Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to prove the following statement.

$$1+\cos{2\pi\over5}+\cos{4\pi\over5}+\cos{6\pi\over5}+\cos{8\pi\over5}=0$$

I have tried to use the sum of angles formula for cosine, but didn't get to a point where I'd be able to show that it is equal to $0$.

share|cite|improve this question
27  
Proving only with a calculator would be harder ;) – ypercubeᵀᴹ Feb 17 at 18:38
2  
Mosquito-nuking: it can be shown that $T_5(x)-\frac1{16}$ has the zeroes $\cos\frac{2\pi k}{5},\;k=0\dots 4$. Since the coefficient of $x^4$ in this polynomial is $0$, the sum in the OP is $0$, by Vieta. – J. M. Feb 18 at 2:33
    
    
Would converting to exponential form help? – Mehrdad Feb 18 at 19:09

11 Answers 11

Let $O = (0,0)$ and $A_i = (\cos 2\pi i/5,\sin 2\pi i/5)$. Then $A_0A_1A_2A_3A_4$ is a regular pentagon with vertices on the unit circle. The sum you've written is the $x$-coordinate of the vector $u = \overrightarrow{OA_0} + \dots \overrightarrow{OA_4}$. If you apply a rotation centred at $O$ with angle $2\pi/5$, the pentagon remains invariant. Therefore $u$ doesn't change when rotated by this angle. That shows that $u = 0$.

share|cite|improve this answer
    
This is the most elegant solution. No need to invoke complex numbers, or really any formulas at all. This is pure geometry. – Steven Gubkin Feb 17 at 23:10
4  
@CiaPan The only vector that is unchanged under a rotation (other than the identity) is the zero vector. – David Feb 18 at 16:58
1  
@CiaPan well, more precisely, rotating the pentagon doesn't just change the order of the terms - it transforms each term into something else. That something else happens to be the next term from the original (pre-rotation) sum, which is why it looks like it's just changing the order. This should be clearer if you imagine doing the rotation on something that's not symmetric; for example pick only two of the terms, $\overrightarrow{OA_0} + \overrightarrow{OA_1}$. When you rotate that, you get $\overrightarrow{OA_1} + \overrightarrow{OA_2}$. Not the same. – David Z Feb 18 at 18:35
1  
@DavidZ More precisely, rotating the pentagon by the given angle $2\pi/5$ transforms each term into the next one term of the LHS sum. So it essentially changes nothing on the LHS. What's important here is what happens on the RHS – and I could not see it until this reply by David. – CiaPan Feb 18 at 21:29
1  
@David You're right, I can see that now. Thank you. – CiaPan Feb 18 at 21:29

Using complex exponential: $$ 1+e^{\frac{2\pi i}{5}}+e^{\frac{4\pi i}{5}}+e^{\frac{6\pi i}{5}}+e^{\frac{8\pi i}{5}}=\frac{(e^{\frac{2\pi i}{5}})^5 -1}{e^{\frac{2\pi i}{5}}-1}=0 $$ and its real part is $0$.

share|cite|improve this answer

Note that $e^{2i\pi/5}=\omega_5$ is the fifth root of unity. Now $$\omega_5+\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5=x$$ so \begin{align} \omega_5x&=\omega_5(\omega_5+\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5)\\ &=\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5+\omega_5^6\\ &=\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5+\omega_5\\ &=x \end{align} so $\omega_5x=x$, and so $x\neq 0$ would imply $\omega_5=1$ which is false. So $x=0$. Now use $$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$ Now take $\theta=\frac{2k\pi}5$ for $k=1,2,3,4,5$ to get $\omega_5^k$. So we get: \begin{align} 0&=\Re(0)\\ &=\Re(\omega_5+\omega_5^2+\omega_5^3+\omega_5^4+\omega_5^5)\\ &=\Re(e^{2i\pi/5}+e^{4i\pi/5}+e^{6i\pi/5}+e^{8i\pi/5}+e^{10i\pi/5})\\ &=\cos(\frac{2\pi}5)+\cos(\frac{4\pi}5)+\cos(\frac{6\pi}5)+\cos(\frac{8\pi}5)+\cos(\frac{10\pi}5)\\ &=\cos(\frac{2\pi}5)+\cos(\frac{4\pi}5)+\cos(\frac{6\pi}5)+\cos(\frac{8\pi}5)+1 \end{align}

share|cite|improve this answer
    
I presume lines 3 and 4 are meant to be understood as equivalent since \omega_5^6\ = \omega_5\ by periodic formula. – CandiedOrange Feb 18 at 10:01
    
@CandiedOrange, yes, that's the step I took. We use $\omega_5^5=1$ – vrugtehagel Feb 18 at 10:05

Let's try this -

Using $$2\cos a \sin b = \sin(a+b) - \sin(a- b)$$ we get:

$$2\cos(2π/5)\sin(2π/5)=\sin(4π/5 )- \sin(0)$$

$$2\cos(4π/5)\sin(2π/5)=\sin(6π/5)-\sin(2π/5)$$

$$2\cos(6π/5)\sin(2π/5)=\sin(8π/5) - \sin(4π/5)$$

$$2\cos(8π/5)\sin(2π/5)=\sin(10π/5) -\sin(6π/5)$ $

$$2\cos(10π/5)\sin(2π/5)=\sin(12π/5) - \sin(8π/5) $$

Adding: \begin{align} &2\sin(2π/5)\{\cos(10π/5)+\cos(2π/5)+\cos(4π/5)+\cos(6π/5)+\cos(8π/5)\}\\ &=2\sin(2π/5)\{1+\cos(2π/5)+\cos(4π/5)+\cos(6π/5)+\cos(8π/5)\} \\ &=\sin(12π/5) - \sin(2π/5)\\ &=2\sin(10π/5)\cos(14π/5)\\&=0\qquad [\;\because \sin a - \sin b = 2\sin(a-b)\cos(a+b)] \end{align}

$$1+\cos(2π/5)+\cos(4π/5)+ \cos(6π/5)+\cos(8π/5)=0$$

share|cite|improve this answer
    
I used it by grouping the last term with the second and the onea in the middle. I disnt manage to do something – Razvan Paraschiv Feb 17 at 16:28
    
@RazvanParaschiv Check my method. I edited my answer. – Win Vineeth Feb 18 at 0:02

Draw a regular pentagon ABCDE with all sides one unit long. Treat these sides as vectors from A to B, then B to C, etc. From the "head to tail" rule for resultants the resultant of all five vectors in the cycle is zero.

Define an "x-axis" along any one of the edge vectors. Work out the components of all five vectors along this axis. Add them up and match to the zero resultant identified above.

Done. And it works for any regular polygon.

share|cite|improve this answer

Here is one additional proof, which although elegant, is certainly overkill. (The easiest proof in my opinion is the ones given above relating the sum to the real part of the sum of the fifth roots of unity).

We know that $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$

We have $1+\cos(2\pi/5)+\cos(4\pi/5)+\cos(6\pi/5)+\cos(8\pi/5)+= \sum\limits_{k=0}^4 \cos(k\cdot 2\pi/5)$

$=\sum\limits_{k=0}^4\frac{e^{k2\pi i/5}+e^{-k2\pi i/5}}{2} = \frac{1}{2}\left(\sum\limits_{k=0}^4(e^{2\pi i/5})^k\right)+\frac{1}{2}\left(\sum\limits_{k=0}^4(e^{-2\pi i/5})^k\right)$

Recognizing each as a geometric sum, we simplify as

$=\frac{1}{2}\left(\frac{1-e^{5\cdot 2\pi i/5}}{1-e^{2\pi i/5}}\right)+\frac{1}{2}\left(\frac{1-e^{-5\cdot 2\pi i/5}}{1-e^{-2\pi i/5}}\right)$

Since $e^{2\pi i} =e^{-2\pi i}= 1$ the numerators of both fractions above simplify to be zero and the sum is zero.

More generally, following the same outline of proof, one can prove the more general statement that:

$$1+\cos(\theta)+\cos(2\theta)+\dots+\cos(n\theta) = \frac{1}{2}+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}$$

or equivalently as

$$=\frac{\sin(\frac{(n+1)\theta}{2})}{\sin(\frac{\theta}{2})}\cos(\frac{n\theta}{2})$$

share|cite|improve this answer

$$z_k:=\cos\left(k\frac{2\pi}5\right)$$ for $k=0,1,2,3,4$ are distinct solutions of $$\cos(5t)=1.$$

As

$$\cos(5t)=16\cos^5(t)- 20\cos^3(t)+ 5\cos(t),$$

they are the roots of

$$z^5-\frac54z^3+\frac5{16}z-\frac1{16}=(z-z_0)(z-z_1)(z-z_2)(z-z_3)(z-z_4).$$

Hence by identifying the coefficients we can deduce that

$$-z_0z_1z_2z_3z_4=-\frac1{16},\\ z_0z_1z_2z_3+z_0z_1z_2z_4+z_0z_1z_3z_4+z_0z_2z_3z_4+z_1z_2z_3z_4=\frac5{16},\\ -z_0z_1z_2-z_0z_1z_3-z_0z_1z_4-z_0z_2z_3-z_0z_2z_4-z_0z_3z_4-z_1z_2z_3-z_1z_2z_4-z_1z_3z_4-z_2z_3z_4=0,\\ z_0z_1+z_0z_2+z_0z_3+z_0z_4+z_1z_2+z_1z_3+z_1z_4+z_2z_3+z_2z_4+z_3z_4=\frac54,\\ -z_0-z_1-z_2-z_3-z_4=0.$$

(Only the last result was requested.)

share|cite|improve this answer
    
Hey it seems promising...but one thing i didnt understand....for k=0,1,2,3,4 are distinct solutions of cos(5t)=1. could you please explain that part? – ZOZ Feb 19 at 9:41
    
@SanchayanDutta: just because $\cos\left(5k\dfrac{2\pi}5\right)=\cos(2k\pi)=1$. – Yves Daoust Feb 19 at 10:06

We may use formulas, mainly double angle formula and Chebyshev Polynomials, to simplify this problem:

$$\cos(2x)=2\cos^2(x)-1$$

$$\cos(4x)=2\cos^2(2x)-1=8\cos^4(x)-8\cos^2(x)+1$$

$$\cos(6x)=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1$$

$$\cos(8x)=64\cos^8(x)-128\cos^6(x)+22\cos^4(x)-16\cos^2(x)+1$$

Use $x=\frac{\pi}5$.

$$1+\cos{2\pi\over5}+\cos{4\pi\over5}+\cos{6\pi\over5}+\cos{8\pi\over5}$$

$$=1+2\cos^2(x)-1+8\cos^4(x)-8\cos^2(x)+1+32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1+64\cos^8(x)-128\cos^6(x)+22\cos^4(x)-16\cos^2(x)+1$$

$$=1-4\cos^2(x)-18\cos^4(x)-96\cos^6(x)+64\cos^8(x)$$

We know that $\cos(\frac{\pi}5)=\frac{1+\sqrt{5}}4$. Plug this in:

$$=1-4\left(\frac{1+\sqrt{5}}4\right)^2-18\left(\frac{1+\sqrt{5}}4\right)^4-96\left(\frac{1+\sqrt{5}}4\right)^6+64\left(\frac{1+\sqrt{5}}4\right)^8$$

Avoiding the messy stuff, I'll leave it to the reader to determine if the above is equal to zero.

share|cite|improve this answer

This is the real part of $1 + \exp{2\pi i/5} + \exp{4\pi i/5} + \exp{6\pi i/5} + \exp{8\pi i/ 5}$. Let $z = \exp {2\pi i/5}$; then this is $1+z+z^2+z^3+z^4 = (z^5-1)/(z-1)$. Since $z^5 = 1$ and $z \not = 1$, you have $(z^5-1)/(z-1) = 0$.

(You simultaneously get the identity $0 + \sin 2\pi/5 + \sin 4\pi/5 + \sin 6\pi/5 + \sin 8\pi/5 = 0$ as well, by taking imaginary parts instead of real parts.)

share|cite|improve this answer
    
The sin identity is more obvious because $\sin 2\pi/5 + \sin 8\pi/5 = 0$ etc. – Neil Feb 19 at 11:03

Place a regular pentagon in a plane with Cartesian coordinates so that one of its sides is parallel to $X$ axis. Notice that when you make each side a vector, consecutive vectors are at angles $$\frac {0\pi} 5, \frac{2\pi} 5, \ldots,\frac{8\pi}5$$ to the $X$ axis, so their cosines are $X$-componenets of respective vectors.
And the chain of polygon's sides is closed, so the vectors' sum is zero, consequently the sum of their $X$-components is zero, too – hence your identity.

share|cite|improve this answer

The given terms are projections of unit radii of a regular pentagon on x-axis. A sum of cyclic vectors or static equilibrium of vectors/forces acting on a point, it sums to zero.

BTW and likewise, the y-axis projection sum

$$ 1+\sin{2\pi\over5}+\sin{4\pi\over5}+\sin{6\pi\over5}+\sin{8\pi\over5} $$

also goes to zero.

Also we have the formula for sum of cosines of $n$ angles in arithmetic progression with common difference $ \beta$

$$ \dfrac{\sin n \beta/2 }{ \sin \beta/2} \cdot \cos \dfrac{\alpha_1 +\alpha_2}{2} $$

which also vanishes.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.