frac{2 \pi}5\right) + \frac12$

Proof without words by Lai Wish you like the solution!

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 $\displaystyle c^2=\cos^2x=\frac{1+\cos x}2$

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 $$2\ (\cos\frac\pi5-\cos\frac{2\pi}5)=1.$$

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.

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 $2\sin18^\circ\cos36^\circ=\sin(18+36)^\circ-\sin(36-18)^\circ=\cos ?-\cos ?$

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 $$\cos\left(\frac\pi5\right)-\cos\left(\frac{2\pi}5\right)=\frac12$$

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

$$\cos(\frac{\pi}5)-\cos(\frac{2\pi}5)=\frac{\phi}2-(2(\frac{\phi}2)^2-1)=\frac{\phi-(\phi+1)+2}{2}=\frac12$$ $$\cos(\frac{\pi}5)\;\cos(\frac{2\pi}5)=\frac{\phi}2\times(\frac{\phi^2}2-1)=\frac{\phi}2\times\frac{\phi-1}2=\frac{\phi^2-\phi}4=\frac14$$