# Prove the following trigonometric identity without a calculator involved [duplicate]

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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 $0$.

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• Proving only with a calculator would be harder ;) – ypercubeᵀᴹ Feb 17 '16 at 18:38
• 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. is a poor mathematician Feb 18 '16 at 2:33
• – lab bhattacharjee Feb 18 '16 at 5:01
• Would converting to exponential form help? – Mehrdad Feb 18 '16 at 19:09

## 11 Answers

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 $u = 0$.

• 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 '16 at 23:10
• @CiaPan The only vector that is unchanged under a rotation (other than the identity) is the zero vector. – David Feb 18 '16 at 16:58
• @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 '16 at 18:35
• @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 '16 at 21:29
• @David You're right, I can see that now. Thank you. – CiaPan Feb 18 '16 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 $0$.

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}

• I presume lines 3 and 4 are meant to be understood as equivalent since \omega_5^6\ = \omega_5\ by periodic formula. – candied_orange Feb 18 '16 at 10:01
• @CandiedOrange, yes, that's the step I took. We use $\omega_5^5=1$ – vrugtehagel Feb 18 '16 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}

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

• 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 '16 at 16:28
• @RazvanParaschiv Check my method. I edited my answer. – Win Vineeth Feb 18 '16 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.

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

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

$$z_k:=\cos\left(k\frac{2\pi}5\right)$$ 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.)

• 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? – user220382 Feb 19 '16 at 9:41
• @SanchayanDutta: just because $\cos\left(5k\dfrac{2\pi}5\right)=\cos(2k\pi)=1$. – Yves Daoust Feb 19 '16 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.

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.)

• The sin identity is more obvious because $\sin 2\pi/5 + \sin 8\pi/5 = 0$ etc. – Neil Feb 19 '16 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.

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.